# Sectional nonassociativity of metrized algebras

Daniel J. F. Fox  Departamento de Matemática Aplicada
Escuela Técnica Superior de Arquitectura
Universidad Politécnica de Madrid
Av. Juan de Herrera 4
28040 Madrid España<EMAIL_ADDRESS>
###### Abstract.

The sectional nonassociativity of a metrized (not necessarily associative or
unital) algebra is defined analogously to the sectional curvature of a pseudo-
Riemannian metric, with the associator in place of the Levi-Civita covariant
derivative. For commutative real algebras nonnegative sectional
nonassociativity is usually called the Norton inequality, while a sharp upper
bound on the sectional nonassociativity of the Jordan algebra of Hermitian
matrices over a real Hurwitz algebra is closely related to the Böttcher-
Wenzel-Chern-do Carmo-Kobayashi inequality. These and other basic examples are
explained, and there are described some consequences of bounds on sectional
nonassociativity for commutative algebras. A technical point of interest is
that the results work over the octonions as well as the associative Hurwitz
algebras.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 The analogy between multiplications and connections
  4. 4 Sectional nonassociativity
  5. 5 Examples of algebras with constant sectional nonassociativity
  6. 6 Examples of bounds on sectional nonassociativity
  7. 7 The Chern-do Carmo-Kobayashi inequality over real Hurwitz algebras
  8. 8 Consequences for commutative algebras of conditions on sectional nonassociativity

##  1\. Introduction

For an algebra $(\mathbb{A},\circ)$ equipped with a metric, meaning a
nondegenerate symmetric bilinear form, that is invariant, meaning $h(x\circ
y,z)=h(x,y\circ z)$ for all $x,y,z\in\mathbb{A}$, the _sectional
nonassociativity_ $\mathscr{K}(x,y)$ of the $h$-nondegenerate subspace
$\operatorname{Span}\\{x,y\\}\subset\mathbb{A}$ is defined in [18] to be

(1.1) $\displaystyle\begin{split}\mathscr{K}(x,y)&=\frac{h(x\circ x,y\circ
y)-h(x\circ y,y\circ x)}{h(x,x)h(y,y)-h(x,y)^{2}}.\end{split}$

(Because $\circ$ need not be commutative or anticommutative, the order of $x$
and $y$ in (1.1) matters.) The sectional nonassociativity is analogous to the
sectional curvature of the Levi-Civita connection of a pseudo-Riemannian
metric, with the multiplication and its associator in place of the covariant
derivative and its curvature. Section 3 describes in detail this analogy
between the multiplication of an algebra an torsion-free connection on a
tangent bundle. The notion itself, and its basic properties, are given in
Section 4.

Quantitative bounds on the sectional nonassociativity of a metrized real
algebra encompass apparently unrelated inequalities such as the Böttcher-
Wenzel-Chern-do Carmo-Kobayashi inequality bounding the norm of a matrix
commutator and the Norton inequality asserting the nonnegativity of the
quadratic expression $h(x\circ x,y\circ y)-h(x\circ y,x\circ y)$ for certain
metrized commutative algebras arising in the theory of vertex operator
algebras whose most famous representative is the $196844$-dimensional Griess
algebra on which the monster finite simple group acts by automorphisms.

It is apparent from (1.1) that, for a commutative algebra, the Norton
inequality may be rephrased as _nonnegative sectional nonassociativity_. On
the other hand, if $(\mathbb{A},[\,\cdot\,,\,\cdot\,],h)$ is a metrized
anticommutative algebra (for example, a Lie or Malcev algebra with $h$ a
multiple of its Killing form), (1.1) takes the form

(1.2) $\displaystyle\mathscr{K}(x,y)$
$\displaystyle=\frac{h([x,y],[x,y])}{h(x,x)h(y,y)-h(x,y)^{2}}.$

If $h$ is positive definite, the right-hand side of (1.2) is bounded from
above by $\frac{h([x,y],[x,y])}{h(x,x)h(y,y)}$ and, by the Cauchy-Schwarz
inequality, this quantity is bounded from above. The Böttcher-Wenzel
inequality asserts that for the algebra $\mathbb{M}(n,\mathbb{C})$ of $n\times
n$ complex matrices metrized by the Frobenius norm this quantity is bounded
from above by $2$ (which is better than the constant obtained by applying
Cauchy-Schwarz), and characterizes the case of equality. The Böttcher-Wenzel
inequality can be interpreted as an upper bound on the sectional
nonassociativity of the Lie algebra of $n\times n$ matrices
$\mathbb{M}(n,\mathbb{C})$ with the usual matrix commutator.

For an anticommutative algebra, (1.2) is obviously nonnegative, with equality
if and only if $[x,y]=0$, so that characterization of the equality case in the
lower bound amounts to characterization of commuting pairs of elements. On the
other hand, for a commutative algebra, (1.1) can take both signs. These
observations suggest looking for similar bounds for other metrized algebras,
for example commutative algebras or other sorts of Lie algebras.

The extreme case of equal upper and lower bounds is that of constant sectional
nonassociativity. This is closely related to the notion of projective
associativity of commutative algebras studied in [18]. Although it is not
solved here, the classification of metrized algebras with constant sectional
nonassociativity appears an accessible problem. However, if anisotropy of the
metric is not assumed, examples abound and it is still not clear what form the
classification will take even in the simplest case of vanishing sectional
nonassociativity. Essentially by definition, associative and alternative
algebras have vanishing sectional nonassociativity. In Section 5, it is shown
that both a quadratic $2$-step nilpotent Lie algebra and a metrized
antiflexible algebra have vanishing sectional nonassociativity, and nontrivial
examples of such algebras are given. In particular, Example 15 exhibits a
Euclidean metrized antiflexible algebra with vanishing sectional
nonassociativity whose underlying Lie algebra is $\mathfrak{so}(3)$ (which has
constant sectional nonassociativity $1$ when metrized by the negative of its
Killing form) and whose underlying commutative algebra has constant negative
sectional nonassociativity. (It gives a similar example with underlying Lie
algebra $\mathfrak{sl}(2,\mathbb{R})$.)

Section 5 describes other examples of algebras having constant sectional
nonassociativity. Example 16 explains that cross product algebras could be
_defined_ as antisymmetric metrized algebras having constant negative
sectional nonassociativity. Example 18 describes a one-parameter family of
pairwise nonisomorphic $3$-dimensional metrized commutative algebras having
constant sectional nonassociativities in $(-\infty,1/4]$.

Section 6 gives examples of algebras for which there can be established bounds
on the sectional nonassociativities. Example 19 recalls a result of Miyamoto
shows that the Griess algebras of OZ vertex operator algebras satisfy the
Norton inequality. In this context it would be interesting to establish upper
bounds. Example 20 shows that symmetric composition algebras such as para-
Hurwitz and Okubo algebras have sectional nonassociativity bounded between
$-1$ and $1$.

The remainder of Section 6 is devoted to the proof of Theorem 6.1, which gives
sharp upper and lower bounds on the sectional nonassociativities of a simple
Euclidean Jordan algebra of rank at least $3$ together with characterization
of the cases of equality in the bounds. These algebras are simply the Jordan
algebras $\mathbb{Herm}(n,\mathbb{h})$ of Hermitian matrices over the real
Hurwitz algebras
$\mathbb{h}\in\\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\\}$, where $n\geq
3$, and, in the case of the octonions, $\mathbb{O}$, attention is restricted
to $n=3$. Theorem 6.1 states that the sectional nonassociativity,
$\mathscr{K}(x,y)$, defined in Definition 4.1, of the subspace spanned by
linearly independent $x,y\in\mathbb{Herm}(n,\mathbb{h})$ satisfies

(1.3) $\displaystyle
0\leq\mathscr{K}(x,y)=\mathscr{K}_{h,\star}(x,y)\leq\tfrac{n}{2},$

and characterizes the subspaces for equality is obtained in the upper and
lower bound. The lower bound is the Norton inequality. The sharp upper bound
is deduced as a consequence of the Chern-do Carmo-Kobayashi inequality for the
commutators of Hermitian matrices. This requires some argument to extend this
inequality to the octonions that is formulated as Lemma 7.2; this extension
may be of independent interest.

These results are of interest from the point of view of applications of the
Böttcher-Wenzel-Chern-do Carmo-Kobayashi inequality. This inequality is
recalled in section 7 and it is shown that the Chern-do Carmo-Kobayashi
inequality holds for $\mathbb{Herm}(n,\mathbb{H})$ and partially for
$\mathbb{Herm}(3,\mathbb{O})$. These refinements, Theorem 7.1 and the variant
Lemma 7.2, valid over a real Hurwitz algebra, are used in the proof of Theorem
6.1 to estimate the sectional nonassociativity of the deunitalization of a
simple Euclidean Jordan algebra. In a sense, this amounts to reinterpreting
the Böttcher-Wenzel-Chern-do Carmo-Kobayashi inequalities as upper bounds on
sectional nonassociativity.

The notion of sectional nonassociativity facilitates quantifications of
nonassociativity both more refined and less exact than nonnegativity or
nonpositivity, and these may in turn suggest refinements of the Böttcher-
Wenzel inequality. More generally, the sectional nonassociativity is
manifestly invariant under isometric algebra automorphisms and its relation
with other similar quantities deserves to be explored more. See Remark 5 for
further comments in this direction.

Not many examples of commutative algebras satisfying the Norton inequality are
known and the description of such algebras is interesting and their
classification, possibly with additional hypotheses, may be a possibly
tractable problem. The main known examples are the Griess algebras of certain
vertex operator algebras [43] and their subalgebras, as is explained next in
Example 19, and, by Theorem 6.1, the simple Euclidean Jordan algebras, as is
shown in Theorem 6.1. (It is known that most of the simple Euclidean Jordan
algebras can also be obtained as Griess algebras of certain vertex operator
algebras [3, 35, 67].) Further examples are the Majorana algebras introduced
by A. A. Ivanov [28, 29] which are defined by a list of axioms including the
Norton inequality. See [64] for a family of examples depending on a parameter
and further references.

More generally, the notion of sectional nonassociativity should help to
organize the study of metrized algebras in the same manner that restriction on
curvature quantities help to organize the study of Riemannian manifolds.
Typically classes of algebras are defined by identities. From the point of
view advocated here this is something like studying symmetric spaces (whose
curvature tensors satisfy algebraic equalities).

Section 8 presents some general consequences for commutative algebras of
bounds on sectional nonassociativity. Combined with the notion of sectional
nonassociativity, results about the eigenvectors and eigenvalues of the
structure tensor can be used to relate the existence and description of
idempotents and square-zero elements to conditions involving sectional
nonassociativity. Since an automorphism of the algebra must permute its
idempotents and square-zero elements, such results yield results about the
size of the automorphism group. For example, Theorem 8.1 shows that a
Euclidean metrized commutative algebra has finite automorphism group if its
sectional nonassociativity is negative or if its sectional nonassociativity is
nonpositive and the algebra contains no trivial subalgebra. In the other
direction, Lemma 8.5 shows that nonnegative sectional nonassociativity
precludes existence of $2$-nilpotents. The flavor of such results conforms
broadly with the geometric analogy. For example, the automorphism group of a
negatively curved Riemann surface is finite, while the (positively curved)
sphere has a large continuous group of automorphisms.

The paper is structured as follows. Section 2 presents basic definitions
related to metrized algebras. Section 3 shows that the analogy between
multiplications and covariant derivatives has substance and presents results
that help motivate the definition of sectional nonassociativity given in
Section 4. Section 4 also describes the most basic properties of the sectional
nonassociativity. Section 5 presents examples having constant sectional
nonassociativity; these include Hurwitz algebras, cross-product algebras,
antiflexible algebras, and some novel three-dimensional algebras. Section 6
describes examples satisfying bounds on sectional nonassociativity
generalizing the Norton inequality; these include symmetric composition
algebras, Griess algebras of OZ vertez operator algebras, and the simple
Euclidean Jordan algebras. The proof of the upper bound on the sectional
nonassociativity of simple Euclidean Jordan algebras given in Theorem 6.1
requires the Böttcher-Wenzel-Chen-do Carmo-Kobayashi inequality for matrices
over a Hurwitz algebra. Section 7 describes this and addresses what happens
over the octonions. Finally, Section 8 presents some general consequences for
commutative algebras of bounds on sectional nonassociativity.

##  2\. Preliminaries

Let $\mathbb{k}$ be a field. Everywhere in the paper there is assumed
$\operatorname{char}\mathbb{k}\notdivides 6$, although sometimes assumptions
on $\operatorname{char}\mathbb{k}$ are stated explicitly for clarity.

An algebra means a finite-dimensional vector space $\mathbb{A}$ over
$\mathbb{k}$ equipped with a bilinear multiplication
$\circ:\mathbb{A}\times\mathbb{A}\to\mathbb{A}$. It need not be associative,
commutative, or unital. The left and right multiplication endomorphisms
$L_{\circ},R_{\circ}:\mathbb{A}\to\operatorname{End}(\mathbb{A})$ are defined
by $L_{\circ}(x)y=x\circ y=R_{\circ}(y)x$.

An algebra is _metrized_ if it is equipped with a _metric_ , meaning a
nondegenerate symmetric bilinear form, $h$, that is _invariant_ (also called
_associative_), meaning that $h(x\circ y,z)=h(x,y\circ z)$ for all
$x,y,z\in\mathbb{A}$.

A _Euclidean_ metrized algebra is a real algebra metrized by a positive
definite metric. (Essentially all results in the paper stated for Euclidean
metrized algebras remain true with a real-closed base field in place of
$\mathbb{R}$, but it would be distracting to state and prove them in this
generality.) That an algebra be Euclidean metrized means that the adjoint of
the left multiplication endomorphism $L_{\circ}(x)$ equals the right
multiplication endomorphism $R_{\circ}(x)$. Consequently, for a Euclidean
metrized commutative algebra the multiplication endomorphisms
$L_{\circ}(x)=R_{\circ}(x)$ are self-adjoint with respect to a positive
definite symmetric bilinear form.

######  Example 1.

A semisimple real Lie algebra $(\mathfrak{g},\circ)$ is metrized by the
negative $h=-B_{\mathfrak{g}}$ of its Killing form
$B_{\mathfrak{g}}=\operatorname{tr}L_{\circ}(x)L_{\circ}(y)$. It is Euclidean
in the above sense if it is moreover compact. More generally, an
anticommutative algebra $(\mathbb{A},\circ)$ is _Malcev_ if

(2.1) $\displaystyle(x\circ y)\circ(x\circ z)=((x\circ y)\circ z)\circ
x+((y\circ z)\circ x)\circ x+((z\circ x)\circ x)\circ y,$
$\displaystyle\text{for all}\,\,x,y,z\in\mathbb{A}.$

for all $x,y,z\in\mathbb{A}$. Any Lie algebra is Malcev. By [56, Theorem
$7.16$] the Killing form
$\tau_{\circ}(x,y)=\operatorname{tr}L_{\circ}(x)L_{\circ}(y)$ of a Malcev
algebra is invariant. The identity (2.1) can be written $L_{\circ}(x\circ
y)L_{\circ}(x)+L_{\circ}(x)L_{\circ}(x\circ
y)=[L_{\circ}(x)^{2},L_{\circ}(y)]$ and tracing this shows
$\tau_{\circ}(x\circ y,x)=\tau_{\circ}(x,y\circ x)$. Polarizing this in $x$
yields the claim.

Lemma 2.1 generalizes the statement that a Lie algebra admitting an invariant
nondegenerate bilinear form has dimension at least $3$.

######  Lemma 2.1.

A two-dimensional metrized algebra $(\mathbb{A},\circ,h)$ is commutative.

###### Proof.

If $u$ and $v$ span $\mathbb{A}$, then $h(u\circ v,u)=h(u,u\circ v)=h(u\circ
u,v)=h(v,u\circ u)=h(v\circ u,u)$ and, similarly, $h(u\circ v,v)=h(v\circ
u,v)$. By the nondegeneracy of $h$ this implies $u\circ v=v\circ u$, which
suffices to show the claim. ∎

The _associator_ of an algebra is the trilinear map
$\otimes^{3}\mathbb{A}^{\ast}\to\mathbb{A}$ defined by

(2.2) $\displaystyle[x,y,z]=\mathscr{A}_{\circ}(x,y)z=(L_{\circ}(x\circ
y)-L_{\circ}(x)L_{\circ}(y))z=(x\circ y)\circ z-x\circ(y\circ z).$

The map $\mathscr{A}_{\circ}(x,y)\in\operatorname{End}(\mathbb{A})$ is the
_(left) associator endomorphism_.

######  Example 2.

A commutative algebra is _Jordan_ if $[x\circ x,y,x]=0$ for all
$x,y,\in\mathbb{A}$. For a Jordan algebra, the bilinear form
$\operatorname{tr}L_{\circ}(x\circ y)$ is invariant, so it is metrized if this
bilinear form is nondegenerate. A real Jordan algebra is Euclidean if it is
formally real, meaning that a sum of squares equals zero if and only if each
of the elements squares is zero. A real Jordan algebra is Euclidean as a
Jordan algebra if and only if $\operatorname{tr}L_{\circ}(x\circ y)$ is
positive definite [30].

######  Example 3.

A _composition algebra_ is an algebra $(\mathbb{A},\circ,q)$ with a
nondegenerate quadratic form $q$ that is multiplicative meaning $q(x\circ
y)=q(x)q(y)$ for all $x,y\in\mathbb{A}$. A _Hurwitz_ algebra is a unital
composition algebra $(\mathbb{A},\circ,e,q)$. By the Hurwitz theorem, a real
Hurwitz algebra is one of the real field $\mathbb{R}$, the complex field
$\mathbb{C}$, the quaternions $\mathbb{H}$, or the octonions $\mathbb{O}$. For
background on composition and Hurwitz algebras see [14, 32, 59].

A Hurwitz algebra carries the nondegenerate symmetric bilinear form
$h(x,y)=q(x+y)-q(x)-q(y)$ obtained from $q$ by linearization. By the
nondegeneracy and multiplicativity of $q$, $h(e,e)=2q(e)=2$. The _standard
involution_ of $(\mathbb{A},\circ,e,q)$ is the algebra antiautomorphism
defined by $\bar{x}=h(x,e)e-x$. Its fixed point set is the subalgebra
generated by $e$ and is isomorphic to $\mathbb{R}$. Taking $z=e$ yields
$h(x,y)e=x\bar{y}+y\bar{x}=h(\bar{x},\bar{y})e$. A Hurwitz algebra equipped
with its standard involution is a metrized algebra with involution in the
sense of [2], meaning that $h$ is _involutively invariant_ in the sense that
$h(x,z\circ\bar{y})=h(x\circ y,z)=h(y,\bar{x}\circ z)$ for all
$x,y,z\in\mathbb{A}$. Note that, except when $\mathbb{h}=\mathbb{k}$, $h$ is
_not_ invariant, so $(\mathbb{h},\circ,h)$ is _not_ metrized as an algebra.

##  3\. The analogy between multiplications and connections

The following analogy between algebras and connections illuminates the point
of view taken here. The multiplication $\circ$ is analogous to a torsion-free
connection on a tangent bundle, and the associator corresponds to (the
negative of) the second covariant derivative. In fact, a torsion-free
connection on a tangent bundle can be regarded as a Lie-admissible product on
the space of vector fields, and in that setting the sectional nonassociativity
defined in Section 4 _is_ simply the sectional curvature. However that setting
differs from that considered here in several ways. First, the space of vector
fields is infinite-dimensional. Second, the covariant derivative is a
$C^{\infty}$-module map in its first argument. A general context encompassing
both finite-dimensional algebras and connections on tangent bundles is
algebroids, but this is not discussed here.

The notions suggested by the multiplication-connection analogy prove
surprisingly robust and provide principles for organizing the zoo of general
algebras different from those that arise from classical points of view. In
particular, they offer possibilities for quantifying nonassociativity that
turn out to be quite useful.

This idea has been discussed in similar terms in [26] in the context of Lie-
admissible algebras and has roots in foundational articles on left-symmetric
algebras [60, 62]. In [8] there are considered algebras satisfying identities
like those treated in this section from the more sophisticated point of view
of Koszul duality of operads. The unrelated duality given by the adjoint
multiplication considered here is motivated by considerations from affine
differential geometry [19]. These ideas are expounded in the more limited
context of metrized commutative algebras in [18, 20].

The analogy is implemented in practice in the following way. The space of
connections on a bundle is an affine space. A reference connection is fixed
and then any other connection is identified with a tensor. Any expression
involving this connection involves some derivatives and some purely algebraic
part expressible in terms of this tensor. The purely algebraic expression make
sense also for the structure tensor of an algebra. For example, the covariant
derivative of a metric $g$ with respect to a torsion-free affine connection
$\nabla$ is $(\nabla_{X}g)(Y,Z)=X(g(Y,Z))-g(\nabla_{X}Y,Z)-g(Y,\nabla_{X}Z)$.
The corresponding expression for a metric $g$ on an algebra
$(\mathbb{A},\circ)$ is $-g(x\circ y,z)-g(y,x\circ z)$. The vanishing of this
expression is analogous to the metric tensor being parallel. This kind of
invariance should probably be properly formulated as a cohomological
condition, but it is not clear how to give sense to this last claim when
working in the generality considered here.

In the analogy, the underlying bracket is regarded as background information
of a topological rather than geometric nature, analogous to the Lie algebra of
vector fields on a smooth manifold.

Applying the connection-algebra analogy to the curvature of an affine
connection yields the following definitions.

Decomposing the product $\circ$ by symmetries leads to two underlying
algebras, from which $(\mathbb{A},\circ)$ can be reconstructed (except when
$\operatorname{char}\mathbb{k}=2$). The _underlying bracket_ (or _commutator_)
of the algebra $(\mathbb{A},\circ)$ is the commutator defined by $[x,y]=x\circ
y-y\circ x$ and $(\mathbb{A},[\,\cdot\,,\,\cdot\,])$ is the _underlying_
antisymmetric algebra. The _underlying commutative algebra_ of an algebra
$(\mathbb{A},\circ)$ is $\mathbb{A}$ equipped with the product $x\odot
y=x\circ y+y\circ x$.

The multiplication $\circ^{t}$ _adjoint_ to $\circ$ is that defined by
$x\circ^{t}y=-y\circ x$. It is the negative of the opposite multiplication. By
definition, that $L_{\circ^{t}}=-R_{\circ}$ and $R_{\circ^{t}}=-L_{\circ}$.
The sign in the definition of the adjoint multiplication $\circ^{t}$ is chosen
so that $\circ$ and the adjoint product $\circ^{t}$ have the same underlying
bracket.

The _curvature_
$\mathsf{r}:\mathbb{A}\otimes\mathbb{A}\to\operatorname{End}(\mathbb{A})$ and
_adjoint curvature_
$\bar{\mathsf{r}}:\mathbb{A}\otimes\mathbb{A}\to\operatorname{End}(\mathbb{A})$
of $(\mathbb{A},\circ)$ are defined by

(3.1)
$\displaystyle\begin{split}\mathsf{r}(x,y)z&=-[x,y,z]+[y,x,z]=\left([L_{\circ}(x),L_{\circ}(y)]-L_{\circ}([x,y])\right)z=(\mathscr{A}_{\circ}(y,x)-\mathscr{A}_{\circ}(x,y))z,\\\
\bar{\mathsf{r}}(x,y)z&=-[z,x,y]+[z,y,x]=\left([R_{\circ}(x),R_{\circ}(y)]+R_{\circ}([x,y])\right)z=(\mathscr{A}_{\circ^{t}}(y,x)-\mathscr{A}_{\circ^{t}}(x,y))z.\end{split}$

The adjoint curvature of $\circ$ is the curvature of the adjoint
multiplication $\circ^{t}$ and vice-versa. For any tensorial object
constructed from a multiplication $\circ$ there is a corresponding tensorial
object constructed in the same way from the adjoint multiplication
$\circ^{t}$, and it is denoted by putting a $\bar{\,}$ over the notation
indicating the object associated with $\circ$. This convention is consistent
with the notation $\mathsf{r}$ and $\bar{\mathsf{r}}$.

The algebra has _self-adjoint curvature_ if $\mathsf{r}=\bar{\mathsf{r}}$ and
_anti-self-adjoint curvature_ if $\mathsf{r}=-\bar{\mathsf{r}}$.

######  Example 4.

An algebra $(\mathbb{A},\circ)$ is _flexible_ if $[x,y,x]=0$ for all
$x,y\in\mathbb{A}$. A commutative or anticommutative algebra is flexible. The
symmetric composition algebras discussed in Example 20 are also flexible. It
is evident from (3.1) that a flexible algebra has self-adjoint curvature.

######  Example 5.

An algebra $(\mathbb{A},\circ)$ is _antiflexible_ if $[x,y,z]=[z,y,x]$ for all
$x,y,z\in\mathbb{A}$. It is evident from (3.1) that an antiflexible algebra
has anti-self-adjoint curvature.

######  Example 6.

An algebra is _left-symmetric_ (also called _pre-Lie_ or _Vinberg_) if
$\mathscr{A}_{\circ}(x,y)=\mathscr{A}_{\circ}(y,x)$ for all
$x,y\in\mathbb{A}$. It is evident from (3.1) that an algebra is left-symmetric
if and only if its curvature $\mathsf{r}$ vanishes. The vanishing of the
adjoint curvature is equivalent to the left symmetry of the adjoint
multiplication, or that the given multiplication is _right-symmetric_.

Any left-symmetric algebra that is not right-symmetric gives an example of an
algebra $(\mathbb{A},\circ)$ not isomorphic to its adjoint
$(\mathbb{A},\circ^{t})$. More generally, an algebra that does not have self-
adjoint curvature is not isomorphic to its adjoint algebra.

######  Example 7.

An awkward aspect of the conventions here is that the underlying bracket of an
anticommutative algebra is twice the original bracket, and this needs to be
remembered when interpreting (3.1).

As a Lie algebra $(\mathfrak{g},[\,\cdot\,,\,\cdot\,])$ is flexible, it has
self-adjoint curvature, and from (3.1) it follows that
$\bar{\mathsf{r}}(x,y)=\mathsf{r}(x,y)=-\operatorname{ad}_{\mathfrak{g}}([x,y])=-[\operatorname{ad}_{\mathfrak{g}}(x),\operatorname{ad}_{\mathfrak{g}}(y)]$.
Consequently any $2$-step nilpotent Lie algebra provides an example of a
nonassociative algebra for which both $\mathsf{r}$ and $\bar{\mathsf{r}}$
vanish identically.

Define $A_{\circ}(x)=L_{[\,\cdot\,,\,\cdot\,]}(x)=L_{\circ}(x)-R_{\circ}(x)$,
so that $A_{\circ}(x)y=[x,y]$, and define
$S_{\circ}(x)=L_{\odot}(x)=L_{\circ}(x)+R_{\circ}(x)$, so that
$S_{\circ}(x)y=x\odot y$. Straightforward computations using (3.1) show that,
in any algebra,

(3.2) $\displaystyle[A_{\circ}(x),S_{\circ}(y)]-S_{\circ}(A_{\circ}(x)y)$
$\displaystyle=\mathsf{r}(x,y)-\bar{\mathsf{r}}(x,y)-[R_{\circ}(x),L_{\circ}(y)]-[R_{\circ}(y),L_{\circ}(x)].$

Decomposing (3.2) into its parts symmetric and antisymmetric in $x$ and $y$
yields the relations

(3.3)
$\displaystyle\begin{split}[A_{\circ}(x),S_{\circ}(y)]+[A_{\circ}(y),S_{\circ}(x)]=-2[R_{\circ}(x),L_{\circ}(y)]-2[R_{\circ}(y),L_{\circ}(x)],\end{split}$

and

(3.4)
$\displaystyle\begin{split}\mathsf{r}(x,y)x-\bar{\mathsf{r}}(x,y)z&=-[x,y,z]+[z,x,y]-[z,y,x]+[y,x,z]\\\
&=\left([L_{\circ}(x),R_{\circ}(z)]+[L_{\circ}(z),R_{\circ}(x)]\right)y-\left([L_{\circ}(y),R_{\circ}(z)]+[R_{\circ}(y),L_{\circ}(z)]\right)x,\\\
\
&=\left(\tfrac{1}{2}[A_{\circ}(x),S_{\circ}(y)]-\tfrac{1}{2}[A_{\circ}(y),S_{\circ}(x)]-S_{\circ}([x,y])\right)z,\end{split}$

for all $x,y,z\in\mathbb{A}$ (the validity of the last equality needs
$\operatorname{char}\mathbb{k}\neq 2$).

######  Lemma 3.1.

If $\operatorname{char}\mathbb{k}\neq 2$, the following conditions on a
$\mathbb{k}$-algebra $(\mathbb{A},\circ)$ are equivalent.

  1. (1)

$[A_{\circ}(x),S_{\circ}(x)]=0$ for all $x\in\mathbb{A}$.

  2. (2)

$[R_{\circ}(x),L_{\circ}(y)]=-[R_{\circ}(y),L_{\circ}(x)]$ for all
$x,y\in\mathbb{A}$.

  3. (3)

$(\mathbb{A},\circ)$ is flexible.

Over a field of any characteristic, a flexible algebra satisfies (1) and (2)
and has self-adjoint curvature.

###### Proof.

By definition, $(\mathbb{A},\circ)$ is flexible if and only if
$[R_{\circ}(x),L_{\circ}(x)]=0$ for all $x\in\mathbb{A}$. By (3.3) this
implies (1) and is implied by (1) if $\operatorname{char}\mathbb{k}\neq 2$.
Polarization shows any flexible algebra satisfies (2), and if
$\operatorname{char}\mathbb{k}\neq 2$, then the validity of (2) implies
$(\mathbb{A},\circ)$ is flexible (taking $y=x$ in (2)). By (2) and (3.4), a
flexible algebra has self-adjoint curvature. ∎

######  Lemma 3.2.

If $\operatorname{char}\mathbb{k}\neq 2$, a $\mathbb{k}$-algebra
$(\mathbb{A},\circ)$ is flexible if and only if it has self-adjoint curvature
and $A_{\circ}(x)$ is a derivation of the underlying commutative algebra
$(\mathbb{A},\odot)$ for all $x\in\mathbb{A}$.

###### Proof.

This is immediate from Lemma 3.1 and (3.2). ∎

######  Example 8.

By [2, Proposition $1$], the right-hand side of (3.4) vanishes for any three
elements in the subspace of a structurable algebra fixed by its involution.

The analogue of the algebraic Bianchi identity for a curvature tensor is the
vanishing of the complete antisymmetrization of $\mathsf{r}(x,y)z$. This need
not occur in general for, as is the case for an affine connection, there is an
obstruction related to compatibility with an underlying Lie bracket.

An algebra $(\mathbb{A},\circ)$ is _Lie-admissible_ if the underlying bracket
$[\,\cdot\,,\,\cdot\,]$ satisfies the Jacobi identity, so that
$(\mathbb{A},[\,\cdot\,,\,\cdot\,])$ is a Lie algebra. That
$(\mathbb{A},\circ)$ be Lie-admissible is equivalent to the vanishing of the
completely antisymmetric tensor

(3.5)
$\displaystyle\begin{split}\mathsf{t}(x_{1},x_{2},x_{3})&=\left(A_{\circ}([x_{1},x_{2}])-[A_{\circ}(x_{1}),A_{\circ}(x_{2})]\right)x_{3}=\mathsf{Cycle}_{1,2,3}[[x_{1},x_{2}],x_{3}]\\\
&=\sum_{\sigma\in
S_{3}}\operatorname{sgn}(\sigma)[x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)}]=-\mathsf{Cycle}_{1,2,3}\mathsf{r}(x_{1},x_{2})x_{3}=-\mathsf{Cycle}_{1,2,3}\bar{\mathsf{r}}(x_{1},x_{2})x_{3}.\end{split}$

(The notation $\mathsf{Cycle}_{1,2,3}$ indicates the sum over the cyclic
permutations of the indicated indices.) The validity of the identity
$-\mathsf{Cycle}_{1,2,3}\mathsf{r}(x_{1},x_{2})x_{3}=\mathsf{Cycle}\left([x_{1},x_{2},x_{3}]-[x_{2},x_{1},x_{3}]\right)=\mathsf{Cycle}[[x_{1},x_{2}],x_{3}]$
in any algebra is noted in [48, Equation $2.19$]. That the vanishing of the
first expression on the second line of (3.5) is equivalent to Lie
admissibility is stated in [40, Example $6$].

Note that $\bar{\mathsf{t}}=\mathsf{t}$. The final two equalities of (3.5)
show that the identity $\mathsf{t}=0$ is analogous to the algebraic Bianchi
identity for a torsion-free connection, so the tensor $\mathsf{t}$ should be
regarded as the algebraic analogue of the torsion of an affine connection.

A straightforward computation shows $\mathsf{t}$ is the self-adjoint part of

(3.6)
$\displaystyle\begin{split}[A_{\circ}(x),L_{\circ}(y)]-L_{\circ}(A_{\circ}(x)y)\end{split}$

which vanishes for all $x,y\in\mathbb{A}$ if and only if $A_{\circ}(x)$ is a
derivation of $(\mathbb{A},\circ)$ for all $x\in\mathbb{A}$.

######  Example 9.

It is immediate from (3.5) that $\mathsf{t}=0$ for an antiflexible algebra, so
an antiflexible algebra is Lie-admissible.

The analogue of the covariant derivative of the curvature tensor,
$(\nabla_{X}R)(Y,Z)$, is
$[L_{\circ}(x),\mathsf{r}(y,z)]-\mathsf{r}(L_{\circ}(x)y,z)-\mathsf{r}(y,L_{\circ}(x)z)$.
The identities (3.7) are analogous to the differential Bianchi identity, and
reinforce the idea that $\mathsf{t}$ should be viewed as something like the
torsion of $\circ$.

######  Lemma 3.3.

In any algebra $(\mathbb{A},\circ)$, for all $x,y,z\in\mathbb{A}$ there hold

(3.7) $\displaystyle\begin{aligned}
\mathsf{Cycle}_{x,y,z}&\left([L_{\circ}(x),\mathsf{r}(y,z)]-\mathsf{r}(L_{\circ}(x)y,z)-\mathsf{r}(y,L_{\circ}(x)z)\right)=L_{\circ}(\mathsf{t}(x,y,z)),\\\
\mathsf{Cycle}_{x,y,z}&\left([R_{\circ}(x),\bar{\mathsf{r}}(y,z)]-\bar{\mathsf{r}}(R_{\circ}(x)y,z)-\bar{\mathsf{r}}(y,R_{\circ}(x)z)\right)=R_{\circ}(\mathsf{t}(x,y,z)).\end{aligned}$

###### Proof.

A computation using (3.1), the Jacobi identity in
$\operatorname{End}(\mathbb{A})$, the identity $[x\circ y,z]-[x,z\circ
y]=[x,y,z]-[z,y,x]$, and (3.5) shows the first identity of (3.7). Taking
$\circ^{t}$ in place of $\circ$ and recalling $\bar{\mathsf{t}}=\mathsf{t}$
yields the second identity of (3.7). ∎

An algebra $(\mathbb{A},\circ)$ is _associator-cyclic_ if it satisfies the
identity

(3.8) $\displaystyle 0=[x,y,z]+[z,x,y]+[y,z,x]=\left(A_{\circ}(x\circ
y)-A_{\circ}(x)L_{\circ}(y)-A_{\circ}(y)R_{\circ}(x)\right)z,$

for all $x,y,z\in\mathbb{A}$. (There seems to be no standard terminology for
algebras satisfying (3.8), and _associator-cyclic_ seems reasonably evocative
terminology.)

######  Example 10.

Commutative algebras and Lie algebras are associator-cyclic, but a general
anticommutative algebra need not be associator-cyclic.

######  Lemma 3.4.

If $\operatorname{char}\mathbb{k}\neq 3$, an associator-cyclic
$\mathbb{k}$-algebra has self-adjoint curvature if and only if it is flexible.

###### Proof.

By Lemma 3.1, a flexible algebra has self-adjoint curvature, so it suffices to
prove that an associator-cyclic algebra with self-adjoint curvature is
flexible. Taking $x=z$ in (3.8) and (3.4) yields
$[x,y,x]=-[x,x,y]-[y,x,x]=-2[x,y,x]$, so that $3[x,y,x]=0$ for all
$x,y\in\mathbb{A}$. If $\operatorname{char}\mathbb{k}\neq 3$, this implies
$(\mathbb{A},\circ)$ is flexible. ∎

Antisymmetrizing (3.8) in $x$ and $y$ shows that an associator-cyclic algebra
is Lie-admissible. Lemma 3.5 shows that these conditions are equivalent for
flexible $\mathbb{k}$-algebras when $\operatorname{char}\mathbb{k}\neq 2$.
That when $\operatorname{char}\mathbb{k}\neq 2$ an algebra is flexible and
Lie-admissible if and only if there holds (2) of Lemma 3.5 is due to Okubo and
Myung in [44, p. $80$] and [50, Theorem $4.1$].

######  Lemma 3.5 ([44, p. $80$], [50, Theorem $4.1$], [45]).

For a $\mathbb{k}$-algebra $(\mathbb{A},\circ)$ the following are equivalent.

  1. (1)

$(\mathbb{A},\circ)$ is flexible and associator-cyclic.

  2. (2)

For all $x\in\mathbb{A}$, $A_{\circ}(x)=L_{\circ}(x)-R_{\circ}(x)$ is a
derivation of $(\mathbb{A},\circ)$.

  3. (3)

$(\mathbb{A},\circ)$ has self-adjoint curvature.

These imply

  1. (4)

$(\mathbb{A},\circ)$ is flexible and Lie-admissible.

If $\operatorname{char}\mathbb{k}\neq 2$, then (4) implies (1) and (2). In
particular, if $\operatorname{char}\mathbb{k}\neq 2$, a flexible algebra is
Lie-admissible if and only if it is associator-cyclic.

###### Proof.

For any algebra there holds

(3.9)
$\displaystyle\begin{split}\mathsf{Cycle}_{x,y,z}[x,y,z]&=\left([A_{\circ}(x),L_{\circ}(y)]-L_{\circ}(A_{\circ}(x)y)\right)z+[L_{\circ}(x),R_{\circ}(z)]y.\end{split}$

It is immediate from (3.9) that if $(\mathbb{A},\circ)$ is flexible and
associator-cyclic, then it satisfies (2). If $(\mathbb{A},\circ)$ satisfies
(2), then
$-[R_{\circ}(x),L_{\circ}(x)]=[A_{\circ}(x),L_{\circ}(x)]=L_{\circ}(A_{\circ}(x)x)=0$
for all $x\in\mathbb{A}$, so $(\mathbb{A},\circ)$ is flexible, and with (3.9)
it follows that $(\mathbb{A},\circ)$ is associator-cyclic. This proves the
equivalence of (1) and (2). The equivalence of (2) and (3) is immediate from
(3.6) and its adjoint relation.

By (3.5),
$\mathsf{t}(x,y,z)=\mathsf{Cycle}_{x,y,z}[x,y,z]-\mathsf{Cycle}_{x,y,z}[y,x,z]$
(alternatively, by antisymmetrizing (3.8) in $x$ and $y$), so an associator-
cyclic algebra is Lie-admissible, and (1) implies (4). If $(\mathbb{A},\circ)$
is Lie-admissible and flexible, then
$0=\mathsf{Cycle}_{x,y,z}([x,y,z]-[y,x,z])=2\mathsf{Cycle}_{x,y,z}([x,y,z])$,
so, if $\operatorname{char}\mathbb{k}\neq 2$, $(\mathbb{A},\circ)$ is
associator-cyclic. ∎

######  Remark 1.

The essential content of Lemma 3.5 is due to Okubo and Myung [44, 45, 50]. It
is formulated here in a manner consistent with the perspective taken here. The
equivalence of (1) and (2) and that these imply (3) is also proved, with
different terminology, in [27] (see also [54, Propositions $1$ and $2$]). (In
[54] an algebra is called _admissible Poisson_ if there holds (2) and _weakly
associative_ if there vanishes the right-hand side of (3.9) (so it satisfies
(3) of Lemma 3.5).)

######  Example 11.

By the Jacobi identity, the associator of a Lie algebra
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,])$ is $[x,y,z]=[y,[x,z]]$ and a Lie
algebra is associator-cyclic. This shows an associator-cyclic, flexible
algebra need not be commutative.

######  Example 12.

By [57] an algebra obtained from a flexible algebra by the Cayley-Dickson
process is flexible. An example of a flexible algebra that is not Lie-
admissible (so not associator-cyclic) is the imaginary octonions
$\operatorname{im}\mathbb{O}$ equipped with the commutator bracket $[x,y]=xy-
yx$. (Although $\operatorname{im}\mathbb{O}$ is Malcev-admissible [45].)

##  4\. Sectional nonassociativity

This section introduces quantitative notions of nonassociativity for metrized
algebras. The sectional nonassociativity was introduced in [18] for metrized
commutative algebras.

In the first part of this section the base field $\mathbb{k}$ has
characteristic not equal to $2$ or $3$. Let $\mathbb{A}$ be an $n$-dimensional
$\mathbb{k}$-vector space with a metric $h$. Define

(4.1) $\displaystyle\begin{aligned}
S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}&=\left\\{\begin{aligned}
a&\in\otimes^{4}\mathbb{A}^{\ast}:\\\
&a(x,y,z,w)=-a(y,x,z,w)=-a(x,y,w,z)=a(z,w,x,y)\,\,\text{for
all}\,\,x,y,z,w\in\mathbb{A}\end{aligned}\right\\},\\\
\mathscr{MC}(\mathbb{A})&=\\{a\in
S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}:a(x,y,z,w)+a(y,z,x,w)+a(z,x,y,w)=0\,\,\text{for
all}\,\,x,y,z,w\in\mathbb{A}\\}.\end{aligned}$

The subspace $\mathscr{MC}(\mathbb{A})$ comprises tensors of metric curvature
tensor type. An element $a\in S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ can
be viewed as the bilinear form on $\mathsf{\Lambda}^{2}\mathbb{A}$ defined on
decomposable elements by $a(x\wedge y,z\wedge w)=4a(x,y,z,w)$, so determines
an associated quadratic form on $\mathsf{\Lambda}^{2}\mathbb{A}$. Endow
$\otimes^{4}\mathbb{A}^{\ast}$ and its subspaces with the metrics given by
complete contraction with the metric $h$. If
$\operatorname{char}\mathbb{k}\neq 3$,
$S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ is isomorphic to the orthogonal
direct sum
$\mathscr{MC}(\mathbb{A})\oplus\mathsf{\Lambda}^{4}\mathbb{A}^{\ast}$, and the
orthogonal projections onto the factors,
$\mathcal{P}:S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}\to\mathscr{MC}(\mathbb{A})$
and
$\mathcal{Q}:S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}\to\mathsf{\Lambda}^{4}\mathbb{A}^{\ast}$
are defined for $a\in S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ by

(4.2) $\displaystyle\begin{aligned}
\mathcal{P}(a)(x,y,z,w)&=\tfrac{1}{3}\left(2a(x,y,z,w)-a(y,z,x,w)-a(z,x,y,w)\right),\\\
\mathcal{Q}(a)(x,y,z,w)&=\tfrac{1}{3}\left(a(x,y,z,w)+a(y,z,x,w)+a(z,x,y,w)\right).\end{aligned}$

The tensors $a$ and $\mathcal{P}(a)$ do not necessarily determine the same
quadratic form on $\mathsf{\Lambda}^{2}\mathbb{A}$, but the quadratic forms
they determine agree on decomposable elements of
$\mathsf{\Lambda}^{2}\mathbb{A}$.

######  Lemma 4.1.

Suppose $\operatorname{char}\mathbb{k}\notdivides 6$. Let $h$ be a metric on
the $\mathbb{k}$-vector space $\mathbb{A}$ and let
$\mathcal{P}:S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}\to\mathscr{MC}(\mathbb{A})$
be the orthogonal projection with respect to the inner products given by
complete contraction with $h$. The quadratic forms determined on
$\mathsf{\Lambda}^{2}\mathbb{A}$ by $a,b\in
S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ coincide on decomposable elements
of $\mathsf{\Lambda}^{2}\mathbb{A}$ if and only if
$\mathcal{P}(a)=\mathcal{P}(b)$.

###### Proof.

For $\omega\in\mathsf{\Lambda}^{2}\mathbb{A}$ and $a\in
S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$,

(4.3) $\displaystyle\begin{split}\langle
a,\omega\otimes\omega\rangle&=\langle\mathcal{P}(a),\omega\otimes\omega\rangle+\langle\mathcal{Q}(a),\omega\otimes\omega\rangle=\langle\mathcal{P}(a),\omega\otimes\omega\rangle+\tfrac{1}{6}\langle\mathcal{Q}(a),\omega\wedge\omega\rangle,\end{split}$

in which $\langle\,\cdot\,,\,\cdot\,\rangle$ denotes the pairing between dual
vector spaces. Because $\omega\wedge\omega=0$ if $\omega$ is decomposable,
this shows that the restrictions to decomposable $2$-forms of the quadratic
forms determined by $a$ and $\mathcal{P}(a)$ coincide, and, moreover, that if
$Q(a)=0$, then these quadratic forms are equal.

On the other hand, that the quadratic forms determined on
$\mathsf{\Lambda}^{2}\mathbb{A}$ by $a,b\in
S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ coincide on decomposable elements
means that $c=a-b\in S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ satisfies
$4c(x,y,x,y)=\langle c,(x\wedge y)\otimes(x\wedge y)\rangle=0$ for all
$x,y\in\mathbb{A}$. Linearizing in $x$ and $y$ and using the symmetries of $c$
yields

(4.4) $\displaystyle 0$
$\displaystyle=c(x,y,z,w)+c(z,y,x,w)+c(x,w,z,y)+c(z,w,x,y)=2c(x,y,z,w)-2c(y,z,x,w)$

for all $x,y,z,w\in\mathbb{A}$. Hence $c(x,y,z,w)=c(y,z,x,w)=c(z,x,y,w)$, and
in (4.2) this yields $\mathcal{P}(a)-\mathcal{P}(b)=\mathcal{P}(c)=0$. ∎

For a metrized algebra $(\mathbb{A},\circ,h)$, it follows straightforwardly
from the invariance of $h$ that

(4.5) $\displaystyle h([x,y,z],w)=-h([y,z,w],x)=h([z,w,x],y)=-h([w,x,y],z),$
$\displaystyle\text{for all}\,\,x,y,z,w\in\mathbb{A}.$

Define $\mathsf{r}^{\flat}(x,y,z,w)=h(\mathsf{r}(x,y)z,w)$ and
$\bar{\mathsf{r}}^{\flat}(x,y,z,w)=h(\bar{\mathsf{r}}(x,y)z,w)$. From (4.5)
and (3.1) it follows that

(4.6)
$\displaystyle\mathsf{r}^{\flat}(x,y,z,w)=-\bar{\mathsf{r}}^{\flat}(x,y,w,z),$
$\displaystyle\text{for all}\,\,x,y,z,w\in\mathbb{A}.$

By (4.6) and (4.5),

(4.7) $\displaystyle\begin{aligned}
&\mathsf{r}^{\flat}(x,y,z,w)+\bar{\mathsf{r}}^{\flat}(x,y,z,w)=\mathsf{r}^{\flat}(x,y,z,w)-\mathsf{r}^{\flat}(x,y,w,z)\\\
&=\mathsf{r}^{\flat}(z,w,x,y)-\mathsf{r}^{\flat}(z,w,y,x)=\mathsf{r}^{\flat}(z,w,x,y)+\bar{\mathsf{r}}^{\flat}(z,w,x,y)\end{aligned}$

so that $\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat}$ is an element of
$S^{2}\mathsf{\Lambda}^{2}\mathbb{A}^{\ast}$ (as it need not satisfy the
algebraic Bianchi identity, it need not be in $\mathscr{MC}(\mathbb{A})$).
This follows also from the alternative expression

(4.8)
$\displaystyle\begin{split}\mathsf{r}^{\flat}(x,y,z,w)&+\bar{\mathsf{r}}^{\flat}(x,y,z,w)\\\
&=-h([x,y],[z,w])+h(y\circ z,w\circ x)-h(x\circ z,w\circ y)+h(z\circ y,x\circ
w)-h(z\circ x,y\circ w),\end{split}$

which is evidently antisymmetric in $x$ and $y$, and in $z$ and $w$, and
symmetric in the pairs $xy$ and $zw$.

######  Lemma 4.2.

Suppose $\operatorname{char}\mathbb{k}\notdivides 6$. A metrized
$\mathbb{k}$-algebra $(\mathbb{A},\circ,h)$ is Lie-admissible if and only if
$\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat}\in\mathscr{MC}(\mathbb{A})$.

###### Proof.

By (3.7), the tensor $\mathsf{m}$ defined by
$\mathsf{m}^{\flat}=3\mathcal{P}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})\in\mathscr{MC}(\mathbb{A})$
satisfies

(4.9)
$\displaystyle\begin{split}\mathsf{m}(x,y)z&=-2(\mathsf{r}(x,y)z+\bar{\mathsf{r}}(x,y)z)+\mathsf{r}(y,z)x+\bar{\mathsf{r}}(y,z)x+\mathsf{r}(z,x)y+\bar{\mathsf{r}}(z,x)y)\\\
&=-3(\mathsf{r}(x,y)z+\bar{\mathsf{r}}(x,y)z)-2\mathsf{t}(x,y,z),\end{split}$

for all $x,y,z\in\mathbb{A}$. The claim follows because $(\mathbb{A},\circ)$
is Lie-admissible if and only if $\mathsf{t}$ vanishes identically. ∎

Let $(\mathbb{A},\circ,h)$ be a metrized algebra of dimension at least $2$. If
$x,y\in\mathbb{A}$ span a two-dimensional subspace
$\mathbb{V}\subset\mathbb{A}$ and $\bar{x}=ax+by$ and $\bar{y}=cx+dy$ span the
same subspace $\mathbb{V}$, then

(4.10) $\displaystyle
h(\bar{x},\bar{x})h(\bar{y},\bar{y})-h(\bar{x},\bar{y})^{2}=(ad-
bc)^{2}\left(h(x,x)h(y,y)-h(x,y)^{2}\right),$ (4.11) $\displaystyle
h(\bar{x}\circ\bar{x},\bar{y}\circ\bar{y})-h(\bar{x}\circ\bar{y},\bar{x}\circ\bar{y})=(ad-
bc)^{2}\left(h(x\circ x,y\circ y)-h(x\circ y,y\circ x)\right),$

from which it follows that (4.11) divided by (4.10) is well-defined and
depends only the subspace $\mathbb{V}$, provided the denominator is non-zero,
which is the case as long $\mathbb{V}$ is $h$-nondegenerate. (A symmetric
bilinear form $h$ on a two-dimensional $\mathbb{k}$-vector space $\mathbb{V}$
is nondegenerate if and only if $h(x,x)h(y,y)-h(x,y)^{2}\neq 0$ for any basis
$\\{x,y\\}$ of $\mathbb{V}$; this is a special case of the statement that a
symmetric bilinear form is nondegenerate if and only if the determinant of its
Gram matrix with respect to any basis is nonvanishing [36, Proposition
I.$1.2$].) These observations show that Definition 4.1 is well made. By (4.5),

(4.12) $\displaystyle h([x,x,y],y)=-h([y,x,x],y),$ $\displaystyle\text{for
all}\,\,x,y\in\mathbb{A}.$

The second equality in (4.13) is (4.12).

######  Definition 4.1.

Let $(\mathbb{A},\circ,h)$ be a metrized $\mathbb{k}$-algebra of dimension at
least $2$. Define the _sectional nonassociativity_ $\mathscr{K}(x,y)$ of the
$h$-nondegenerate subspace $\operatorname{Span}\\{x,y\\}\subset\mathbb{A}$ to
be

(4.13)
$\displaystyle\begin{split}\mathscr{K}(x,y)&=\mathscr{K}_{\circ,h}(x,y)=-\tfrac{1}{2}\frac{h(\mathsf{r}(x,y)x,y)+h(\bar{\mathsf{r}}(x,y)x,y)}{h(x\circ
x,y\circ y)-h(x,y)^{2}}\\\
&=\frac{h([x,x,y],y)}{h(x,x)h(y,y)-h(x,y)^{2}}=\frac{-h([y,x,x],y)}{h(x,x)h(y,y)-h(x,y)^{2}}=\frac{h(x\circ
x,y\circ y)-h(x\circ y,y\circ x)}{h(x,x)h(y,y)-h(x,y)^{2}}.\end{split}$

The proviso that $\operatorname{Span}\\{x,y\\}$ be $h$-nondegenerate is
automatic when $h$ is anisotropic.

By definition, the sectional nonassociativity is invariant under isometric
algebra isomorphisms of metrized algebras.

The dependence of $\mathscr{K}=\mathscr{K}_{\circ,h}$ on $\circ$ and $h$ is
indicated with subscripts when helpful. For $r,s\in\dot{\mathbb{R}}$,
$\mathscr{K}_{s\circ,rh}=s^{2}r^{-1}\mathscr{K}_{\circ,h}$, where $s\circ$
means the multiplication $sx\circ y$.

Given an algebra $(\mathbb{A},\circ)$, with underlying anticommutative and
commutative products $[\,\cdot\,,\,\cdot\,]$ and $\odot$, a straightforward
computation shows that that $h$ is invariant with respect to both of
$[\,\cdot\,,\,\cdot\,]$ and $\odot$, consequently the sectional
nonassociativities $\mathscr{K}_{[\,\cdot\,,\,\cdot\,],h}(x,y)$ and
$\mathscr{K}_{\odot,h}(x,y)$ are defined. From the identity

(4.14) $\displaystyle\begin{aligned}
{}_{\odot}&+[x,y,z]_{[\,\cdot\,,\,\cdot\,]}=[x,y,z]_{\odot}+[[x,y],z]+[[y,z],x]\\\
&=[x,y,z]_{\odot}-[[z,x],y]+\mathsf{t}_{\circ}(x,y,z)=2[x,y,z]_{\circ}-2[z,y,x]_{\circ},\end{aligned}$

(subscripts indicate the dependence on the different multiplications) together
with (4.12) there follows

(4.15) $\displaystyle
h([x,x,y]_{\odot},y)+h([x,x,y]_{[\,\cdot\,,\,\cdot\,]},y)=4h([x,x,y]_{\circ},y),$

and with (4.13) this yields

(4.16) $\displaystyle\begin{aligned}
4\mathscr{K}_{\circ,h}(x,y)&=\mathscr{K}_{\odot,h}(x,y)+\mathscr{K}_{[\,\cdot\,,\,\cdot\,],h}(x,y)\\\
&=\frac{h(x\odot x,y\odot y)-h(x\odot y,y\odot
x)}{h(x,x)h(y,y)-h(x,y)^{2}}+\frac{h([x,y],[x,y])}{h(x,x)h(y,y)-h(x,y)^{2}}.\end{aligned}$

It follows that for a Euclidean metrized algebra,
$4\mathscr{K}_{\circ,h}(x,y)\geq\mathscr{K}_{\odot,h}(x,y)$ for all linearly
independent $x$ and $y$, with equality if and only if $x$ and $y$ commute.

For $h\in S^{2}\mathbb{A}^{\ast}$, define $h\owedge
h\in\mathscr{MC}(\mathbb{A})$ by

(4.17) $\displaystyle(h\owedge h)(x,y,z,w)=h(x,z)h(y,w)-h(x,w)h(y,z),$
$\displaystyle\text{for all}\,\,x,y,z,w\in\mathbb{A}.$

By definition $(h\owedge h)(x,y,x,y)=|x|^{2}|y|^{2}-h(x,y)^{2}$.

######  Lemma 4.3.

Let $\mathbb{k}$ be an infinite field. A metrized $\mathbb{k}$-algebra
$(\mathbb{A},\circ,h)$ of dimension $n\geq 2$ has constant sectional
nonassociativity $c\in\mathbb{k}$ if and only if the quadratic forms
determined on $\mathsf{\Lambda}^{2}\mathbb{A}$ by
$-\tfrac{1}{2}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})$ and $c(h\owedge
h)$ coincide on decomposable elements of $\mathsf{\Lambda}^{2}\mathbb{A}$.

###### Proof.

That the coincidence of
$-\tfrac{1}{2}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})$ and $c(h\owedge
h)$ on decomposable elements of $\mathsf{\Lambda}^{2}\mathbb{A}$ implies
constant sectional nonassociativity is immediate from the definition (4.13).

The Grassmannian $Gr(2,\mathbb{A})$ of two-dimensional subspaces of
$\mathbb{A}$ is a projective variety via the Plücker embedding
$Gr(2,\mathbb{A})\to\mathbb{P}(\mathsf{\Lambda}^{2}\mathbb{A})$ sending
$\mathbb{k}\\{x,y\\}\in Gr(2,\mathbb{A})$ to the span $\mathbb{k}\\{x\wedge
y\\}$. The homogeneous functions
$H,K:\mathsf{\Lambda}^{2}\mathbb{A}\to\mathbb{k}$ defined by $H(x\wedge
y)=h(x,x)h(y,y)-h(x,y)^{2}$ and $K(x\wedge
y)=-\tfrac{1}{2}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})(x,y,x,y)=h(x\circ
x,y\circ y)-h(x\circ y,y\circ x)$ are quadratic polynomials that restrict to
the image of the Plücker embedding of the Grassmannian $Gr(2,\mathbb{A})$ of
two-dimensional subspaces of $\mathbb{A}$. If $\mathbb{k}$ is infinite, then
the intersection of any two nonempty Zariski open subsets of
$Gr(2,\mathbb{A})$ is nonempty. That the sectional nonassociativity be
constant equal to $c$ is the same as assuming $cH-K$ vanishes on
$Gr(2,\mathbb{A})\setminus\\{H=0\\}$. As the complement
$Gr(2,\mathbb{A})\setminus\\{H=0\\}$ is infinite, if $cH-K$ vanishes on
$Gr(2,\mathbb{A})\setminus\\{H=0\\}$, it vanishes on $Gr(2,\mathbb{A})$. ∎

######  Lemma 4.4.

Let $(\mathbb{A},\circ,h)$ be a metrized algebra over an infinite field of
characteristic not equal to $2$ or $3$.

  1. (1)

$(\mathbb{A},\circ,h)$ has constant sectional nonassociativity $c$ if and only
if $\mathcal{P}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})=2ch\owedge h$.

  2. (2)

If $(\mathbb{A},\circ,h)$ is Lie-admissible, it has constant sectional
nonassociativity $c$ if and only if
$\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat}=2ch\owedge h$.

  3. (3)

If $(\mathbb{A},\circ,h)$ is Lie-admissible and flexible, it has constant
sectional nonassociativity $c$ if and only if
$\mathsf{r}=\bar{\mathsf{r}}=ch\owedge h$.

###### Proof.

By definition of $\mathscr{K}(x,y)$ and Lemma 4.1,

(4.18) $\displaystyle\begin{split}(h\owedge
h)(x,y,x,y)\mathscr{K}(x,y)&=-\tfrac{1}{2}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})(x,y,x,y)=-\tfrac{1}{2}\mathcal{P}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})(x,y,x,y).\end{split}$

If there holds (1), then the last term of (4.18) equals $c(h\owedge
h)(x,y,x,y)$, so (4.18) shows $\mathscr{K}(x,y)=c$. On the other hand, by
Lemma 4.3, that $\mathscr{K}(x,y)=c$ for all linearly independent
$x,y\in\mathbb{A}^{\ast}$ means that the quadratic forms determined on
$\mathsf{\Lambda}^{2}\mathbb{A}$ by
$-\tfrac{1}{2}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})$ and $c(h\owedge
h)$ coincide on decomposable elements of $\mathsf{\Lambda}^{2}\mathbb{A}$. By
Lemma 4.1 this implies (1). If $\circ$ is Lie-admissible, then, by Lemma 4.2,
$\mathcal{P}(\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat})=\mathsf{r}^{\flat}+\bar{\mathsf{r}}^{\flat}$,
so in this case (1) is equivalent to (2). If $\circ$ is moreover flexible,
then $\mathsf{r}=\bar{\mathsf{r}}$, so (2) is equivalent to (3). ∎

######  Example 13.

Evidently the sectional nonassociativity of a metrized associative algebra is
identically zero. More generally, the same is true for an alternative algebra.
An algebra $(\mathbb{A},\circ)$ is _alternative_ if $[x,x,y]=0$ and
$[x,y,y]=0$ for all $x,y\in\mathbb{A}$. By (4.13), the sectional
nonassociativity of a metrized alternative algebra is identically zero. This
can also be seen as a consequence of the theorem of E. Artin (see [58, p.
$18$]) that shows that the subalgebra generated by any two elements of an
alternative algebra is associative.

For example, any Hurwitz algebra is alternative, so has vanishing sectional
nonassociativity.

This observation illustrates the limitations of the analogy between the
associator and the curvature tensor, for the octonions are alternative but not
associative.

In the rest of this section it is supposed the base field is $\mathbb{R}$.

A Euclidean metrized algebra $(\mathbb{A},\circ,h)$ has positive, nonnegative,
zero, etc. sectional nonassociativity if the given qualifier is valid for
every two-dimensional subspace of $\mathbb{A}$.

For a Euclidean metrized algebra, by the Cauchy-Schwarz inequality, the
numbers

(4.19)
$\displaystyle\mathscr{U}(\mathbb{A},\circ,h)=\sup_{x,y\in\mathbb{A}:x\wedge
y\neq 0}\mathscr{K}_{h}(x,y),$
$\displaystyle\mathscr{L}(\mathbb{A},\circ,h)=\inf_{x,y\in\mathbb{A}:x\wedge
y\neq 0}\mathscr{K}_{h}(x,y),$

are finite and invariant under isometric automorphisms of
$(\mathbb{A},\circ,h)$. Their estimation for any well delineated class of
metrized algebras is a basic problem.

Such an estimate is most interesting when the invariant metric $h$ is
determined in some canonical manner by the algebra structure, as is the case
for the Killing form
$\tau_{\circ}(x,y)=\operatorname{tr}L_{\circ}(x)L_{\circ}(y)$ of a Lie or
Malcev algebra or the form $\operatorname{tr}L_{\circ}(x\circ y)$ on a Jordan
algebra.

Lemma 4.5 shows that taking direct sums of Euclidean metrized commutative
algebras preserves conditions such as nonpositive and nonnegative sectional
nonassociativity.

For readability, in the proofs of Lemmas 4.5 and 4.6, there are written
$|x|^{2}=h(x,x)$, $\langle x,y\rangle=h(x,y)$.

######  Lemma 4.5.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra and
suppose that $\mathbb{A}=\mathbb{A}_{1}\oplus\mathbb{A}_{2}$ is a
decomposition into $h$-orthogonal ideals. There hold:

(4.20)
$\displaystyle\mathscr{L}(\mathbb{A},\circ,h)\geq\min\\{\mathscr{L}(\mathbb{A}_{1},\circ,h),\mathscr{L}(\mathbb{A}_{2},\circ,h),0\\},$
$\displaystyle\mathscr{U}(\mathbb{A},\circ,h)\leq\max\\{\mathscr{U}(\mathbb{A}_{1},\circ,h),\mathscr{U}(\mathbb{A}_{2},\circ,h),0\\}.$

In particular, if $(\mathbb{A}_{1},\circ,h)$ and $(\mathbb{A}_{2},\circ,h)$
have nonpositive (resp. nonnegative) sectional nonassociativity, then
$(\mathbb{A},\circ,h)$ has nonpositive (resp. nonnegative) sectional
nonassociativity.

###### Proof.

For any $x_{1},y_{1}\in\mathbb{A}_{2}$ and $x_{2},y_{2}\in\mathbb{A}_{2}$, by
the Cauchy-Schwarz inequality,

(4.21) $\displaystyle\begin{split}|x\wedge y|^{2}&-|x_{1}\wedge
y_{1}|^{2}-|x_{2}\wedge
y_{2}|^{2}=|x_{1}|^{2}|y_{2}|^{2}+|x_{2}|^{2}|y_{1}|^{2}-2\langle
x_{1},y_{1}\rangle\langle x_{2},y_{2}\rangle=\\\
&=\left(|x_{1}||y_{2}|-|x_{2}||y_{1}|\right)^{2}+2\left(|x_{1}||y_{1}||x_{2}||y_{2}|-\langle
x_{1},y_{1}\rangle\langle x_{2},y_{2}\rangle\right)\geq 0,\end{split}$

with equality if and only if $x=x_{1}+x_{2}$ and $y=y_{1}+y_{2}$ are linearly
dependent. Let $x_{i}\in\mathbb{A}_{i}$ be the $h$-orthogonal projection of
$x\in\mathbb{A}$. By (4.21), $r_{i}=\tfrac{|x_{i}\wedge y_{i}|^{2}}{|x\wedge
y|^{2}}$ is contained in $[0,1]$, and, by the orthogonality of the ideals
$\mathbb{A}_{1}$ and $\mathbb{A}_{2}$, it follows that for linearly
independent $x,y\in\mathbb{A}$,

(4.22) $\displaystyle\begin{aligned}
m_{1}r_{1}+m_{2}r_{2}\leq\mathscr{K}_{\mathbb{A},h}(x,y)=\mathscr{K}_{\mathbb{A}_{1},h}(x_{1},y_{1})r_{1}+\mathscr{K}_{\mathbb{A}_{2},h}(x_{2},y_{2})r_{2}\leq
M_{1}r_{1}+M_{2}r_{2}.\end{aligned}$

Applying to (4.22) the inequalities $\min\\{a,b,c\\}\leq
ar_{1}+br_{2}+c(1-r_{1}-r_{2})\leq\max\\{a,b,c\\}$, valid for any
$r_{1},r_{2}\in[0,1]$, with $(a,b,c)=(m_{1},m_{2},0)$ or
$(a,b,c)=(M_{1},M_{2},0)$ yields (4.20). ∎

It is more difficult to relate the sectional nonassociativity of a tensor
product to the sectional nonassociativities of its factors, but in some cases
something can be said.

The tensor product of metrized algebras $(\mathbb{A}_{i},\circ_{i},h_{i})$,
$i\in\\{1,2\\}$, is the vector space $\mathbb{A}_{1}\otimes\mathbb{A}_{2}$
with the product defined by $(a_{1}\otimes b_{1})\circ(a_{2}\otimes
b_{2})=(a_{1}\circ_{1}b_{1})\otimes(a_{1}\circ_{2}b_{2})$ and extending
bilinearly, and the metric defined by $h(a_{1}\otimes b_{1},a_{2}\otimes
b_{2})=h_{1}(a_{1},b_{1})h_{2}(a_{1},b_{2})$ and extending bilinearly.

######  Lemma 4.6.

For $i\in\\{1,2\\}$ let $(\mathbb{A}_{i},\circ_{i},h_{i})$ be Euclidean
metrized commutative algebras. For $a_{i},\bar{a}_{i}\in\mathbb{A}_{i}$, if
$\mathscr{K}_{\mathbb{A}_{1},h_{1}}(a_{1},\bar{a}_{1})$ and
$\mathscr{K}_{\mathbb{A}_{2},h_{2}}(a_{2},\bar{a}_{2})$ are both nonnegative,
then $\mathscr{K}_{\mathbb{A}_{1}\otimes\mathbb{A}_{2},h_{1}\otimes
h_{2}}(a_{1}\otimes a_{2},\bar{a}_{1}\otimes\bar{a}_{2})$ is nonnegative.

###### Proof.

Straightforward computations show

(4.23) $\displaystyle\begin{aligned} |(a_{1}\otimes
a_{2})\wedge(\bar{a}_{1}\otimes\bar{a}_{2})|^{2}&=|a_{1}\wedge\bar{a}_{1}|^{2}|a_{2}\wedge\bar{a}_{2}|^{2}+|a_{2}\wedge\bar{a}_{2}|^{2}\langle
a_{1},\bar{a}_{1}\rangle^{2}+|a_{1}\wedge\bar{a}_{1}|^{2}\langle
a_{2},\bar{a}_{2}\rangle^{2},\\\
\mathscr{K}_{\mathbb{A}_{1}\otimes\mathbb{A}_{2},h_{1}\otimes
h_{2}}(a_{1}\otimes a_{2},\bar{a}_{1}\otimes\bar{a}_{2})&\tfrac{|(a_{1}\otimes
a_{2})\wedge(\bar{a}_{1}\otimes\bar{a}_{2})|^{2}}{|a_{1}\wedge\bar{a}_{1}|^{2}|a_{2}\wedge\bar{a}_{2}|^{2}}=\mathscr{K}_{\mathbb{A}_{1},h_{1}}(a_{1},\bar{a}_{1})\mathscr{K}_{\mathbb{A}_{2},h_{2}}(a_{2},\bar{a}_{2})\\\
&+\tfrac{|a_{1}\circ_{1}\bar{a}_{1}|^{2}}{|a_{1}\wedge\bar{a}_{1}|^{2}}\mathscr{K}_{\mathbb{A}_{2},h_{2}}(a_{2},\bar{a}_{2})+\tfrac{|a_{2}\circ_{2}\bar{a}_{2}|^{2}}{|a_{2}\wedge\bar{a}_{2}|^{2}}\mathscr{K}_{\mathbb{A}_{1},h_{1}}(a_{1},\bar{a}_{1}),\end{aligned}$

from which the claim follows. ∎

##  5\. Examples of algebras with constant sectional nonassociativity

This section presents examples of metrized algebras having constant sectional
nonassociativity.

######  Example 14.

By Example 7 a $2$-step nilpotent Lie algebra satisfies
$\mathsf{r}=\bar{\mathsf{r}}=0$. If such a Lie algebra is equipped with an
invariant metric, then its sectional nonassociativity vanishes identically. In
general a $2$-step nilpotent Lie algebra need not admit an invariant metric.
For example, the Heisenberg Lie algebra of any odd dimension admits no
invariant metric. However, there are $2$-step nilpotent Lie algebras that do
admit invariant metrics. G. Ovando [52] has given some examples. For example
the semidirect product of a Lie algebra with its dual carries a split
signature invariant metric, and if the original Lie algebra is $2$-step
nilpotent, its semidirect product with its dual is $2$-step nilpotent as well.

However, these examples obtained from $2$-step nilpotent real Lie algebras
cannot be Euclidean - the invariance of the metric and the $2$-step nilpotence
of $\mathfrak{g}$ imply that $[\mathfrak{g},\mathfrak{g}]$ is a nontrivial
isotropic subspace.

######  Example 15.

By Example 5 an antiflexible algebra has anti-self-adjoint-curvature, so a
metrized antiflexible algebra has vanishing sectional nonassociativity. The
following example of a three-dimensional antiflexible algebra that is not
power-associative is a modification (by unimportant scalar factors) of one
given by F. Kosier in [34, p. $303$]. Suppose
$\operatorname{char}\mathbb{k}\neq 2$ and equip $\mathbb{k}^{3}$ with the
product $x\circ y=(2x_{1}y_{1}+x_{2}y_{3},2x_{1}y_{2},2x_{3}y_{1})$ where
$x=(x_{1},x_{2},x_{3})$ and $y=(y_{1},y_{2},y_{3})$. The two possible third
powers of $(0,1,1)$ are different, so the algebra is not power-associative.

On $(\mathbb{k}^{3},\circ)$, the metric $h$ defined by
$h(x,y)=x_{1}y_{1}+\tfrac{1}{2}(x_{2}y_{3}+x_{3}y_{2})$ is invariant, for
$h(x\circ
y,z)=2x_{1}y_{1}z_{1}+x_{2}y_{3}z_{1}+x_{1}y_{2}z_{3}+x_{3}y_{1}z_{2}$ is
preserved by cyclic permutations of $x$, $y$, and $z$. A computation shows
$[x,y,z]=(0,2x_{2}y_{3}z_{2},-2x_{3}y_{2}z_{3})$, which is evidently symmetric
in $x$ and $z$. It follows that $h([x,x,y],y)=0$, and $(\mathbb{A},\circ,h)$
has vanishing sectional nonassociativity.

By Example 9, $(\mathbb{A},\circ)$ is Lie admissible as can also be checked
using that its underlying bracket is
$[x,y]=(x_{2}y_{3}-x_{3}y_{2},2(x_{1}y_{2}-x_{2}y_{1}),2(x_{3}y_{1}-x_{1}y_{3}))$.
The underlying Lie algebra has Killing form equal to $8h$ and is isomorphic to
$\mathfrak{sl}(2,\mathbb{k})$ for the elements $p=(1,0,0)$, $q=(0,1,0)$, and
$r=(0,0,1)$ are an $\mathfrak{sl}(2)$-triple satisfying $[p,q]=r$, $[r,p]=2p$,
and $[r,q]=-2q$.

This example can be modified to yield a Euclidean example by taking
$\mathfrak{so}(3,\mathbb{k})$ in place of $\mathfrak{sl}(2,\mathbb{k})$.
Define a product $\star$ on $\mathbb{k}^{3}$ by $x\star
y=(x_{2}y_{3},x_{3}y_{1},x_{1}y_{2})$. The underlying anticommutative bracket
is the usual vector cross product $x\times
y=(x_{2}y_{3}-x_{3}y_{2},x_{3}y_{1}-x_{1}y_{3},x_{1}y_{2}-x_{2}y_{1})$ on
$\mathbb{k}^{3}$, while the underlying commutative product is its symmetric
analogue $x\odot
y=(x_{2}y_{3}+x_{3}y_{2},x_{3}y_{1}+x_{1}y_{3},x_{1}y_{2}+x_{2}y_{1})$ (which
is isomorphic to the $3$-dimensional case of the algebra of Example 17). That
the products $\star$, $\times$, and $\odot$ are invariant with respect to the
metric $g(x,y)=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}$ follows from the cyclic
symmetry of $g(x\star y,z)=x_{1}y_{2}z_{3}+x_{2}y_{3}z_{1}+x_{3}y_{1}z_{2}$ in
$x$, $y$, and $z$. The associators satisfy

(5.1) $\displaystyle\begin{aligned}
{}_{\star}&=((x_{3}z_{3}-x_{2}z_{2})y_{1},(x_{1}z_{1}-x_{3}z_{3})y_{2},(x_{2}z_{2}-x_{1}z_{1})y_{3}),\\\
[x,y,z]_{\times}&=(x\times z)\times y=-[x,y,z]_{\odot},\end{aligned}$

from which it is apparent that $\star$ is antiflexible and it follows that
$g([x,x,y]_{\star},y)=0$, so that the Euclidean metrized algebra
$(\mathbb{k}^{3},\star,g)$ has vanishing sectional nonassociativity, and
$-g([x,x,y]_{\odot},y)=g([x,x,y]_{\times},y)=g(x,x)g(y,y)-g(x,y)^{2}$ so that
the Euclidean metrized algebras $(\mathbb{k}^{3},\odot,g)$ and
$(\mathbb{k}^{3},\times,g)$ have constant sectional nonassociativities $-1$
and $1$, respectively. The last statement is particular cases of Examples 16
and 17.

######  Example 16.

A metrized algebra $(\mathbb{A},\circ,h)$ is a (twofold) _cross product
algebra_ if it satisfies:

  1. (1)

The multiplication $\circ$ is antisymmetric.

  2. (2)

The quantity $h(x\circ y,x\circ y)$ equals the Gram determinant
$h(x,x)h(y,y)-h(x,y)^{2}$.

It follows from (1) and the invariance of $h$ that the product $x\circ y$ is
$h$-orthogonal to $x$ and $y$, $h(x\circ y,x)=0$. In [13] and [9] it is shown
that a real cross product algebra exists only in dimensions $3$ and $7$, in
which case it is determined uniquely up to isomorphism by the signature of
$h$, the only possibilities being the positive definite case, and the case
with negative inertial index one greater than the positive inertial index.

######  Theorem 5.1.

Let $\mathbb{k}$ be an infinite field such that
$\operatorname{char}\mathbb{k}\neq 2$. A metrized algebra
$(\mathbb{A},\circ,h)$ with antisymmetric multiplication is a cross product
algebra if and only if it has constant sectional nonassociativity $1$.

###### Proof.

The antisymmetry of $\circ$ means that (4.13) takes the form (1.2) It follows
that a cross product algebra has constant sectional nonassociativity $1$. On
the other hand, if a metrized algebra $(\mathbb{A},\circ,h)$ with
antisymmetric multiplication has constant sectional nonassociativity $1$ then
$h(x\circ y,x\circ y)=h(x,x)h(y,y)-h(x,y)^{2}$ whenever $x$ and $y$ span an
$h$-nondegenerate subspace. Because $\mathbb{k}$ is infinite, arguing as in
the proof of Lemma 4.3, this holds if and only if $h(x\circ y,x\circ
y)=h(x,x)h(y,y)-h(x,y)^{2}$ for all $x$ and $y$, so that
$(\mathbb{A},\circ,h)$ is a cross product algebra. ∎

######  Corollary 5.1.

Suppose $\operatorname{char}\mathbb{k}\neq 2$. A metrized algebra
$(\mathbb{A},\circ,h)$ with anisotropic $h$ satisfies $h(x\circ y,x\circ
y)=h(x,x)h(y,y)-h(x,y)^{2}$ for all $x,y\in\mathbb{A}$ if and only if
$(\mathbb{A},\circ,h)$ is a cross product algebra.

###### Proof.

To show the forward direction it suffices to show $\circ$ is antisymmetric. If
$h(x\circ y,x\circ y)=h(x,x)h(y,y)-h(x,y)^{2}$ for all $x,y\in\mathbb{A}$,
then $h(x\circ x,x\circ x)=0$ for all $x\in\mathbb{A}$. Because $h$ is
anisotropic, this implies $x\circ x=0$ for all $x\in\mathbb{A}$. Because
$\operatorname{char}\mathbb{k}\neq 2$, this implies $\circ$ is antisymmetric.
The reverse implication is immediate. ∎

Hence a cross product algebra can be _defined_ as a metrized algebra with
antisymmetric multiplication and having sectional nonassociativity $1$ on
every $h$-nondegenerate subspace. Moreover, it is not necessary to suppose
antisymmetry if the metric is anisotropic and
$\operatorname{char}\mathbb{k}\neq 2$.

######  Example 17.

An example of a commutative algebra having constant negative sectional
nonassociativity is the following. Consider $\mathbb{R}^{n}$ with the
coordinatewise product $x\cdot y=\sum_{i=1}^{n}x_{i}y_{i}$. Define
$\ell:\mathbb{R}^{n}\to\mathbb{R}$ by $\ell(x)=\sum_{i=1}^{n}$ and define a
modified product $x\circ y=\tfrac{n+1}{n-1}x\cdot
y-\tfrac{1}{n-1}(\ell(x)y+\ell(y)x)$. Then $(\mathbb{R}^{n},\circ)$ is a
commutative algebra metrized by its Killing form

(5.2)
$\displaystyle\tau_{\circ}(x,y)=\operatorname{tr}L_{\circ}(x)L_{\circ}(y)=\tfrac{n+1}{n-1}\ell(x\cdot
y)-\tfrac{1}{n-1}\ell(x)\ell(y),$

which is positive definite. With this metric, this algebra satisfies the
reverse Norton inequality; in fact it has constant negative sectional
nonassociativity. See [18] for details.

######  Example 18.

This example describes a one-parameter family of pairwise nonisomorphic
$3$-dimensional Euclidean metrized commutative algebras having constant
sectional nonassociativity.

For $\epsilon\in[0,\infty)$, define a commutative algebra
$(\mathbb{U}_{\epsilon},\circ)$ as the $3$-dimensional $\mathbb{R}$-vector
space $\mathbb{U}_{\epsilon}$ with basis $\\{f_{0},f_{1},f_{2}\\}$ and
multiplication $\circ$ given by

(5.3) $\displaystyle\begin{aligned} &f_{0}\circ f_{0}=f_{0},&&f_{0}\circ
f_{1}=(\tfrac{1}{2}-\epsilon)f_{1},&&f_{0}\circ
f_{2}=(\tfrac{1}{2}+\epsilon)f_{2},&\\\ &f_{1}\circ
f_{1}=(\tfrac{1}{2}-\epsilon)f_{0},&&f_{1}\circ f_{2}=0,&&f_{2}\circ
f_{2}=(\tfrac{1}{2}+\epsilon)f_{0}.\end{aligned}$

For $x=x_{0}f_{0}+x_{1}f_{1}+x_{2}f_{2}$ the matrix of $L_{\circ}(x)$ with
respect to the ordered basis $\\{f_{0},f_{1},f_{2}\\}$ is

(5.4)
$\displaystyle\begin{pmatrix}x_{0}&(\tfrac{1}{2}-\epsilon)x_{1}&(\tfrac{1}{2}+\epsilon)x_{2}\\\
(\tfrac{1}{2}-\epsilon)x_{1}&(\tfrac{1}{2}-\epsilon)x_{0}&0\\\
(\tfrac{1}{2}+\epsilon)x_{2}&0&(\tfrac{1}{2}+\epsilon)x_{0}\end{pmatrix},$

A calculation using

(5.5) $\displaystyle\begin{aligned} x\circ
y=&(x_{0}y_{0}+(\tfrac{1}{2}-\epsilon)x_{1}y_{1}+(\tfrac{1}{2}+\epsilon)x_{2}y_{2})f_{0}+(\tfrac{1}{2}-\epsilon)(x_{0}y_{1}+x_{1}y_{0})f_{1}+(\tfrac{1}{2}+\epsilon)(x_{0}y_{2}+x_{2}y_{0})f_{2},\end{aligned}$

shows that the metric $h(x,y)=x_{0}y_{0}+x_{1}y_{1}+x_{2}y_{2}$ satisfies

(5.6) $\displaystyle h(x\circ
y,z)=x_{0}y_{0}z_{0}+(\tfrac{1}{2}-\epsilon)(x_{1}y_{1}z_{0}+x_{1}y_{0}z_{1}+x_{0}y_{1}z_{1})+(\tfrac{1}{2}+\epsilon)(x_{2}y_{2}z_{0}+x_{2}y_{0}z_{2}+x_{0}y_{2}z_{2}).$

Because this is completely symmetric in $x$, $y$, and $z$ it shows $h$ is
invariant with respect to $\circ$. From (5.4) and (5.5) it follows that the
matrix of $\mathscr{A}_{\circ}(x,y)=L_{\circ}(x\circ
y)-L_{\circ}(x)L_{\circ}(y)$ is

(5.7)
$\displaystyle\begin{split}&(\tfrac{1}{4}-\epsilon^{2})\begin{pmatrix}x_{1}y_{1}+x_{2}y_{2}&-x_{0}y_{1}&-x_{0}y_{2}\\\
-x_{1}y_{0}&x_{0}y_{0}+x_{2}y_{2}&-x_{1}y_{2}\\\
-x_{2}y_{0}&-x_{2}y_{1}&x_{0}y_{0}+x_{1}y_{1}\end{pmatrix}=(\tfrac{1}{4}-\epsilon^{2})\left(h(x,y)\operatorname{Id}_{\mathbb{U}_{\epsilon}}-x\otimes
h(y,\,\cdot\,)\right),\end{split}$

Note that
$\operatorname{tr}\mathscr{A}_{\circ}(x,y)=2(1/4-\epsilon^{2})h(x,y)$, so that
$h$ is determined by $\circ$. From (5.7) it follows that
$\mathscr{K}(x,y)=\tfrac{1}{4}-\epsilon^{2}$ for all linearly independent $x$
and $y$ in $\mathbb{U}_{\epsilon}$, so that $(\mathbb{U}_{\epsilon},\circ,h)$
has constant sectional nonassociativity $\tfrac{1}{4}-\epsilon^{2}$.

######  Lemma 5.1.

If $\epsilon\neq 1/2$, the algebra $(\mathbb{U}_{\epsilon},\circ)$ is simple.

###### Proof.

By (5.4) and (5.5),

(5.8) $\displaystyle\det
L_{\circ}(x)=x_{0}(x_{0}^{2}+(\tfrac{1}{2}-\epsilon)x_{1}^{2}+(\tfrac{1}{2}+\epsilon)x_{2}^{2})=\tfrac{1}{4}\operatorname{tr}L_{\circ}(x)\operatorname{tr}L_{\circ}(x\circ
x).$

If $\mathbb{I}\subset\mathbb{U}_{\epsilon}$ is a nontrivial ideal so is its
$h$-orthogonal complement $\mathbb{I}^{\perp}$, so without loss of generality
it can be assumed $\mathbb{I}$ is one-dimensional and generated by
$z\in\mathbb{U}_{\epsilon}$. Consequently there is $\lambda\in\mathbb{k}$ such
that $z\circ z=\lambda z$. Since
$L_{\circ}(z)\mathbb{U}_{\epsilon}\subset\mathbb{I}$, $L_{\circ}(z)$ has rank
at most $1$ and $\operatorname{tr}L_{\circ}(z)=\lambda$. By (5.8), $0=\det
L_{\circ}(z)=(\lambda/4)(\operatorname{tr}L_{\circ}(z))^{2}=\lambda
z_{0}^{2}$. Were $z_{0}\neq 0$, then $\lambda=0$ and $z\circ z=0$. By (5.5),
$z\circ z=0$ if and only if $(1-2\epsilon)z_{1}z_{0}=0$,
$(1+2\epsilon)z_{2}z_{0}=0$, and
$z_{0}^{2}+(1/2-\epsilon)z_{1}^{2}+(1/2+\epsilon)z_{2}^{2}=0$. Because
$2\epsilon\notin\\{\pm 1\\}$, if $z_{0}\neq 0$, the first two equations imply
$z_{1}=0$ and $z_{2}=0$, which in the third equation yield the contradiction
$z_{0}^{2}=0$.

Consequently, $z_{0}=0$ and there are $a,b\in\mathbb{R}$ such that
$z=af_{1}+bf_{2}$ and $\lambda af_{1}+\lambda bf_{2}=\lambda z=z\circ
z=((\tfrac{1}{2}-\epsilon)a^{2}+(\tfrac{1}{2}+\epsilon)b^{2})f_{0}$. Because
$2\epsilon\notin\\{\pm 1\\}$, were $\lambda\neq 0$ this would imply $a=0=b$, a
contradiction. If $\lambda=0$ then
$(\tfrac{1}{2}-\epsilon)a^{2}+(\tfrac{1}{2}+\epsilon)b^{2}=0$. If either of
$a$ or $b$ is $0$ then so is the other. Were $ab\neq 0$, then it would follow
from (5.4) that $L_{\circ}(z)=L_{\circ}(af_{1}+bf_{2})$ has rank $2$,
contradicting that $L_{\circ}(z)$ has rank at most $1$. ∎

######  Lemma 5.2.

If $\epsilon\neq 1/2$, then the nontrivial idempotents and square-zero
elements in $(\mathbb{U}_{\epsilon},\circ)$ have forms indicated in Table 1.

Element $x$ | Condition | Type | $h(x,x)$ |  Spectrum $L_{\circ}(x)$
---|---|---|---|---
$f_{0}$ | none | Idempotent | $1$ | $\\{1,\tfrac{1}{2}(1\pm 2\epsilon)\\}$
$\tfrac{1}{1-2\epsilon}(f_{0}\pm\omega f_{1})$ | $\omega^{2}=\tfrac{4\epsilon}{2\epsilon-1}$ | Idempotent | $\tfrac{1-6\epsilon}{(1-2\epsilon)^{3}}$ | $\\{1,\tfrac{1}{2}\tfrac{1+2\epsilon}{1-2\epsilon}\\}$
$\tfrac{1}{1+2\epsilon}(f_{0}\pm\omega f_{2})$ | $\omega^{2}=\tfrac{4\epsilon}{1+2\epsilon}$ | Idempotent | $\tfrac{1+6\epsilon}{(1+2\epsilon)^{3}}$ | $\\{1,\tfrac{1}{2}\tfrac{1-2\epsilon}{1+2\epsilon}\\}$
$\theta(f_{1}\pm\tfrac{\omega}{1+2\epsilon}f_{2})$ |  $\theta\in\dot{\mathbb{k}}$, $\omega^{2}=4\epsilon^{2}-1$ | Square-zero | $\tfrac{4\epsilon\theta^{2}}{1+2\epsilon}$ | $\\{0,\pm\theta\sigma\\},\sigma^{2}=\epsilon(2\epsilon-1)$
Table 1. Idempotents and square-zero elements of
$(\mathbb{U}_{\epsilon},\circ)$ when $1\neq 1/2$.

###### Proof.

From (5.5) it follows that $x\in\mathbb{U}_{\epsilon}$ is an idempotent if and
only if it solves

(5.9) $\displaystyle x_{0}$
$\displaystyle=x_{0}^{2}+\tfrac{1}{2}(1-2\epsilon)x_{1}^{2}+\tfrac{1}{2}(1+2\epsilon)x_{2}^{2},$
(5.10) $\displaystyle x_{1}$ $\displaystyle=(1-2\epsilon)x_{0}x_{1},$ (5.11)
$\displaystyle x_{2}$ $\displaystyle=(1+2\epsilon)x_{0}x_{2}.$

Combining (5.10) and (5.11) shows
$(1-2\epsilon)x_{0}x_{1}x_{2}=x_{1}x_{2}=(1+2\epsilon)x_{0}x_{1}x_{2}$.
Because $2\epsilon\neq\pm 1$ this implies $x_{0}x_{1}x_{2}=0$. If $x_{0}=0$,
then, by (5.10) and (5.11), there vanish both $x_{1}$ and $x_{2}$, so $x=0$.
Suppose $x_{0}\neq 0$. Then $x_{1}x_{2}=0$. By (5.9) it cannot be that both
$x_{1}$ and $x_{2}$ vanish, so either $x_{2}=0$ or $x_{1}=0$. These two cases
yield the idempotents listed in Table 1, provided there exist solutions to
$\omega^{2}=\tfrac{4\epsilon}{2\epsilon-1}$ and
$\omega^{2}=\tfrac{4\epsilon}{1+2\epsilon}$ in $\mathbb{k}$.

From (5.5) it follows that $x\in\mathbb{U}_{\epsilon}$ is square-zero if and
only if it solves

(5.12) $\displaystyle 0$
$\displaystyle=x_{0}^{2}+\tfrac{1}{2}(1-2\epsilon)x_{1}^{2}+\tfrac{1}{2}(1+2\epsilon)x_{2}^{2},$
(5.13) $\displaystyle 0$ $\displaystyle=(1-2\epsilon)x_{0}x_{1},$ (5.14)
$\displaystyle 0$ $\displaystyle=(1+2\epsilon)x_{0}x_{2}.$

Were $x_{0}\neq 0$, then, by (5.13) and (5.14), there would vanish both
$x_{1}$ and $x_{2}$, but in (5.12) this yields a contradiction, so $x_{0}=0$.
Upon substituting $x_{0}=0$ in (5.12), it follows that the square-zero
elements are as indicated in the last line of Table 1. To see that
$L_{\circ}(x)$ is diagonalizable proceed as follows. If
$(1-2\epsilon)a^{2}+(1+2\epsilon)b^{2}=0$ admits a solution with
$a,b\in\mathbb{k}$, then necessarily $ab\neq 0$, and
$4\epsilon^{2}-1=\left(\tfrac{(1+2\epsilon)b}{a}\right)^{2}$, so there is
$\omega\in\mathbb{k}$ such that $4\epsilon^{2}-1=\omega^{2}$ and
$b=\pm\tfrac{\omega}{1+2\epsilon}a$. In this case write
$z_{\pm}=f_{1}\pm\tfrac{\omega}{1+2\epsilon}f_{2}$. It is straightforward to
check that $z_{+}\circ z_{-}=(1-2\epsilon)f_{0}$, $z_{+}\circ
f_{0}=\tfrac{1}{2}z_{+}-\epsilon z_{-}$, and $z_{-}\circ f_{0}=-\epsilon
z_{+}+\tfrac{1}{2}z_{-}$. If there is $\sigma\in\mathbb{k}$ such that
$\sigma^{2}=\epsilon(2\epsilon-1)$, then $\pm\sigma
f_{0}+\tfrac{1}{2}z_{+}-\epsilon z_{-}$ are eigenvectors of $L_{\circ}(z_{+})$
with eigenvalues $\pm\sigma$ and $\pm\sigma f_{0}-\epsilon
z_{+}+\tfrac{1}{2}z_{-}$ are eigenvectors of $L_{\circ}(z_{-})$ with
eigenvalues $\pm\sigma$. ∎

######  Corollary 5.2.

For $\epsilon,\bar{\epsilon}\in[0,\infty)$, $(\mathbb{U}_{\epsilon},\circ)$
and $(\mathbb{U}_{\bar{\epsilon}},\circ)$ are not isomorphic if
$\epsilon\neq\bar{\epsilon}$.

###### Proof.

The algebra $(\mathbb{U}_{1/2},\circ)$ is associative, isomorphic to a direct
sum of a trivial ideal spanned by $f_{1}$ and a two-dimensional ideal spanned
by $f_{0}$ and $f_{2}$ that is isomorphic to the quadratic algebra
$\mathbb{R}[t]/(t^{2}-1)$. When $\epsilon\neq 1/2$,
$(\mathbb{U}_{\epsilon},\circ)$ is not power associative. For example, for
$x=f_{1}+f_{2}$, $((x\circ x)\circ x)\circ
x=(\tfrac{1}{2}+2\epsilon^{2})f_{0}$ while $((x\circ x)\circ(x\circ
x))=f_{0}$. Consequently, for $\epsilon\neq 1/2$,
$(\mathbb{U}_{\epsilon},\circ)$ is not isomorphic to the associative algebra
$(\mathbb{U}_{1/2},\circ)$.

An algebra isomorphism maps idempotents to idempotents and preserves the
spectra of the corresponding left multiplication endomorphisms. By Lemma 5.2,
the algebra $(\mathbb{U}_{0},\circ)$ has a unique nontrivial idempotent and
only one one-dimensional subalgebra, while for all $\epsilon>0$,
$(\mathbb{U}_{\epsilon},\circ)$ has at least three distinct idempotents, so,
for $\epsilon>0$, $(\mathbb{U}_{\epsilon},\circ)$ is not isomorphic to
$(\mathbb{U}_{0},\circ))$.

Suppose $\epsilon,\bar{\epsilon}\in(0,1/2)\cup(1/2,\infty)$ and
$\phi:\mathbb{U}_{\epsilon}\to\mathbb{U}_{\bar{\epsilon}}$ is an algebra
isomorphism. By Lemma 5.2, for $\omega^{2}=\tfrac{4\epsilon}{1+2\epsilon}$,
$\tfrac{1}{1+2\epsilon}\phi(f_{0}\pm\omega
f_{2})\in\mathbb{Idem}(\mathbb{U}_{\bar{\epsilon}},\circ)$ whose left
multiplication endomorphisms have the same spectrum
$\\{1,1/2,\tfrac{1-2\epsilon}{1+2\epsilon}\\}$. Because
$\tfrac{1-2\epsilon}{1+2\epsilon}\neq 1$, it follows from Lemma 5.2 that
either
$\tfrac{1-2\epsilon}{1+2\epsilon}=\tfrac{1-2\bar{\epsilon}}{1+2\bar{\epsilon}}$
or
$\tfrac{1-2\epsilon}{1+2\epsilon}=\tfrac{1+2\bar{\epsilon}}{1-2\bar{\epsilon}}$.
Because $f(t)=\tfrac{1-2t}{1+2t}$ decreases monotonically on $[0,\infty)$, the
first case implies $\bar{\epsilon}=\epsilon$. The second case implies
$\epsilon+\bar{\epsilon}=0$, which is impossible because $\epsilon$ and
$\bar{\epsilon}$ are both positive. ∎

In summary, $(\mathbb{U}_{\pm\epsilon},\circ,h)$ is a one-parameter family of
pairwise nonisomorphic Euclidean commutative metrized algebras having strictly
positive constant sectional nonassociativity for $\epsilon\in[0,1/2)$ and
strictly negative constant sectional nonassociativity for
$\epsilon\in(1/2,\infty)$ and which are simple when $\epsilon\neq 1/2$.

##  6\. Examples of bounds on sectional nonassociativity

A Euclidean metrized algebra $(\mathbb{A},\circ,h)$ has nonnegative sectional
nonassociativity if and only if

(6.1) $\displaystyle 0\leq h([x,x,y],y)=h(x\circ x,y\circ y)-h(x\circ y,y\circ
x),$ $\displaystyle\text{for all}\,\,x,y\in\mathbb{A}.$

That is, the associator endomorphism
$\mathscr{A}_{\circ}(x,x)=[x,x,\,\cdot\,]$ is nonnegative definite for all
$x\in\mathbb{A}$. For metrized commutative algebras (6.1) is known as the
_Norton inequality_ because S. P. Norton showed that it holds for the Griess
algebra of the monster finite simple group [12, 46].

######  Example 19.

A _vertex operator algebra_ (VOA) over a field $\mathbb{k}$ of characteristic
zero is a $\mathbb{k}$-vector space $\mathbb{V}$ equipped with a linear
_state-field correspondence_
$Y:\mathbb{V}\to(\operatorname{End}\mathbb{V})[[z,z^{-1}]]$ taking values in
endomorphism-valued formal power series and written
$Y(a,z)=\sum_{n\in\mathbb{Z}}a_{(n)}z^{-n-1}\in(\operatorname{End}\mathbb{V})[[z,z^{-1}]]$,
and two distinguished vectors, the _vacuum vector_ $\mathbb{1}\in\mathbb{V}$
and the _conformal vector_ $\omega\in\mathbb{V}$, that together satisfy some
axioms that can be found in [21, 22, 31].

The axioms include that the endomorphisms $L_{m}=\omega_{(m+1)}$ satisfy
$[L_{m},L_{n}]=(m-n)L_{m+n}+\tfrac{m^{3}-m}{12}c\delta_{m+n,0}$ for some
$c\in\mathbb{k}$ called the _central charge_ of the VOA, and that the operator
$L_{0}=\omega_{(1)}$ is semisimple and there is a direct sum decomposition
$\mathbb{V}=\oplus_{n=0}^{\infty}\mathbb{V}^{n}$ where $\mathbb{V}^{n}$ is the
eigenspace of $L_{0}$ with eigenvalue $n$, which is moreover finite-
dimensional for all $n\geq 0$. These force $\mathbb{1}\in\mathbb{V}^{0}$ and
$\mathbb{V}^{i}_{(n)}\mathbb{V}^{j}\subset\mathbb{V}^{i+j-n-1}$.

A VOA is OZ (short for _one-zero_) if $\mathbb{V}^{0}$ is generated by
$\mathbb{1}$ and $\mathbb{V}^{1}=\\{0\\}$. For an OZ VOA, the multiplication
$\star$ defined by $a\star b=a_{(1)}b$ for $a,b\in\mathbb{V}^{2}$ makes
$\mathbb{V}^{2}$ into a commutative algebra and the symmetric bilinear form
$g$ defined by $g(a,b)\mathbb{1}=a_{(3)}b$ for $a,b\in\mathbb{V}^{2}$ is
invariant with respect to $\star$. This is explained in detail in [25, section
$3.5$] and [41, Section $8$]. By definition,
$L_{\star}(\omega)=2\operatorname{Id}_{\mathbb{A}}$ and
$g(\omega,\omega)=c/2$. Hence $e=\omega/2$ is a unit in $\mathbb{V}^{2}$ such
that $g(e,e)=c/8$. The triple $(\mathbb{V}^{2},\star,g)$ is called the _Griess
algebra_ of the OZ VOA.

By a theorem of Miyamoto [43, Theorem $6.3$], the Griess algebra of a real OZ
VOA having a positive definite invariant bilinear form satisfies the Norton
inequality, so has nonnegative sectional nonassociativity.

Because, as mentioned in the introduction, the simple Euclidean Jordan
algebras can be obtained as Griess algebras of VOAs, it follows from the
theorem of Miyamoto that they have nonnegative sectional nonassociativity.
Theorem 6.1 indicates a direct proof (it is not hard) and the characterization
of equality.

Nonpositive sectional nonassociativity is equivalent to the _reverse Norton
inequality_ , which is (6.1) with the inequality reversed. Example 17 gives an
example of algebras satisfying the reverse Norton inequality.

Example 20 shows the sectional nonassociativities of symmetric composition
algebras can take both signs and shows sharp upper and lower bounds for them.

######  Example 20.

A composition algebra is _symmetric_ if the symmetric bilinear form
$h(x,y)=q(x+y)-q(x)-q(y)$ is invariant. Basic results about symmetric
composition algebras are given in [32, Chapter $34$]. Any Hurwitz algebra
$(\mathbb{A},\circ,q,e)$ has an associated _para-Hurwitz_ algebra
$(\mathbb{A},\diamond,q)$ that is a symmetric composition algebra. Its
multiplication is the isotope of $\circ$ determined by the canonical
involution and so defined by $x\diamond y=\bar{x}\circ\bar{y}$. It is not
unital, but the unit $e$ of the original Hurwitz algebra is a distinguished
idempotent for which $L_{\diamond}(e)=R_{\diamond}(e)$ is the canonical
involution of the original Hurwitz algebra.

Symmetric composition algebras have been classified by A. Elduque and H. C.
Myung in [14, 15, 16, 32]; not quite all examples are given by the para-
Hurwitz and forms of what are known as Okubo or Petersen algebras. In
particular, a symmetric composition algebra has dimension in $\\{1,2,4,8\\}$,
it is alternative, it is nonunital if $\dim\mathbb{A}>1$, and it is
commutative if and only if $\dim\mathbb{A}\leq 2$.

######  Lemma 6.1.

Let $(\mathbb{A},\circ,q)$ be a real symmetric composition algebra having
associated bilinear form $h$. If $h$ is positive definite, the sectional
nonassociativity of $(\mathbb{A},\circ,h)$ satisfies $|\mathscr{K}(x,y)|\leq
1$ for all linearly independent $x,y\in\mathbb{A}$. There holds
$\mathscr{K}(x,y)=1$ if and only if $x\circ x$ and $y\circ y$ are linearly
dependent, and there holds $\mathscr{K}(x,y)=-1$ if and only if $x\circ
y=y\circ x$.

###### Proof.

Linearizing the relation $q(x\circ y)=q(x)q(y)$ shows that $h$ satisfies

(6.2) $\displaystyle h(x\circ y,w\circ z)+h(w\circ y,x\circ
z)=h(x,w)h(y,z)=h(y\circ x,z\circ w)+h(z\circ x,y\circ w).$

Taking $z=x$ and $w=y$ in (6.2) yields

(6.3) $\displaystyle h(x\circ y,y\circ x)+h(x\circ x,y\circ y)=h(x,y)^{2},$

while the multiplicativity of $q$ yields $2|x\circ
y|^{2}=|x|^{2}|y|^{2}=2|y\circ x|^{2}$. Together these yield

(6.4) $\displaystyle|x|^{2}|y|^{2}-\langle x,y\rangle^{2}=2|x\circ
y|^{2}-h(x\circ y,y\circ x)-h(x\circ x,y\circ y).$

There results

(6.5) $\displaystyle\begin{split}\mathscr{K}(x,y)&=\tfrac{h(x\circ x,y\circ
y)-h(x\circ y,y\circ x)}{|x|^{2}|y|^{2}-\langle
x,y\rangle^{2}}=-1+\tfrac{2\left(|x\circ y|^{2}-h(x\circ y,y\circ
x)\right)}{|x|^{2}|y|^{2}-\langle x,y\rangle^{2}}=-1+\tfrac{|x\circ y-y\circ
x|^{2}}{|x|^{2}|y|^{2}-\langle x,y\rangle^{2}}\geq-1,\end{split}$

where the second equality follows from $|x\circ y|=|y\circ x|$. Equality holds
if and only if $x\circ y=y\circ x$. On the other hand, (6.3) yields

(6.6) $\displaystyle\begin{split}\mathscr{K}(x,y)&=\tfrac{h(x\circ x,y\circ
y)-h(x\circ y,y\circ x)}{|x|^{2}|y|^{2}-\langle
x,y\rangle^{2}}=\tfrac{2h(x\circ x,y\circ
y)-h(x,y)^{2}}{|x|^{2}|y|^{2}-\langle x,y\rangle^{2}}\leq\tfrac{2|x\circ
x||y\circ y|-h(x,y)^{2}}{|x|^{2}|y|^{2}-\langle x,y\rangle^{2}}=1,\end{split}$

the inequality by the Cauchy-Schwarz inequality, and the last equality because
$2|x\circ x|^{2}_{h}=|x|^{4}$. Equality holds if and only if $x\circ x$ and
$y\circ y$ are linearly dependent. ∎

Lemma 6.1 implies that a para-Hurwitz algebra has sectional nonassociativities
of both signs provided its dimension is at least $4$. For a para-Hurwitz
algebra $(\mathbb{A},\diamond,q)$ with underlying Hurwitz algebra
$(\mathbb{A},\circ,q,e)$, there holds $h(x,e)=0$ if and only if $x\diamond
x=-\tfrac{1}{2}h(x,x)e$ and $x$ is not a multiple of $e$. Consequently, by
Lemma 6.1, there holds $\mathscr{K}_{h,\diamond}(e,x)=-1$ if $h(x,e)=0$ and
$\mathscr{K}_{h,\diamond}(x,y)=1$ if $h(x,e)=0$, $h(y,e)=0$, and $x$ and $y$
are linearly independent. The latter case cannot occur for a two-dimensional
para-Hurwitz algebra, but does occur for para-quaternions or para-octonions.
On the other hand, it follows that a two-dimensional para-Hurwitz algebra has
constant sectional nonassociativity $-1$.

The complex Okubo algebra is $\mathbb{B}=\mathfrak{sl}(3,\mathbb{C})$ equipped
with the multiplication

(6.7) $\displaystyle\begin{aligned} x\star y&=\omega
xy-\omega^{2}yx+\tfrac{\omega-\omega^{2}}{3}h(x,y)I=-\tfrac{1}{2}[x,y]+\tfrac{\omega-\omega^{2}}{2}\left(xy+yx+\tfrac{2}{3}h(x,y)I\right).\end{aligned}$

where juxtaposition indicates the ordinary matrix product, $[x,y]=xy-yx$ is
the usual matrix commutator, $h(x,y)=-\operatorname{tr}(xy)$, and $\omega$ is
a nontrivial cube root of unity (so $\omega+\omega^{2}=-1$). The product (6.7)
or an equivalent product was studied independently by Okubo [47, 49, 51],
Petersson [53], and by Laquer [37] (as one of a family of products on
$\mathfrak{su}(n)$).

From (6.7) it is apparent that $[x,y]_{\star}=-[x,y]$ so that
$(\mathbb{B},\star)$ is Lie-admissible, with underlying Lie algebra isomorphic
to $\mathfrak{sl}(3,\mathbb{C})$ with the opposite of the usual Lie bracket.
The underlying symmetric product $x\odot
y=(\omega-\omega^{2})(xy+yx+\tfrac{2}{3}h(x,y)I)$ is a rescaling of the usual
trace-free Jordan product.

Straightforward computations show $[x,y,z]_{\star}=[y,[z,x]]-h(x,y)z+h(z,y)x$,
so that $(x\star y)\star x=\tfrac{1}{2}h(x,x)y=x\star(y\star x)$, showing
$(\mathbb{B},\star)$ is a symmetric composition algebra. Its sectional
nonassociativity satisfies

(6.8)
$\displaystyle\mathscr{K}_{\mathbb{B},\star,h}(x,y)=\tfrac{h([x,x,y]_{\star},y)}{h(x,x)h(y,y)-h(x,y)^{2}}=\tfrac{h([x,y],[x,y])}{h(x,x)h(y,y)-h(x,y)^{2}}-1.$

The real linear map $\sigma\in\operatorname{End}_{\mathbb{R}}(\mathbb{B})$
defined by $\sigma(x)=-\bar{x}^{t}$ is an order two isometric automorphism of
$(\mathbb{B},\star)$. Its fixed point set is $\mathbb{A}=\mathfrak{su}(3)$,
and so $(\mathbb{A},\star)$ is a subalgebra metrized by the Euclidean inner
product $h(x,y)=\tfrac{1}{2}\operatorname{tr}(\bar{x}^{t}y+\bar{y}^{t}x)$. The
algebra $(\mathbb{A},\star,h)$ is the compact real form of the Okubo algebra.
From (6.8) it follows that $-1\leq\mathscr{K}_{\mathbb{A},\star,h}(x,y)$ with
equality if and only if $[x,y]=0$. As explained in Remark 4, from the
Böttcher-Wenzel-Chern-do Carmo-Kobayashi inequality, Theorem 7.1, it follows
that $|[x,y]|^{2}\leq 2|x|^{2}|y|^{2}$ for $x,y\in\mathfrak{su}(3)$, so, from
(6.8) it follows that $1\geq\mathscr{K}_{\mathbb{A},\star,h}(x,y)$; the case
of equality is characterized in Theorem 7.1.

The remainder of this section is devoted to the formulation and proof of
Theorem 6.1, which gives sharp bounds on sectional nonassociativity of simple
Euclidean Jordan algebras.

Let $\mathbb{M}(n,\mathbb{h})$ denote the $n\times n$ matrices over the
Hurwitz algebra $\mathbb{h}$. The subspace of $n\times n$ Hermitian matrices
over $\mathbb{h}$ is the subspace
$\mathbb{Herm}(n,\mathbb{h})=\\{x\in\mathbb{M}(n,\mathbb{h}):\bar{x}^{t}=x\\}$
comprising the fixed points of the conjugate transpose $x\to\bar{x}^{t}$. The
space $\mathbb{Herm}(n,\mathbb{h})$ is a commutative algebra with unit the
identity matrix, $I$, when equipped with the symmetrized matrix product
$x\star y=\tfrac{1}{2}(xy+yx)$, where juxtaposition, as in the expression
$xy$, indicates the matrix product in $\mathbb{Herm}(n,\mathbb{h})$. The
algebra $(\mathbb{Herm}(n,\mathbb{h}),\star)$ is Jordan if $\mathbb{h}$ is
associative or $\mathbb{h}=\mathbb{O}$ and $n=3$ [30, Section III.$1$, Theorem
$1$]. The algebra $(\mathbb{Herm}(2,\mathbb{h}),\star)$ is isomorphic to the
$(2+\dim\mathbb{h})$-dimensional Jordan algebra of a quadratic form often
called a _spin factor_ ; this follows from the observation that the
$\star$-square of a trace-free element is a multiple of the unit.

The matrix trace $\operatorname{tr}:\mathbb{M}(n,\mathbb{h})\to\mathbb{h}$ is
a $\mathbb{k}$-linear map satisfying
$\operatorname{tr}(\bar{x}^{t})=\operatorname{tr}(\bar{x})=\overline{\operatorname{tr}(x)}$
and so also $\mathsf{t}(\operatorname{tr}(x))=\operatorname{tr}(x+\bar{x})$.
The $\mathbb{k}$-bilinear form $h$ on $\mathbb{M}(n,\mathbb{h})$ defined by
$h(x,y)=\tfrac{1}{2n}\mathsf{t}(\operatorname{tr}(\bar{x}^{t}y))$ is symmetric
(even when $\mathbb{h}$ is noncommutative). Its restriction to
$\mathbb{Herm}(n,\mathbb{h})$ satisfies
$h(x,y)=\tfrac{1}{2n}\mathsf{t}(\operatorname{tr}(xy))=\tfrac{1}{n}\mathsf{t}(\operatorname{tr}(x\star
y))$. The dimension dependent constant factor is a normalization.

The _nucleus_ $\mathscr{N}(\mathbb{A},\circ)$ of an algebra
$(\mathbb{A},\circ)$ is the set of elements $a\in\mathbb{A}$ such that there
vanish the associators $[a,x,y]$, $[x,a,y]$, and $[y,a,x]$ for all
$x,y\in\mathbb{A}$. It is straightforward to see that if $(\mathbb{A},\circ)$
is commutative, then
$\mathscr{N}(\mathbb{A},\circ)=\\{a\in\mathbb{A}:[L_{\circ}(a),L_{\circ}(x)]=0\,\,\text{for
all}\,\,x\in\mathbb{A}\\}$. If a unital Jordan algebra $(\mathbb{A},\circ)$
has nondegenerate trace form $\operatorname{tr}L_{\circ}(x\circ y)$ and
$\sigma\in S^{2}\mathbb{A}^{\ast}$ is an invariant symmetric bilinear form,
then there is $z\in\mathscr{N}(\mathbb{A},\circ)$ such that
$\sigma(x,y)=\operatorname{tr}L_{\circ}((z\circ x)\circ y)$ for all
$x,y\in\mathbb{A}$ [33, Chapter $3$, Theorem $10$].

######  Lemma 6.2.

Suppose $\operatorname{char}\mathbb{k}\notdivides 2n$. Let $\mathbb{h}$ be a
$d$-dimensional Hurwitz $\mathbb{k}$-algebra and let $N=n+dn(n-1)/2$. The
symmetric bilinear form $h$ on $\mathbb{Herm}(n,\mathbb{h})$ defined by
$h(x,y)=\tfrac{1}{2n}\mathsf{t}(\operatorname{tr}(xy))=\tfrac{1}{n}\mathsf{t}(\operatorname{tr}(x\star
y))$ is nondegenerate and invariant and satisfies
$\operatorname{tr}L_{\star}(x,y)=Nh(x,y)$ for all
$x,y\in\mathbb{Herm}(n,\mathbb{h})$.

###### Proof.

A consequence of the involutive invariance of the metric $h$ on $\mathbb{h}$
is that the linear map $\mathsf{t}:\mathbb{h}\to\mathbb{R}$ defined by
$\mathsf{t}(x)=x+\bar{x}$ satisfies

(6.9)
$\displaystyle\mathsf{t}((xy)z)=h((xy)z,e)=h(xy,\bar{z})=h(x,\bar{z}\bar{y})=h(x(yz),e)=\mathsf{t}(x(yz)),$
$\displaystyle\text{for}\,\,x,y,z\in\mathbb{h}.$

From (6.9) it follows that
$\mathsf{t}(\operatorname{tr}((xy)z))=\operatorname{tr}((xy)z+\bar{z}(\bar{y}\bar{x}))=\operatorname{tr}(x(yz)+(\bar{z}\bar{y})\bar{x})=\mathsf{t}(\operatorname{tr}(x(yz))$
for $x,y,z\in\mathbb{M}(n,\mathbb{h})$, and there follows

(6.10) $\displaystyle\begin{split}2nh(x\star
y,z)&=2\mathsf{t}(\operatorname{tr}(x\star
y)z)=\mathsf{t}(\operatorname{tr}((xy)z))+\mathsf{t}(\operatorname{tr}((yx)z))\\\
&=\mathsf{t}(\operatorname{tr}(x(yz)))+\mathsf{t}(\operatorname{tr}(z(yx))=\mathsf{t}(\operatorname{tr}(x(yz)))+\mathsf{t}(\operatorname{tr}((zy)x))\\\
&=\mathsf{t}(\operatorname{tr}(x(yz)))+\mathsf{t}(\operatorname{tr}(x(zy)))=2\mathsf{t}(\operatorname{tr}(x(y\star
z))=2nh(x,y\star z),\end{split}$

showing $h$ is invariant on $\mathbb{Herm}(n,\mathbb{h})$. Using a basis of
$\mathbb{Herm}(n,\mathbb{h})$ as in [30, pp. $125-126$] is straightforward to
check that $h$ is nondegenerate.

From the invariance of $h$ is follows that there is
$z\in\mathscr{N}(\mathbb{Herm}(n,\mathbb{h}),\star)$ such that
$h(x,y)=\operatorname{tr}L_{\star}((z\star x)\star y)$ for all
$x,y\in\mathbb{Herm}(n,\mathbb{h})$. The _center_ of a Jordan algebra is the
subset of its nucleus comprising elements that commute with all other
elements. By a theorem of A. A. Albert [1], the nucleus of a simple Jordan
$\mathbb{k}$-algebra ($\operatorname{char}\mathbb{k}\neq 2$) equals its
center. Consequently, since $(\mathbb{Herm}(n,\mathbb{h}),\star)$ is simple,
its nucleus equals its center. Because $(\mathbb{Herm}(n,\mathbb{h}),\star)$
is central simple, its center is its subalgebra, isomorphic to $\mathbb{k}$,
spanned by its unit [30, Chapter $5$, Section $7$]. It follows that $z$ is a
multiple of the unit, so $h$ is a multiple of
$\operatorname{tr}L_{\star}(\,\cdot\,\star\,\cdot\,)$. Because $h(e,e)=1$ and
$\operatorname{tr}L_{\star}(e\star
e)=\dim\mathbb{Herm}(n,\mathbb{h})=N=n+dn(n-1)/2$ where $d=\dim\mathbb{h}$, it
follows that $\operatorname{tr}L_{\star}(\,\cdot\,\star\,\cdot\,)=Nh$. ∎

The simple Euclidean Jordan algebras are classified; a modern exposition is
[17]. A simple Euclidean Jordan algebra of rank $n\geq 4$ is isomorphic to
$\mathbb{Herm}(n,\mathbb{h})$ for $\mathbb{h}$ an associative Hurwitz algebra
and a simple Euclidean Jordan algebra of rank $n=3$ is isomorphic to
$\mathbb{Herm}(n,\mathbb{h})$ for $\mathbb{h}$ a Hurwitz algebra. The rank $2$
case consists of the spin factor algebras; as mentioned these include
$\mathbb{Herm}(2,\mathbb{h})$.

In the statement of Theorem 6.1, $e_{ij}$ denotes the matrix with a $1$ in the
$ij$ entry and $0$ in other entries.

######  Theorem 6.1.

Let $\mathbb{h}$ be a real Hurwitz algebra. Equip
$(\mathbb{Herm}(n,\mathbb{h}),\star)$ with the invariant metric
$h(x,y)=N^{-1}\operatorname{tr}L_{\star}(x\star
y)=n^{-1}\operatorname{tr}(x\star y)$ where
$N=\dim\mathbb{Herm}(n,\mathbb{h})$ and $n\geq 2$ is the rank of
$\mathbb{Herm}(n,\mathbb{h})$. For linearly independent
$x,y\in\mathbb{Herm}(n,\mathbb{h})$, the sectional nonassociativity of
$(\mathbb{Herm}(n,\mathbb{h}),\star,h)$ satisfies

(6.11) $\displaystyle
0\leq\mathscr{K}(x,y)=\mathscr{K}_{h,\star}(x,y)\leq\tfrac{n}{2},$

for all linearly independent $x,y\in\mathbb{Herm}(n,\mathbb{h})$.

  1. (1)

If $\mathbb{h}$ is associative, equality holds in the lower bound of (6.11) if
and only if $x$ and $y$ commute with respect to the matrix product; in
particular, if $x$ and $x\star x$ are linearly independent, then
$\mathscr{K}(x,x\star x)=0$.

  2. (2)

Equality holds in the upper bound of (6.11) if and only if $x$ and $y$ are
simultaneously equivalent under
$\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}),\star)$ to scalar multiples of
$e_{11}-e_{nn}$ and $e_{1n}+e_{n1}$.

###### Proof.

For $x,y\in\mathbb{M}(n,\mathbb{h})$, let $[x,y]=xy-yx$. By (4.14), the
associators of $\star$ and the matrix product $\cdot$ satisfy

(6.12)
$\displaystyle\begin{split}4[x,y,z]_{\star}&=2[x,y,z]_{\cdot}-2[z,y,x]_{\cdot}+[x,[y,z]]+[z,[x,y]].\end{split}$

If $\mathbb{h}$ is associative, $(\mathbb{M}(n,\mathbb{h}),\cdot)$ is
associative, so (6.12) becomes simply $4[x,y,z]_{\star}=[y,[x,z]]$. Hence

(6.13) $\displaystyle\begin{split}h(x\star x,z\star z)&-h(x\star z,z\star
x)=h([x,x,z]_{\star},z)=\tfrac{1}{4}h([x,[x,z]],z)=\tfrac{1}{4n}\operatorname{tr}([x,[x,z]]\star
y)\\\
&=\tfrac{1}{8n}\operatorname{tr}\left([x,[x,z]]z+z[x,[x,z]]\right)=-\tfrac{1}{4n}\operatorname{tr}[x,z][x,z]=\tfrac{1}{4n}\operatorname{tr}\overline{[x,z]}^{t}[x,z].\end{split}$

When $\mathbb{h}$ is associative, an automorphism of
$(\mathbb{Herm}(n,\mathbb{h}),\star)$ preserves the ordinary matrix product
and commutator bracket on $\mathbb{M}(n,\mathbb{h})$, but I do not know if
this is true when $\mathbb{h}=\mathbb{O}$, so in this case the following
alternative argument is given (it works for associative $\mathbb{h}$ too).

Recall that the principal axis theorem states that every element of
$\mathbb{Herm}(n,\mathbb{h})$ (where $n\leq 3$ if $\mathbb{h}=\mathbb{O}$) is
equivalent via an element of
$\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}),\star)$ to a diagonal matrix.
For $\mathbb{h}\in\\{\mathbb{R},\mathbb{C},\mathbb{H}\\}$ this is well known.
For $\mathbb{h}=\mathbb{O}$ this is [23, Theorem $5.1$]; see also [17, Theorem
V.$2.5$].

By the principal axis theorem there are
$g\in\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}),\star))$ and diagonal
$\lambda\in\mathbb{Herm}(n,\mathbb{h})$ such that $x=g\lambda$. Let
$y=g^{-1}z$. Since $\lambda$ is Hermitian its elements are real, so
$L_{\cdot}(\lambda\cdot\lambda)=L_{\cdot}(\lambda)^{2}$ and
$R_{\cdot}(\lambda\cdot\lambda)=R_{\cdot}(\lambda)^{2}$, and hence
$[\lambda,\lambda,y]_{\cdot}=[y,\lambda,\lambda]_{\cdot}=0$. In (6.12) this
yields

(6.14) $\displaystyle
4[x,x,z]_{\star}=4g[\lambda,\lambda,y]_{\star}=g[\lambda,[\lambda,y]].$

As in (6.13) this yields

(6.15) $\displaystyle\begin{split}h&(x\star x,z\star z)-h(x\star z,z\star
x)=h([x,x,z]_{\star},z)=\tfrac{1}{4}h(g[\lambda,[\lambda,y]],gy)=\tfrac{1}{4}h([\lambda,[\lambda,y]],y)\\\
&=\tfrac{1}{4n}\operatorname{tr}([\lambda,[\lambda,y]]\star
y)=\tfrac{1}{8n}\operatorname{tr}\left([\lambda,[\lambda,y]]y+y[\lambda,[\lambda,y]]\right)=-\tfrac{1}{4n}\operatorname{tr}[\lambda,y][\lambda,y]=\tfrac{1}{4n}\operatorname{tr}\overline{[\lambda,y]}^{t}[\lambda,y],\end{split}$

where, in the case $\mathbb{h}=\mathbb{O}$, the penultimate equality follows
from the invariance $\operatorname{tr}([x,y]z)+\operatorname{tr}(y[x,z])=0$
for $x,y,z\in\mathbb{Herm}(n,\mathbb{h})$ proved in [23, section $4.4$]. By
(6.13) and (6.15), $\mathscr{K}(x,y)\geq 0$. If $\mathbb{h}$ is associative,
by (6.13), equality holds if and only if $[x,y]=0$; in particular, $[x\star
x,x]=[x,x,x]_{\star}=0$ because $\star$ is commutative, so if $x$ and $x\star
x$ are linearly independent, then $\mathscr{K}(x,x\star x)=0$.

Since $\operatorname{tr}\bar{x}^{t}x$ equals the Frobenius norm on
$\mathbb{Herm}(n,\mathbb{h})$, by Lemma 7.2, if $\mathbb{h}$ is associative

(6.16) $\displaystyle h(x\star x,z\star z)-h(x\star z,z\star
x)=\tfrac{1}{4n}\operatorname{tr}\overline{[x,z]}^{t}[x,z]\leq\tfrac{n}{2}\left(|x|^{2}_{h}|z|^{2}_{h}-h(x,z)^{2}\right),$

which shows the upper bound in (6.11). The characterization of the equality
case in Lemma 7.2 yields the characterization of the equality case in the
upper bound of (6.11). If $\mathbb{h}=\mathbb{O}$, then by Lemma 7.2,

(6.17)
$\displaystyle\tfrac{1}{4n}\operatorname{tr}\overline{[\lambda,y]}^{t}[\lambda,y]\leq\tfrac{n}{2}\left(|\lambda|^{2}_{h}|y|^{2}_{h}-h(\lambda,y)^{2}\right)=\tfrac{n}{2}\left(|x|^{2}_{h}|z|^{2}_{h}-h(x,z)^{2}\right),$

(where $n=3$) and in (6.15) this shows the upper bound in (6.11). The
characterization of the equality case in Lemma 7.2 again yields the
characterization of the equality case in the upper bound of (6.11). ∎

##  7\. The Chern-do Carmo-Kobayashi inequality over real Hurwitz algebras

The Frobenius bilinear form
$f(x,y)=\operatorname{Re}\operatorname{tr}\bar{x}^{t}y=\tfrac{1}{2}\operatorname{tr}(\bar{x}^{t}y+\bar{y}^{t}x)$
on $\mathbb{M}(n,\mathbb{h})$ is symmetric [17, proposition V.$2.1$] and
positive definite (this is true for $\mathbb{h}=\mathbb{O}$ even if $n>3$).

######  Lemma 7.1.

Let $\mathbb{h}$ be a real Hurwitz algebra. Let $\mathbb{M}(n,\mathbb{h})$ be
metrized by $f(x,y)=\operatorname{Re}\operatorname{tr}\bar{x}^{t}y$. The
number

(7.1)
$\displaystyle\mathscr{W}=\mathscr{W}(\mathbb{M}(n,\mathbb{h}))=\sup\left\\{\tfrac{|[x,y]|^{2}}{|x|^{2}|y|^{2}}:x,y\in\mathbb{M}(n,\mathbb{h})\setminus\\{0\\}\right\\}$

is finite and

(7.2)
$\displaystyle|[x,y]|^{2}\leq\mathscr{W}(|x|^{2}|y|^{2}-f(x,y)^{2})\leq\mathscr{W}|x|^{2}|y|^{2},$
$\displaystyle\text{for all}\,\,x,y\in\mathbb{M}(n,\mathbb{h}).$

If $|[x,y]|^{2}\leq\mathscr{W}(|x|^{2}|y|^{2}-f(x,y)^{2})$ then $p=|y|x-|x|y$
and $q=|y|x+|x|y$ satisfy $|[p,q]|^{2}=\mathscr{W}|p|^{2}|q|^{2}$.

###### Proof.

The Cauchy-Schwarz inequality implies directly
$\mathscr{W}(\mathbb{M}(n,\mathbb{h}))\leq 4$, which shows
$\mathscr{W}(\mathbb{M}(n,\mathbb{h}))$ is finite but is not optimal. The
following argument from the proof of [65, Theorem $2.2$] shows that the
apparently weaker inequality
$|[x,y]|^{2}\leq\mathscr{W}(\mathbb{M}(n,\mathbb{h}))|x|^{2}|y|^{2}$ implies
the first inequality of (7.2). If either $x$ or $y$ equals $0$ there is
nothing to show, so suppose $x\neq 0$ and $y\neq 0$. Take $p=|y|x-|x|y$ and
$q=|y|x+|x|y$. Then $[p,q]=2|x||y|[x,y]$, and straightforward computation
shows

(7.3)
$\displaystyle\begin{split}|[x,y]|^{2}=\tfrac{1}{4|x|^{2}|y|^{2}}|[p,q]|^{2}\leq\tfrac{1}{2|x|^{2}|y|^{2}}|p|^{2}|q|^{2}=2(|x|^{2}|y|^{2}-f(x,y)^{2}).\end{split}$

If there is equality in the first inequality of (7.2), then (7.3) shows that
$|[p,q]|^{2}=\mathscr{W}|p|^{2}|q|^{2}$. ∎

The Böttcher-Wenzel inequality was first proved for real symmetric matrices,
with a characterization of the equality
$|[x,y]|^{2}_{f}=2|x|_{f}^{2}|y|_{f}^{2}$, as [11, Lemma $1$]. It was proved
for $\mathbb{M}(n,\mathbb{C})$ in full generality by Z. Lu in [38] and as
stated in the proof of [7, Theorem $3.1$]. A nice exposition of the proof of
(7.4), following [4], is in [66, chapter $9$]. For variants and discussion see
[4, 6, 7, 10, 38, 39, 66] among many references.

######  Theorem 7.1 ([7, 11, 38]).

There holds $\mathscr{W}(\mathbb{M}(n,\mathbb{C}))=2$. That is, for the inner
product $f(x,y)=\operatorname{Re}\operatorname{tr}\bar{x}^{t}y$,

(7.4) $\displaystyle|[x,y]|^{2}\leq 2(|x|^{2}|y|^{2}-f(x,y)^{2})\leq
2|x|^{2}|y|^{2},$ $\displaystyle\text{for}\,\,x,y\in\mathbb{M}(n,\mathbb{C}).$

  1. (1)

If either $x,y\in\mathbb{Herm}(n,\mathbb{C})$ or $x,y\in\mathfrak{su}(n)$,
then $|[x,y]|^{2}=2|x|^{2}|y|^{2}$ if and only if $x$ and $y$ are
simultaneously unitarily conjugate to scalar multiples of $e_{11}-e_{nn}$ and
$e_{1n}+e_{n1}$.

  2. (2)

If either $x,y\in\mathbb{Herm}(n,\mathbb{C})$ or $x,y\in\mathfrak{su}(n)$,
then $|[x,y]|^{2}=2(|x|^{2}|y|^{2}-f(x,y)^{2})$ if and only if $x$ and $y$
either are linearly dependent over $\mathbb{C}$ or are simultaneously
unitarily conjugate to scalar multiples of $e_{11}-e_{nn}$ and
$e_{1n}+e_{n1}$.

###### Proof.

Characterizations of equality in (7.4) are given in various references, for
instance [7]. Here these characterizations are needed when
$x,y\in\mathbb{Herm}(n,\mathbb{C})$. Since multiplication by $i$ is an
isometric real linear isomorphism from $\mathfrak{su}(n)$ to
$\mathbb{Herm}(n,\mathbb{C})$, the validity of either of (1) or (2) for $x$
and $y$ both Hermitian or both anti-Hermitian implies its validity for the
other class. The characterization of equality in (1) for Hermitian matrices is
proved as it is in [11] for $\mathbb{Herm}(n,\mathbb{R})$; see the proof of
Lemma 7.2 below. Suppose there holds
$|[x,y]|^{2}=2(|x|^{2}|y|^{2}-f(x,y)^{2})$ in (7.4) for linearly independent
$x,y\in\mathbb{Herm}(n,\mathbb{C})$. Then (7.3) shows that the Hermitian
matrices $p$ and $q$ satisfy $|[p,q]|^{2}=2|p|^{2}|q|^{2}$, so, by (1), are
simultaneously unitarily conjugate to (nonzero) scalar multiples of
$e_{11}-e_{nn}$ and $e_{1n}+e_{n1}$. Since this implies that
$|p|^{2}=|q|^{2}$, from $x=\tfrac{p+q}{2|x|}$ and $y=\tfrac{q-p}{2|y|}$ it
follows that $f(x,y)=0$, so, by (7.3), there holds
$|[x,y]|^{2}=2|x|^{2}|y|^{2}$. By (1), $x$ and $y$ are simultaneously
unitarily conjugate to scalar multiples of $e_{11}-e_{nn}$ and
$e_{1n}+e_{n1}$. This proves (2). ∎

The inequality (7.4) is not valid over $\mathbb{H}$ and $\mathbb{O}$. Taking
$x=iI$ and $y=jI\in\mathbb{M}(n,\mathbb{H})$ shows
$\mathscr{W}(\mathbb{M}(n,\mathbb{h}))\geq 4$ when
$\mathbb{h}\in\\{\mathbb{H},\mathbb{O}\\}$. In [24, Theorem $3.1$] it is shown
that $\mathscr{W}(\mathbb{M}(n,\mathbb{H}))=4$ and the equality case is
characterized. Nonetheless, the proof of (7.4) given in [11] for
$\mathbb{Herm}(n,\mathbb{R})$ adapts for $\mathbb{Herm}(n,\mathbb{h})$ over
any real Hurwitz algebra
$\mathbb{h}\in\\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\\}$. Over
$\mathbb{R}$ or $\mathbb{C}$, because a Hermitian matrix is unitarily
diagonalizable, it suffices to prove the claim in the case one of the matrices
is diagonal. This proof requires the prinicipal axis theorem, which is valid
over $\mathbb{H}$ and $\mathbb{O}$, and one has to check that the
noncommutativity of $\mathbb{H}$ and $\mathbb{O}$ does not effect the rest of
the argument. This works over $\mathbb{H}$, but the reduction via
diagonalization faces an obstacle over $\mathbb{O}$, namely that, for
$x,y\in\mathbb{Herm}(n,\mathbb{h})$, the inequality (7.4) is not manifestly
invariant with respect to the action of
$\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}),\star)$. Over associative
$\mathbb{h}$, $\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}))$ preserves the
commutator of the matrix product on $\mathbb{M}(n,\mathbb{h})$ restricted to
$\mathbb{Herm}(n,\mathbb{h})$, but when $\mathbb{h}=\mathbb{O}$, in which case
$\operatorname{Aut}(\mathbb{Herm}(3,\mathbb{O}))$ is the compact form of the
simple real Lie group of type $F_{4}$ [58], it is not clear to the author
whether this is true. However the partial result, that (7.4) holds when $x$ is
diagonal, is true, and this suffices for the application to sectional
nonassociativity in the proof of Theorem 6.1.

######  Lemma 7.2.

Let $\mathbb{h}$ be a real Hurwitz algebra. Let
$f(x,y)=\operatorname{Re}\operatorname{tr}\bar{x}^{t}y$ on
$\mathbb{M}(n,\mathbb{h})$.

  1. (1)

If $\mathbb{h}$ is associative, for all $x,y\in\mathbb{Herm}(n,\mathbb{h})$
there holds

(7.5) $\displaystyle|[x,y]|^{2}\leq 2(|x|^{2}|y|^{2}-f(x,y)^{2})\leq
2|x|^{2}|y|^{2}.$

  2. (2)

If $\mathbb{h}=\mathbb{O}$ and $n=3$, (7.5) holds for all
$y\in\mathbb{Herm}(3,\mathbb{O})$ and all _diagonal_
$x\in\mathbb{Herm}(3,\mathbb{O})$.

  3. (3)

The equality $|[x,y]|^{2}=2|x|^{2}|y|^{2}$ holds in (7.5) (where, if
$\mathbb{h}=\mathbb{O}$ and $n=3$, $x$ is assumed diagonal) if and only if $x$
and $y$ are simultaneously equivalent via an automorphism of the Jordan
algebra $(\mathbb{Herm}(n,\mathbb{h}),\star)$ to real multiples of
$e_{11}-e_{nn}$ and $e_{1n}+e_{n1}$.

  4. (4)

The equality $|[x,y]|^{2}=2(|x|^{2}|y|^{2}-f(x,y)^{2})$ holds in (7.5) (where,
if $\mathbb{h}=\mathbb{O}$ and $n=3$, $x$ is assumed diagonal) if and only if
$x$ and $y$ are either linearly dependent over $\mathbb{h}$ or are
simultaneously equivalent via an automorphism of the Jordan algebra
$(\mathbb{Herm}(n,\mathbb{h}),\star)$ to real multiples of $e_{11}-e_{nn}$ and
$e_{1n}+e_{n1}$.

###### Proof.

Suppose $x\in\mathbb{Herm}(n,\mathbb{h})$ is diagonal, where $n=3$ if
$\mathbb{h}=\mathbb{O}$. Since $x$ is Hermitian, its entries are real. Let
$x_{i}=x_{ii}$. Then $[x,y]_{ij}=x_{i}y_{ij}-y_{ij}x_{j}=(x_{i}-x_{j})y_{ij}$,
the last equality because $x_{j}$ is real, so in the center of $\mathbb{h}$.
Now the proof goes through as in [11]:

(7.6)
$\displaystyle\begin{split}|[x,y]|^{2}&=\sum_{i,j=1}^{n}(x_{i}-x_{j})^{2}|y_{ij}|^{2}\leq
2\sum_{i,j=1}^{n}(x_{i}^{2}+x_{j}^{2})|y_{ij}|^{2}\leq
2\sum_{k=1}^{n}x_{k}^{2}\sum_{i,j=1}^{n}|y_{ij}|^{2}=2|x|^{2}|y|^{2}.\end{split}$

This proves $|[x,y]|^{2}\leq 2|x|^{2}|y|^{2}$ for diagonal $x$. By the
principal axis theorem, any element of $(\mathbb{Herm}(n,\mathbb{h}),\star)$
is equivalent via an automorphism of $(\mathbb{Herm}(n,\mathbb{h}),\star)$ to
a diagonal matrix. Since an automorphism of
$(\mathbb{Herm}(n,\mathbb{h}),\star)$ preserves the trace, it is isometric,
and, when $\mathbb{h}$ is associative it preserves the commutator of the
matrix product on $\mathbb{M}(n,\mathbb{h})$ restricted to
$\mathbb{Herm}(n,\mathbb{h})$, when $\mathbb{h}$ is associative to prove
$|[x,y]|^{2}\leq 2|x|^{2}|y|^{2}$ in general it suffices to prove it when $x$
is diagonal. By Lemma 7.1, in either case $|[x,y]|^{2}\leq 2|x|^{2}|y|^{2}$
implies (7.5).

The characterization of the equality case (3) goes through as in the proof of
[11, Lemma $1$]. Precisely, if there is equality in (7.6), then

(7.7) $\displaystyle 0=4\sum_{i=1}^{n}x_{i}^{2}|y_{ii}|^{2}+2\sum_{i\neq
j}(x_{i}+x_{j})^{2}|y_{ij}|^{2}=2\sum_{k=1}^{n}x_{k}^{2}\sum_{i=1}^{n}|y_{ii}|^{2}+2\sum_{i\neq
j}(x_{i}+x_{j})^{2}|y_{ij}|^{2}.$

If $x$ is nonzero, this implies $y_{ii}=0$ for $1\leq i\leq n$,
$x_{i}+x_{j}=0$ when $y_{ij}\neq 0$ for $i\neq j$, and that at exactly two of
$x_{1},\dots,x_{n}$ are nonzero. Suppose $i\neq j$ are the indices such that
$x_{i}=-x_{j}\neq 0$. It follows that $y_{kl}=0$ if $\\{k,l\\}\neq\\{i,j\\}$.
Hence $x$ and $y$ are scalar multiples of $e_{ii}-e_{jj}$ and $e_{ij}+e_{ji}$,
respectively. Since $\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}))$ contains
a subgroup acting as permutations on the diagonal subalgebra of
$\mathbb{Herm}(n,\mathbb{h})$, it can be assumed that $i=1$ and $j=n$. The
proof of the equality case (4) follows from (3) exactly as in the proof of (2)
of Theorem 7.1. ∎

######  Remark 2.

Although the same difficulty related to diagonalization occurs in the proof of
Theorem 6.1 as occurs in the proof of Lemma 7.2, the end result in Theorem 6.1
is stated in terms manifestly invariant with respect to
$\operatorname{Aut}(\mathbb{Herm}(n,\mathbb{h}))$, and as a result takes the
same form whether or not $\mathbb{h}$ is associative. This observation
suggests that the upper bound on sectional nonassociativity, which is closely
related to the commutator bound, is the more natural bound to consider, at
least from the algebraic point of view, although it could also simply reflect
a technical deficiency in the proof of Lemma 7.2.

######  Remark 3.

It would be useful to know whether (7.5) is true for
$\mathbb{Herm}(3,\mathbb{O})$. This seems likely. More, generally, it would be
interesting to know whether (7.5) is true for $\mathbb{Herm}(n,\mathbb{O})$
for $n>3$. This would follow from a principal axis theorem and a
characterization of the automorphism group, but it appears that neither is
known. See [68] for what is known in this regard.

######  Remark 4.

For a compact semisimple real Lie algebra $\mathfrak{g}$ with Killing form
$B_{\mathfrak{g}}$ the number

(7.8)
$\displaystyle\mathscr{W}(\mathfrak{g})=\mathscr{U}(\mathfrak{g},[\,\cdot\,,\,\cdot\,],-B_{\mathfrak{g}})=\sup_{x,y\in\mathfrak{g}:x\wedge
y\neq
0}\frac{-B_{\mathfrak{g}}([x,y],[x,y])}{B_{\mathfrak{g}}(x,x)B_{\mathfrak{g}}(y,y)-B_{g}(x,y)^{2}}$

is positive and finite, by the Cauchy-Schwarz inequality. Its value is a basic
automorphism invariant of $(\mathfrak{g},[\,\cdot\,,\,\cdot\,])$. Its
calculation for particular $\mathfrak{g}$ can be viewed as a refinement of the
Böttcher-Wenzel inequality. On the other hand, for some compact simple real
Lie algebras $\mathscr{W}(\mathfrak{g})$ can be estimated using the Böttcher-
Wenzel inequality. For $\mathfrak{su}(n)$ and $\mathfrak{so}(n)$,
$B_{\mathfrak{su}(n)}(x,y)=-2nf(x,y)$ and
$B_{\mathfrak{so}(n)}(x,y)=-(n-2)f(x,y)$ where
$f(x,y)=\operatorname{tr}\bar{x}^{t}y$, so in these cases it follows from the
Böttcher-Wenzel inequality, Theorem 7.1, that

(7.9) $\displaystyle\mathscr{W}(\mathfrak{so}(n))\leq\tfrac{2}{n-2},$
$\displaystyle\mathscr{W}(\mathfrak{su}(n))\leq\tfrac{1}{n}.$

For the special case of $\mathfrak{so}(n)$, viewed as antisymmetric matrices,
equipped with the Frobenius norm $f$, the essentially equivalent quantity
$\sup_{x,y\in\mathfrak{g}:x\wedge y\neq 0}\tfrac{|[x,y]|_{f}}{|x|_{f}|y|_{f}}$
was shown to equal $\sqrt{2}$ if $n\geq 4$ in [5, Theorem $6$]. By Lemma 7.3,
these estimates yield sharp numerical bounds on the sectional
nonassociativites of the subspaces spanned by decomposable elements in
algebras such as $\mathfrak{so}(m)\otimes\mathfrak{so}(n)$,
$\mathfrak{so}(m)\otimes\mathfrak{su}(n)$, and
$\mathfrak{su}(m)\otimes\mathfrak{su}(n)$.

Lemma 7.3 shows a partial lower bound on the sectional nonassociativities of
the tensor product of Lie algebras.

######  Lemma 7.3.

For compact semisimple real Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$,
let $\tau_{\circ}$ be the Killing form of the tensor product algebra
$(\mathfrak{g}\otimes\mathfrak{h},\circ)$, defined by
$\tau_{\circ}(a,b)=\operatorname{tr}L_{\circ}(a)L_{\circ}(b)$ for
$a,b\in\mathfrak{g}\otimes\mathfrak{h}$. The sectional nonassociativity of the
subspace of $(\mathfrak{g}\otimes\mathfrak{h},\circ,\tau_{\circ})$ spanned by
decomposable elements $a_{1}\otimes b_{1},a_{2}\otimes
b_{2}\in\mathfrak{g}\otimes\mathfrak{h}$ satisfies

(7.10) $\displaystyle 0\geq\mathscr{K}(a_{1}\otimes b_{1},a_{2}\otimes
b_{2})\geq-\mathscr{W}(\mathfrak{g})\mathscr{W}(\mathfrak{h}).$

###### Proof.

Because $\mathfrak{g}$ and $\mathfrak{h}$ are compact, their Killing forms,
$B_{\mathfrak{g}}$ and $B_{\mathfrak{h}}$, satisfy
$-B_{\mathfrak{g}}([a_{1},a_{2}],[a_{1},a_{2}])\geq 0$ and
$-B_{\mathfrak{h}}([b_{1},b_{2}],[b_{1},b_{2}])\geq 0$. The upper bound in
(7.10) follows from

(7.11) $\displaystyle\begin{split}\mathscr{K}(a_{1}\otimes b_{1},a_{2}\otimes
b_{2})&=-\tfrac{B_{\mathfrak{g}}([a_{1},a_{2}],[a_{1},a_{2}])B_{\mathfrak{h}}([b_{1},b_{2}],[b_{1},b_{2}])}{B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}B_{\mathfrak{h}}(b_{1},b_{2})^{2}}\leq
0,\end{split}$

where the positivity of the denominator follows from the Cauchy-Schwarz
inequality. On the other hand, by definition of $\mathscr{W}(\mathfrak{g})$
and $\mathscr{W}(\mathfrak{h})$,

(7.12) $\displaystyle\begin{split}-&\tfrac{\mathscr{K}(a_{1}\otimes
b_{1},a_{2}\otimes
b_{2})}{\mathscr{W}(\mathfrak{g})\mathscr{W}(\mathfrak{h})}=\tfrac{1}{\mathscr{W}(\mathfrak{g})\mathscr{W}(\mathfrak{h})}\tfrac{\left(-B_{\mathfrak{g}}([a_{1},a_{2}],[a_{1},a_{2}])\right)\left(-B_{\mathfrak{h}}([b_{1},b_{2}],[b_{1},b_{2}])\right)}{B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}B_{\mathfrak{h}}(b_{1},b_{2})^{2}}\\\
&\leq\tfrac{\left(B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}\right)\left(B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{h}}(b_{1},b_{2})^{2}\right)}{B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}B_{\mathfrak{h}}(b_{1},b_{2})^{2}}\\\
&=1-\tfrac{B_{\mathfrak{g}}(a_{1},a_{2})^{2}\left(B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{h}}(b_{1},b_{2})^{2}\right)+B_{\mathfrak{h}}(b_{1},b_{2})^{2}\left(B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}\right)}{B_{\mathfrak{g}}(a_{1},a_{1})B_{\mathfrak{g}}(a_{2},a_{2})B_{\mathfrak{h}}(b_{1},b_{1})B_{\mathfrak{h}}(b_{2},b_{2})-B_{\mathfrak{g}}(a_{1},a_{2})^{2}B_{\mathfrak{h}}(b_{1},b_{2})^{2}}\leq
1,\end{split}$

which shows the lower bound in (7.10). ∎

Note that Lemma 7.3 does not imply that $\mathfrak{g}\otimes\mathfrak{h}$ has
nonpositive sectional nonassociativity. In fact, for any compact simple real
Lie algebra $\mathfrak{g}$, $\mathfrak{so}(3)\otimes\mathfrak{g}$ has
sectional nonassociativities of both signs [18].

######  Remark 5.

Although this does not seem to be realized widely, for anti-Hermitian matrices
the inequality (7.4) is closely related to Vinberg’s results on invariant
norms on compact simple Lie algebras in [63] applied in the special case of
$\mathfrak{su}(n)$. Precisely, Vinberg shows that, for an invariant norm,
$||\,\cdot\,||$ on a compact simple Lie algebra $\mathfrak{g}$ the quantity
$\theta(x)=\sup_{0\neq y\in\mathfrak{g}}\tfrac{||[x,y]||}{||y||}$ does not
depend on the choice of invariant norm, and equals the spectral norm of
$\operatorname{ad}_{\mathfrak{g}}(x)$. Since, by definition,
$\theta([x,y])\leq\theta(x)\theta(y)$, taking $\mathfrak{g}$ to be a compact
simple Lie algebra of matrices, e.g. antisymmetric or anti-Hermitian matrices,
and taking $||\,\cdot\,||$ to be the Frobenius norm there results
$|[x,y]|_{f}\leq\theta(x)|y|_{f}$. It suffices then to check that the spectral
norm of $\operatorname{ad}_{\mathfrak{g}}(x)$ is no greater than
$\sqrt{2}|x|_{f}$. In the case $\mathfrak{g}=\mathfrak{su}(n)$ this can be
shown as follows. Since any anti-Hermitian matrix is unitarily conjugate to a
diagonal matrix and the spectral norms of the adjoint representations of
unitarily conjugate matrices are the same, it suffices to consider the case of
diagonal $x\in\mathfrak{su}(n)$. In this case the nonzero eigenvalues of
$\operatorname{ad}_{\mathfrak{su}(n)}(x)$ have the form
$\lambda_{i}-\lambda_{j}$ where $\lambda_{1},\dots,\lambda_{n}$ are the
diagonal elements of $x$. Since $|\lambda_{i}-\lambda_{j}|^{2}\leq
2|\lambda|^{2}\leq 2|x|^{2}_{f}$, where $\lambda$ is the element of $x$ with
the greatest modulus, the spectral norm of
$\operatorname{ad}_{\mathfrak{su}(n)}(x)$ is no greater than
$\sqrt{2}|x|_{f}$.

##  8\. Consequences for commutative algebras of conditions on sectional
nonassociativity

This section describes some general structural consequences for commutative
algebras of assumptions on the sectional nonassociativity.

Consequences of nonnegative sectional nonassociativity for commutative
algebras can be found in [12, section $17$] and in [42], where they are
applied to the study of maximal associative subalgebras of the Griess algebra.
By the analogy advocated here, associative subalgebras are analogous to flat
submanifolds of a nonnegatively curved Riemannian manifold. Here such results
are cast in a somewhat more general setting.

An element $e\in\mathbb{A}$ is _idempotent_ if $e\circ e=e$ and _square-zero_
if $e\circ e=0$. Let $\mathbb{Idem}(\mathbb{A},\circ)=\\{0\neq
e\in\mathbb{A}:e\circ e=e\\}$ and $\mathbb{Nil}_{2}(\mathbb{A},\circ)=\\{0\neq
z\in\mathbb{A}:z\circ z=0\\}$.

######  Lemma 8.1.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra.

  1. (1)

If $a+ib\in\mathbb{Nil}_{2}(\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C},\circ)$
and the $\mathbb{R}$-span of $a$ and $b$ is two-dimensional, then

(8.1) $\displaystyle\mathscr{K}(a,b)=\tfrac{|a\circ a|^{2}_{h}}{|a\wedge
b|^{2}}\geq 0,$

with equality if and only if $a$ and $b$ span a trivial subalgebra of
$(\mathbb{A},\circ)$.

  2. (2)

If $a+ib\in\mathbb{Idem}(\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C},\circ)$ and
the $\mathbb{R}$-span of $a$ and $b$ is two-dimensional, then

(8.2) $\displaystyle\mathscr{K}(a,b)=\tfrac{1}{4}\tfrac{4|b\circ
b|^{2}_{h}+|b|^{2}_{h}}{|a\wedge b|^{2}}>0.$

###### Proof.

That $a+ib\in\mathbb{Nil}_{2}(\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C},\circ)$
is equivalent to the equations $a\circ a=b\circ b$ and $a\circ b=0$, and in
(4.13) these yield (8.1). If equality holds in (8.1), then $b\circ b=a\circ
a=0$, so $a$ and $b$ generate a trivial subalgebra. Suppose
$a+ib\in\mathbb{Idem}(\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C},\circ)$, so
that $a\circ a-b\circ b=a$ and $2a\circ b=b$. Then

(8.3) $\displaystyle\begin{split}h(a\circ a,b\circ b)&-h(a\circ b,a\circ
b)=h(b\circ b+a,b\circ b)=|b\circ b|^{2}_{h}+h(b,a\circ
b)-\tfrac{1}{4}|b|^{2}=|b\circ b|^{2}_{h}+\tfrac{1}{4}|b|^{2}_{h},\end{split}$

and in (4.13) this yields (8.2). In (8.2), equality cannot hold because it
would imply $b=0$, contrary to hypothesis. ∎

######  Theorem 8.1.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra.

  1. (1)

If $(\mathbb{A},\circ,h)$ has negative sectional nonassociativity, then
$\operatorname{Aut}(\mathbb{A},\circ,h)$ is finite.

  2. (2)

If $(\mathbb{A},\circ,h)$ has nonpositive sectional nonassociativity, either
$\operatorname{Aut}(\mathbb{A},\circ,h)$ is finite or $(\mathbb{A},\circ,h)$
contains a trivial subalgebra of dimension $1$ or $2$ (the possibilities are
not exclusive).

###### Proof.

For $k\geq 3$, let $\theta$ be a multilinear $k$-form on a complex vector
space $\mathbb{V}$. By a theorem of H. Suzuki [61, Theorem B], either the
group of linear automorphisms of $\theta$ is finite or there is a nonzero
vector $v\in\mathbb{V}$ such that $\theta(v,\dots,v,w)=0$ for all
$w\in\mathbb{V}$. Regard $\mu(x,y,z)=h(x\circ y,z)$ as a trilinear form on
$\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C}$. An orthogonal automorphism of
$(\mathbb{A},\circ,h)$ preserves $\mu$, and extends to an automorphism of
$\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C}$. By Suzuki’s theorem if
$\operatorname{Aut}(\mathbb{A},\circ,h)$ is not finite there exists
$a+ib\in\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C}$ such that
$h((a+ib)\circ(a+ib),w)=0$ for all
$w\in\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C}$. By the nondegeneracy of $h$,
$a+ib\in\mathbb{Nil}_{2}(\mathbb{A}\otimes_{\mathbb{R}}\mathbb{C},\circ)$.
Note that it is not asserted that $a$ and $b$ are linearly independent over
$\mathbb{R}$. The conclusion follows from (1) of Lemma 8.1. ∎

For a Euclidean metrized commutative algebra $(\mathbb{A},\circ,h)$, the
endomorphism $L_{\circ}(e)$ is $h$-self-adjoint for any $e\in\mathbb{A}$. If
$e\in\mathbb{Idem}(\mathbb{A},\circ)$, then $L_{\circ}(e)$ preserves $\langle
e\rangle^{\perp}=\\{y\in\mathbb{A}:h(e,y)=0\\}$, for if $h(e,y)=0$ then
$h(L_{\circ}(e)y,e)=h(y,e\circ e)=h(y,e)=0$. Define the _orthogonal spectrum_

(8.4)
$\displaystyle\operatorname{Spec}^{\perp}(e)=\\{\lambda\in\mathbb{k}:\text{there
is}\,\,x\in\mathbb{k}\\{e\\}^{\perp}\,\,\text{such
that}\,\,L_{\circ}(e)x=\lambda x\\},$

so that $\operatorname{Spec}(e)=\operatorname{Spec}^{\perp}(e)\cup\\{1\\}$.
Because $L_{\circ}(e)$ preserves $\langle e\rangle^{\perp}$ it has an
eigenvector in $\langle e\rangle^{\perp}$. Because $h(e,e)>0$, this
eigenvector is not a multiple of $e$, so in this case
$\operatorname{Spec}^{\perp}(e)$ is not empty.

######  Lemma 8.2.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra of
dimension $n\geq 2$. If $0\neq e\in\mathbb{Idem}(\mathbb{A},\circ)$, then
$4\mathscr{K}(e,x)\leq|e|_{h}^{-2}$ for all $x\in\mathbb{A}$ such that
$x\wedge e\neq 0$, with equality if and only if
$x-h(e,e)^{-1}h(e,x)e\in\ker(L_{\circ}(e)-\tfrac{1}{2}\operatorname{Id}_{\mathbb{A}})$.

###### Proof.

Let $e\in\mathbb{Idem}(\mathbb{A},\circ)$. Because $\mathscr{K}(e,x)$ depends
only on the subspace spanned by $e$ and $x$, in computing $\mathscr{K}(e,x)$
it may be supposed that $h(e,x)=0$. Write $x=\sum_{i=1}^{n-1}x_{i}$ where
$x_{1},\dots,x_{n-1}\in\langle e\rangle^{\perp}$ are $h$-orthogonal
eigenvectors of $L_{\circ}(e)$ with (not necessarily distinct) eigenvalues
$\lambda_{1},\dots,\lambda_{n-1}$. There results

(8.5) $\displaystyle\begin{split}\mathscr{K}(e,x)&=\frac{h(e\circ e,x\circ
x)-h(e\circ x,e\circ x)}{|e|^{2}_{h}|x|^{2}_{h}}=\frac{h(e\circ x,x)-h(e\circ
x,e\circ
x)}{|e|^{2}_{h}|x|^{2}_{h}}=|e|^{-2}_{h}\sum_{i=1}^{n-1}\lambda_{i}(1-\lambda_{i})\frac{|x_{i}|^{2}}{|x|^{2}_{h}}.\end{split}$

Because $\sum_{i=1}^{n-1}|x_{i}|^{2}=|x|^{2}$, (8.5) exhibits
$\mathscr{K}(e,x)$ as a convex combination, so implies

(8.6)
$\displaystyle\begin{split}|e|^{-2}_{h}\min\\{\lambda_{i}(1-\lambda_{i}):1\leq
i\leq
n-1\\}\leq\mathscr{K}(e,x)\leq|e|^{-2}_{h}\max\\{\lambda_{i}(1-\lambda_{i}):1\leq
i\leq n-1\\}.\end{split}$

As $\lambda(1-\lambda)\leq 1/4$ for all $\lambda\in\mathbb{R}$, (8.6) yields
$4\mathscr{K}(e,x)\leq|e|_{h}^{-2}$. There holds equality if and only if
$\lambda_{i}=1/2$ for $1\leq i\leq n-1$, in which case
$x=\sum_{i=1}^{n-1}x_{i}$ is an eigenvector of $L_{\circ}(e)$ with eigenvalue
$1/2$. ∎

######  Lemma 8.3.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra of
dimension $n\geq 2$. Let $0\neq e\in\mathbb{Idem}(\mathbb{A},\circ)$.

  1. (1)

If there is $b\in\mathbb{R}$ such that $\mathscr{K}(\mathbb{U})\leq b$ for all
two-dimensional subspaces $\mathbb{U}\subset\mathbb{A}$ containing $e$, then
$\operatorname{Spec}^{\perp}(e)\subset(-\infty,\tfrac{1-\sqrt{B}}{2}]\cup[\tfrac{1+\sqrt{B}}{2},\infty)$,
where $B=\max\\{1-4b|e|_{h}^{2},0\\}$. In particular, if
$(\mathbb{A},\circ,h)$ has nonpositive sectional nonassociativity, then
$\operatorname{Spec}^{\perp}(e)\cap(0,1)=\emptyset$.

  2. (2)

If there is $b\in\mathbb{R}$ such that $\mathscr{K}(\mathbb{U})\geq b$ for all
two-dimensional subspaces $\mathbb{U}\subset\mathbb{A}$ containing $e$, then
$b\leq 1/(4|e|^{2})$ and
$\operatorname{Spec}^{\perp}(e)\subset[\tfrac{1-\sqrt{B}}{2},\tfrac{1+\sqrt{B}}{2}]$
for $B=1-4b|e|_{h}^{2}\geq 0$. In particular, if $(\mathbb{A},\circ,h)$ has
nonnegative sectional nonassociativity, then
$\operatorname{Spec}^{\perp}(e)\subset[0,1]$.

###### Proof.

In the setting of (1), if $0\neq y\in\mathbb{A}$ is an eigenvector of
$L_{\circ}(e)$ with eigenvalue $\lambda$ and orthogonal to $e$, then, by
(8.5), $\lambda(1-\lambda)=|e|^{2}_{h}\mathscr{K}(e,y)\leq b|e|_{h}^{2}$, so
$\lambda^{2}-\lambda+b|e|_{h}^{2}\geq 0$, which forces $\lambda$ to be in the
indicated range. In the setting of (2), the same argument shows that
$\lambda^{2}-\lambda+b|e|^{2}_{h}\geq 0$, which forces $1-4b|e|_{h}^{2}\geq 0$
and forces $\lambda$ to be in the indicated range. ∎

Next there are given some results showing that nonnegative sectional
nonassociativity precludes the existence of square-zero elements.

######  Lemma 8.4.

If a nontrivial metrized algebra $(\mathbb{A},\circ,h)$ satisfies
$\mathbb{A}\circ\mathbb{A}=\mathbb{A}$ then its multiplication is faithful. In
particular, the multiplication of a metrized semisimple commutative algebra
$(\mathbb{A},\circ,h)$ is faithful.

###### Proof.

If $z\in\ker L_{\circ}$, then, by the invariance of $h$,
$0=h(L_{\circ}(z)x,y)=h(z,x\circ y)$ for all $x,y\in\mathbb{A}$. This shows
$\ker L$ is contained in the orthocomplement
$(\mathbb{A}\circ\mathbb{A})^{\perp}$ of $\mathbb{A}\circ\mathbb{A}$. In
particular, if $\mathbb{A}\circ\mathbb{A}=\mathbb{A}$, then $\ker
L_{\circ}\subset(\mathbb{A}\circ\mathbb{A})^{\perp}=\mathbb{A}^{\perp}=\\{0\\}$,
so $L_{\circ}$ is injective. If $\mathbb{A}=\oplus_{i=1}^{k}\mathbb{A}_{i}$ is
a direct sum of simple ideals, then
$\mathbb{A}_{i}\circ\mathbb{A}_{j}=\\{0\\}$ if $i\neq j$, and, since
$\mathbb{A}_{i}$ is simple,
$\mathbb{A}_{i}\circ\mathbb{A}_{i}=\mathbb{A}_{i}$, so
$\mathbb{A}\circ\mathbb{A}=\sum_{i}\mathbb{A}_{i}\circ\mathbb{A}_{i}=\sum_{i}\mathbb{A}_{i}=\mathbb{A}$,
and $L_{\circ}$ is injective. By the invariance of $h$,
$h(R_{\circ}(y)x,z)=h(x,L_{\circ}(y)z)$ for all $x,y,z\in\mathbb{A}$, so were
$R_{\circ}$ not injective, $L_{\circ}$ would not be injective. ∎

######  Lemma 8.5.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra. Let
$0\neq z\in\mathbb{Nil}_{2}(\mathbb{A},\circ)$. For $0\neq y\in\mathbb{A}$ not
in the span of $z$, $\mathscr{K}(z,y)\leq 0$, with equality if and only if
$z\circ y=0$. In particular, $\mathbb{Nil}_{2}(\mathbb{A},\circ)=\\{0\\}$ if
there holds either of the following conditions:

  1. (1)

$(\mathbb{A},\circ,h)$ has positive sectional nonassociativity.

  2. (2)

$(\mathbb{A},\circ,h)$ has nonnegative sectional nonassociativity and is
semisimple.

###### Proof.

Let $0\neq z\in\mathbb{Nil}_{2}(\mathbb{A},\circ)$. For $y\in\mathbb{A}$
linearly independent of $z$, there holds $\mathscr{K}(z,y)|z\wedge
y|^{2}=-|L_{\circ}(z)y|^{2}_{h}\leq 0$, with equality if and only if $z\circ
y=0$. In particular, $(\mathbb{A},\circ,h)$ cannot have positive sectional
nonassociativity, and if $(\mathbb{A},\circ,h)$ has nonnegative sectional
nonassociativity, then $L_{\circ}(z)y=0$ for all $y\in\mathbb{A}$, so
$L_{\circ}(z)=0$. By Lemma 8.4, if $L_{\circ}(z)=0$, then $(\mathbb{A},\circ)$
is not semisimple. ∎

A commutative algebra is _exact_ if $\operatorname{tr}L_{\circ}(x)=0$ for all
$x\in\mathbb{A}$. Exactness is a condition analogous to unimodularity of a Lie
algebra. Evidently, an exact commutative algebra is not unital.

######  Lemma 8.6.

A Euclidean metrized commutative algebra $(\mathbb{A},\circ,h)$ with
nonnegative sectional nonassociativity that is positive on some two-
dimensional subspace is not exact.

###### Proof.

By [55, Lemma $(1.1)$, Proposition $(1.6)$], there exists
$e\in\mathbb{Idem}(\mathbb{A},\circ)\cup\mathbb{Nil}_{2}(\mathbb{A},\circ)$.
Because $L_{\circ}(e)$ is self-adjoint, there is an $h$-orthogonal basis
$\\{e,z_{1},\dots,z_{n-1}\\}$ of $\mathbb{A}$ such that $z_{1},\dots,z_{n-1}$
are eigenvectors of $L_{\circ}(e)$ with (not necessarily distinct) eigenvalues
$\lambda_{1},\dots,\lambda_{n-1}\in\mathbb{R}$. By (2) of Lemma 8.3,
nonnegativity of the sectional nonassociativity implies $\lambda_{i}\in[0,1]$
for $1\leq i\leq n-1$. It follows that
$\operatorname{tr}L_{\circ}(e)=\epsilon+\sum_{i=1}^{n-1}\lambda_{i}\geq\epsilon$
where $\epsilon$ is $1$ or $0$ as $e\in\mathbb{Idem}(\mathbb{A},\circ)$ or
$e\in\mathbb{Nil}_{2}(\mathbb{A},\circ)$. If there is some two-dimensional
subspace on which the sectional nonassociativity is positive, then there is
nonzero $z=\sum_{i=1}^{n-1}\alpha_{i}z_{i}\in\mathbb{A}$ such that
$\mathscr{K}(e,z)>0$. Then $0<\mathscr{K}(e,z)|e\wedge z|^{2}=h(e,z\circ
z)-h(e\circ z,e\circ
z)=\sum_{i=1}^{n-1}\lambda_{i}(1-\lambda_{i})\alpha_{i}^{2}h(z_{i},z_{i})$.
Since $\lambda_{i}\in[0,1]$ this implies that $\lambda_{i}>0$ for at least one
index $i\in\\{1,\dots,n-1\\}$. Consequently,
$\operatorname{tr}L_{\circ}(e)=\epsilon+\sum_{i=1}^{n-1}\lambda_{i}>\epsilon\geq
0$, so $(\mathbb{A},\circ)$ is not exact. ∎

In a commutative $\mathbb{R}$-algebra $(\mathbb{A},\circ)$ the set
$\mathbb{Q}(\mathbb{A},\circ)=\\{x\circ x:x\in\mathbb{A}\\}$ is a closed cone,
the _cone of squares_. By definition,
$\mathbb{Idem}(\mathbb{A},\circ)\subset\mathbb{Q}(\mathbb{A},\circ)$.

######  Lemma 8.7.

Let $(\mathbb{A},\circ,h)$ be a Euclidean metrized commutative algebra having
nonnegative sectional nonassociativity. The set

(8.7)
$\displaystyle\mathbb{C}(\mathbb{A},\circ,h)=\\{x\in\mathbb{A}:L_{\circ}(x)\,\,\text{is
nonnegative}\\}$

is a closed convex cone containing $\mathbb{Q}(\mathbb{A},\circ)$ for all
$x\in\mathbb{A}$ and consequently containing
$\mathbb{Idem}(\mathbb{A},\circ)$.