version 2.5

An inside view of the tensor product ††${}^{{}^{\displaystyle\star}}$ To
appear in Particles, Fields and Topology, Celebrating A.P. Balachandran,
edited by T.R. Govindrajan, Guiseppe Marmo, V. Parameswaran Nair, Denjoe
O’Connor, Sarada Rajeev and Sachindeo Vaidya (World Scientific, Singapore)

Rafael D. Sorkin

Perimeter Institute, 31 Caroline Street North, Waterloo ON, N2L 2Y5 Canada

and

Raman Research Institute, C.V. Raman Avenue, Sadashivanagar, Bangalore – 560
080 India

and

School of Theoretical Physics, Dublin Institute for Advanced Studies, 10
Burlington Road, Dublin 4, Ireland

and

Department of Physics, Syracuse University, Syracuse, NY 13244-1130, U.S.A.

address for email<EMAIL_ADDRESS>

Abstract

Given a vector-space $~{}V~{}$ which is the tensor product of vector-spaces
$A$ and $B$, we reconstruct $A$ and $B$ from the family of simple tensors
$a{\otimes}b$ within $V$. In an application to quantum mechanics, one would be
reconstructing the component subsystems of a composite system from its
unentangled pure states. Our constructions can be viewed as instances of the
category-theoretic concepts of functor and natural isomorphism, and we use
this to bring out the intuition behind these concepts, and also to critique
them. Also presented are some suggestions for further work, including a hoped-
for application to entanglement entropy in quantum field theory.

Keywords and phrases: tensor product, tensor structure, simple vector,
subsystem, unentangled, intrinsic definition, square-construction, functor

Contents

1\. Introduction

2\. Posing the problem

3\. The space S of simple vectors in $V$

4\. How to recover $V_{1}$ and $V_{2}$ up to scale

5\. The analysis and synthesis of a tensor product

Using “pointed vector spaces”

6\. Categorical matters (exposition and criticism)

A small illustration: topology remembers spin

A functorial gloss on our constructions

7\. Questions; further developments; connection to quantum field theory

This article is dedicated to my friend, A.P. Balachandran, on the occasion of
his 85th birthday. With his knack for discerning concrete implications of
abstract mathematical relationships, maybe he’ll think of an unexpected use in
physics for the conception of tensor product proposed herein!

1\. Introduction

It would hardly be possible to review all the ways in which tensors enter into
physics. General Relativity and Quantum Field Theory would not exist without
certain individual tensors or tensor-fields, like the Lorentzian metric, the
Riemann curvature, or the stress-energy tensor, but it is perhaps in abstract
quantum theory where the concept of tensor-product itself, and of the
corresponding product-space is most prominent. The reason, of course, is that
insofar as one deals with state-spaces of quantal “systems”, the tensor-
product furnishes the construction that combines the state-spaces of two or
more subsystems into that of the larger compound system or “whole”.

Given this role of tensor-product, it could be unsettling that aside from its
dimension, the resulting state-space (call it $V$) appears to remember nothing
about the constituent spaces whose product it is. For example, because
$12=4\times 3=2\times 6$, a given state-space of dimension 12 that arose by
combining spin $3/2$ with spin $1$ might equally well be describing a
composite of spins $1/2$ and $5/2$. Thus arises the following mathematical
question.

Suppose that a certain vector space $V$ is the tensor product of spaces
$V_{1}$ and $V_{2}$. What extra information do you need in order to recover
$V_{1}$ and $V_{2}$ from $V$? Or to put the question another way: What does it
mean for $V$ to carry the structure of a tensor product space?

To bring this question into sharper focus, imagine that instead of being a
tensor-product, $V$ were a direct sum, as it would be for example if $V_{1}$
described an ionized Hydrogen atom, while $V_{2}$ described the same atom in
unionized form. Then the appropriate state-space would be
$V=V_{1}\oplus{V_{2}}$; and the same equation would describe other mutually
exclusive alternatives, like an alpha particle being inside vs. outside a
nucleus, or a molecule being ortho-hydrogen vs. para-hydrogen. To our question
about tensor products, the counterpart in such cases would be: What does it
mean for $V$ to carry the structure of a direct sum? Here however the answer
is simple. One only needs to indicate $V_{1}$ and $V_{2}$ as subspaces of $V$.
These two “parts” of $V$ are contained bodily within the whole, and every
$v\in V$ is uniquely a sum, $v=v_{1}+v_{2}$. A “direct sum structure” for $V$,
in other words, is nothing but a pair of complementary subspaces of $V$.

Why isn’t the case of a tensor-product space $V$ equally straightforward? To
appreciate what’s different, let $V=V_{1}\otimes V_{2}$ be a tensor product
space. The first thing one can notice is that $V_{1}$ and $V_{2}$ are no
longer contained within $V$ in any obvious way. Moreover, the analog of
$v=v_{1}+v_{2}$ in a direct sum, would be $v=v_{1}\otimes v_{2}$, but how
could one express this within $V$, given that vector-spaces are by definition
endowed with a notion of sum but not of product! And lacking the operation
$\otimes$ within $V$, how could you recover $V$ from $V_{1}$ and $V_{2}$, even
if two such subspaces of $V$ could be identified? The development that follows
will answer these questions and show that the space S of simple vectors in $V$
— elements of $V$ of the form $v=v_{1}\otimes v_{2}$ — can play the structural
role that the complementary subspaces played in the case of direct sum.

Although our questions are physically inspired, they are purely mathematical,
and the answers we will come upon must surely be well known in some circles,
even if they haven’t shown up in the literature I’m familiar with. Nor do I
know whether the constructions we will explore have any deeper physical
ramifications. They do, however, offer a more intrinsic way to conceive of the
tensor product (an “inside view” as one might say); and a more intrinsic
conception, more often than not, illuminates and deepens one’s intuition. As
anyone who has taught a course in Relativity, Differential Geometry, or
“Mathematical Methods” can attest, there’s something about the concept of
tensor that intuition finds hard to grasp. Perhaps it’s no coincidence then
that quantum entanglement, the mathematics of which is that of tensor product
spaces, also seems counter-intuitive to so many people.

In this situation, the “square construction” on which our development will
rest, offers a complementary way to think about tensor products, a way that
starts not with the individual factor-spaces, $V_{1}$ and $V_{2}$, but with
the product space $V$ itself. Inasmuch as this more analytical approach shifts
the main emphasis onto the simple vectors within $V$, and inasmuch as these
simple vectors in a quantum context are precisely the unentangled state-
vectors, our development arguably makes better contact with physical intuition
than the more formal definitions one usually encounters. At the very least, it
offers an alternative to more familiar ways of approaching the topic. After
all, the more ways one has to think about a subject, the better the prospect
that at least one of them will be able to provide the key to the deeper
understanding that one is seeking.

The main ingredients of our constructions are presented in Sections 2-5 below.
Sections 2 and 3 are preparatory, while the constructions themselves appear in
Sections 4 and 5.

In Section 6 we ask whether the results of Sections 4 and 5 can be regarded as
fully capturing the structure of $V$ as a tensor product; and we show in some
detail how to make this type of question precise in terms of the category-
theoretic concepts of functor and natural transformation, to which we provide
a brief introduction. In this connection, we also point out a certain
shortcoming of the functor concept itself.

In Section 7 we suggest some extensions of our constructions to tensor
products of three or more spaces, or to products of a single space with
itself, in which case symmetry-conditions come into play (bosonic, fermionic,
or nonabelian). We also speculate that a suitable generalization of the notion
of simple vector to infinite dimensions could help to clear away the
mathematical obstructions (involving type-III von Neumann algebras, where
tensor products as normally defined are not available) that prevent one from
understanding entanglement-entropy (with its need for a cutoff) in terms of
reduced density matrices.

In what follows, we will assume, unless otherwise specified, that all vector
spaces are real and finite dimensional. Nothing would change if we replaced
the field ℝ with the field ℂ, but it’s convenient for exposition to pick one
or the other and stick with it. We will also assume without special mention
that the spaces $V_{1}$ and $V_{2}$ are distinct from each other.

2\. Posing the problem

Our question asks for a more intrinsic definition of tensor product, or as we
worded it above: What could play the role of a tensor-product structure for
$V$? We will contemplate three possible answers to this question, and in doing
so we will always assume that the two spaces of which $V$ is a product are
distinct from each other. Often this will not matter, but sometimes it would
make a difference, notably in definition (1) of the paragraph after next, and
then farther below in the “Second answer” to our main question.

Before suggesting answers to our main question, however, it seems advisable to
dwell for a moment on the more common definitions of tensor product. Different
authors favor different ones, and the answers to our questions will tend to
take on different forms, depending on which definition one has in view. What,
then, are some of the popular definitions [1] of the space
$V_{1}{\otimes}V_{2}\,$ and of the tensors therein?

(1) an element of $V_{1}{\otimes}V_{2}\,$ is a numerical matrix whose entries
depend on a choice of bases for $V_{1}$ and $V_{2}$ and transform in a certain
way when these bases are changed. (This might be the oldest definition.)

(2) an element of $V_{1}{\otimes}V_{2}\,$ is a linear mapping between two
vector spaces, for example a linear mapping from $V_{2}^{*}$ to $V_{1}$, where
$V_{2}^{*}$ is the dual space of $V_{2}\,$

(3) an element of $V_{1}{\otimes}V_{2}\,$ is an equivalence class of formal
sums of symbols, $\alpha{\otimes}\beta\,$, where $\alpha\in V_{1}$ and
$\beta\in V_{2}$

(4) “the” space $V_{1}{\otimes}V_{2}$ is any solution of a certain “universal
mapping problem” involving bilinear functions from $V_{1}\times V_{2}$ to an
arbitrary vector space $Y$. (Thereby a bilinear function from $V_{1}\times
V_{2}$ to $Y$ induces a unique linear mapping from $V_{1}{\otimes}V_{2}$ to
$Y$.)

Remark Notice that definition (1) refers to independently chosen bases for
$V_{1}$ and $V_{2}$. Were we to assume that $V_{1}$ and $V_{2}$ were literally
the same space, only a single basis would come into play. Tensors in this vein
are common in GR and differential geometry, with $V_{1}=V_{2}$ being the
tangent space to a point of spacetime. In quantum mechanics on the other hand,
distinct factor-spaces are typical, albeit not in the case of
indistinguishable particles.

In finite dimensions, all these definitions are provably equivalent. In
infinite dimensions one must distinguish between the so-called algebraic
tensor product and various topological tensor products, not all of which are
the same in general. For our purposes, entering into those subtleties would be
too much of a distraction. [2] Instead, we will work throughout in finite
dimensions. In the jargon of category theory, each of these definitions yields
a “functor” from pairs of vector spaces to vector spaces, and the statement
that (in finite dimensions) they are all equivalent asserts that between any
two of the functors there is an invertible “natural transformation”. Now back
to our main question and some possible answers to it.

First possible answer. The most direct and obvious answer, but at the same
time the least informative, is that a tensor-product structure for $V$ is an
isomorphism between $V$ and a space of the form $V_{1}{\otimes}V_{2}$. This is
a good start, but it has the drawback that the auxiliary spaces $V_{1}$ and
$V_{2}$ are not derived from $V$, with the consequence that different choices
of them would strictly speaking define different tensor-product structures for
$V$. We could address this difficulty by forming equivalence-classes under
isomorphisms of $V_{1}$ and $V_{2}$, but let’s instead continue on to the
second and third proposals.

Second answer. The second possible answer to our main question, already more
concrete and “intrinsic”, hearks back to definition (1) in the above list. In
order to represent an element of $V=V_{1}{\otimes}V_{2}$ as a numerical
matrix, one needs a basis of $V$ whose members are themselves organized into a
matrix. Specifically, if a list of vectors $e_{j}\in V_{1}$ furnishes a basis
for $V_{1}$ and a second list of vectors $f_{k}\in V_{2}$ furnishes a basis
for $V_{2}$, then the products $e_{j}{\otimes}f_{k}$ furnish a basis for
$V_{1}{\otimes}V_{2}$ whose members array themselves in a rectangular matrix
with rows labeled by $j$ and columns by $k$. In such a basis the matrix
representing a simple vector $v_{1}{\otimes}v_{2}$ will be a matrix-product of
the form, column-vector $\times$ row-vector. (Such a “special basis” is
precisely an isomorphism between $V$ and ${\hbox{\openface
R}}^{m}{\otimes}{\hbox{\openface R}}^{n}$, where $m$ and $n$ are the
respective dimensions of $V_{1}$ and $V_{2}$.)

We could thus answer that a tensor-product structure for $V$ is a basis for
$V$ organized into a rectangular matrix. Unfortunately this won’t quite do,
because many other bases will define the same product-structure. First of all,
one might swap rows with columns, which amounts to writing
$V_{2}{\otimes}V_{1}$ instead of $V_{1}{\otimes}V_{2}$. This does nothing.
However one can also replace each of the two bases by some other basis for the
same space, which doesn’t change the spaces $V_{1}$ or $V_{2}$ themselves but
only their representations. We are thus led to identify a tensor-product
structure for $V$ as an equivalence-class of bases, parameterized by
$G=GL(n_{1})\times GL(n_{2})$ where $n_{i}=\dim(V_{i})$. But even this is not
quite correct since it is always possible to rescale the basis for $V_{1}$ by
some factor, while rescaling the basis for $V_{2}$ in the opposite way. Since
this doesn’t affect the resulting basis for $V_{1}{\otimes}V_{2}$, we conclude
that $G$ is really the quotient group $GL(n_{1}){\times}GL(n_{2})\,/\,GL(1)$.
Here of course $GL(n)$ is the group of invertible $n\times n$ matrices.

The fact that true group is $GL(n_{1})\times GL(n_{2})\,/\,GL(1)$ and not
simply $GL(n_{1})\times GL(n_{2})$ seems a detail, but it is actually telling
us something that will show up again in our deliberations below. From a
tensor-product structure for $V$, we cannot fully reconstruct the factor-
spaces $V_{1}$ and $V_{2}$; we can obtain them only up to a joint scaling
ambiguity.

Remark In a classical (non-quantum) context, a composite system would be
described by a cartesian product, $A\times B$. In that case, the counterpart
of a rectangular basis would just be (for discrete spaces $A$ and $B$) a
rectangular list of elements of $A\times B$, and the story would more or less
end there. ††⋆ The counterpart of $G$ would be the product of the permutation
groups of $A$ and $B$ Tensor products are more subtle than cartesian products,
however, and there’s a third possible answer to our question which is still
more concrete and intrinsic than an equivalence class of bases.

Third answer. The third possible answer to our question, and the one which the
rest of this paper will explore, is that the tensor-product structure for
$V=V_{1}{\otimes}V_{2}$ can be taken to be the subspace S of simple vectors:

${\hbox{\german S}}=\left\\{\alpha\otimes\beta\,|\,\alpha\in V_{1},\beta\in
V_{2}\right\\}$ $None$

As we will see, there exist explicit constructions that take you from
${\hbox{\german S}}\subseteq V$ to (copies of) $V_{1}$ and $V_{2}$, and thence
back to $V$.

Remark In quantum language, S would be the set of unentangled state-vectors.
Obviously there’s something special about them, but mathematically they are
only the first in a hierarchy of successively more generic tensors, those of
ranks 2, 3, etc, where the “rank” of $v$ is the minimum number of simple
vectors of which it is a sum (quantum mechanically the number of terms in a
Schmidt decomposition of $v$.) ††† Algebraic Geometry has given the name
“Segre variety”, not quite to S itself, but to the set of rays in S

3\. The space S of simple vectors in $V$

Since it is related quadratically to $V_{1}$ and $V_{2}$, the set S of simple
vectors obviously will not be a linear subspace of $V$ in general, but it will
be foliated by two families of linear subspaces, which we will denote by
${\cal{M}}_{1}$ and ${\cal{M}}_{2}$.

Before demonstrating this, let us deal with two trivial cases that don’t fit
easily into the general pattern, In the most trivial case, both $V_{1}$ and
$V_{2}$ are one-dimensional: $\dim(V_{1})=\dim(V_{2})=1$. Both are then copies
of ℝ, as also is $V=V_{1}\otimes V_{2}$. In this case every $v\in V$ is
plainly a simple vector, and so S is all of $V$. Conversely, given that
$\dim(V)=1$, and since we know in general that $\dim V_{1}\otimes V_{2}=\dim
V_{1}\times\dim V_{2}$, we know immediately that both $V_{1}$ and $V_{2}$ are
isomorphic to $V$ itself. In a reconstruction of $V_{1}$ and $V_{2}$, we can
thus do no better than to take both to be copies of $V$, and this suffices.
The only small subtlety shows up when, having passed from $V$ to $V_{1}$ and
$V_{2}$, we seek to reconstruct $V$ as $V_{1}\otimes V_{2}=V\otimes V$, but
have to face the fact that although $V$ is isomorphic to $V\otimes V$, the
isomorphism is not canonical. ††♭ In some sense this is just “dimensional
analysis”. If the elements of $V$ were “lengths”, then those of $V\otimes V$
would be squared lengths.

In the second trivial case, $\dim V_{1}>1$ while $\dim V_{2}=1$ (or vice
versa). Here again S is trivially all of $V$. (By definition any $v\in V$ is a
sum of terms of the form $a\otimes b$ for $a\in V_{1}$ and $b\in V_{2}$, but
since all nonzero $b$ are proportional to each other, all the $b$ can be taken
equal, whence $v=a_{1}\otimes b+a_{2}\otimes
b+\cdots=(a_{1}+a_{2}\cdots)\otimes b\in{\hbox{\german S}}$.) It follows that
$V=V_{1}\otimes V_{2}\simeq V_{1}\otimes{\hbox{\openface R}}=V_{1}$ (where
‘$\simeq$’ signifies isomorphic-to). Conversely, whenever ${\hbox{\german
S}}=V$, we can construct spaces $V_{1}$ and $V_{2}$ by taking $V_{1}$ to be
$V$ and $V_{2}$ to be any one-dimensional vector space, for example the
subspace of $V$ given by ${\hbox{\openface R}}\,b_{0}$, where $b_{0}$ is any
fixed vector ††⋆ The reason for this particular choice will become clear soon.
Notice also that we could of course have exchanged the roles of $V_{1}$ and
$V_{2}$. in $V$. The now-familiar scaling-ambiguity corresponds then to the
undetermined normalization of $b_{0}$.

Notice in these two examples that $V_{1}$ was identified with a maximal linear
subspace of S. Although completely trivial in the two examples, this
observation will be the basis of our reconstruction of $V_{1}$ and $V_{2}$ in
the generic case. In seeking to understand the linear subspaces of S, we will
need a few “obvious” facts about tensor products which we will now review in
the spirit of definition (3) mentioned in the previous section.

Recall then that any tensor $T\in V_{1}{\otimes}V_{2}$ is a sum of simple
tensors, i.e. a sum of products of vectors from $V_{1}$ with vectors from
$V_{2}$:

$T\,=\,\sum_{j}\,a_{j}\otimes\,b_{j}$ $None$

To fully characterize the space $V_{1}{\otimes}V_{2}$, however, one needs to
specify which such sums are equal to which others, or equivalently which
expressions $\,T\,$ equal the zero tensor. Intuitively the answer is that
$T=0$ iff it is forced to vanish by the combining rules for the symbols
$a{\otimes}b$ together with the linear dependences among the vectors of
$V_{1}$ and $V_{2}$. This criterion is implicit in the aforementioned
definition (4), but it is more useful to express it algorithmically.

Rule Provided that the vectors $a_{j}$ in (2) are linearly independent, $T=0$
if and only if all of the $b_{j}$ vanish.

(Obviously the same rule will hold true if we exchange the roles of $a_{j}$
and $b_{j}$.) As stated, the rule wants the $a_{j}$ to be linearly
independent. If they are not, then some of them can be expressed as linear
combinations of the others, and one should do this before applying the rule.
Thus, an algorithm for deciding whether $T=0$ consists in first writing any
redundant $a_{j}$ in terms of the others, second expanding out the resulting
expression to put $T$ into the form (2), and third applying the rule as
stated.

As a trivial consequence of this rule, we learn that $a{\otimes}b$ is nonzero
if both $a$ and $b$ are. In stating the following further consequences, we
will interpret $a\propto b$ to mean that either $a=\lambda b$ or $b=\lambda
a$, $\lambda\in{\hbox{\openface R}}$.

Lemma 1. If $\alpha$ and $\beta$ are nonzero then
$\alpha{\otimes}\beta=\alpha^{\prime}{\otimes}\beta^{\prime}$ $\Rightarrow$
$\alpha^{\prime}\propto\alpha$ and $\beta^{\prime}\propto\beta$.

Proof Were $\alpha^{\prime}$ not proportional to $\alpha$, they would be
linearly independent. Our “Rule” would then imply that
$\alpha{\otimes}\beta-\alpha^{\prime}{\otimes}\beta^{\prime}$ could not
vanish. Therefore $\alpha^{\prime}\propto\alpha$, and by symmetry
$\beta^{\prime}\propto\beta$.

Lemma 2. Let $\alpha\otimes\beta\in{\hbox{\german S}}$ and
$\alpha^{\prime}\otimes\beta^{\prime}\in{\hbox{\german S}}$ be nonzero simple
vectors. If their sum is also simple then either
$\alpha^{\prime}\propto\alpha$ or $\beta^{\prime}\propto\beta$.

Proof (by contradiction). Assume that $\alpha^{\prime}\not\propto\alpha$ and
$\beta^{\prime}\not\propto\beta$. The four terms $\alpha{\otimes}\beta$,
$\alpha^{\prime}{\otimes}\beta^{\prime}$, $\alpha{\otimes}\beta^{\prime}$,
$\alpha^{\prime}{\otimes}\beta$ are then (by a simple application of the Rule)
linearly independent. By hypothesis
$\alpha\otimes\beta+\alpha^{\prime}\otimes\beta^{\prime}=\gamma{\otimes}\delta$
for some $\gamma$ and $\delta$. Appealing once again to the Rule, and
remembering that $\alpha^{\prime}$ is independent of $\alpha$, we conclude
that $\gamma$ must be a linear combination of $\alpha$ and $\alpha^{\prime}$;
similarly $\delta$ must be a linear combination of $\beta$ and
$\beta^{\prime}$. But then $\gamma{\otimes}\delta$, when expanded out, could
not contain the required terms, $\alpha\otimes\beta$ and
$\alpha^{\prime}\otimes\beta^{\prime}$ without also containing terms in
$\alpha\otimes\beta^{\prime}$ and $\alpha^{\prime}\otimes\beta$.

Returning now to the analysis of S, and recalling that we have already
disposed of the possibility that either $\dim{V_{1}}=1$ or $\dim V_{2}=1$, we
can assume for now that $\dim{V_{1}}\geq 2$ and $\dim V_{2}\geq 2$, this being
where the typical structure of S reveals itself, namely that of the two
foliations ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$ already alluded to but not yet
defined. For the time being, we will define ${\cal{M}}_{1}$ and
${\cal{M}}_{2}$ as follows. Soon, we will define them intrinsically (meaning
directly from $V$ and S alone), whereupon (3) and (4) will shed their status
as definitions and become theorems. The members of ${\cal{M}}_{1}$ will be the
subsets of S of the form $V_{1}{\otimes}\beta$ for $\beta\in V_{2}$, and
likewise for ${\cal{M}}_{2}$:

${\cal{M}}_{1}=\left\\{V_{1}{\otimes}\,\beta\,|\,\beta\in V_{2}\right\\}$
$None$ ${\cal{M}}_{2}=\left\\{\alpha\,{\otimes}V_{2}\,|\,\alpha\in
V_{1}\right\\}$ $None$

(Here of course our notation means that, e.g,
$V_{1}{\otimes}\,\beta=\left\\{\alpha{\otimes}\,\beta\,|\,\alpha\in
V_{1}\right\\}$.)

We want to prove first, that every $M$ in either ${\cal{M}}_{1}$ or
${\cal{M}}_{2}$ is a maximal linear subspace of S; second that ${\cal{M}}_{1}$
and ${\cal{M}}_{2}$ exhaust the maximal linear subspaces of S; third that
$M,N\in{\cal{M}}_{1}$ and $M\not=N$ $\Rightarrow M\cap N=\\{0\\}$ (and
likewise for ${\cal{M}}_{2}$); and fourth that
$M\in{\cal{M}}_{1},N\in{\cal{M}}_{2}\Rightarrow\dim(M\cap N)=1$.

Why are the members of ${\cal{M}}_{1}$ (for example) maximal linear subspaces
of S? That $M=V_{1}{\otimes}\,\beta$ is a linear subspace is obvious, but why
is it maximal? Well, any simple vector not in $M$ must take the form
$\alpha^{\prime}{\otimes}\beta^{\prime}$ with $\beta^{\prime}$ independent of
$\beta$. Choose also an $\alpha\in V_{1}$ that is independent of
$\alpha^{\prime}$ (which is always possible since $\dim{V_{1}}>1$), and notice
that $\alpha{\otimes}\beta\in{M}$. If we could adjoin
$\alpha^{\prime}{\otimes}\beta^{\prime}$ to $M$ then
$\alpha^{\prime}{\otimes}\beta^{\prime}+\alpha{\otimes}\beta$ would also have
to be in $M$, and therefore simple, contrary to Lemma 2 above. ††† What we are
effectively proving could be reduced to a lemma to the effect that every
linear subspace of S has the form $W{\otimes}\beta$ or $\alpha{\otimes}W$, for
some vector-subspace $W$ of $V_{1}$ or $V_{2}$

And why does every maximal linear subspace of S have to belong to either
${\cal{M}}_{1}$ or ${\cal{M}}_{2}$? Well, let $M$ be such a subspace, and let
$\alpha{\otimes}\beta\in M$. Certainly $\alpha{\otimes}\beta$ alone is not
maximal (it belongs to $V_{1}{\otimes}\,\beta$, for example), so let
$\alpha^{\prime}{\otimes}\beta^{\prime}$ be an independent member of $M$. By
the same lemma either $\alpha$ and $\alpha^{\prime}$ are proportional or
$\beta$ and $\beta^{\prime}$ are proportional, say the latter. Then as we just
saw, every other member of $M$ must also take the form $\gamma{\otimes}\beta$
for some $\gamma\in V_{1}$, in other words $M\subseteq
V_{1}{\otimes}\,\beta\in{\cal{M}}_{1}$, whence $M=V_{1}{\otimes}\,\beta$ since
$M$ is maximal.

Third, if $M,N\in{\cal{M}}_{1}$ are unequal then $M=V_{1}{\otimes}\beta$ and
$N=V_{1}{\otimes}\beta^{\prime}$ with $\beta$ independent of $\beta^{\prime}$.
Hence any $v\in M\cap N$ must satisfy
$v=\alpha{\otimes}\beta=\alpha^{\prime}{\otimes}\beta^{\prime}$ for some
$\alpha$ and $\alpha^{\prime}$. But by Lemma 1, this is impossible unless
$v=0$.

Fourth, if $M\in{\cal{M}}_{1},N\in{\cal{M}}_{2}$ then $M=V_{1}{\otimes}\beta$,
$N=\alpha{\otimes}V_{2}$ for some $\alpha\in V_{1},\beta\in V_{2}$. If $v\in
M\cap N$ then by definition,
$v=\alpha^{\prime}{\otimes}\beta=\alpha{\otimes}\beta^{\prime}$ for some
$\alpha^{\prime}\in V_{1},\beta^{\prime}\in V_{2}$. The lemma just cited then
informs us that $\alpha^{\prime}\propto\alpha$ and
$\beta^{\prime}\propto\beta$, whence
$v=\alpha^{\prime}{\otimes}\beta\propto\alpha{\otimes}\beta$. In other words
$M\cap N$ is the 1-dimensional subspace, $\,{\hbox{\openface
R}}\,\alpha{\otimes}\beta$

The essential feature we have discovered is that any two members of different
foliations meet in a ray (a one-dimensional subspace of $V$) and any two
distinct members of the same foliation are disjoint. This lets us determine
the foliations ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$ simply from a knowledge of
${\hbox{\german S}}\subseteq V$, without any further recourse to how $V$ arose
as a tensor product: if $M$ and $N$ are elements of the set ${\cal{M}}$ of all
maximal linear subspaces of S, then they belong to the same foliation if and
only if they are disjoint, and this criterion is guaranteed to produce exactly
two disjoint subsets of ${\cal{M}}$, which we can label as ${\cal{M}}_{1}$ and
${\cal{M}}_{2}$. Henceforth, we will adopt this intrinsic definition of
${\cal{M}}_{1}$ and ${\cal{M}}_{2}$, which we can record in the following two
maps that associate with each simple vector in $V$ the two maximal linear
subspaces of S to which it belongs.

Definition Let $v\in{\hbox{\german S}}$. Then $\pi_{1}(v)$ [resp.
$\pi_{2}(v)$] is the unique maximal linear subspace of type ${\cal{M}}_{1}$
[resp. ${\cal{M}}_{2}$] that contains $v$.

Equations, (3)-(4), are hereby no longer definitions but theorems which apply
whenever we can exhibit vector-spaces $V_{1}$ and $V_{2}$ such that
$V=V_{1}{\otimes}V_{2}$.

With these observations, we have taken a first step in recovering the tensor
product structure of $V$ from S. In fact, one sees from (3) and (4) that each
$M_{1}$ in ${\cal{M}}_{1}$ is a copy of $V_{1}$ and each $M_{2}$ in
${\cal{M}}_{2}$ is a copy of $V_{2}$. In the following section, we will build
on our knowledge of ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$ to recover fully the
ray-spaces associated to $V_{1}$ and $V_{2}$, and then to recover $V_{1}$ and
$V_{2}$ themselves up to scale.

4\. How to recover $V_{1}$ and $V_{2}$ up to scale

Our ultimate aim is to find a construction that, relying on nothing more than
the set S of simple vectors in $V$, will resolve the latter into its two
factors (as uniquely as possible), and then to discover how to rebuild $V$ as
the tensor product of these factors. This will take place in Section 5, and
not everything from the present section will be needed there. If you are
reading these lines, you might thus want to skip over the present section in
order to appreciate the great simplicity of the final constructions. On the
other hand, the present section, as well as providing much of the background
for Section 5, will also show how, in becoming aware of the two spaces
${\cal{M}}_{1}$ and ${\cal{M}}_{2}$, we have already recovered from S the rays
of $V_{1}$ and $V_{2}$, which in a quantum context means we have already
recovered, if not the respective subsystems themselves, then at least their
“pure states”.

To appreciate this fact, recall that when $V=V_{1}{\otimes}V_{2}$, any member
$M$ of ${\cal{M}}_{2}$ can be expressed in the form (4). But the subspace
$M=\alpha\,{\otimes}V_{2}$ determines and is determined by the ray,
${\hbox{\openface R}}\,\alpha\subseteq V_{1}$. The points of ${\cal{M}}_{2}$
are thus in bijective correspondence with the rays of $V_{1}$, and likewise
for ${\cal{M}}_{1}$ and $V_{2}$. Introducing the notation ${\bf{P}}{V}$ for
the projective space formed from the rays of any vector-space $V$, we can
therefore assert that

${\bf{P}}V_{1}={\cal{M}}_{2}\qquad\hbox{and}\qquad{\bf{P}}V_{2}={\cal{M}}_{1}$
$None$

Of course, there’s more to it than this, because so far, we have only
introduced ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$ as sets without further
structure. In order to fully corroborate the claim that
${\bf{P}}V_{1}={\cal{M}}_{2}$, we need to present ${\cal{M}}_{2}$ as the set
of rays of some intrinsically defined vector space, this being one way to
equip it with a projective structure. In the course of doing so, we will also
see how to get our hands on $V_{1}$ itself up to scale.

Let’s first see the procedure per se and then return to see more fully why it
works. To get started, select arbitrarily any $M\in{\cal{M}}_{1}$ and let $P$
be the restriction of $\pi_{2}$ to $M$. It is not hard to see that
$P:M\to{\cal{M}}_{2}$ sets up a one-to-one correspondence between the rays in
$M$ and the points of ${\cal{M}}_{2}$. By definition, if $v\in M$ then
$P(v)=\pi_{2}(v)$ is the unique maximal linear subspace in ${\cal{M}}_{2}$
that contains $v$; being linear, it also contains the entire ray,
${\hbox{\openface R}}\,v$. Furthermore, $P$ is trivially surjective because
for any $N\in{\cal{M}}_{2}$, $M\cap N$ is (as observed earlier) a ray $\ell$
in $M$ that gets mapped by $P$ to $N$ itself. This also proves that $P$ is
injective (on the rays of $M$) because any other ray in $M$ that was mapped to
$N$ by $P$ would by definition have to lie in $M\cap N$ and therefore coincide
with $\ell$.

The mapping, $P:M\to{\cal{M}}_{2}$, is what we were looking for, but it
remains to demonstrate that any other $M^{\prime}\in{\cal{M}}_{1}$ would have
induced the same projective structure on ${\cal{M}}_{2}$. For this, it
suffices to find a linear isomorphism between $M$ and $M^{\prime}$ that
commutes with the corresponding projections. In other words, with $P^{\prime}$
taken to be the restriction of $\pi_{2}$ to $M^{\prime}$, we should seek an
isomorphism $f:M\to M^{\prime}$ such that $P=P^{\prime}\circ f$. Such an $f$
would induce an isomorphism $\hat{f}:{\bf{P}}M\to{\bf{P}}M^{\prime}$, and it
is actually easier to characterize this isomorphism intrinsically than to
exhibit $f$ itself. Let us therefore define $\hat{f}$ first, and only then
consider how to lift it to a linear map $f$. It turns out that the Ansatz,

$\hat{f}(M\cap N)=M^{\prime}\cap N$ $None$

(where $N$ is an arbitrary element of ${\cal{M}}_{2}$) does what is needed. In
particular, if $M^{\prime\prime}$ is a third element of ${\cal{M}}_{1}$, then
the isomorphisms $M\to M^{\prime}\to M^{\prime\prime}$ defined by (6)
obviously compose consistently.

Our remaining task is to lift the just-constructed mapping,
$\hat{f}:{\bf{P}}M\to{\bf{P}}M^{\prime}$, to a linear function, $f:M\to
M^{\prime}$. Given that for $v\in M$, the mapping $\hat{f}$ already determines
the ray in $M^{\prime}$ to which $v$ should go, the only further input needed
to define $f(v)$ is its normalization. Although this seems a tiny bit of extra
information, the construction via which we will obtain it is surprisingly
intricate. In fact, it is not really needed for present purposes; all we
really need to know is that a linear lift $f$ exists, which could be proven
more easily. If nevertheless we take the trouble to construct $f$ explicitly,
it is because doing so will introduce us to a certain type of “simple-square”
that will play an important role in the next section.

Fix spaces $M,M^{\prime}\in{\cal{M}}_{1}$ as above, and let $\ell_{0}$ be any
ray in $M$, with $\ell_{0}^{\prime}=\hat{f}(\ell_{0})$ being the corresponding
ray in $M^{\prime}$, as given by (6). We know that $f$ will take any point in
$\ell_{0}$ to some point in $\ell_{0}^{\prime}$. Given now some arbitrarily
chosen reference vector, $v_{0}\in\ell_{0}$, we need to decide which vector in
$\ell_{0}^{\prime}$ will be $f(v_{0})$, and it turns out that this decision
determines $f$ fully. Let $v_{0}^{\prime}$ be the vector selected to be
$f(v_{0})$. The problem then is to determine $f(v)$ when $v$ belongs to some
other ray $\ell\subseteq M$. That is, we need to figure out where $f(v)$ lies
along the ray $\ell^{\prime}=\hat{f}(\ell)$.

This problem admits a generic case and a couple of special cases. In the
generic case, $v_{0}$, $v_{0}^{\prime}$, and $v$ are all linearly independent.
Consider then an arbitrary $v^{\prime}\in\ell^{\prime}$ and the square

$\pmatrix{a&b\cr&\cr c&d\cr}=\pmatrix{v_{0}&v_{0}^{\prime}\cr&\cr
v&v^{\prime}\cr}$ $None$

whose elements belong to the rays

$\pmatrix{\ell_{0}&\ell_{0}^{\prime}\cr&\cr\ell&\ell^{\prime}\cr}\ .$ $None$

By construction (cf. (6)),

$\pi_{1}a=\pi_{1}c\,,\quad\pi_{1}b=\pi_{1}d\,,\quad\pi_{2}a=\pi_{2}b\,,\quad\pi_{2}c=\pi_{2}d\,.$
$None$

Consequently, the two row-sums and the two column-sums belong to S (i.e. all
four sums are simple vectors in $V$), but what about the overall sum,
$a+b+c+d\,$? In the answer to this question lies the key to our construction
of $f$. In fact (as we will prove shortly) this sum meets S for precisely one
point $v^{\prime}$ in the ray $\ell^{\prime}$, and by setting
$f(v)=v^{\prime}$ we define $f$ uniquely on the ray $\ell$. Doing the same for
the other rays in $M$, we will obtain a function $f:M\to M^{\prime}$ which is
linear, unique up to a multiplicative prefactor, and whose action on rays is
by definition that of $\hat{f}$.

So much for the generic case. Before turning to the special cases, observe
that just from (9) alone, we can write the rays in (8) as

$\pmatrix{M\cap N&M^{\prime}\cap N\cr&\cr M\cap N^{\prime}&M^{\prime}\cap
N^{\prime}\cr}\ ,$ $None$

where $M=\pi_{1}a=\pi_{1}c$, $N=\pi_{2}a=\pi_{2}b$,
$M^{\prime}=\pi_{1}b=\pi_{1}d$, $N^{\prime}=\pi_{2}c=\pi_{2}d$. The generic
case just treated corresponded to an array (10) in which the subspaces,
$\,M,M^{\prime},N,N^{\prime}\,$, were all distinct, and correspondingly the
vectors, $\,a,b,c,d\,$, in (7) were linearly independent. The special cases we
still need to treat are those in which $M=M^{\prime}$ or $N=N^{\prime}$.

Consider first the special case where $N=N^{\prime}$, or equivalently,
$\ell=\ell_{0}$. Here we know the answer trivially because $v=\lambda v_{0}$
for some scalar $\lambda$, whence $v^{\prime}=f(v)=f(\lambda v_{0})=\lambda
f(v_{0})=\lambda v_{0}^{\prime}$. The square in (7) thus assumes the form,

$\pmatrix{a&b\cr&\cr\lambda a&\lambda b\cr}$ $None$

The other special case is that where $M=M^{\prime}$, or equivalently (since,
as we know, any two elements of ${\cal{M}}_{1}$ are either equal or disjoint),
$\ell_{0}^{\prime}=\ell_{0}$. Here $f$ is just mapping $M$ to itself, an
obvious solution for which would be to take $f$ to be the identity map.
However, we could equally well take it to be a multiple of the identity by a
scalar $\mu$, in which case our square would take on the appearance,

$\pmatrix{a&\mu a\cr&\cr c&\mu c\cr}\ ,$ $None$

a form that follows immediately from (11) by symmetry. For completeness, let
us also record the doubly special case where $M=M^{\prime}$ and $N=N^{\prime}$
both hold, leading to a square of the design,

$\pmatrix{a&\mu a\cr&\cr\lambda a&\lambda\mu a\cr}\ ,$ $None$

as one sees by combining (11) with (12). All these special cases, (11) – (13),
can be obtained from the generic case by forming limits. Amalgamating these
special cases with the generic one, we arrive at the following definition.

Definition A square (or simple square) is a matrix $\pmatrix{a&b\cr c&d}$ of
simple vectors which satisfy (9), and which in the generic case satisfy
$a+b+c+d\in{\hbox{\german S}}$, or in the special cases take on one of the
forms (11)–(13).

The reason for separating the generic from special cases in the definition is
that $a+b+c+d\in{\hbox{\german S}}$ suffices in the generic case, but not in
the special cases. Of course, it holds in the latter cases too, albeit it is
trivial there. It’s also worth noting that given any simple square, one can
multiply any row or column by a scalar without invalidating its status as a
square. And for completeness, let us recall from above that the two row-sums
and the two column-sums also belong to S.

This completes the description of our procedure for defining $f$. In order to
understand why it works, let’s “look behind the curtain” to see what our
squares amount to when expressed in terms of vectors in $V_{1}{\otimes}V_{2}$.
(This should also help to illuminate the rather abstract development we have
been following in this section.) Recall that the four rays in (8) can also be
written as the intersecting subspaces exhibited in (10). Now by equations (3)
and (4), $M=V_{1}{\otimes}\beta_{0}$ for some $\beta_{0}\in V_{2}$, while
$N=\alpha_{0}{\otimes}V_{2}$ for some $\alpha_{0}\in V_{1}$, and similarly
$M^{\prime}=V_{1}{\otimes}\beta$, $N^{\prime}=\alpha{\otimes}V_{2}$, for some
$\alpha$ and $\beta$. Without loss of generality we can therefore write (7) in
the form,

$\pmatrix{a&b\cr&\cr
c&d\cr}=\pmatrix{\alpha_{0}\,\otimes\,\beta_{0}&\alpha_{0}\,\otimes\,\beta\cr&\cr\alpha\,\otimes\,\beta_{0}&\lambda\,\alpha\,\otimes\,\beta\cr}$
$None$

where $\lambda$ is some unknown coefficient of proportionality. This form
makes it plain that the row- and column-sums are indeed simple, for example
$a+b=\alpha_{0}{\otimes}(\beta_{0}+\beta)$. As for the overall sum, $a+b+c+d$,
it will be the simple vector, $(\alpha_{0}+\alpha){\otimes}(\beta_{0}+\beta)$
provided that $\lambda=1$. Were $\lambda\not=1$ on the other hand, the same
sum would equal
$(\alpha_{0}+\alpha){\otimes}(\beta_{0}+\beta)+(\lambda-1)\alpha{\otimes}\beta$,
which according to Lemma 2, could be simple only if $\alpha$ were proportional
to $\alpha_{0}$ or $\beta$ were proportional to $\beta_{0}$, meaning we’d be
back in one of the special cases we disposed of earlier.

In summary, consider a square of simple vectors belonging to rays of the form
exhibited in (10) with $M,M^{\prime}\in{\cal{M}}_{1}$ and
$N,N^{\prime}\in{\cal{M}}_{2}$, and assume we are in the generic case where
$M\not=M^{\prime}$, $N\not=N^{\prime}$. On condition that the sum of all four
simple vectors is itself simple, any three of them determine the fourth
uniquely. The vectors must in that case “secretly” take the form (14) with
$\lambda=1$:

$\pmatrix{\alpha_{0}\,\otimes\,\beta_{0}&\alpha_{0}\,\otimes\,\beta\cr&\cr\alpha\,\otimes\,\beta_{0}&\alpha\,\otimes\,\beta\cr}$
$None$

Our special cases correspond to $\alpha_{0}\propto\alpha$ and/or
$\beta_{0}\propto\beta$, and they also fit the pattern (15), which accordingly
represents the universal form that a square assumes when one views it “from
behind the curtain”.

Returning to the task of lifting $\hat{f}:{\bf{P}}M\to{\bf{P}}M^{\prime}$ to a
linear isomorphism, $f:M\to M^{\prime}$, we can now see that the construction
of $f(v)$ following equation (9) does indeed do the job, because it maps
$M=V_{1}{\otimes}\beta_{0}$ to $M^{\prime}=V_{1}{\otimes}\beta$ by carrying
$\alpha{\otimes}\beta_{0}\in M$ to $\alpha{\otimes}\beta\in M^{\prime}$, a
correspondence which is plainly linear when $\alpha$ varies. Of course, the
fact that $f$ is a lift of $\hat{f}$ cannot determine it uniquely, because any
multiple of a lift is another lift. It’s thus no accident that our
construction involved a free choice of reference vectors, $v_{0}$ and
$v_{0}^{\prime}$. A different choice, however, could only alter $f$ by an
overall factor, as follows from the general fact that any two linear
isomorphisms that induce the same mapping on rays must agree up to scale. ††♭
Proof. Call the maps $f$ and $g$ and let $x$ and $y$ be any two independent
vectors in their domain with $z=x+y$. By assumption $g(x)=\lambda f(x)$ and
$g(y)=\mu f(y)$, and we want to prove that $\mu=\lambda$. By rescaling either
$f$ or $g$ if necessary, we can assume that $\lambda=1$. But then
$f(z)=f(x)+f(y)$ would lie in a different ray from $g(z)=f(x)+\mu f(y)$ unless
$\mu=1$ as well.  For the same reason, we don’t need to check our isomorphisms
$f$ for coherence. Given that they cohere on
${\bf{P}}M\to{\bf{P}}M^{\prime}\to{\bf{P}}M^{\prime\prime}$, as we already
know they do, they must also cohere up to scale on $M\to M^{\prime}\to
M^{\prime\prime}$, and that’s the best we can do.

Taking as input solely the set S of simple vectors in $V$, we have now
identified with one another the members of ${\cal{M}}_{1}$ via isomorphisms
which are unique up to scale. On one hand we used this to derive from
$V$-cum-S a canonically given projective space that is naturally isomorphic to
${\bf{P}}V_{1}$ (the space of “pure states of system-1” in a quantal
interpretation). On the other hand, these same identifications produce a
vector space that is naturally isomorphic to $V_{1}$ itself, albeit only
modulo a scaling ambiguity. The same procedure applied to ${\cal{M}}_{2}$
rather than ${\cal{M}}_{1}$ would obviously recover ${\bf{P}}V_{2}$ and
$V_{2}$ in the same sense. Our task now is to complete the story by re-
building $V$ as the tensor product of the two vector spaces just constructed.

5\. The analysis and synthesis of a tensor product

Our previous work has already led us to pay close attention to the maximal
linear subspaces of ${\hbox{\german S}}\subseteq V=V_{1}\otimes V_{2}$. Let us
now select two such spaces, $W_{1}\in{\cal{M}}_{1}$ and
$W_{2}\in{\cal{M}}_{2}$, and then select further a vector $w_{0}\in W_{1}\cap
W_{2}$ to serve as their common “base point”. ††⋆ In Sections 3 and 4 we
usually used the letters $M$ and $N$ to denote maximal linear subspaces of S.
The notation, $W_{1}$, $W_{2}$, here is chosen to emphasize the parallelism
with $V_{1}$, $V_{2}$. We have then

$W_{1}=\pi_{1}w_{0}\ ,\qquad W_{2}=\pi_{2}w_{0}\ .$

We want to demonstrate that $V$ can be construed as the tensor-product of
these two spaces.

To that end, and basing ourselves on the concept of “square” introduced in
Section$\,4$, we will introduce a new bilinear product,
$\overline{\otimes}:W_{1}\times W_{2}\to V$, as follows. For $w_{i}\in W_{i}\
(i=1,2)$, let us define $\,w_{1}\overline{\otimes}w_{2}\,$ to be the solution
of the following square:

$\pmatrix{w_{0}&w_{2}\cr&\cr w_{1}&w_{1}\overline{\otimes}\,w_{2}\cr}$ $None$

In other words, $w=w_{1}\overline{\otimes}\,w_{2}$ must satisfy the
conditions,

$\,w\in\pi_{2}w_{1}\,\cap\,\pi_{1}w_{2}\,$
$w_{0}+w_{1}+w_{2}+w\,\in\,{\hbox{\german S}}\ .$

As we have seen, these conditions determine $w_{1}\overline{\otimes}\,w_{2}$
uniquely in the generic case where $w_{1}\notin W_{2}$ and $w_{2}\notin
W_{1}$. In the special cases where this is not true, a scaling ambiguity
remains. To supplement (16) for such cases, we can stipulate that
$w_{0}\overline{\otimes}w_{0}=w_{0}$, and more generally that
$w_{0}\overline{\otimes}w_{2}=w_{2}$ and $w_{1}\overline{\otimes}w_{0}=w_{1}$.
These rules ††† In the previous section we already introduced rules for the
special cases; the rules stated here are simply their instances for the
situation at hand. If we have restated them here, it is only in order to make
the definition of $\overline{\otimes}$ more self-contained.  render
$w_{1}\overline{\otimes}\,w_{2}$ unique. For example if $w_{1}\in W_{2}$ then
$w_{1}\in W_{1}\cap W_{2}$, whence $w_{1}=\lambda w_{0}$ since, as always,
$\dim(W_{1}\cap W_{2})=1$. Therefore $w_{1}\overline{\otimes}\,w_{2}=(\lambda
w_{0})\overline{\otimes}\,w_{2}=\lambda w_{2}\,$, exactly as in (11).

We learned in the previous section [following eq. (15)] that
$\overline{\otimes}$ would be bilinear when defined in this manner. ††♭ One
can also deduce the bilinearity of $\overline{\otimes}$ directly from the
definition (16), if one proves first the following useful lemma: The set of
first rows $(a,b)$ which make a square with a fixed second row $(c,d)$ is
closed under addition and scalar multiplication; and similarly for columns
instead of rows. Closure under scalar multiplication we already noticed, and
closure under sum can be deduced from the general square-form (15). Taken
together, the row and column assertions suffice to prove linearity of
$\overline{\otimes}$ in both arguments. Therefore (compare definition (2) of
tensor-product in Section 2) it induces a linear map
$\Phi:W_{1}{\otimes}W_{2}\to V$. In fact $\Phi$ is an isomorphism. To see
this, let’s go back to the representation of $V$ as $V_{1}{\otimes}V_{2}$, and
write $w_{0}=\alpha_{0}{\otimes}\beta_{0}$, $w_{1}=\alpha{\otimes}\beta_{0}$,
$w_{2}=\alpha_{0}{\otimes}\beta$. Then as one sees by comparing (15) with
(16), $w_{1}\overline{\otimes}w_{2}=\alpha{\otimes}\beta$. This means first of
all that the simple vectors $\alpha{\otimes}\beta\in{\hbox{\german S}}$
coincide with the vectors of the form $w_{1}\overline{\otimes}w_{2}$ for some
$w_{i}\in W_{i}\ (i=1,2)$. Consequently, we can build up a basis of $V$ by
choosing a basis $\left\\{e_{j}\,|\,j=1\cdots\dim W_{1}\right\\}$ for $W_{1}$,
and a similar basis $\left\\{f_{k}\,|\,k=1\cdots\dim W_{2}\right\\}$ for
$W_{2}$, and then taking our basis-elements to be
$e_{jk}=e_{j}\overline{\otimes}\,f_{k}$. That these $e_{jk}$ constitute a
basis for $V$ follows from the fact that the $e_{j}$ (respectively the
$f_{k}$) have the form $\alpha_{j}{\otimes}\beta_{0}$ (resp.
$\alpha_{0}{\otimes}\beta_{k}$), whereby the $\alpha_{j}$ (resp. $\beta_{k}$)
constitute a basis of $V_{1}$ (resp. $V_{2}$) if and only if the $e_{j}$
(resp. $f_{k}$) constitute a basis of $W_{1}$ (resp. $W_{2}$), and furthermore
$e_{jk}=e_{j}\overline{\otimes}\,f_{k}=\alpha_{j}{\otimes}\beta_{k}$.

The upshot is that a “special basis” for $V$ (i.e. a basis of vectors
$\alpha_{j}{\otimes}\beta_{k}$) is the same thing as a pair of bases for
$W_{1}$ and $W_{2}$, modulo the familiar $GL(1)$ ambiguity that one can
rescale the $W_{1}$-basis by $\lambda$ if one simultaneously rescales the
$W_{2}$-basis by $1/\lambda$. Recall now from Section 2 that our “second
possible answer” to what constitutes a tensor-product structure for $V$ was
“an equivalence-class T of special bases for $V$”. We have thus demonstrated
that from ${\hbox{\german S}}\subseteq V$ one can derive uniquely a tensor-
product structure in that sense. Conversely, given such a structure T, we
immediately obtain S from it as the union of all of the members of the special
bases that comprise T. To the extent that S is a simpler and more natural
object than an equivalence class of special bases (and is also more intrinsic
to $V$), we have reason to maintain that in S we have an answer to the
question, “What does it mean for $V$ to be a tensor product?”.

Using “pointed vector spaces”

The above construction began with an arbitrarily chosen “base-vector”
$w_{0}\in V$ such that $W_{1}=\pi_{1}w_{0}$ and $W_{2}=\pi_{2}w_{0}$. The
ambiguity inherent in such a choice does not impugn our demonstration of the
equivalence, ${\hbox{\german T}}\leftrightarrow{\hbox{\german S}}$, but it
does mean that in the procession,
$(V_{1},V_{2})\,\to\,V\hbox{-cum-}{\hbox{\german S}}\,\to\,(W_{1},W_{2})$, a
different choice of $w_{0}$ would produce a different pair of spaces, $W_{1}$,
$W_{2}$. If desired, one could arrange for $W_{1}$ and $W_{2}$ to be unique by
working with “pointed vector spaces”, i.e. by equipping $V_{1}$ and $V_{2}$
with distinguished “base-points”, $\alpha_{0}\in V_{1}$ and $\beta_{0}\in
V_{2}$, and then taking $v_{0}=\alpha_{0}{\otimes}\beta_{0}\in
V_{1}{\otimes}V_{2}$ to be the base-point of $V$. Our construction above (with
$w_{0}$ taken to be $v_{0}$) would then recover the pairs ($V_{1},\alpha_{0}$)
and ($V_{2},\beta_{0}$) essentially uniquely from the triple
($V,{\hbox{\german S}},v_{0}$).

Remark Interestingly, the “histories Hilbert spaces” H that play a role in
Quantum Measure Theory [3] automatically come with distinguished vectors
$|\Omega\rangle\in{\hbox{\german H}}$, where $\Omega$ represents the full
history-space (the unit of the corresponding event-algebra). However it is
generally false for coupled subsystems that H for the composite system is the
Hilbert-space tensor product of the H’s for the subsystems. (Even when the
vectorspace dimensions match, the norms in general will not.)

6\. Categorical matters (and a shortcoming of the functor concept)

From a given vector space $V$ one can form new spaces, like the dual-space
$V^{*}$ or the double dual $V^{**}$. With two vector spaces, there are other
possibilities, including their direct sum, their tensor product, and so forth.
Although a vector space formed in one of these ways will be isomorphic to
infinitely many other vector spaces, its “inner constitution” will in general
be distinctive, with the result that it will to a certain extent “remember
where it came from”. One may say then that it carries the structure of a dual
space, a direct sum, or a tensor product, as the case may be. In each instance
one can try to identify concretely where this extra information resides, and
for a vector-space $V$ that arose as a tensor product, our discussion has
pointed to the set S of simple vectors within $V$ as the pertinent structure.
Adopting a notation that keeps track of S, we may say that from an ordered
pair $(V_{1},V_{2})$ of vector spaces, there arises via tensor-product the
ordered pair $(V,{\hbox{\german S}})$.

A question then is to what extent the transformation
$(V_{1},V_{2})\to(V,{\hbox{\german S}})$ is reversible. How perfectly does $V$
remember where it came from, or to ask this another way, how well can we
reconstruct $V_{1}$ and $V_{2}$, given $V$ and S? When we dealt with pointed
spaces, we discovered that $(V_{1},V_{2})$ could “in essence” be recovered
fully. But in the unpointed case, it appeared that although $(V,{\hbox{\german
S}})$ is determined by $(V_{1},V_{2})$, the latter could be recovered from the
former only up to some sort of $GL(1)$ ambiguity. This suggests that in the
pointed case a vector space carrying the structure of a tensor product is in
some sense equivalent to the factor spaces from which it arose, whereas in the
un-pointed case there is only partial equivalence.

But what concept of equivalence is implicitly animating these expectations?
Simple isomorphism will not do, being too narrow in one way (because by
definition structures of different types like $(V_{1},V_{2})$ and
$(V,{\hbox{\german S}})$ cannot be isomorphic) and too broad in another way
(because, for example, any two vector spaces of equal dimension are
isomorphic). Maybe one can put the underlying thought into words by saying
that “$A$ and $B$ are equivalent if $B$ can be constructed from $A$ and vice
versa.” I am not sure that mathematics knows any framework which really does
justice to this thought, but perhaps the category-theoretical concepts of
functor and natural isomorphism come closest to providing one, and so it seems
worth considering how they apply to the question at hand. We will take this up
momentarily, but first let’s see a very simple illustration of how S is able
to remember “where $V$ came from”.

A small illustration: topology remembers spin

As a small illustration of how the set S of simple tensors encodes the
structure of $V$ as a tensor product, let us return to the example of
$spin$-$3/2$ ${\otimes}$ $spin$-$1$ versus $spin$-$1/2$ ${\otimes}$
$spin$-$5/2$. To distinguish these two possible provenances of $V$, one from
the other, it is enough to pay attention to the topology of S, for example its
dimensionality. Taking into account that an element of S has by definition the
form $\alpha\otimes\beta$, and that $\alpha$ and $\beta$ are unique modulo the
obvious $GL(1)$ ambiguity, we can observe that the (complex) dimensionality of
S is one less than the sum of the dimensionalities of the factor spaces. In
our examples this yields for $\dim({\hbox{\german S}})$ the respective values,
$4+3-1=6$ for 3/2$\,\otimes\,$1 and $2+6-1=7$ for 1/2$\,\otimes\,$5/2. In
fact, it is easy to verify that this simple test works in general. If we know
that $V$ arose from the combination of two spins, then the topological
dimension of S determines fully what those spins were. Of course (and as we
have now seen in great detail) the same information can be deduced with a bit
more work from the dimensionalities of the maximal linear subspaces of S, or
from the dimensionalities of ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$ as
“foliations” of S.

A functorial gloss on our constructions

Now back to categories, functors, and natural isomorphisms. A category is
basically a collection of spaces of a given type (its “objects”) and of
structure-preserving mappings between these spaces (its “morphisms”). For our
purposes it will be best to limit the latter to isomorphisms, i.e. to require
them to be invertible. A functor between two categories, I and II, is a kind
of black box that converts the objects and morphisms of category-I to objects
and morphisms of category-II while preserving composition of morphisms.
Conceptually, it is telling you that you can build spaces and mappings of type
II from spaces and mappings of type I (but unfortunately it is not telling you
how to do so.)

The two categories of interest to us here can be denoted as
$\hbox{VEC}\times\hbox{VEC}$ and TVEC, where the former is the category of
pairs $(V_{1},V_{2})$ and the latter ††⋆ The “T” in TVEC is meant to suggest
the word “tensor” is the category of pairs $(V,{\hbox{\german S}}_{V})$. A
morphism in $\hbox{VEC}\times\hbox{VEC}$ will thus be a pair of linear
isomorphisms, while a morphism in TVEC will be a linear isomorphism between
vector spaces that preserves their respective subsets S. When our spaces are
pointed, all these isomorphisms will of course also need to preserve the
respective base-points. Let us now describe some of our constructions in terms
of functors between $\hbox{VEC}\times\hbox{VEC}$ and TVEC, concentrating for
the time being exclusively on the pointed case.

The first functor of interest, which we will designate as
${\otimes}:\hbox{VEC}\times\hbox{VEC}\to\hbox{TVEC}$, is that induced by the
tensor product itself. It takes a pair of vector spaces $(A,B)$ to their
tensor product, $V=A{\otimes}B$ equipped with its space ${\hbox{\german
S}}_{V}$ of simple vectors $\alpha{\otimes}\beta$, and it takes a pair $(f,g)$
of (invertible) linear functions between vector spaces to their tensor product
$f{\otimes}g$. Conversely, given an object $(V,{\hbox{\german
S}}_{V},v_{0})\in\hbox{TVEC}$ (where I’ve now indicated the base-point $v_{0}$
explicitly), we saw how to locate within $V$ the subspaces
$W_{1}=\pi_{1}(v_{0})$ and $W_{2}=\pi_{2}(v_{0})$, which were certain maximal
linear subsets of ${\hbox{\german S}}_{V}$. Thereby, we in effect defined a
second functor, $D:\hbox{TVEC}\to\hbox{VEC}\times\hbox{VEC}$, that goes in the
direction opposite to ${\otimes}$, and for which $D(V,{\hbox{\german
S}}_{V},v_{0})=((W_{1},v_{0}),(W_{2},v_{0}))$. Of course one has not defined a
functor fully until one tells how it acts on morphisms, but that is self-
evident for $D$. A morphism in TVEC from $(V,{\hbox{\german S}}_{V},v_{0})$ to
$(V^{\prime},{\hbox{\german S}}_{V}^{\prime},v_{0}^{\prime})$ is nothing but
an invertible linear function, $f:V\to V^{\prime}$, such that
$f[{\hbox{\german S}}_{V}]={\hbox{\german S}}_{V}^{\prime}$ and
$f(v_{0})=v_{0}^{\prime}$. Such an $f$ induces immediately a pair of
(basepoint preserving) functions $f_{1}:W_{1}{\to}W_{1}^{\prime}$ and
$f_{2}:W_{2}{\to}W_{2}^{\prime}$, and so $D(f)=(f_{1},f_{2})$.

Now what of the expectation that ${\otimes}$ and $D$ are in essence each
other’s inverses? Were that literally true, we would be able to express it by
writing ${\otimes}\circ D=1$ and $D\circ{\otimes}=1$, but unfortunately both
equations are, strictly speaking, false. Consider first the composed functor,
$D\circ{\otimes}$. What happens when we apply it to the pair of pointed vector
spaces $((A,\alpha_{0}),(B,\beta_{0}))$? Tracing through the definitions, we
find ${\otimes}((A,\alpha_{0}),(B,\beta_{0}))=(A{\otimes}B,{\hbox{\german
S}}_{A{\otimes}B},\alpha_{0}{\otimes}\beta_{0})$, and then
$D(A{\otimes}B,{\hbox{\german
S}}_{A{\otimes}B},\alpha_{0}{\otimes}\beta_{0})=((W_{1},w_{1}),(W_{2},w_{2}))$,
where $W_{1}=\pi_{1}(\alpha_{0}{\otimes}\beta_{0})=A{\otimes}\beta_{0}$,
$W_{2}=\pi_{2}(\alpha_{0}{\otimes}\beta_{0})=\alpha_{0}{\otimes}B$, and
$w_{1}=w_{2}=\alpha_{0}{\otimes}\beta_{0}$. In other words,

$(D\circ{\otimes})((A,\alpha_{0}),(B,\beta_{0}))=((A{\otimes}\beta_{0},\alpha_{0}{\otimes}\beta_{0}),(\alpha_{0}{\otimes}B,\alpha_{0}{\otimes}\beta_{0}))$
$None$

While $(D\circ{\otimes})((A,\alpha_{0}),(B,\beta_{0}))$ is thus not exactly
identical with $((A,\alpha_{0}),(B,\beta_{0}))$, there is between them an
obvious correspondence,
$((A,\alpha_{0}),(B,\beta_{0}))\longleftrightarrow(D\circ{\otimes})((A,\alpha_{0}),(B,\beta_{0}))$,
given by the linear isomorphisms,

$\alpha\,\longleftrightarrow\,\alpha\,\otimes\beta_{0}\qquad\hbox{and}\qquad\beta\,\longleftrightarrow\,\alpha_{0}\,\otimes\beta$
$None$

The bijection (18) is an instance of what is called a natural isomorphism
between functors, and so category theory gives us a precise way to express
that $D$ is effectively a right-inverse of ${\otimes}$ by saying that
${\otimes}\circ D$ is “naturally isomorphic” to the identity-functor, a
relationship which we will write as

${\otimes}\circ D\cong 1\ .$ $None$

It is evident from its definition that the correspondence (18) establishes an
isomorphism between two objects in $\hbox{VEC}\times\hbox{VEC}$. If read from
left to right, it is a mapping
$\Psi:((A,\alpha_{0}),(B,\beta_{0}))\to(D\circ\otimes)((A,\alpha_{0}),(B,\beta_{0}))$,
but what is it that earns $\Psi$ the title, “natural”, thereby authorizing the
use of the symbol $\cong$ in (19)? It is that $\Psi$ also induces the correct
correspondence between morphisms by converting $(f_{1},f_{2})$ into
$(D\circ\otimes)(f_{1},f_{2})$. This is self-evident when one unpacks the
definitions (cf. (20) below), but even without unpacking the definitions, we
could have been assured that $\Psi$ was natural, if we had reflected that it
was defined intrinsically, utilizing nothing more than the structures
displayed in (17). Indeed, I think it would be fair to say that this
possibility of being constructed from intrinsic information without the
intervention of any arbitrary choices is what best expresses the intuitive
meaning of “naturality”.

The distinction between plain isomorphism $\simeq$ and natural isomorphism
$\cong$ is perhaps most familiar in the example of dual vector-spaces, where
both $V^{**}$ and $V^{*}$ are isomorphic to $V$, but only the isomorphism
between $V^{**}$ and $V$ is natural. Given an element $v\in V$ one can define
$v^{**}\in V^{**}$ by the equation, $v^{**}(f)=f(v)$, where $f\in V^{*}$. On
the other hand, there is no way to pass deterministically from $v$ to an
element $f\in V^{*}$ without the aid of a basis for $V$, or a metric, or some
such auxiliary structure.

Remark A natural isomorphism sets up an equivalence between functors in much
the same way as a similarity transformation sets up an equivalence between
group representations. If $R_{1}$ and $R_{2}$ are representations of the group
$G$ related by the similarity transformation $S$, then
$SR_{1}(g)S^{-1}=R_{2}(g)$, or equivalently $SR_{1}(g)=R_{2}(g)S$. Now if we
replace $R_{1}$ and $R_{2}$ by functors $F_{1}$ and $F_{2}$, and the arbitrary
group-element $g$ by an arbitrary morphism $f$, we obtain the condition for a
family of invertible morphisms $S$ to define a natural isomorphism between
$F_{1}$ and $F_{2}$, namely $SF_{1}(f)=F_{2}(f)S$. Often this last equation is
represented by drawing the commutative diagram,

$\matrix{F_{1}X&{F_{1}f\atop\rightarrow}&F_{1}Y\cr
S_{X}\downarrow\quad&&\quad\downarrow S_{Y}\cr F_{2}X&{\rightarrow\atop
F_{2}f}&F_{2}Y\cr}$

where $X$ and $Y$ are any objects in the category and $f:X\to Y$ is any
morphism between them. When, as in our case, $F_{1}$ is the identity functor,
the diagram for $S:1\to F$ simplifies to

$\matrix{X&{f\atop\rightarrow}&Y\cr S\downarrow\quad&&\quad\downarrow S\cr
FX&{\rightarrow\atop Ff}&FY\cr}$ $None$

One sees in this case that the functor $F$ must be a bijection between the
morphisms $f$ and the morphisms $Ff$; and conversely, the fact that $F$ is
such a bijection captures to a large extent everything that the equation
$F\cong 1$ means.

Having established that $D\circ{\otimes}\cong 1$, let us now try to
demonstrate the complementary equivalence, ${\otimes}\circ D\cong 1$.
Following the same steps as before, let us apply the functor ${\otimes}\circ
D$ to the object $(V,{\hbox{\german S}}_{V},v_{0})\in\hbox{TVEC}$, obtaining
first $D(V,{\hbox{\german S}}_{V},v_{0})=((W_{1},v_{0}),(W_{2},v_{0}))$ and
then
${\otimes}((W_{1},v_{0}),(W_{2},v_{0}))=(W_{1}{\otimes}W_{2},{\hbox{\german
S}}_{W_{1}{\otimes}W_{2}},v_{0}{\otimes}v_{0})$, which taken together tell us
that

$({\otimes}\circ D)(V,{\hbox{\german
S}}_{V},v_{0})=(W_{1}{\otimes}W_{2},{\hbox{\german
S}}_{W_{1}{\otimes}W_{2}},v_{0}{\otimes}v_{0})$ $None$

Can we exhibit a natural isomorphism equating
$(W_{1}{\otimes}W_{2},{\hbox{\german
S}}_{W_{1}{\otimes}W_{2}},v_{0}{\otimes}v_{0})$ to $(V,{\hbox{\german
S}}_{V},v_{0})$, and therefore ${\otimes}\circ D$ to the identity functor? To
this question we already have the answer in the form of the isomorphism,
$\Phi:W_{1}{\otimes}W_{2}\to V$, which we constructed earlier with the aid of
the intrinsically defined product $\overline{\otimes}$, and for which
$\Phi(w_{1}{\otimes}w_{2})=(w_{1}\overline{\otimes}w_{2})$. As with $\Psi$
before, it is straightforward to verify that $\Phi$ is natural, as indeed it
had to be, given its intrinsic nature. Therefore ${\otimes}\circ D\cong 1$.

We have now proven that the composition of ${\otimes}$ with $D$ in either
order is naturally isomorphic to the identity. Thus category theory, by
introducing the concept of natural isomorphism $\cong$ as a replacement for
strict equality, has given us a way to make precise (and then to verify) the
informal claims that, in the pointed case, the functor ${\otimes}$ is
invertible and that $D$ is its inverse.

Turn now to the unpointed case and to our expectation that it will not be
possible to recover the pair $(A,B)$ from $(A{\otimes}B,{\hbox{\german S}})$
when the spaces involved are not equipped with base-points. Can we also
corroborate this expectation within the categorical framework? Stated
formally, the question is whether there exists a functor
$D:\hbox{TVEC}\to\hbox{VEC}\times\hbox{VEC}$ which is a “left inverse” to
${\otimes}$ in the sense that $D\circ{\otimes}\cong 1$. In fact, it’s easy to
see that no such functor can exist. Were $D$ such a functor then, as we
noticed in connection with (20), the mapping, $f\mapsto(D\circ\otimes)f$,
would have to be invertible for morphisms,
$f:(A,B)\to(A^{\prime},B^{\prime})$, of the category
$\hbox{VEC}\times\hbox{VEC}$, where $f$ is by definition a pair $(g,h)$ of
individual morphisms in VEC. This, however, is clearly impossible because the
functor ${\otimes}$ (and therefore its composition with $D$ if the latter
existed) fails to be injective, since it maps both $f=(g,h)$ and
$\tilde{f}=(\lambda g,h/\lambda)$ to the single morphism $g{\otimes}h=(\lambda
g){\otimes}(h/\lambda)$. In other words, ${\otimes}$ acting on morphisms is
not injective but many-to-one, the “many” being parametrized by a non-zero
scalar $\lambda$ which embodies the same $GL(1)$ ambiguity we met with
earlier. This confirms that the equation $D\circ{\otimes}\cong 1$ can have no
solution, and a fortiori that the functor ${\otimes}$ is not invertible.

A somewhat simpler example of the same nature occurs in connection with the
attempt to represent a spinor geometrically. Starting from a 2-component Weyl
spinor $\zeta$, for example, one can derive algebraically a so-called null
flag $F$, which consists of a lightlike vector together with a half-plane
matched to the vector. [4] But because vectors are quadratically related to
spinors, both $\zeta$ and $-\zeta$ give rise to the same flag $F$, whence one
can recover the spinor from the flag only up to an unknown sign. (This loss of
information was inevitable, because spinors change sign after rotating through
$2\pi$, whereas vectors do not.) To couch these relationships in categorical
language, one could introduce a category of spinor-spaces and a category of
spaces of null-flags and a functor $\phi$ from the former to the latter. Like
${\otimes}$ above, $\phi$ would not be invertible, because it would be $2\to
1$ on morphisms. At best, one might be able to devise, as a kind of right
inverse to $\phi$, a “functor manqué” or “functor up to sign” going from flag-
spaces to spinor spaces. Its existence would proclaim that, although not fully
a geometrical object, a spinor is nevertheless “geometrical up to sign”.

Remark Despite its utility, the concept of functor does not necessarily
illuminate the connection between its inputs and its outputs as fully as one
might have expected it to do, because unlike a morphism, it is blind to the
individual elements of the spaces on which it acts; by definition it does not
“look inside”. Thus if $\phi$ is a functor and $X$ a space (or a mapping), and
if $Y$ is the space (or mapping) that results when $\phi$ acts on $X$, then
the equation $Y=\phi(X)$ tells us that $Y$ is in some sense built from $X$,
but it tells us nothing concretely about how $Y$ is built from $X$. ††† Could
Bourbaki’s concept [5] of “deduction of structures” come any closer to doing
this? For example $X$ could be a spinor-space, and $\phi$ the above functor.
Then $Y=\phi(X)$ would be the space comprised of all the null flags derived
from the spinors comprising $X$. But if $\zeta\in X$ were an individual spinor
in $X$, and if $F$ were the individual flag derived from $\zeta$, the rules
governing functors would not allow us to write “$F=\phi(\zeta)$”, even though
it might seem natural to do so, and even though we know perfectly well what we
would mean by it!

7\. Questions; further developments; connection to quantum field theory

In conclusion, let me mention a few questions and possible further
developments suggested by the above considerations.

The most important of the constructions introduced in Sections 4 and 5 revolve
around the “foliations” ${\cal{M}}_{1}$ and ${\cal{M}}_{2}$, the corresponding
mappings $\pi_{1}$ and $\pi_{2}$, the square concept, and the product
$\overline{\otimes}$ which results from these via the definition (16).

An obvious question that one might ask is how these distinctive ingredients
generalize to the tensor product of three or more vector spaces. One could of
course just treat a threefold product like $A\otimes{}B\otimes C$ as an
iterated pairwise product like $(A{\otimes}B)\otimes C$, but a more symmetric
construction ought to be possible, and one might expect it to uncover some new
structures that are not visible in connection with simple pairwise products
like $A\otimes{}B$.

One might also wonder whether there was anything of interest to be learned
from the study of the various symmetry types that become possible when two or
more of the factor-spaces are equal to each other. For example when
$V\subseteq{A}\otimes A$ is the subspace of symmetric tensors, two natural
analogs of S as used above would be the set of tensors of the form,
$\alpha\otimes\beta+\beta\otimes\alpha$, or even more simply, of the form
$\alpha\otimes\alpha$. Or for the anti-symmetric tensor product of $A$ with
itself, the set of tensors of the form,
$\alpha\wedge\beta=\alpha\otimes\beta-\beta\otimes\alpha$, would be a natural
analog of S. To what extent, and in what form, could one repeat the above
discussion with one of these subsets replacing S? And still more generally,
what might be an analog of S belonging to the non-abelian symmetry-types
(those corresponding to more general Young tableaux) that arise as subspaces
of higher products, ${A}\otimes A\cdots\otimes A$, and which are neither
“bosonic” nor “fermionic”?

Another sub-case of obvious interest is that where the vector spaces are
equipped with metrics, in particular where they are Hilbert spaces. One might
then expect orthonormality to play a role, but would any additional,
unexpected features of interest show up?

Our discussion so far has proceeded in finite dimensions. If we want to
generalize it to infinite-dimensional vector spaces, a whole raft of further
questions will appear, some of which concern the definition of tensor product
itself. Clearly, the subset S of simple tensors $\alpha\otimes\beta$ within
$V=A\otimes{B}$ can be defined without difficulty, but will our constructions
based on it also go through as before? Will they still let us recover $A$ and
$B$, and will they still lead us as in Sections 4 and 6 (say in the pointed
case) to a functor $D$ inverse to $\otimes$? In all of this, what consequences
might flow from ambiguities in the definition of $\otimes$? When $A$ and $B$
are Hilbert spaces, $A\otimes{B}$ qua Hilbert space is unambiguous, but when
they are only Banach spaces (normed vector spaces), many different spaces
$A\otimes{B}$ have been defined [2]. One may wonder then whether S will still
be able to “remember” which specific choice of $\otimes$ went into the
creation of $V$.

Among infinite-dimensional vector spaces, the Hilbert spaces have a special
significance for quantum theories. Although the ambiguity in defining
$A\otimes B$ is not an issue when $A$ and $B$ are Hilbert spaces, it can
happen in connection with quantum field theory that the notion of tensor
product itself seems to be transcended. If one divides a Cauchy surface into
two complementary regions, then naively one would expect the overall Hilbert
space H of the field-theory to be the tensor product of Hilbert spaces
associated with the two regions, just as happens with composite systems in
ordinary quantum mechanics. Unfortunately, this would conflict with the fact
that the operator algebras associated with the two regions (technically with
their domains of dependence in spacetime) are known (for free fields) to be of
“type III”, this being intimately linked to the infinite entanglement-entropy
between the two regions. One still has operator subalgebras for the regions
(so-called coupled factors), but these subalgebras cannot be interpreted as
acting on the separate factors of a tensor-product. One thus confronts
something like a tensor product of operator-algebras that does not derive from
a tensor-product decomposition of the underlying Hilbert space H. (Adopting
the language of “quantum systems”, one might say that one is dealing with
“subsystems which possess observables but lack state-vectors”.) ††♭ This is
not quite the same as saying that a type-III factor lacks pure states. As most
commonly defined in the theory of operator-algebras, a pure state on A is an
extreme point in the convex set of normalized positive linear functionals on
A. It is known that such pure states exist copiously, and one could thus
entertain them as generalized state-vectors, since in finite dimensions,
state-vector = pure state. However when A is a type-III factor, its pure
states seem to be mathematically pathological (perhaps even “ineffable”), and
one could plausibly regard them as unphysical. See [6].

In the absence of a tensor-product structure for H, the notion of simple-
vector is not defined, and therefore neither is our subset ${\hbox{\german
S}}\subseteq{\hbox{\german H}}$. Nevertheless, one might hope that some
generalization of simple vector, and some corresponding subset of H, could
serve a similar function. Simple vectors are tensors of rank 1, but the
tensors of ranks, 2, 3, 4, etc. also respond to the tensor-product structure
of $V$. Could it be that suitable analogs of the spaces of such tensors (or
better, of the tensors of finite “co-rank” in some suitable sense) are able to
capture the structure of coupled factors of types-III or II, and in so doing
shed light on features like the area-law for entanglement-entropy? Especially
salient in this connection is the “spatiotemporal cutoff” needed to render the
entropy finite [7]. Physically, such a cutoff needs to be frame-independent
(locally Lorentz invariant), and it seems suggestive that an analog of S, if
it could be defined, would not obviously need to refer to any arbitrarily
chosen reference-frame.

As a first approach to some of these questions, one could ask in finite
dimensions how to relate the operator-algebra framework to that of the present
paper. Indeed, one might have thought to identify a tensor-product-structure
for $V$, not with the family S of simple vectors in $V$, but with a pair of
commuting operator-subalgebras which generate the algebra $L(V)$ of all linear
operators on $V$ and which have in common only the multiples of the identity-
operator (like Murray-von Neumann coupled factors but without any
specialization to complex numbers or self-adjointness). The advantage of such
an alternative approach would be that the algebras $L(A)$ and $L(B)$ reappear
bodily in $L(A\otimes B)$, whereas the spaces $A$ and $B$ themselves need to
be excavated from $A\otimes B$ more painfully, as we have seen in great detail
above. (Quantally speaking, the “observables” of a subsystem carry over to the
composite system, whereas the “states” do not. But see the remark below.) Its
disadvantage would be that an algebra of operators in $V$ is a considerably
more complicated beast than the simple subset ${\hbox{\german
S}}\subseteq{V}$. Be that as it may, it’s clear that the two viewpoints are
related. For example, an operator acting only on $A$ (an operator in
$L(A)\otimes{\bf{1}}$), or only on $B$, will automatically be an operator that
preserves S, suggesting how one might derive $L(A)$ and $L(B)$ from S.

Remark In the context of Quantum Measure Theory [3], the histories-hilbert-
space associated to a subsystem actually does reappear as a true subspace of
the histories-hilbert-space of the full system, the reason being that an event
in a subsystem is ipso facto an event in the full system. Moreover this
subspace carries a distinguished “base point”, as remarked in Section 5. When,
in addition, the overall quantum-measure is the product-measure (as for “non-
interacting subsystems in a product-state”), the full histories-hilbert-space
is the tensor product of the subspaces, and the aforementioned advantage of an
approach via operator algebras disappears.

Let us return, finally, to finite dimensions and to the cone S of simple
vectors within $V$, on which most of our work has been based. We have seen how
S endows $V$ with the structure of a product space, but we did not provide (or
even ask for) a simple criterion that would let us recognize whether a given
subset S could actually play the role assigned to it. That is, we did not
provide necessary and sufficient conditions for there to exist an isomorphism
mapping $V$ to a space $A\otimes B$ that would map S to the set of tensors of
the form $\alpha\otimes\beta$. One trivially adequate criterion is that the
re-constructions undertaken in Section 5 should succeed, and in particular
that the building up of the squares should never encounter an obstacle. But
one might wish for criteria that were more self-contained and more simply
stated. Given that the rays in S constitute a “Segre variety”, one might hope
that the Algebraic-Geometry literature would contain something of this sort.

Alternatively, rather than seeking axioms for S, one might instead seek axioms
for the squares themselves, i.e. axioms for quadruples of vectors in $V$. A
tensor-product structure for $V$ would then be a set of quadruples satisfying
these axioms.

This research was supported in part by NSERC through grant RGPIN-418709-2012.
This research was supported in part by Perimeter Institute for Theoretical
Physics. Research at Perimeter Institute is supported by the Government of
Canada through Industry Canada and by the Province of Ontario through the
Ministry of Economic Development and Innovation.

References

[1] For more detail on these definitions, see:

Shlomo Sternberg, Lectures on Differential Geometry (Englewood Cliffs, N.J:
Prentice-Hall, 1964)

Tracy Y. Thomas, Concepts from Tensor Analysis and Differential Geometry
(Academic Press, 1961), pages 7ff

Saunders Mac Lane, Homology (Academic Press, 1963), pages 138ff

Chapter 1 of Sternberg contains examples of most of the definitions, including
implicitly the definition of $V\otimes W$ as the space of bilinear mappings of
$V^{*}\times W^{*}$ into ℝ. The time-honoured definition of a tensor in terms
of transformation laws for its components is presented in Thomas.

[2] A. Grothendieck, “Produits tensoriels topologiques et espaces nucléaires”,
Memoirs of the American Mathematical Society–16 : (1955)

[3] Rafael D. Sorkin, “Quantum Mechanics as Quantum Measure Theory”, Mod.
Phys. Lett. A–9 (No. 33) : 3119-3127 (1994), gr-qc/9401003,
http://www.pitp.ca/personal/rsorkin/some.papers/80.qmqmt.pdf

Fay Dowker and Rafael D. Sorkin, “An intrinsic causality principle in
histories-based quantum theory: a proposal” (to appear)

Fay Dowker, Steven Johnston, Sumati Surya, “On extending the Quantum Measure”,
J. Phys. A–43 : 505305 (2010) arXiv:1007.2725 [gr-qc]

[4] Roger Penrose, “Structure of space-time”, in Battelle Rencontres: 1967
Lectures in Mathematics and Physics (edited by Cecile M. DeWitt and John A.
Wheeler) (New York: Benjamin, 1968)

[5] N. Bourbaki, Theory of Sets  (Paris: Hermann, 1968), Chapter IV

[6] Bruce Blackadar (2017), Operator Algebras: Theory of C*-Algebras and von
Neumann Algebras, https://packpages.unr.edu/media/1224/cycr.pdf See III.2.2.3
and the remarks in III.2.2.15

[7] Rafael D. Sorkin, “On the Entropy of the Vacuum Outside a Horizon”, in B.
Bertotti, F. de Felice and A. Pascolini (eds.), Tenth International Conference
on General Relativity and Gravitation (held Padova, 4-9 July, 1983),
Contributed Papers, vol. II, pp. 734-736 (Roma, Consiglio Nazionale Delle
Ricerche, 1983),
http://www.pitp.ca/personal/rsorkin/some.papers/31.padova.entropy.pdf
http://arxiv.org/abs/1402.3589

Rafael D. Sorkin, “The Statistical Mechanics of Black Hole Thermodynamics”, in
R.M. Wald (ed.) Black Holes and Relativistic Stars, (U. of Chicago Press,
1998), pp. 177-194 gr-qc/9705006
http://www.pitp.ca/personal/rsorkin/some.papers/92.chandra.pdf