We study the geometry of the Fano schemes $\fano{k}{\SD[r]{n}}$ of the projective variety $\SD[r]{n}$ defined by the $r\times r$ minors of a symmetric $n\times n$ matrix filled with indeterminates.
These schemes are fine moduli spaces parameterizing $(k+1)$-dimensional linear spaces of symmetric matrices of rank less than $r$.
We prove that the schemes $\fano{k}{\SD[r]{n}}$ can have generically non-reduced components, characterize their irreducibility and connectedness, and give results on their smoothness.
Our approach to connectedness also applies to Fano schemes of rectangular matrices as well as alternating matrices and answers a question of Ilten and Chan.
Furthermore, we give a complete description of $\fano{1}{\SD[r]{n}}$ and show that when $r=n$, the Fano schemes of lines have the expected dimension.
As an application, we provide geometric arguments for several previous results concerning spaces of symmetric matrices of bounded rank.

§ INTRODUCTION

§.§ Motivation

Let $\KK$ be an algebraically closed field of characteristic not equal to 2.
For an integer $n\geq 3$, consider the $\KK$-vector space $S_n$ of symmetric $n \times n$ matrices with entries in $\KK$.
Fixing an integer $3\leq r\leq n$, we are interested in studying those subspaces $Q\subset S_n$ such that every matrix in $Q$ has rank less than $r$.

Denote by $\SD[r]{n}$ the closed subscheme of $\PP_{\KK}^{\binom{n+1}{2}-1}$ cut out by the $r\times r$ minors of the generic symmetric matrix

\begin{equation}
\label{eq:gen-sym-matrix}
\begin{pmatrix}
x_{1,1} & x_{1,2} & \cdots & x_{1,n} \\
x_{1,2} & x_{2,2} &        & x_{2,n} \\
\vdots  &         & \ddots & \vdots  \\
x_{1,n} & x_{2,n} & \cdots & x_{n,n}
\end{pmatrix},
\end{equation}

filled with $\binom{n+1}{2}$ independent forms $x_{i,j}$.
The variety $\SD[r]{n}$ is irreducible and has dimension $n(r-1)-\binom{r-1}{2}-1$ (see <cit.>).
In this article we study the Fano scheme $\fano{k}{\SD[r]{n}}$ that parameterizes those $k$-planes in $\PP_{\KK}^{\binom{n+1}{2}-1}$ that lie on $\SD[r]{n}$.
This is a subscheme of the Grassmannian $\Gr\left(k+1,\binom{n+1}{2}\right)$ whose closed points correspond to $(k+1)$-dimensional linear subspaces $Q\subset S_n$ containing matrices of rank less than $r$.
See <cit.> for details on Fano schemes.

Our original motivation for studying the Fano schemes $\fano{k}{\SD[r]{n}}$ was to generalize the work of Ilten and Chan <cit.> on Fano schemes $\fano{k}{D_{m,n}^r}$ of rectangular matrices of bounded rank to the symmetric setting.
Their systematic use of the dimensions of tangent spaces to $\fano{k}{D_{m,n}^r}$ at select points characterizes the irreducibility and smoothness of $\fano{k}{D_{m,n}^r}$, and gives results on the components and connectedness of these schemes.
Their techniques will prove helpful in our study of $\fano{k}{\SD[r]{n}}$, although we will encounter new challenges in the symmetric case.
The difficulty is due to the existence of generically non-reduced components which prevents us from leveraging the full power of tangent space dimensions.

The next reason to study $\fano{k}{\SD[r]{n}}$ is to provide geometric proofs for several previous results concerning subspaces of $S_n$ of bounded rank.
One motivation for these results is a classical theorem due to Kronecker and Weierstrass that gives a canonical form for a pencil of singular matrices (see <cit.>).
Even though the classification of subspaces of $S_n$ of bounded rank in linear dimension at least 3 is open, there are partial results.
One is Meshulam's theorem <cit.> giving the maximum dimension of a subspace of symmetric (and alternating) matrices with rank less than $r$.
In Proposition <ref> we give a short proof of this theorem using $\fano{k}{\SD[r]{n}}$.
Actually, our method treats subspaces of symmetric, alternating and rectangular matrices of bounded rank simultaneously and gives a unified geometric proof of Meshulam's theorem for symmetric and alternating matrices and of Flanders' theorem <cit.> for rectangular matrices.

Another result we will reprove (in Corollary <ref>) is Loewy and Radwan's theorem <cit.> that classifies subspaces of $S_n$ of bounded rank having the maximum dimension in Meshulam's result.
Loewy and Radwan's proof, although lengthy, uses elementary methods in linear algebra and graph theory and has the advantage that it imposes few restrictions on the ground field $\KK$.
We will give a shorter geometric proof for an algebraically closed field $\KK$ by studying the Fano schemes $\fano{k}{\SD[r]{n}}$.

Our third motivation concerns the classification of trilinear forms, i.e. elements $\omega\in \KK^{k+1}\otimes \KK^m\otimes \KK^n$ up to the action of $\gl{k+1}\times\gl{m}\times\gl{n}$.
Classifying the orbits in general is an open problem but particular cases have been studied (see e.g.  <cit.> and  <cit.>).
A closely related problem is the classification of elements in $\KK^{k+1}\otimes\Sym^2(\KK^n)$, or equivalently $\Hom\left((\KK^{k+1})^*, \Sym^2(\KK^n)\right)$, up to $\gl{k+1}\times\gl{n}$.

With respect to the standard bases, an element $\omega:(\KK^{k+1})^*\to \Sym^2(\KK^n)$ is an $n\times n$ matrix filled with linear forms in the coordinates of $(\KK^{k+1})^*$.
We may assume, after possibly quotienting by the kernel, that $\omega$ is injective.
Up to $\gl{k+1}$, $\omega$ then corresponds to the pair $(k+1,Q)$, where $Q$ is the image of $\omega$, a $(k+1)$-dimensional subspace of $n\times n$ symmetric matrices.
The Fano scheme $\fano{k}{\SD[r]{n}}$ is the moduli space of those $Q$ having elements with rank less than $r$.

The study of linear spaces of singular symmetric matrices is also related to the study of quadrics.
A subspace $Q\subset S_n$ gives a linear family of quadrics.
The linear family $Q$ is called singular if every quadric in $Q$ is given by a singular matrix, otherwise $Q$ is non-singular.
Pencils of quadrics have been extensively studied.
A theorem due to Weierstrass and Segre classifies the $\gl{n}$-orbits of non-singular pencils of quadrics via a combinatorial object called the Segre symbol (see <cit.>).
In <cit.> the author studies singular pencils of quadrics of constant rank.
Orbits of nets of conics have been classified in <cit.>.
Our Fano schemes parameterize linear spaces (not necessarily pencils) of quadrics of bounded rank.
We hope that this study leads to a better understanding of these linear spaces.

Finally, this study provides a family of examples for general Hilbert schemes.
For a projective scheme $Y\subset \PP_{\KK}^n$, the general Hilbert scheme $\mathrm{Hilb}_f(Y)$ is the fine moduli space parameterizing those closed subschemes of $Y$ that have a fixed Hilbert polynomial $f$.
When $Y=\PP_{\KK}^n$, we get the classical Hilbert scheme $\mathrm{Hilb}_f(\PP_{\KK}^n)$ which is known to be connected <cit.>.
But for general Hilbert schemes, connectedness is not guaranteed.
The Fano schemes $\fano{k}{Y}$ are particular examples of general Hilbert schemes.
In <cit.>, the authors show that the Fano schemes $\fano{k}{D_{m,n}^r}$ that parameterize $k$-planes of the space of $m\times n$ matrices of rank less than $r$ is not always connected.
We will see that the Fano scheme $\fano{k}{\SD[r]{n}}$ is also not always connected (see Corollary <ref>) and we will even encounter generically non-reduced components (see Proposition <ref>).
Our investigation will provide a family of examples of general Hilbert schemes that arise naturally but exhibit such geometric pathologies.

§.§ Main results

The general linear group $\gl{n}$ over the field $\KK$ acts on the space of symmetric matrices $S_n$ through conjugation $(B, M) \mapsto B\cdot M=BMB^t$, for every $B\in\gl{n}$ and $M\in S_n$.
Here $B^t$ denotes the transpose of $B$.
This action naturally induces an action of $\gl{n}$ on $\fano{k}{\SD[r]{n}}$.

An example of a subspace of $S_n$ whose every matrix has rank less than $r$ is a $\gl{n}$-translate of a subspace of matrices with $1+n-r$ zero rows.
This is a particular example of what is called a compression space, first introduced in <cit.>.

Let $V,W$ be $\KK$-vector spaces of dimensions $n$ and $m$ respectively and fix $r\leq m,n$.
For an integer $0\leq s\leq r-1$, a linear subspace $Q\subset\Hom(V,W)$ is an $s$-compression space if every element of $Q$ maps a fixed $(s+1+n-r)$-dimensional subspace of $V$ into a fixed $s$-dimensional subspace of $W$.

For an $s$-compression space $Q$, every element in $Q$ has rank less than $r$ since every $r$-dimensional subspace of $V$ intersects the fixed $(s+1+n-r)$-dimensional subspace in dimension at least $s+1$ and thus every $r$-dimensional subspace of $V$ meets the kernel of every element in $Q$ non-trivially.
After a suitable choice of bases for $V$ and $W$, an $s$-compression space, considered as a subspace of all $m \times n$ matrices, is of the block form

\begin{equation}
\label{eq:standard-rect-zero-pattern}
\begin{pNiceArray}{c|cc}[first-row,first-col,margin]
& r-s-1 & & s+1+n-r\\
& & & \\
s & * & &* \\
\hline
& &&   \\
&&   &\\
\end{pNiceArray}.
\end{equation}

A symmetric (or alternating) compression space is a compression space consisting of symmetric (or alternating) matrices.
We denote the maximum dimension of an $s$-compression space by $\kappa(s)$ (see Definition <ref>).

Our first result is a complete characterization of the connectedness of $\fano{k}{\SD[r]{n}}$.
Our approach also treats the connectedness of Fano schemes of rectangular matrices of bounded rank $\fano{k}{D_{m,n}^r}$, and Fano schemes of alternating matrices of bounded rank $\fano{k}{\AD[r]{n}}$ (see §<ref> for the exact definitions of $D_{m,n}^r$ and $\AD[r]{n}$).

Let $X$ be one of the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$ where $r\leq m\leq n$ and $k$ are integers such that $X$ is non-empty.
The compression spaces form a closed subscheme $\mathrm{\textbf{C}}\subset X$ and the natural map $\pi_0(\mathrm{\textbf{C}})\rightarrow\pi_0\left(X\right)$ on the sets of connected components of $\textbf{C}$ and $X$  is bijective.
In particular, $X$ is connected if and only if $\mathrm{\textbf{C}}$ is.

We will prove this in §<ref> and there we will combinatorially describe the connectedness of the compression subscheme $\mathrm{\textbf{C}}\subset X$ and obtain the following characterization of the connectedness of $X$.

Let $X$ be as in Theorem <ref>.

\[
\mathbf{V}=\left\{ s\in\ZZ \, | \, 0\leq s\leq s_{\max},\,\mathrm{and}\, k\leq\kappa(s) \right\},
\]

\[
\begin{cases}
\left\lfloor \frac{r-1}{2} \right\rfloor & \text{if } X=\fano{k}{\mathrm{SD}_n^r}\, \text{ or }\, \fano{k}{\mathrm{Pf}_n^r},\\
r-1 & \text{if } X=\fano{k}{D_{m,n}^r}.
\end{cases}
\]

Assume $s_1,s_2,\ldots,s_{\ell}$ are the distinct elements of $\mathbf{V}$ in ascending order.
Then $X$ is disconnected if and only if

* $\ell\geq 2$, and

* At least two terms of the sequence

\[
\]

are less than $k$, where $g(\{s,s'\})=\kappa(s')-(s+1+n-r)(s'-s)$ for $s<s'$.

Corollary <ref> completes the partial characterization of the connectedness of $\fano{k}{D_{m,n}^r}$ by Ilten and Chan <cit.>.
In loc. sit. the authors ask whether there exist $m,n,r$ such that $\fano{k}{D_{m,n}^r}$ is connected but its compression subscheme is disconnected <cit.>.
Theorem <ref> answers this question in the negative.

The next result (proved in §<ref>) characterizes the irreducibility of $\fano{k}{\SD[r]{n}}$.
We note that by Meshulam's theorem <cit.>, the scheme $\fano{k}{\SD[r]{n}}$ is non-empty if and only if

\[
k\leq\max\left\{ \kappa(0),\kappa(s_{\max}) \right\},
\]

where $s_{\max}=\lfloor\frac{r-1}{2}\rfloor$.
We will give a short proof of this fact in §<ref>.

The Fano scheme $\fano{k}{\SD[r]{n}}$ is irreducible if and only if

\[
\max\left\{\kappa(1),\kappa(s_{\max})\right\}< k\leq\kappa(0),
\]

\[
\max\left\{\kappa(0),\kappa(s_{\max}-1)\right\}< k\leq\kappa(s_{\max}).
\]

In the particular case of the Fano scheme of lines $\fano{1}{\SD[r]{n}}$ we can describe all the irreducible components.
We say a subspace of $S_n$ is a nested $s$-compression space if it is a $\gl{n}$-translate of a subspace of the form

\begin{equation}
\label{eq:standard-zero-pattern}
\begin{pNiceArray}{ccccc}[first-row,first-col]
&&&&& \\
s&&*&&*&* \\
& &  &&& \\
& &&  && \\
\end{pNiceArray}.
\end{equation}

We will see that the nested $s$-compression spaces form a closed subscheme of $\fano{k}{\SD[r]{n}}$ which we denote by $\compk{k}{s}$.
We will prove the following result in §<ref>.

The subschemes $\compk{1}{s}$ for $s=0, 1,\ldots,\lfloor\frac{r-1}{2}\rfloor$ are distinct, and are exactly the irreducible components of $\fano{1}{\SD[r]{n}}$.
Each component $\compk{1}{s}$ has dimension $nr+(s-1)(n-r)-5$ and their intersection is non-empty.
In particular when $r=n$, the components are equidimensional and of the expected dimension.

It is known that if the Fano scheme $\fano{k}{Y}$ of a general hypersurface $Y\subset \PP^{N}$ of degree $d$ is non-empty, then it has the expected dimension $(k+1)(N-k)-\binom{k+d}{k}$.
When the Fano scheme of a hypersurface, e.g. $\fano{k}{\SD[n]{n}}$, has the expected dimension, it is given by the vanishing of a section of a certain vector bundle on the Grassmannian.
In particular, its class in the Chow ring of the Grassmannian can be determined using tools from intersection theory (see <cit.>).

For higher values of $k$, a full characterization of the irreducible components of $\fano{k}{\SD[r]{n}}$ is difficult since non-compression components appear.
Although we suspect that $\compk{k}{s}$ forms an irreducible component for every $k$, we are able to prove this only when $k=1$,  $k=\kappa(s)$, or when $k$ is arbitrary but $s=\frac{r-1}{2}$ (see Corollary <ref> and Proposition <ref>).
One obstacle in proving $\compk{k}{s}$ is an irreducible component is that it could be generically non-reduced as the following result shows.

When the closed subscheme $\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$ is an irreducible component (e.g. when $k=1$ or $k=\kappa(s)$ or $s=\frac{r-1}{2}$), it is generically reduced if and only if $s=\frac{r-1}{2}$.

We will prove this in Section <ref>.
In particular, all components of $\fano{1}{\SD[r]{n}}$ (except one component when $r$ is odd) are generically non-reduced.
Obviously the scheme $\fano{k}{\SD[r]{n}}$ is not smooth when it has a generically non-reduced component.
We will show in Proposition <ref> that when $r$ is odd and $k\leq\kappa(\frac{r-1}{2})$, the scheme has smooth points.
Based on the above corollary and the characterization of connectedness, we formulate a conjecture on the smoothness of $\fano{k}{\SD[r]{n}}$ (see Conjecture <ref>).

§.§ Related work and organization

The study of linear spaces of matrices with bounded rank has attracted attention in the last sixty years.
Dieudonné <cit.> showed that a subspace of singular $n\times n$ matrices has dimension at most $n(n-1)$ and those with this maximum dimension must be $0$- or $(n-1)$-compression spaces.
Flanders generalized this result to rectangular matrices of bounded rank and Atkinson, Lloyd, Beasley and de Seguins Pazzis <cit.> improved Flanders' classification to dimensions near the maximal one.

For symmetric matrices, Meshulam <cit.> proved the equivalent of Flanders' result and determined the maximum dimension of a subspace of $S_n$ of bounded rank.
Loewy and Radwan <cit.> showed subspaces having the maximum dimension are compression spaces and de Seguins Pazzis <cit.> generalized this classification to subspaces near the maximal dimension.
For a study of subspaces of $S_n$ of fixed rank, see e.g. <cit.>.

Fano schemes have been studied in numerous contexts.
Fano <cit.> studied the family of lines on a cubic hypersurface having a finite number of singularities.
Altman and Kleiman <cit.> gave a modern treatment of these moduli spaces.
There have also been extensive studies on Fano schemes of hypersurfaces <cit.> and complete intersections <cit.>.

In <cit.>, the authors study the Fano schemes $\fano{k}{D_{m,n}^r}$ and $\fano{k}{P_{m,n}^r}$ where $D_{m,n}^r$ and $P_{m,n}^r$ are subschemes of $\PP_{\KK}^{mn-1}$ given by the vanishing of the $r\times r$ determinants, respectively permanents, of the generic $m\times n$ matrix.
They give results on irreducibility, smoothness, connectedness and components of these schemes.

This paper is organized as follows.
In §<ref> we review preliminary results on compression spaces, ending it with a short proof of Flanders' and Meshulam's theorems.
In §<ref> we characterize the connectedness of the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$, and $\fano{k}{D_{m,n}^r}$ and prove Theorem <ref> and Corollary <ref>.
In §<ref> we will prove results on irreducible components of $\fano{k}{\SD[r]{n}}$ and give a geometric proof for Loewy and Radwan's theorem.
In §<ref> we will prove Theorem <ref> on the Fano scheme of lines and characterize the irreducibility of $\fano{k}{\SD[r]{n}}$ by proving Theorem <ref>.
In §<ref> we study the smoothness of the Fano schemes $\fano{k}{\SD[r]{n}}$ via tangent space dimensions, discuss the existence of generically non-reduced components, and prove Proposition <ref>.

§.§ Acknowledgements

This paper was written on the unceded traditional territories of the Coast Salish peoples of the Tsleil-Waututh, Squamish, Kwikwetlem, and Musqueam nations.
I would like to thank Michel Brion and Allen Knutson for helpful discussions and Nathan Ilten for suggesting this problem and being a great mentor to me.
The author was supported by NSERC.

§ PRELIMINARIES

§.§ The varieties $\SD[r]{n}$, $\AD[r]{n}$ and $D_{m,n}^r$

The main focus of this paper is the Fano schemes $\fano{k}{\SD[r]{n}}$ of symmetric matrices of bounded rank.
However, in §<ref>, we will treat the connectedness of these schemes together with that of Fano schemes of alternating matrices $\fano{k}{\AD[r]{n}}$ and rectangular matrices $\fano{k}{D_{m,n}^r}$ of bounded rank.
In this section we define $\AD[r]{n}, D_{m,n}^r$ and related terminology.
When considering alternating or rectangular matrices, $\KK$ can be of any characteristic (even 2).

An alternating matrix is a skew-symmetric matrix with zeros on the diagonal.
Denote by $A_n$ the vector space of $n\times n$ alternating matrices over $\KK$.
For integers $2<r\leq n$, where $r$ is even, we define $\AD[r]{n}$ to be the subscheme of $\PP^{\binom{n}{2}-1}$ defined by the vanishing of the $r\times r$ Pfaffians of the generic alternating matrix

\begin{equation}
\label{eq:gen-alter-matrix}
\begin{pmatrix}
0        & x_{1,2}  & \cdots & x_{1,n} \\
-x_{1,2} & 0        & \cdots & x_{2,n} \\
\vdots   &          & \ddots & \vdots  \\
-x_{1,n} & -x_{2,n} & \cdots & 0
\end{pmatrix}.
\end{equation}

The group $\gl{n}$ acts on $A_n$ by $BMB^t$ for every $B\in \gl{n}$ and $M\in A_n$.
The matrix $BMB^t$ is obtained by applying to $M$ a sequence of row-column operations.
A row-column operation is an elementary row operation followed by the same operation on the columns of the matrix.
This action induces an action on $\fano{k}{\AD[r]{n}}$.

For an even integer $r$, the $r\times r$ Pfaffians of (<ref>) generate a prime ideal (see <cit.>) that cuts out the locus of alternating matrices of rank at most $r-2$ (see <cit.>).
We choose Pfaffians over minors because the ideal generated by minors of size $r$ (or $r-1$) is not prime.

Next, fix integers $2\leq r\leq m\leq n$.
Let $M_{m,n}$ be the vector space of all $m\times n$ matrices over $\KK$ and define $D_{m,n}^r$ to be the subscheme of $\PP^{mn-1}$ given by the vanishing of the $r\times r$ minors of the generic $m\times n$ matrix

\begin{equation}
\label{eq:generic-m-by-n}
\begin{pmatrix}
x_{1,1} & x_{1,2} & \cdots & x_{1,n} \\
x_{2,1} & x_{2,2} & \cdots & x_{2,n} \\
\vdots  &         & \ddots & \vdots  \\
x_{m,1} & x_{m,2} & \cdots & x_{m,n}
\end{pmatrix}.
\end{equation}

The group $\gl{m}\times\gl{n}$ acts on $M_{m,n}$, and hence on $\fano{k}{D_{m,n}^r}$, via $B_1 MB_2^t$ for every $(B_1,B_2)\in \gl{m}\times\gl{n}$ and $M\in M_{m,n}$.

A linear space of matrices $Q$ can be represented non-uniquely as a matrix with linear forms as entries by picking a basis in $Q$ (see Example <ref>).
We will freely switch between the subspace and the matrix perspective.
For example the expressions $A\in Q,$ and $BQB^t$ are both valid, the latter expressing that we are acting on a chosen matrix that represents $Q$.
The expression $B\cdot Q$ always means $BQB^t$.

Fix $3\leq r\leq n$ and let $0\leq s\leq \lfloor\frac{r-1}{2}\rfloor$ (we take $r$ to be even when considering alternating matrices).
The standard symmetric (respectively alternating) $s$-compression space is the subspace of $S_n$ (respectively of $A_n$) of the block form (<ref>) where the entries in the star blocks vary independently over $\KK$, subject to the condition that the matrix be symmetric (respectively alternating).

Next fix $2\leq r\leq m\leq n$ and let $0\leq s\leq r-1$.
The standard rectangular $s$-compression space is the subspace of $M_{m,n}$ of the block form (<ref>) where the star blocks vary independently over $\KK$.

Define $\kappa(s,r)$ to be the projective dimension of the standard $s$-compression space

\begin{equation}
\label{eq:kappa}
\kappa(s,r) =
\begin{cases}
s(s+1+n-r)+\binom{r-s}{2}-1 & \text{for symmetric matrices},\\
s(s+1+n-r)+\binom{r-s-1}{2}-1 & \text{for alternating matrices},\\
mn-(s+1+n-r)(m-s)-1 & \text{for rectangular matrices}.
\end{cases}
\end{equation}

We usually use the notation $\kappa(s)$ when $r$ is known from the context.

For $\fano{k}{\SD[r]{n}}$, the case $r=2$ is trivial since a subspace of symmetric matrices of rank at most 1 is necessarily the span of $u^t u$ for a row vector $u\in\KK^n$

§.§ Bilinear forms

An $n\times n$ matrix $A$ over $\KK$ defines a bilinear form $\beta_A:\KK^n\times \KK^n\to \KK$ via $\beta_A(u,w)=u A w^t$ for row vectors $u,w\in\KK^n$.
A change of basis on $\KK^n$ corresponds to changing the matrix $A$ to $BAB^t$ for some $B\in\gl{n}$.
We say $u,w\in \KK^n$ are orthogonal with respect to $A$ and write $u\perp_A w$ if $\beta_A(u,w)=0$.
For a set $Q$ of matrices, $u\perp_Q w$ means $u\perp_A w$ for every $A\in Q$.
Similarly, for $U,W\subset \KK^n$, by $U\perp_Q W$ we mean $u\perp_Q w$ for every $u\in U, w\in W$.
The next proposition characterizes compression spaces in terms of bilinear forms.

A linear space of $n\times n$ matrices $Q$ over $\KK$ is an $s$-compression space if and only if there exist subspaces $U,W\subset \KK^n$ of dimensions $s+1+n-r$ and $n-s$ respectively, such that $U\perp_Q W$.

A matrix $A\in Q$ defines a map $L_A:\KK^n\to (\KK^n)^*$ by $v\mapsto (x\mapsto\beta_A(x,v))$ for every $v\in \KK^n$.
The matrix of $L_A$ with respect to the standard basis on $\KK^n$ and the dual basis on $(\KK^n)^*$ is $A$.
We also have $u\perp_A w$ if and only if $L_A(u)$ vanishes on $w$.

Suppose first that $Q$ is an $s$-compression space.
By definition, there exist subspaces $U\subset \KK^n$ and $U'\subset (\KK^n)^*$ of dimensions $s+1+n-r$ and $s$, respectively such that $L_A(U)\subset U'$ for every $A\in Q$.
Let $W=\ann(U')=\{w\in \KK^n:f(w)=0,\, \text{for every}\, f\in U'\}$.
Then $\dim W=n-s$ and $U\perp_Q W$.
Conversely, suppose $U$ and $W$ have dimensions $s+1+n-r$ and $n-s$ and $U\perp_Q W$.
Take $U'=\ann(W)$, the subspace of $(\KK^n)^*$ vanishing on $W$.
Then $L_A(U)\subset U'$ for every $A\in Q$.
Because $\dim U'=s$, we deduce that $Q$ is an $s$-compression space.

An immediate corollary is that when we consider $s$-compression spaces of symmetric or alternating matrices, we may assume $0\leq s\leq\lfloor\frac{r-1}{2}\rfloor$.

Every $s$-compression space $Q\subset S_n$ or $A_n$ is also an $(r-s-1)$-compression space.

This follows from Proposition <ref> by changing the roles of $U$ and $W$.

We now describe compression spaces of $S_n$ and $A_n$ in terms of the standard ones.

Let $0\leq s\leq \lfloor\frac{r-1}{2} \rfloor$.
Then every symmetric or alternating $s$-compression space is a subspace of a $\gl{n}$-translate of the standard $s'$-compression space for some $0\leq s'\leq s$.

Let $Q\subset S_n$ or $A_n$ be an $s$-compression space.
By Proposition <ref>, there are subspaces $U,W\subset \KK^n$ of dimensions $s+1+n-r$ and $n-s$ respectively, such that $U\perp_Q W$.
Define $s'=(\dim U\cap W) -(n-r+1)$.
Note that $0\leq s'\leq s$.
Since $\dim U\cap W=s'+1+n-r$, we have $\dim (U+W)=n-s'$ and $(U\cap W) \perp_Q (U+W)$.
Choose a basis $\mathcal{B}_1$ for $U\cap W$ and extend it to a basis $\mathcal{B}_2$ for $U+W$.
Then extend $\mathcal{B}_2$ to a basis $\mathcal{B}$ for $\KK^n$.
With respect to a suitable ordering of the elements in $\mathcal{B}$, every matrix in $Q$ has zero blocks similar to the standard $s'$-compression space in (<ref>).

We distinguish the case where the orthogonal subspaces in Proposition <ref> satisfy the additional property that one contains the other.

Let $Q\subset S_n$ or $A_n$ be a linear subspace and let $0\leq s\leq \lfloor\frac{r-1}{2}\rfloor$.
We say $Q$ is a nested $s$-compression space if there exist subspaces $U\subset W\subset\KK^n$ of linear dimensions $s+1+n-r$ and $n-s$ respectively, such that $U\perp_Q W$.

Nested $s$-compression spaces are exactly subspaces of $\gl{n}$-translates of the standard $s$-compression space.
On the other hand, by Corollary <ref>, an $s$-compression space is a nested $s'$-compression space for some $0\leq s'\leq s$ but $s'$ need not be the same as $s$.
For example, let $n=r=3$.
The subspace of $S_n$ given by

\[
z_0 & 0   & 0 \\
0   & z_1 & 0 \\
0   & 0   & 0
\end{pmatrix},
\]

is both a $0$- and a $1$-compression space but only a nested $0$-compression space.
Because if it were a nested $1$-compression space, then there would be a subspace $U\subset \KK^3$ of dimension 2 that is totally isotropic with respect to every matrix in $Q$.
If $u=\alpha_1 e_1+\alpha_2 e_2+\alpha_3 e_3$ is an isotropic vector where $e_i$ are the standard basis vectors of $\KK^3$, then $\beta_Q(u,u)=\alpha_1 ^2 z_0 + \alpha_2^2 z_1=0$ for all $z_0,z_1\in \KK$.
Hence $u\in \mathrm{span}\{e_3\}$, a contradiction to $\dim U=2$.

Being a (nested) $s$-compression space is a closed property.
For example for symmetric matrices, the locus of points $([Q],[U,W])\in\Gr(k+1,S_n)\times\flag{s+1+n-r,n-s}{\KK^n}$ with $U\perp_Q W$ is closed by inspecting the affine charts of the factors.
Here $\flag{\ell_1,\ell_2}{\KK^n}$ denotes the flag variety parametrizing flags $U\subset W$ in $\KK^n$ with dimensions $\ell_1, \ell_2$, respectively.
The image of the projection of this locus to $\Gr(k+1,S_n)$ is closed and is the locus of nested $s$-compression spaces.

When $X=\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$, we denote the closed subscheme of $X$ consisting of nested $s$-compression spaces by $\compk{k}{s}$.
When $X=\fano{k}{D_{m,n}^r}$, the notation $\compk{k}{s}$ means the closed subscheme consisting of $s$-compression spaces, which is consistent with the notation of <cit.>.

§.§ Fixed points

We will use the Borel subgroup $G_B(n)\subset\gl{n}$ consisting of all invertible upper triangular matrices and the unipotent subgroup $G_U(n)$ consisting of those matrices in $G_B(n)$ with ones on the diagonal.
We say a point of $\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$ is a Borel fixed point, respectively a unipotent fixed point, if it is fixed under the action of $G_B(n)$, respectively $G_U(n)$.
For the action of $\gl{n}\times\gl{m}$ on $\fano{k}{D_{m,n}^r}$, we define the Borel subgroup and the unipotent subgroup to be $G_B(n)\times G_B(m)$ and $G_U(n)\times G_U(m)$, respectively.
This section gives several properties of the unipotent fixed points of the Fano schemes $\fano{k}{\SD[r]{n}}, \fano{k}{\AD[r]{n}}$ and $\fano{k}{D_{m,n}^r}$ and fully characterizes the Borel fixed points.

The utility of fixed points of $G_B(n)$ is due to Borel's fixed point theorem <cit.> which states that the action of $G_B(n)$ on a non-empty complete variety (e.g. every $G_B(n)$-stable closed subscheme of our Fano schemes) has a fixed point.
On the other hand, Horrocks's theorem <cit.> states in our setting that the set of connected components of the unipotent fixed locus of the scheme is in bijection with the set of connected components of the scheme.

Assume integers $m,n$ are fixed.
We denote by $e_{(i,j)}$ the $m\times n$ matrix with 1 in the $(i,j)$ entry and zero elsewhere.
Next assume $m=n$.
Set $e_{\{i,j\}}=e_{(i,j)}+e_{(j,i)}$ when $i\neq j$ and $e_{\{i,j\}}=e_{(i,j)}$ when $i=j$.
Also, for $i\neq j$, set $e_{\langle  i,j\rangle}=e_{(i,j)}-e_{(j,i)}$.

Let $\mathcal{V}$ be one of the vector spaces $S_n, A_n$ or $M_{m,n}$.
Suppose $[P]\in\Gr(k+1,\mathcal{V})$ is a unipotent fixed point, where $P=[p_{i,j}]$ is a $(k+1)$-dimensional subspace of $\mathcal{V}$ written as a symmetric, alternating or rectangular matrix whose entries $p_{i,j}$ are linear forms in $k+1$ free variables.

* If $p_{i,j}=0$ for some $i,j$ (we also require $i\neq j$ if $\mathcal{V}=A_n$), then $p_{i',j'}=0$ for every $i',j'$ with $i'\geq i$ and $j'\geq j$.

* If $e_{\{i,j\}}$, respectively $e_{\langle i,j \rangle}$ or $e_{(i,j)}$, is in $P$ for some $i,j$, then $e_{\{i',j'\}}$, respectively $e_{\langle i',j' \rangle}$ or $e_{(i',j')}$, is also in $P$ for every $i',j'$ with $i'\leq i$ and $j'\leq j$ (and $i'\neq j'$ when $\mathcal{V}=A_n$).

* Assume $p_{i,j}\neq 0$ for some $i,j$.
Then for all $i',j'$ with $i'< i$, and $j'< j$ (and $i'\neq j'$ when $\mathcal{V}=A_n$), we have $e_{\{i',j'\}}\in P$ when $\mathcal{V}=S_n$, $e_{\langle  i',j'\rangle}\in P$ when $\mathcal{V}=A_n$ and $e_{(i',j')}\in P$ when $\mathcal{V}=M_{m,n}$ .

See Example <ref> for what a unipotent fixed point looks like.
We prove the lemma for the case $\mathcal{V}=S_n$.
The other cases are proved similarly with minor adjustments.

Part <ref>: It is enough to show the claim for $(i',j')=(i,j+1)$ and $(i',j')=(i+1,j)$.
Assume to the contrary that for one of these two, $p_{i',j'}\neq 0$.
We consider two cases.
The first case is when $i\neq j$, i.e. $p_{i,j}$ is not on the diagonal.
We may assume $i< j$.
If $p_{i,j+1}\neq 0$, take $u=I_n+e_{(j,j+1)}\in G_U(n)$ where $I_n$ is the identity matrix of size $n$.
Then $uPu^t$ is obtained from $P$ by adding row/column $j+1$ to row/column $j$.
We see that $[uPu^t]$ is not equal to $[P]$ because the $(i,j)$ entry in $uPu^t$ is nonzero, a contradiction.
Similarly, if $p_{i+1,j}\neq 0$, we take $u=I_n+e_{(i,i+1)}$ and get $[uPu^t]\neq [P]$.

The second case is when $i=j$, i.e. $P$ is as in Figure <ref>.
Take $u=I_n+e_{(i,i+1)}$, then $uPu^t$ will be as in Figure <ref>.
If $p_{i+1,i+1}+2p_{i,i+1}\neq 0$, then certainly $[uPu^t]\neq[P]$, a contradiction to $[P]$ being a unipotent fixed point.
On the other hand, if $p_{i+1,i+1}+2p_{i,i+1}= 0$, then we take $u'=I_n+\frac{1}{2}e_{(i,i+1)}$ and then the $(i,i+1)$-entry of $u'P(u')^t$ is zero, while the same entry in $P$ is nonzero.

\begin{pNiceArray}{ccccc}[first-row,first-col]
&&&&& \\
i&&&0&p_{i,i+1}& \\
i+1&&&p_{i,i+1}&p_{i+1,i+1}& \\
\end{pNiceArray}

\begin{pNiceArray}{ccccc}[first-row,first-col]
&&&&& \\
i&&&p_{i+1,i+1}+2p_{i,i+1}&p_{i,i+1}+p_{i+1,i+1}& \\
i+1&&&p_{i,i+1}+p_{i+1,i+1}&p_{i+1,i+1}& \\
\end{pNiceArray}

The matrices in the proof of Lemma <ref> part <ref>

Part <ref>: It is enough to prove the statement for $(i',j')=(i-1,j)$ and $(i',j')=(i,j-1)$.
We split the argument into two cases: $i\neq j$ and $i=j$.
If $i\neq j$, we may assume $i<j$.
Take $u=I_n+e_{(i-1,i)}$, then $e_{\{i-1,j\}}=ue_{\{i,j\}}u^t-e_{\{i,j\}}\in P$.
A similar argument with $u=I_n+e_{(j-1,j)}$ proves $e_{\{i,j-1\}}\in P$.
If $i=j$, take $u=I_n+e_{(i-1,i)}$ and define $v = ue_{\{i,i\}}u^t-e_{\{i,i\}}\in P$.
Then $e_{\{i-1,i\}}=\frac{1}{2}(3v-uvu^t)\in P$.

Part <ref>:
Without loss of generality we let $i\leq j$ and by increasing $i$ and $j$ (if necessary) we may assume $p_{i'',j''}=0$ for every $i'',j''$ such that $(i'',j'')\neq(i,j)$, $i''\geq i$, and $j''\geq j$.
We show $e_{\{i-1,j-1\}}\in P$.
The result then follows from part <ref>.
By assumption $P$ is of the form

\begin{equation*}
\begin{pNiceArray}{ccccc}[first-row,first-col]
i&&p_{i,j}&0&\cdots&0 \\
\end{pNiceArray}.
\end{equation*}

Take $u=I_n+e_{(i-1,i)}$, then the subspace $P'= uPu^t-P$ of $P$ has the form

\begin{equation*}
\begin{pNiceArray}{ccccccc}[first-row,first-col]
&&&&&\vdots&& \\
&&&&&0&& \\
i-1&&&&&p_{i,j}&& \\
i&&&&&0&\ldots&0 \\
\end{pNiceArray}.
\end{equation*}

By taking $u'=I_n+e_{(j-1,j)}$ we have

\[
\alpha p_{i,j}e_{\{i-1,j-1\}}=u'P'(u')^t-P'\subset P,
\]

where $\alpha=1$ if $i<j$ and $\alpha=2$ if $i=j$.
Because $p_{i,j}\neq0$, we get $e_{\{i-1,j-1\}}\in P$.

\begin{pmatrix}
*      &         &     & *      & P_1 \\
*      &         & *   & P_2    & 0      \\
*      &         & P_3 & 0      & 0      \\
\vdots & \iddots &     & \vdots & \vdots \\
P_t    & \cdots  & 0   & 0      & 0
\end{pmatrix}

\begin{pNiceMatrix}[r,left-margin=0.6em,right-margin=0.6em]
&                     &   &   & \Block[draw]{2-2}{} & * \\
&                     &   &   & *                   & * \\
& \Block[draw]{3-3}{} &   & * & \mathbf{0}          & 0 \\
&                     & * & * &     0                & 0 \\
& *                   & * &   &     0                & 0 \\
\Block[draw]{1-1}{}* & \mathbf{0}  & 0 & 0 & 0      & 0
\end{pNiceMatrix}

Partitioning $P'$ in Proposition <ref>: (a) general case (b) the three blocks $P_1,P_2, P_3$ of sizes 2, 3 and 1 in a $6\times 6$ matrix determined by the marked zeros (in bold) on the second anti-diagonal

The following statements hold.

* Every unipotent fixed point of $\fano{k}{\SD[r]{n}}$ and $\fano{k}{\AD[r]{n}}$ is a subspace of the form (<ref>) for some $s$.

* Every unipotent fixed point of $\fano{k}{D_{m,n}^r}$ is a subspace of the form (<ref>) for some $s$.

In particular, every unipotent fixed point of $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$ is an $s$-compression space for some $s$.

Let $[P]$ be a unipotent fixed point of $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$ where $P=[p_{i,j}]$ is a symmetric, alternating or rectangular matrix whose entries are linear forms in $k+1$ free variables.
Let $P'$ be the upper left $r\times r$ submatrix of $P$.
We show that at least one entry on the anti-diagonal of $P'$ is zero.
By Lemma <ref> part <ref>, all entries to the right and bottom of this entry in $P$ will be zero and therefore for some $s$, $P$ will have zero blocks as in (<ref>) when we are considering $\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$ and as in (<ref>) when considering $\fano{k}{D_{m,n}^r}$ (note that in the case of $\fano{k}{\AD[r]{n}}$, $r$ is even and the diagonal and anti-diagonal of $P'$ do not intersect).

To show an entry on the anti-diagonal of $P'$ is zero, we use the action of the unipotent subgroup to produce from $P$ a matrix $Q$ filled with linear forms such that as vector spaces, $Q$ is a subspace of $P$ and as a matrix with linear forms, the upper left $r\times r$ minor of $Q$ is the product of the anti-diagonal entries of $P'$.
Since this minor must be zero, then at least one entry on the anti-diagonal of $P'$ is zero.

In a square matrix of size $n$, we call the entries $(i,j)$ with $i+j=n+2$, the second anti-diagonal.
We claim we can partition the matrix $P'$ as in Figure <ref>, where $t\geq 1$ and each $P_i$ is a square matrix such that every entry on its second anti-diagonal is non-zero (in the alternating case, the middle $P_i$ block will have a zero on the intersection of its second anti-diagonal with the diagoanl of $P'$, but that will not be a problem).
Mark the zero entries on the second anti-diagonal of $P'$ (in the alternating case, the zero entry on the intersection of the diagonal and the second anti-diagonal is not marked).
Assume the marked entries are $(1+\ell_i,n+1-\ell_i)$ for $1\leq i\leq t-1$ where $\ell_1,\ldots,\ell_{t-1}$ are in ascending order.
Defining $\ell_0=0, \ell_t=n$, we set the size of $P_i$ to be $\ell_i-\ell_{i-1}$ for $1\leq i\leq t$.
See Figure <ref> for an example.

By construction every entry on the second anti-diagonal of each $P_i$ is non-zero (except the entry on the intersection of the diagonal and the second anti-diagonal of $P'$ in the alternating case).
By Lemma <ref> part <ref>, the marked zeros force every entry below and to the right of the $P_i$ blocks to be zero.
We want to change the entries in $P_i$ so that the entries above the anti-diagonal are zero.
Then the determinant in each block will be the product of the entries on the anti-diagonal.

By applying Lemma <ref> part <ref> to every entry on the second anti-diagonal of every $P_i$, we can subtract from $P$ suitable multiples of those $e_{\{i,j\}}$, $e_{\langle i,j \rangle}$ or $e_{(i,j)}$  in $P$ (depending on whether we are working with symmetric, alternating, or rectangular matrices respectively) in such a way that in each $P_i$ block, every entry above the anti-diagonal is zero.
Let $Q$ be the resulting matrix with $Q'$ its upper left $r\times r$ submatrix and denote by $Q_i$ the submatrix of $Q$ corresponding to $P_i$.
Then as subspaces, $Q$ is contained in $P$ and as a matrix with linear forms we have

\[
0=\det Q'=\pm\prod_{\ell=1}^t \det Q_{\ell}=\pm\prod_{\ell=1}^t p_{\ell,n-\ell+1}.
\]

This shows an entry on the anti-diagonal of $P$ is zero, completing the proof.

Next we characterize Borel fixed points of the Grassmannian and our Fano schemes.

Assume $\mathcal{V}$ is one of the vector spaces $S_n$, $A_n$ or $M_{m,n}$ and let $Q\subset \mathcal{V}$ be a linear subspace of dimension $k+1$ written as a matrix in linear forms.
Then $[Q]\in\Gr\left(k+1,\mathcal{V}\right)$ is a Borel fixed point if and only if it satisfies the following conditions:

* Every non-zero entry of $Q$ varies freely when $\mathcal{V}=M_{m,n}$, and every non-zero entry on and above the main diagonal varies freely when $\mathcal{V}=S_n$ or $A_n$,

* For every zero entry $(i,j)$ in $Q$ (we also require $i\neq j$ if $\mathcal{V}=A_n$), all entries $(i',j')$ with $i'\geq i$ and $j'\geq j$ are also zero.

See Example <ref> for a matrix satisfying these conditions.
Suppose $Q=[q_{i,j}]$ satisfies the two conditions of the statement.
To prove the claim for $\mathcal{V}=M_{m,n}$, let $A=[a_{i,j}]\in G_B(m)$ and $A'=[a'_{i,j}]\in G_B(n)$ be two upper-triangular matrices.
If $q_{i,j}=0$ for some $(i,j)$, then the $(i,j)$ entry of $AQ(A')^t$ is zero too because

\[
\sum_{\substack{1\leq \ell\leq m\\ 1\leq \ell'\leq n}} a_{i,\ell} q_{\ell,\ell'} a'_{j,\ell'} =
\sum_{\substack{i\leq \ell\leq m\\ j\leq \ell'\leq n}} a_{i,\ell} q_{\ell,\ell'} a'_{j,\ell'} = 0,
\]

where the first equality is due to $A,A'$ being upper triangular and the second equality is due to condition <ref>.

Since non-zero entries of $Q$ vary freely, $Q$ has $k+1$ non-zero entries.
This means the number of non-zero entries of $AQ(A')^t$ is at most $k+1$.
But $AQ(A')^t$ is also a subspace of $\mathcal{V}$ of dimension $k+1$.
Therefore, the number of non-zero entries of $AQ(A')^t$ is exactly $k+1$ and they are linearly independent and are positioned exactly the same as non-zero entries of $Q$.
Thus $Q$ and $AQ(A')^t$ are equal as subspaces of $\mathcal{V}$.
The proof for $\mathcal{V}=S_n$ or $A_n$ is similar except that we take $A'=A$ and consider only non-zero entries on and above the diagonals of $Q$ and $AQA^t$.

Conversely, assume $Q$ is fixed under the action of the Borel subgroup.
Since every Borel fixed point is a unipotent fixed point, by Lemma <ref> part <ref>, the matrix $Q$ satisfies condition <ref>.
To show $Q$ satisfies <ref>, it is enough to prove if $q_{i,j}\neq 0$ for some $i,j$, then the subspace $Q$ contains $e_{\{i,j\}}$, $e_{\langle i,j \rangle}$ or $e_{(i,j)}$ depending on whether $\mathcal{V}=S_n$, $A_n$ or $M_{m,n}$ respectively.

Assume $\mathcal{V}=S_n$ or $A_n$.
First we consider the case $i\neq j$.
For $\alpha\in\KK^*$ and $1\leq \ell\leq n$, define $E_{n,\ell,\alpha}$ to be the diagonal $n\times n$ matrix with entry $(\ell,\ell)$ equal to $\alpha$ and ones elsewhere on the diagonal.
When $\mathcal{V}=S_n$, for every $\alpha,\beta\in\KK^*$ we act on the matrix $Q$ to get the matrix

\[
(\beta-1)(\alpha-1)q_{i,j}e_{\{i,j\}} =
E_{n,j,\beta}\cdot\left(E_{n,i,\alpha}\cdot Q- Q\right)-\left(E_{n,i,\alpha}\cdot Q- Q\right),
\]

representing a subspace of $Q$, which gives $e_{\{i,j\}}\in Q$ (in the above, the dot refers to the $\gl{n}$ action on the chosen matrix of $Q$).
When $\mathcal{V}=A_n$, a similar equation with $e_{\{i,j\}}$ replaced by $e_{\langle i,j \rangle}$ holds showing $e_{\langle i,j \rangle}\in Q$.

If $i=j$, then necessarily $\mathcal{V}=S_n$ and the matrix  $E_{n,i,\alpha}\cdot Q-Q$ represents a subspace of $Q$ and has nonzero entries only in the $i$-th row (and column) and in particular the $(i,i)$ entry equals $(\alpha^2-1)q_{i,i}$.
Using the proof of the claim for entries not on the diagonal, we can zero out all the entries in the $i$-th row (and column) except the entry on the diagonal.
This shows $e_{\{i,i\}}\in Q$.

Finally, for $\mathcal{V}=M_{m,n}$ we use a similar computation on the matrix $Q$ to get

\[
(\beta-1)(\alpha-1)q_{i,j}e_{(i,j)} =
(I_m,E_{n,j,\beta})\cdot\left( (E_{m,i,\alpha},I_n)\cdot Q-Q \right)-\left( (E_{m,i,\alpha},I_n)\cdot Q-Q \right),
\]

representing a subspace of $Q$, hence $e_{(i,j)}\in Q$.

Let $X$ be one of the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$.
A Borel fixed point $[Q]$ of the Grassmannian lies on $X$ if and only if for some $s$, $Q$ is a subspace of the form (<ref>) when $X=\fano{k}{D_{m,n}^r}$ and of the form (<ref>) when $X=\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$.
In particular, Borel fixed points of $X$ are compression spaces.

If for some $s$, $Q$ is a subspace of the standard $s$-compression space, then clearly $[Q]\in X$.
Conversely, if $[Q]$ is a Borel fixed point of $X$, then it is unipotent fixed and by Proposition <ref>, $Q$ has the required form.

The following are two examples of unipotent fixed points of $\Gr(5,S_4)$.
The point on the right is Borel fixed and also lies on $\fano{4}{\SD[4]{4}}$.

\[
\begin{pmatrix}
z_0 & z_1 & z_2 & z_3 \\
z_1 & z_4 & z_3 & 0   \\
z_2 & z_3 & 0   & 0   \\
z_3 & 0   & 0   & 0
\end{pmatrix},\qquad
\begin{pmatrix}
z_0 & z_1 & z_2 & z_3 \\
z_1 & z_4 & 0   & 0   \\
z_2 & 0   & 0   & 0   \\
z_3 & 0   & 0   & 0
\end{pmatrix}.
\]

Using Borel fixed points, we can characterize when the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ and $\fano{k}{D_{m,n}^r}$ are non-empty.
Flanders <cit.> proved this for rectangular matrices (see also <cit.> for the case $m=n=r$) and Meshulam <cit.> gave a proof for the symmetric and alternating cases.
We give a unified geometric proof for symmetric, alternating, and rectangular matrices.
Our argument is similar to the proof of Ilten and Chan <cit.> for $\fano{k}{D_{m,n}^r}$ that uses fixed points of a torus action.

Let $X$ be one of the Fano schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$.
Then $X$ is non-empty if and only if

\[
k\leq\max\left\{\kappa(0),\kappa\left( s_{\max} \right)\right\},
\]

where $\kappa(s)$ is as in (<ref>), $s_{\max}=\left\lfloor\frac{r-1}{2}\right\rfloor$ when $X=\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$ and $s_{\max}=r-1$ when $X=\fano{k}{D_{m,n}^r}$.

If $k$ satisfies the inequality, then $X$ contains a point that is an $s$-compression space for $s=0$ or $s=s_{\max}$ and is non-empty.
Conversely, if $X$ is non-empty, by Borel's fixed point theorem <cit.>, it contains a Borel fixed point which by Corollary <ref> is a subspace of the standard $s$-compression space for some $0\leq s\leq s_{\max}$.
This implies $k\leq\kappa(s)$ and since $\kappa(s)$ is convex in $s$, the integer $k$ will satisfy the inequality of the statement.

§ CONNECTEDNESS

In this section we characterize when the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ and $\fano{k}{D_{m,n}^r}$ are connected.
First we prove Theorem <ref> by using the unipotent fixed points.

We saw in Remark <ref> that $\mathrm{\textbf{C}}$ is closed.
Let $Y$ be the closed subscheme of $X$ consisting of unipotent fixed points.
By Proposition <ref>, we have the inclusions $Y\rightarrow\mathrm{\textbf{C}}\rightarrow X$, inducing the maps

\begin{equation}
\label{eq:maps-connected-components}
\pi_0(Y)\rightarrow\pi_0(\mathrm{\textbf{C}})\rightarrow\pi_0\left( X \right).
\end{equation}

The closed subscheme $\mathrm{\textbf{C}}$ is Borel-stable.
By Borel's fixed point theorem <cit.>, every connected component of $\mathrm{\textbf{C}}$ contains a Borel fixed point which is unipotent fixed.
The first map in (<ref>) is thus surjective.
By Horrocks' theorem <cit.>, the set of connected components of the unipotent fixed points is in bijection with the set of connected components of the scheme, i.e. the composition in (<ref>) is bijective and therefore the second map is bijective.

Our goal is to obtain a combinatorial description of connectedness for the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ and $\fano{k}{D_{m,n}^r}$ and prove Corollary <ref>.
We will associate each of these schemes with a graph whose set of connected components is in bijection with the set of connected components of the scheme.

Let $X$ be one of the schemes $\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$ where $r\leq m\leq n$ and $k$ are integers and set

\[
\begin{cases}
\left\lfloor \frac{r-1}{2} \right\rfloor & \text{if } X=\fano{k}{\mathrm{SD}_n^r}\, \text{ or }\, \fano{k}{\mathrm{Pf}_n^r},\\
r-1 & \text{if } X=\fano{k}{D_{m,n}^r}.
\end{cases}
\]

Define $\mathcal{\tilde{G}}_X=(\mathbf{V},\mathbf{E})$ to be the complete simple graph on the vertex set $\mathbf{V}=\{s\in\ZZ:0\leq s\leq s_{\max}\}$ equipped with a function $g:\mathbf{V}\cup \mathbf{E}\to \ZZ_{>0}$ that labels the vertices and edges of $\mathcal{\tilde{G}}_X$ as follows.
For a vertex $s\in \mathbf{V}$ we define $g(s)=\kappa(s)$ and for the edge $\{s,s'\}\in \mathbf{E}$ with $s<s'$ we set

\begin{equation}
\label{eq:edge-label}
\end{equation}

We define the graph $\mathcal{G}_X$ associated to $X$ to be the subgraph of $\mathcal{\tilde{G}}_X$ obtained by deleting those vertices and edges whose labels are less than $k$.

Let $X=\fano{k}{\SD[6]{6}}$.
Figure <ref> shows the graph $\mathcal{G}_X$ and its labels for three values of $k$.

[node distance=14mm, thick, main/.style = draw, circle]
[main] (0) $0$;
[above] at (0.north) 14;
[main] (1) [below right of=0] $1$;
[below] at (1.south) 11;
[main] (2) [above right of=1] $2$;
[above] at (2.north) 11;
(0) – node[midway, below left, pos=.5] 10 (1);
(0) – node[midway, above,  pos=.5] 9 (2);
(1) – node[midway, below right, pos=.5] 9 (2);

[node distance=14mm, thick, main/.style = draw, circle]
[main] (0) $0$;
[above] at (0.north) 14;
[main] (1) [below right of=0] $1$;
[below] at (1.south) 11;
[main] (2) [above right of=1] $2$;
[above] at (2.north) 11;
(0) – node[midway, below left, pos=.5] 10 (1);

[node distance=14mm, thick, main/.style = draw, circle]
[main] (0) 0;
[above] at (0.north) 14;
(1) [below of=0] ;

The graph $\mathcal{G}_X$ for $X=\fano{k}{\SD[6]{6}}$ and three values of $k$.

When $X=\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$, in Definition <ref>, the label of the vertex $s\in \mathbf{V}$ denotes the highest integer $k$ such that the scheme $X$ contains a Borel fixed point that is a nested $s$-compression space.
Similarly, by the next lemma, the label for the edge $\{s,s'\}$ denotes the highest integer $k$ such that $X$ contains a Borel fixed point that is both a nested $s$- and a nested $s'$-compression space or equivalently $\compk{k}{s}$ intersects $\compk{k}{s'}$.
A similar remark holds when $X=\fano{k}{D_{m,n}^r}$ with the expression nested compression space substituted by compression space.

If $X=\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$ and $[Q]\in X$ is a Borel fixed point that is a nested $s$-compression space, then $Q$ is a subspace of the standard $s$-compression space.
The statement also holds for $X=\fano{k}{D_{m,n}^r}$ when $[Q]$ is a Borel fixed $s$-compression space.

Let $X$ be $\fano{k}{\SD[r]{n}}$ or $\fano{k}{\AD[r]{n}}$.
By assumption there are subspaces $U\subset W\subset \KK^n$ of dimensions $s+1+n-r$ and $n-s$ with $U\perp_Q W$.
Moreover, by Proposition <ref>, we can write $Q$ as a matrix in linear forms such that   the $(i,j)$ entry is either zero, or a free variable, say $z_{i,j}$.

Let $\mathcal{B}_U$ be a $(s+1+n-r)\times n$ matrix with entries in $\KK$ whose rows form a basis for $U$ with respect to the standard basis of $\KK^n$.
Define a similar $(n-s)\times n$ matrix $\mathcal{B}_W$ for $W$.
We may assume that $\mathcal{B}_U$ and $\mathcal{B}_W$ are in row-echelon form.
Let $I_U\subset\{1,\ldots,n\}$ be the indices of columns containing leading 1's in $\mathcal{B}_U$ and similarly define $I_W$ for $\mathcal{B}_W$.
The sets $I_U$ and $I_W$ have sizes $s+1+n-r$ and $n-s$ respectively.
Assume the elements of $I_U$ and $I_W$ in ascending order are

\[
I_U = \{i_1,\ldots,i_{s+1+n-r}\},\qquad I_W = \{i'_1,\ldots,i'_{n-s}\}.
\]

We claim the $(i,i')$ entry in $Q$ is zero for every $i\in I_U$ and $i'\in I_W$.
By construction, there are elements $u=e_i+\sum_{j>i}\alpha_j e_j\in U$ and $w=e_{i'}+\sum_{j>i'}\beta_j e_j\in W$ for suitable scalars $\alpha_j,\beta_j$.
If the $(i,i')$ entry in $Q$ is not zero, the term $z_{i,i'}$ appears in $uQw^t$, contrary to $u\perp_Q w$.

Next, we show that for every $i\geq r-s$ and $i'\geq s+1$, the $(i,i')$ entry in $Q$ is zero which puts $Q$ in the form (<ref>).
When $X=\fano{k}{\SD[r]{n}}$, apply Proposition <ref> part <ref> to the zero entry $(i_1,i'_1)$ in $Q$.
Because the dimensions of $U$ and $W$ imply $i_1\leq r-s$ and $i'_1\leq s+1$, we have the required zero blocks in $Q$.

When $X=\fano{k}{\AD[r]{n}}$, if $i_1\neq i'_1$, then the same argument as in the symmetric case applies.
Suppose now that $i_1=i'_1\leq s+1$.
If $i'_2\leq s+1$, we can apply Proposition <ref> to the zero entry $(i_1,i'_2)$ because $i_1\neq i'_2$ and we are done.
If $i'_2>s+1$, then $\dim W=n-s$ gives $i'_2=s+2$.
On the other hand, $r$ is even in the alternating case and $\dim U\leq \dim W$ yields $i'_2=s+2\leq r-s$.
Another application of Proposition <ref> to the zero entry $(i_1,i'_2)$ in $Q$ gives that for every $i\geq s+1$ and $i'\geq r-s$, the $(i,i')$ entry in $Q$ is zero, putting this matrix in the desired form.
The proof is the same for $X=\fano{k}{D_{m,n}^r}$ by viewing an $m\times n$ matrix $A\in Q$ as a bilinear map $\KK^m\times\KK^n\to\KK$ via $(u,w)\mapsto uAw^t$.

The set of connected components of a scheme $X=\fano{k}{\SD[r]{n}}$, $\fano{k}{\AD[r]{n}}$ or $\fano{k}{D_{m,n}^r}$ is in bijection with the set of connected components of its associated graph $\mathcal{G}_X$.
In particular, $X$ is connected if and only if $\mathcal{G}_X$ is a connected graph.

Let $\mathrm{\textbf{C}}$ be the subscheme of $X$ consisting of all compression spaces, i.e. $\mathrm{\textbf{C}}=\bigcup_s \compk{k}{s}$.
By definition $\mathcal{G}_X$ has a node for each $s$ such that $\compk{k}{s}\neq\emptyset$ and by Remark <ref>, two nodes corresponding to $s,s'$ are adjacent if and only if $\compk{k}{s}$ intersects $\compk{k}{s'}$.
This means the connected components of the graph $\mathcal{G}_X$ are in bijection with the connected components of $\mathrm{\textbf{C}}$.
Now use Theorem <ref>.

Next, we prove Corollary <ref> that characterizes when the graph $\mathcal{G}_X$, and as a result $X$, is connected.

By Lemma <ref>, the scheme $X$ is disconnected if and only if the graph $\mathcal{G}_X$ is.
The two conditions of the statement are equivalent to the fact that the graph $\mathcal{G}_X$ has at least two vertices and the cycle $s_1-s_2-\cdots-s_{\ell}-s_1$ is missing two edges.
We need to show $\mathcal{G}_X$ is disconnected if and only if these two conditions on $\mathcal{G}_X$ hold.

One direction is clear.
Assume that $\mathcal{G}_X$ has at least two vertices and the cycle is missing two edges.
If $\ell=2$, $\mathcal{G}_X$ is obviously disconnected, so assume $\ell\geq 3$.
Consider the graph $\tilde{\mathcal{G}}_X$ in Definition <ref>.
Build the subgraph $\mathcal{G}'_X$ of $\tilde{\mathcal{G}}_X$ by removing those edges whose labels are less than $k$.
This is similar to building $\mathcal{G}_X$ from $\tilde{\mathcal{G}}_X$ except that we are not deleting any vertices.
This makes $\mathcal{G}_X$ a subgraph of $\mathcal{G}'_X$.

We claim the following statements hold for $\mathcal{G}'_X$ (we use $\sim$ for the relation of two vertices being adjacent):

* If $i\not\sim i+1$ then $s\not\sim i+1$ for $0\leq s\leq i$,

* If $i\not\sim i+1$ then $i\not\sim s$ for $i+1\leq s\leq s_{\max}$.

The following are immediate consequences of the above two statements:

* If $i\not\sim i+1\not\sim i+2$, then $i+1$ is an isolated vertex (i.e. not adjacent to any other vertex),

* If $i\not\sim i+1$ and $j\not\sim j+1$ for $0\leq i<j<s_{\max}$, then every vertex $s$ with $i+1\leq s \leq j$ is an isolated
vertex: the function $g(\{s,s+1\})$ is convex in $s$ and by assumption $g(\{i,i+1\}),g(\{j,j+1\})<k$.
This shows $g(\{s,s+1\})<k$ or equivalently $s\not\sim s+1$ for $i\leq s \leq j$.
Now use statement <ref>.

* If $i\not\sim i+1$ for some $0\leq i<s_{\max}$ and $s_{\max}\not\sim 0$, then there are no edges between $\mathbf{V}_1=\{0,\ldots,i\}$ and $\mathbf{V}_2=\{i+1,\ldots,s_{\max}\}$: by statements <ref> and <ref> we have $0\not\sim i+1$ and $i\not\sim s_{\max}$, i.e. $k$ is larger than all of

\begin{equation}\label{eq:four-points}
g(\{0,i+1\}), g(\{i,i+1\}), g(\{i,s_{\max}\}), g(\{0,s_{\max}\}).
\end{equation}

We claim $k>g(\{s,s'\})$ for all $0\leq s\leq i$ and $i+1\leq s'\leq s_{\max}$, which is equivalent to no edges between $\mathbf{V}_1$ and $\mathbf{V}_2$.
Momentarily think of the polynomial defining $g(\{s,s'\})$ as a function that accepts a pair of real numbers $(s,s')$ with domain $s<s'$.
This function is  convex since its Hessian is positive semidefinite.
In particular when four points in the domain are the vertices of a quadrilateral (such as those in (<ref>)), then the value of the function at every convex combination of the vertices, e.g. $(s,s')$ with $s,s'\in\ZZ_{\geq0}$, $0\leq s\leq i$, and $i+1\leq s'\leq s_{\max}$, is at most the maximum of the values at the vertices.

Assuming these claims hold, we prove that $\mathcal{G}_X$ is disconnected.
By assumption, at least two edges are missing from $\{s_1,s_2\},\{s_2,s_3\},\ldots,\{s_{\ell-1},s_{\ell}\},\{s_{\ell},s_1\}$.
There are two cases: if $\{s_i,s_{i+1}\}$ and $\{s_j,s_{j+1}\}$ are missing in $\mathcal{G}_X$ for some $1\leq i<j<\ell$, then $\{s_i,s_i +1\}$ and $\{s_j,s_j +1\}$ are also missing in $\mathcal{G}'_X$.
This is because either $s_{i+1}=s_i+1$ or $s_{i+1}>s_i+1$.
If the latter holds, then the vertex $s_i+1$ is not present in $\mathcal{G}_X$, meaning that $k>g(\{s_i+1,s\})$ for all $s\neq s_i+1$, i.e. $s_i+1$ is isolated in $\mathcal{G}'_X$.
A similar argument works for $\{s_j,s_j+1\}$.
By statement <ref>, $s_j$ is isolated in $\mathcal{G}'_X$ and hence in $\mathcal{G}_X$.

The second case is when $\{s_i,s_{i+1}\}$ and $\{s_{\ell},s_1\}$ are missing in $\mathcal{G}_X$ for some $1\leq i<\ell$.
Then $\{s_i,s_{i}+1\}$ and $\{s_{\max},0\}$ are missing in $\mathcal{G}'_X$ and by statement <ref>, there are no edges between $\{0,\ldots,s_i\}$ and $\{s_i +1,\ldots,s_{\max}\}$ in $\mathcal{G}'_X$ and therefore there are no edges between $\{s_1,\ldots,s_i\}$ and $\{s_{i+1},\ldots,s_{\ell}\}$ in $\mathcal{G}_X$, completing the proof.

Proof of statement <ref>: by assumption $k>g(\{i,i+1\})$.

\[
g(\{i,i+1\})-g(\{0,i+1\})= (n-r)i\geq 0
\]

and $g(\{s,i+1\})$ is convex in $s$ when $s<i+1$, we deduce $k>g(\{s,i+1\})$ for $0\leq s\leq i$, i.e. $s\not\sim i+1$.

Proof of statement <ref>: if $i+1=s_{\max}$, there is nothing to prove, so let $i+2\leq s_{\max}$.
In the symmetric case $s_{\max}\leq \frac{r-1}{2}$, and we have $i+2\leq\frac{r-1}{2}$.

\[
\Delta=g(\{i,i+1\})-g(\{i,\frac{r-1}{2}\})=\frac{1}{8}\left( r-2i-5 \right)\left( r-2i-3 \right)\geq 0,
\]

which together with $k>g(\{i,i+1\})$ and convexity of $g(\{s,s'\})$ yields $k>g(\{i,s\})$ for $s\geq i+1$, i.e. $i\not\sim s$.
In the alternating and rectangular cases, $s_{\max}=\frac{r-2}{2}$ and $s_{\max}=r-1$ respectively, and a similar argument using $\Delta=g(\{i,i+1\})-g(\{i,s_{\max}\})$ goes through.

§ ON THE COMPONENTS

Starting with this section, we will solely work with the Fano schemes $\fano{k}{\SD[r]{n}}$.
The main result of this section is that when $k=\kappa(s)$, the closed subscheme $\compk{\kappa(s)}{s}$ is an irreducible component.
This subscheme is the image of

\begin{equation}
\label{eq:map-from-flag-variety}
\flag{s+1+n-r,n-s}{\KK^n}\to\fano{\kappa(s)}{\SD[r]{n}},
\end{equation}

that sends a flag $U\subset W\subset\KK^n$ to the point $[Q]$ where $Q\subset S_n$ is the subspace of all symmetric bilinear forms with $U \perp_Q W$.

Let $0\leq s\leq\lfloor\frac{n-1}{2}\rfloor$ and $k=\kappa(s)$ and let $[Q]\in\fano{k}{\SD[r]{n}}$ be the unique Borel fixed nested $s$-compression space.
There is an affine open neighborhood of $[Q]$ such that every closed point in this neighborhood is in the $\gl{n}$ orbit of $[Q]$ and hence a nested $s$-compression space.

Let $Q_{n,s}\subset S_n$ be the standard s-compression space in (<ref>) where the star blocks are populated with $\kappa(s)+1$ independent variables $z_{i,j}$.
There is an affine chart of the Grassmannian $\Gr\left(\kappa(s)+1, S_n\right)$ corresponding to $[Q_{n,s}]$ that consists of all points $[P_{n,s}]$ where $P_{n,s}\subset S_n$ is of the form $Q_{n,s}$ except that the zero entries are replaced by linear forms $c_{i,j}$ (see <cit.>).
It is enough to prove that if the $r\times r$ minors of $P_{n,s}$ are identically zero, then $P_{n,s}$ is in the $\gl{n}$ orbit of $Q_{n,s}$.

We start with the case $r=n$.
We perform a sequence of row-column operations to reduce the $c_{i,j}$ entries in $P_{n,s}$ to zero.
This proves $P_{n,s}$ is in the $\gl{n}$-orbit of $Q_{n,s}$.
Partition $P_{n,s}$ into blocks with block sizes as in (<ref>), then:

\begin{pmatrix}
z_{1,1}        & \cdots & \bm{z_{1,n-1}} & c_{1,n}   \\
\vdots         & \bm{\iddots} & \vdots         & \vdots    \\
\bm{z_{1,n-1}} & \cdots & z_{n-1,n-1}    & c_{n-1,n} \\
c_{1,n}        & \cdots & c_{n-1,n}      & c_{n,n}
\end{pmatrix}

\begin{pmatrix}
z_{1,1} & z_{1,2} & z_{1,3} & z_{1,4} &\bm{z_{1,5}} \\
z_{1,2} & z_{2,2} & \bm{z_{2,3}} &         &         \\
z_{1,3} & \bm{z_{2,3}} & z_{3,3} &         &         \\
z_{1,4} &         &         &         &         \\
z_{1,5}   &         &         &         &
\end{pmatrix}

Marked entries (in bold) in the proof of Proposition <ref>; the entries not shown in (b) are linear forms $c_{i,j}$.

* Mark the entries on the anti-diagonal of the middle $(r-2s-1)\times(r-2s-1)$ block as well as the entries on the anti-diagonal of $P_{n,s}$ that are in the top right $s\times(s+1+n-r)$ block.
These entries are all independent variables $z_{i,j}$ for suitable $i$ and $j$.
Label these entries as $1,2,\ldots$ starting from the bottom left and going up and to the right (see Figure <ref>).

* For each $i=1,\ldots,n$ do the following for the marked entry with label $i$ (the first one is the $(r-s-1,s+1)$ entry): perform row-column operations to remove this marked entry from the $c_{i,j}$ entries below it in that column.
After a change of variables and relabeling, we may assume these operations do not change any of the $z_{i,j}$ entries in the $*$ blocks and at each step we do not disturb the elimination we did for the previous marked entry.

We call the above sequence of operations, simplifying the matrix $P_{n,s}$ and the resulting matrix simplified.
We claim if $P_{n,s}$ is simplified, then all the $c_{i,j}$ entries are reduced to zero and we prove this by inducting on $n$.

It is easy to prove the base case $n=2$ (and necessarily $s=0$).
Assume $n>2$ and the claim holds for simplified matrices of size smaller than $n$.
First we consider the case $s=0$ (see Figure <ref>).
Modulo $z_{1,1},z_{1,2},\ldots,z_{1,n-1}$, we have $\det P_{n,0}=c_{1,n}^2\det \tilde{P}=0$, where $\tilde{P}$ is the generic symmetric matrix obtained by deleting rows and columns $\{1,n\}$ of $P_{n,0}$.
This gives $c_{1,n}=0$, i.e. $c_{1,n}$ involves only the $z_{i,j}$ in the first row except $z_{1,n-1}$ (in simplifying $P_{n,0}$ we removed $z_{1,n-1}$ from $c_{1,n}$).

Modulo $z_{1,1}, z_{1,2},\ldots,z_{1,n-2}$, we have

\begin{equation}\label{eq:det-base-case}
0=\det P_{n,0}=
\begin{pmatrix}
z_{2,2}   & \cdots & z_{2,n-2}   & c_{2,n}   \\
\vdots    & \ddots & \vdots      & \vdots    \\
z_{2,n-2} & \cdots & z_{n-2,n-2} & c_{n-2,n} \\
c_{2,n}   & \cdots & c_{n-2,n}   & c_{n,n}
\end{pmatrix},
\end{equation}

where the matrix on the right is simplified.
Using the induction hypothesis gives $c_{2,n},\ldots,c_{n-2,n},c_{n,n}$ are zero (modulo $z_{1,1},\ldots,z_{1,n-2}$) and thus do not involve $z_{1,n-1}$.
We conclude that $z_{1,n-1}$ appears only in entries $(1,n-1)$ and $(n-1,1)$.
Therefore in the expansion of $P_{n,0}$, the coefficient of $z_{1,n}^2$ is exactly the determinant on the right-hand side of (<ref>) which must vanish (this time not modulo any variables).
Another application of the induction hypothesis gives $c_{2,n},\ldots,c_{n-2,n},c_{n,n}$ are indeed zero.

It remains to show $c_{1,n}=c_{n-1,n}=0$.
If a term in the expansion of $\det P_{n,0}$ involves entry $(1,n-1)$, then due to the zeros in the last column and row, this term has to involve $c_{n-1,n}$ in the last column and $c_{1,n}$ in the last row.
Combining all such terms gives $\pm z_{1,n-1}c_{n-1,n}c_{1,n}\det P'$ where $P'$ is obtained from $P_{n,0}$ by deleting rows and columns $\{1,n-1,n\}$.
Due to the symmetry of $P_{n,0}$, the terms in $\det P_{n,0}$ involving $z_{1,n-1}$ with exponent 1 are $\pm 2z_{1,n-1}c_{n-1,n}c_{1,n}\det P'$.
Because $\det P'\neq 0$, we get $c_{n-1,n}c_{1,n}=0$.
Since one of $c_{n-1,n}$ or $c_{1,n}$ is zero, by expanding $\det P_{n,0}$ along the last column and using Proposition <ref>, we get the other one is also zero.

Next, assume $s\geq 1$.
Call the $s$ marked entries in the upper right $s\times (s+1+n-r)$ block of $P_{n,s}$ pivots  (see Figure <ref>).
Fix $1\leq \ell\leq s$ and consider the pivot $z_{\ell,n-\ell+1}$ in row $\ell$.
Modulo the variables in row $\ell$ of $P_{n,s}$ except the pivot we have

\[
0=\det P_{n,s}=z_{\ell,n-\ell+1}^2\det P_{n-2,s-1},
\]

where $P_{n-2,s-1}$ is the matrix obtained from $P_{n,s}$ by deleting rows and columns $\{\ell,n-\ell+1\}$.
Because $P_{n-2,s-1}$ is simplified, by the induction hypothesis, its $c_{i,j}$ entries are zero modulo the variables in the pivot row except the pivot itself.
This shows that these $c_{i,j}$ entries in $P_{n-2,s-1}$ do not involve the pivot and since $P_{n,s}$ is simplified, the only entries involving the pivot $z_{\ell,n-\ell+1}$ are the entries $(\ell,n-\ell+1)$ and $(n-\ell+1,\ell)$.
Hence, the coefficient of $z_{\ell,n-\ell+1}^2$ in $\det P_{n,s}$ is $\det P_{n-2,s-1}$ which must be zero (this time not modulo any variables) and another application of the induction hypothesis gives that the $c_{i,j}$ entries in $P_{n-2,s-1}$ are indeed zero (see Figure <ref> for an example).

\begin{pmatrix}
\cdots & z_{1,n-s}   & z_{1,n-s+1}      &              & \bm{z_{1,n}} \\
&    \vdots         &       \vdots           & \bm{\iddots} &     \vdots         \\
\cdots  & z_{s,n-s}   & \bm{z_{s,n-s+1}} &              &     z_{s,n}         \\
\cdots & c_{s+1,n-s} & c_{s,n-s+1}      &              & c_{s,n}      \\
&  \vdots     &    \vdots              &              & \vdots
\end{pmatrix}
pivots in $P_{n,s}$ (in bold)

\begin{pmatrix}
\cdots & z_{1,n-2}   & z_{1,n-1}      & \bm{z_{1,n}} \\
\cdots & z_{2,n-2}   & \bm{z_{2,n-1}} & z_{2,n}      \\
& 0           & c_{3,n-1}      & 0            \\
& 0           & \vdots         & 0            \\
\cdots & c_{n-1,n-2} & c_{n-1,n-1}    & c_{n-1,n}    \\
& 0           & c_{n-1,n}      & 0
\end{pmatrix}
using pivot in row $\ell=2$ of $P_{n,2}$

For pivot in row $\ell$, $c_{i,j}=0$ for $i,j\neq n-\ell+1$.

The above argument shows that for $1\leq\ell\leq s$ the $c_{i,j}$ not in column $n-\ell+1$ and not in row $n-\ell+1$ are zero.
Depending on $s$, the number of pivots, there are three cases:

* If $s\geq 3$, then for every $c_{i,j}$ there is a pivot such that the argument gives $c_{i,j}=0$ because there exists $1\leq \ell\leq s$ with $i,j\neq n-\ell+1$.

* If $s=2$, the argument gives all $c_{i,j}$ are zero except possibly $c_{n-1,n}$ (see Figure <ref>).
Since $z_{1,n}$ is a pivot, the only entries involving $z_{1,n}$ are $(1,n)$ and $(n,1)$.
Therefore, in $\det P_{n,s}$, combining the terms involving $z_{1,n}$ with exponent 1 gives $\pm 2z_{1,n}z_{2,n-2}c_{n-1,n}\det P'$ where $P'$ is the submatrix of $P_{n,s}$ obtained by deleting rows $\{1,2,n\}$ and columns $\{n-2,n-1,n\}$.
By Proposition <ref>, $\det P'\neq0$ implying $c_{n-1,n}=0$.

* If $s=1$, then we have only $z_{1,n}$ as pivot and the argument gives all $c_{i,j}$ are zero except possibly those in the last row and column.
We claim $c_{n-1,n}=0$.
Since none of the $c_{i,j}$ involve $z_{1,n}$, the terms in $\det P_{n,s}$ involving $z_{1,n}$ with exponent 1 are $\pm 2z_{1,n}c_{n-1,n}\det P'$, where $P'$ is obtained from $P_{n,s}$ by deleting rows $\{1,n\}$ and columns $\{n-1, n\}$.
By Proposition <ref>, $\det P'\neq 0$ and we get $c_{n-1,n}=0$.
Expanding $\det P_{n,s}$ along column $n-1$ gives

\[
0=\det P_{n,s}=z_{1,n-1}^2 \det
\begin{pmatrix}
z_{2,2}   & \cdots & z_{2,n-2}   & c_{2,n}   \\
\vdots    & \ddots & \vdots      & \vdots    \\
z_{2,n-2} & \cdots & z_{n-2,n-2} & c_{n-2,n} \\
c_{2,n}   & \cdots & c_{n-2,n}   & c_{n,n}
\end{pmatrix}.
\]

Since the above matrix is simplified, the induction hypothesis gives $c_{i,n}=0$.

Finally, we prove the case $r<n$ using the statement for $r=n$.
Simplify the matrix $P_{n,s}$ and partition it in the following way

\[
\begin{pNiceArray}{cccc}[first-row,first-col,margin]
& s       & r-2s-1 & n-r+1 & s\\
s      &  Z_1    &  Z_2   &  Z_3  & Z_4\\
&         &        &       &    \\
r-2s-1 &  Z_2^t  &  Z_5   &   C_1 & C_2\\
&         &        &       &    \\
n-r+1  &  Z_3^t  &  C_1^t &   C_3 & C_4\\
&         &        &       &    \\
s      &  Z_4^t  &  C_2^t & C_4^t & C_5
\end{pNiceArray},
\]

where the $Z_i$ blocks are filled with free variables $z_{i,j}$ and the entries in the $C_i$ blocks are linear forms $c_{i,j}$.
We claim $C_i$ are all zero.
For each $r-s\leq \ell \leq n-s$, we consider the principal $r\times r $ minor of $P_{n,s}$ involving rows and columns $\{1,\ldots,r-s-1,\ell,n-s+1,\ldots,n\}$.
This minor is the determinant of a simplified matrix of the form $P_{r,s}$ and vanishes.
By what we proved for $r=n$, we get that the $c_{i,j}$ entries in these minors are zero.
This means all $C_i$ blocks are zero except possibly $C_3$ (only diagonal entries of $C_3$ appear in these minors).

To show $C_3$ is zero, pick the entry $c_{i,j}$ in $C_3$.
Then consider the $r\times r$ minor given by the first $r-s-1$ rows and columns, the last $s$ rows and columns, and the $i$-th row and $j$-th column.
This minor is $\pm c_{i,j}(\det Z_4)^2\det Z_5$ which must vanish and since $\det Z_4\det Z_5\neq 0$, we get $c_{i,j}=0$.
This completes the proof.

For $0\leq s\leq\lfloor\frac{n-1}{2}\rfloor$, the closed subscheme $\comp{s}$ is isomorphic in its reduced structure to the flag variety $\flag{s+1+n-1,n-s}{\KK^n}$ and is an irreducible component of $\fano{\kappa(s)}{\SD[r]{n}}$.
In particular, $\comp{s}$ is the $\gl{n}$ orbit of the standard $s$-compression space and does not intersect other components (if any) of the scheme.

The nested $s$-compression spaces form an irreducible closed subscheme $\compk{\kappa(s)}{s}$ of $\fano{\kappa(s)}{\SD[r]{n}}$ as it is the image of the morphism in (<ref>).
Denote by $\phi$ the composition of $\fano{\kappa(s)}{\SD[r]{n}}\to\Gr(\kappa(s),S_n)$ with the map in (<ref>).
Then $\phi$ is $\gl{n}$-equivariant and bijective onto its image.

A standard argument using affine charts of the flag variety and the Grassmannian shows the differential $d\phi$ is injective at the flag that maps to the standard $s$-compression space $[Q]$.
Because $\gl{n}$ acts transitively on $\flag{s+1+n-r,n-s}{\KK^n}$, $d\phi$ is injective everywhere, hence $\phi$ is a closed embedding.
This implies $\comp{s}$ is isomorphic in its reduced structure to $\flag{s+1+n-r,n-s}{\KK^n}$.
In particular it is the orbit of any single point, say $[Q]$, and does not intersect other components.

The point $[Q]$ is the unique Borel fixed point in $\compk{\kappa(s)}{s}$.
If $\compk{\kappa(s)}{s}$ is contained properly in an irreducible component, then any Zariski open neighborhood of $[Q]$ must contain a point that is not a nested $s$-compression space which contradicts Proposition <ref>.
Therefore, $\compk{\kappa(s)}{s}$ is an irreducible component.

Next, we prove Loewy and Radwan's theorem <cit.> as a corollary.

Let $Q\subset S_n$ be a subspace whose elements have rank less than $r$ and assume $Q$ is of maximal dimension.
Then $Q$ is a $\gl{n}$-translate of either the standard $0$-compression space or the standard $\lfloor\frac{r-1}{2}\rfloor$-compression space.

Let $k+1$ be the linear dimension of $Q$ and set $s_{\max}=\lfloor\frac{r-1}{2}\rfloor$.
By assumption $k$ is the maximum integer such that $\fano{k}{\SD[r]{n}}\neq \emptyset$.
By Meshulam's result (Proposition <ref>), $k=\max \{\kappa(0),\kappa(s_{\max})\}$.
If $k=\kappa(0)$ (or $k=\kappa(s_{\max})$), then Corollary <ref> gives that $\compk{k}{0}$ (respectively $\compk{k}{s_{\max}}$) is an irreducible component not intersecting other components and is the $\gl{n}$-orbit of the standard $0$-compression space (respectively $\left(s_{\max}\right)$-compression space).

Since $\kappa(s)$ is convex in $s$, for $0< s <s_{\max}$ we have $k>\kappa(s)$ and thus every Borel fixed point is either a $0$- or $\left(s_{\max}\right)$-compression space.
By Borel's theorem <cit.>, every irreducible component must contain a Borel fixed point.
We deduce that the Fano scheme $\fano{k}{\SD[r]{n}}$ is the disjoint union of $\compk{k}{0}$ and $\compk{k}{s_{\max}}$.

Let $\mathcal{U}(s)$ be the restriction of the universal bundle on $\fano{\kappa(s)}{\SD[r]{n}}$ to $\comp{s}$.
For $k<\kappa(s)$, the subscheme $\compk{k}{s}$ is the image of the natural morphism

\begin{equation}
\label{eq:map-from-Grassmann-bundle}
\Gr\left(k+1,\mathcal{U}(s)\right)\to\fano{k}{\SD[r]{n}},
\end{equation}

from the Grassmann bundle $\Gr(k+1,\mathcal{U}(s))$ that sends a $(k+1)$-dimensional subspace of $\KK^{\binom{n+1}{2}}$ to its corresponding point on the Fano scheme.

We suspect that the subscheme $\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$ is also an irreducible component when $k<\kappa(s)$.
We will prove this for $k=1$ in §<ref> and for $r=2s+1$ in §<ref>.
However, our techniques do not apply to arbitrary $k,s$.

§ FANO SCHEME OF LINES

In this section we give a complete description of the irreducible components of $\fano{1}{\SD[r]{n}}$.
Using this description, we will characterize when $\fano{k}{\SD[r]{n}}$ is irreducible.
To prove the main results, we need three lemmas.

Fix integers $n\geq2$ and $0<s\leq\frac{n}{2}$.
For a general 2-dimensional linear subspace $Q\subset S_n$, there are no linear subspaces $U,W\subset\KK^n$ of dimensions $s$ and $n-s$ such that $U\subset W$ and $U\perp_Q W$.

Define the incidence variety $Y\subset\flag{s,n-s}{\KK^n}\times\Gr(2,S_n)$ by

\[
Y=\left\{ \big( [U,W], [P] \big)\in\flag{s,n-s}{\KK^n}\times\Gr(2,S_n) \,|\, U\perp_{P} W\right\}.
\]

Consider the diagram

\[
\begin{tikzcd}
Y \arrow[r, "\pi_1"] \arrow[d, "\pi_2"'] & \flag{s,n-s}{\KK^n} \\
\Gr(2,S_n)                                      &
\end{tikzcd}

where $\pi_1$ and $\pi_2$ are the projections of $\flag{s,n-s}{\KK^n}\times\Gr(2,S_n)$ onto its factors.

We show that the restriction of $\pi_2$ to $Y$ is not surjective by comparing the dimensions of $Y$ and $\Gr(2,S_n)$.

Fix a point $[U,W]\in\flag{s,n-s}{\KK^n}$, where $U\subset W$ are subspaces of $\KK^n$ of dimensions $s$ and $n-s$ respectively.
The $\gl{n}$ action on $\KK^n$ induces a natural action of $\gl{n}$ on $\flag{s,n-s}{\KK^n}$

and on $Y$.
The restriction of $\pi_1$ to $Y$ is $\gl{n}$-equivariant.
We may thus assume

\[
U = \spn\{e_{n-s+1},\ldots,e_n\},\qquad W = \spn\{e_{s+1}, \ldots,e_n\},
\]

where $e_i$ denotes the $i$-th standard basis vector in $\KK^n$.
The fiber of $\pi_1$ over $[U,W]$ consists of all 2-dimensional subspaces of $\tilde{P}\subset S_n$ where $\tilde{P}$ is the subspace of all symmetric matrices of the form

\[
\begin{pNiceArray}{cccccc}[first-row,first-col]
&& &&n-2s&s&\\
&&&&&& \\
&&*&&*&*& \\
& &  &&&& \\
& &&  &&& \\
\end{pNiceArray}.
\]

\[
\dim\pi_1^{-1}([U,W]) =\dim\Gr(2, \tilde{P})=2\left( \binom{n+1}{2}-\left( \binom{s+1}{2}+s(n-2s) \right) -2 \right).
\]

The dimension of $Y$ is then

\[
\dim Y=\dim\flag{s,n-s}{\KK^n}+\dim\pi_1^{-1}([U,W])=n^2+n-s-4.
\]

We now see that

\[
\dim \Gr(2,S_n)-\dim Y = s>0.
\]

Since $\pi_2$ is not surjective and its image is closed, the claim is proved.

Let $r=2s+1$ and assume $[Q]\in\fano{k}{\SD[r]{n}}$ is a nested $t$-compression space only for $t=s$.

* There exists a unique flag $U\subset W\subset\KK^n$ with $\dim U=s+1+n-r$, $\dim W=n-s$, and $U\perp_Q W$.

* If a flag $U'\subset W'\subset \KK^n$ satisfies $\dim U'+\dim W'\geq 2n-r+1$ and $U'\perp_Q W'$, then $\dim U'+\dim W'= 2n-r+1$ and $U'\subset W'$ is the unique flag in part <ref>.

Part <ref>: the existence of the flag $U\subset W$ follows from the definition of $[Q]$.
To show uniqueness, assume $U_1\subset W_1$ is another flag satisfying the conditions.
We have $\dim U\cap U_1\geq \dim U+\dim U_1 - n = 1+n-r$.
Let $p\geq 0$ be such that $\dim U\cap U_1=p+1+n-r$.
It follows $\dim (U+U_1)=n-p$.
Since $U\cap U_1\perp_Q U+U_1$, $[Q]$ is a nested $p$-compression space, forcing $p=s$ and therefore $U_1=U$.

Next, we show $W_1=W$.
If $s=0$, then $W=W_1=\KK^n$.
If $s>0$ and $W\neq W_1$, then $\dim W+W_1> n-s$.
Choose subspaces $U_2\subset U$ and $W_2\subset W+W_1$ with $U_2\subset W_2$, $\dim U_2=(s-1)+1+n-r$ and $\dim W_2=n-(s-1)$.
Then $U_2\perp_Q W_2$ turns $[Q]$ into a nested $(s-1)$-compression space, a contradiction.

Part <ref>: the inequality implies $\dim U'\geq 1+n-r$.
Let $p\geq0$ be such that $\dim U'=p+1+n-r$.
Then $\dim W'\geq n-p$, and $U'\perp_Q W'$ makes $[Q]$ a nested $p$-compression space forcing $p=s$.
If $\dim W'>n-s$, then similar to the proof of part <ref>, $[Q]$ becomes a nested $(s-1)$-compression space.
We conclude $\dim W'=n-s$.
Now use part <ref>.

Let $0<n'< n$ and assume $[Q]\in\Gr(k+1,S_n)$ is given by an $n\times n$ matrix of the form
P & 0\\
0 & P'
\end{pmatrix}$
where $P$ and $P'$ are square blocks of size $n'$ and $n-n'$, respectively, with linear forms as entries.
Let $U=\spn\{e_1,\ldots,e_{n'}\}$ and $W=\spn\{e_{n'+1},\ldots,e_n\}$ and define $\pi_U$ and $\pi_W$ to be projections of $\KK^n$ onto $U$ and $W$.
For a flag $U'\subset W'\subset\KK^n$ satisfying $U'\perp_Q W'$,

* $U'\cap U \perp_P \pi_U(W')$ and $\pi_W(U')\perp_{P'} W'\cap W$,

* if $\dim U'+\dim W'\geq n+\ell+1$ for some $0\leq \ell<n$, then $\dim U'\cap U+ \dim\pi_U(W')\geq n'+\ell+1$ or $\dim \pi_W(U')+ \dim W'\cap W\geq n-n'$,

* if $\dim U'+\dim W'= n+\ell+1$ for some $0\leq \ell<n$, we have $\dim U'\cap U+ \dim\pi_U(W')= n'+\ell+1$ if and only if $\dim \pi_W(U')+ \dim W'\cap W= n-n'$.

Part <ref>: let $u\in U'\cap U$ and $v\in\pi_U(W')$.
Then $v=w'-w''$ for some $w'\in W'$ and $w''\in \pi_W(W')$.
Since $U'\perp_Q W'$ and $U\perp_Q W$ we have $u\perp_Q w'$ and $u\perp_Q w''$, hence $u\perp_Q v$.
This shows $U'\cap U \perp_Q \pi_U(W')$ and since $U'\cap U,\pi_U(W')\subset U$, they are also orthogonal with respect to $P$.
The proof of $\pi_W(U')\perp_{P'} W'\cap W$ is similar.

Part <ref>: By the exact sequence $0\to U'\cap U\to U' \xrightarrow{\pi_W} W$ we have $\dim U'=\dim U'\cap U+ \dim \pi_W(U')$ and similarly $\dim W'=\dim\pi_U(W')+ \dim W'\cap W$.
Assume to the contrary that $\dim U'\cap U+ \dim\pi_U(W')< n'+\ell+1$ and $\dim \pi_W(U')+ \dim W'\cap W< n-n'$.
Adding these inequalities yields

\[
\left(\dim U'\cap U+ \dim \pi_W(U')\right)+ \left(\dim\pi_U(W')+ \dim W'\cap W\right) < n+\ell+1.
\]

But the left-hand side is $\dim U'+\dim W'$, a contradiction to $\dim U'+\dim W'\geq n+\ell+1$.

To prove <ref>, for each equality in the if and only if statement, subtract it from $\dim U'+\dim W'= n+\ell+1$.

We now prove Theorem <ref> as part of the next two results.

The following holds for the Fano scheme of lines $\fano{1}{\SD[r]{n}}$.

* The subschemes $\compk{1}{s}$ for $s=0, 1,\ldots,\lfloor\frac{r-1}{2}\rfloor$ are distinct, and are exactly the irreducible components of $\fano{1}{\SD[r]{n}}$.
Moreover, $\bigcap_s \compk{1}{s}\neq\emptyset$.

* Let $\,\mathcal{U}(s)$ be the restriction of the universal bundle on $\fano{\kappa(s)}{\SD[r]{n}}$ to $\comp{s}$.
The fiber of the natural map in (<ref>) (with $k=1$) over a general point of $\compk{1}{s}$ is finite.

Part <ref>: by the Kronecker canonical form of a pencil of singular matrices <cit.>, every line on $\SD[r]{n}$ is a compression space, hence a nested one by Corollary <ref>.
Therefore, the irreducible closed subschemes $\compk{1}{s}$ cover $\fano{1}{\SD[r]{n}}$.
For every $0\leq s\leq \lfloor\frac{r-1}{2}\rfloor$, we show $\compk{1}{s}$ has a point not contained in $\bigcup_{s'\neq s}\compk{1}{s'}$ which proves that $\compk{1}{s}$ is an irreducible component.

Set $r'=2s+1$ and $n'=r'+n-r$ and assume we have a pencil $[P]\in\fano{1}{\mathrm{SD}^{r'}_{n'}}$ with $P$ an $n'\times n'$ matrix over $\KK[z_0,z_1]_1$ that is a nested $t$-compression space only for $t=s$.
Also, take $P'$ to be a generic symmetric matrix of size $n-n'$ over $\KK[z_0,z_1]_1$ and define the $n\times n$ matrix

\[
P & 0  \\
0 & P'
\end{pmatrix}.
\]

Define $U = \spn\{e_1,\ldots,e_{n'}\}$ and $W = \spn\{e_{n'+1},\ldots,e_n\}$, so that $\KK^n=U\oplus W$ and the matrices $P$ and $P'$ are pencils of bilinear forms on $U$ and $W$ respectively.
We claim $[Q]\in\fano{1}{\SD[r]{n}}$ and is a nested $t$-compression space if and only if $t=s$.

To show $Q$ is a nested $s$-compression space, let $U_1\subset W_1$ be a flag in $U$ with $U_1\perp_P W_1$ making $P$ into a nested $s$-compression space.
By definition $\dim U_1=s+1+n'-r'$ and $\dim W_1=n'-s$.
Then the flag $U_1\subset W_1+W$ in $\KK^n$ satisfies $U_1\perp_Q W_1+W$ and makes $Q$ into a nested $s$-compression space because $\dim U_1=s+1+n-r$ and $\dim W_1+W=n-s$.
Here we used $n-r=n'-r'$.

Conversely, assume $Q$ is a nested $t$-compression space and $U'\subset W'\subset \KK^n$ is a flag satisfying $\dim U'=t+1+n-r$, $\dim W'=n-t$ and $U'\perp_Q W'$.
Our goal is to show $t=s$.
By Lemma <ref> part <ref>, with $\ell=n-r=n'-r'$, we have $U'\cap U\perp_P \pi_U(W')$ and $\pi_W (U')\perp_{P'} W'\cap W$.
Since $P'$ is not singular, $\dim \pi_W (U')+ \dim W'\cap W\leq \dim W=n-n'$.
By Lemma <ref> parts <ref> and <ref>, the orthogonal flag $U'\cap U\subset \pi_U(W')$ in $U$ satisfies $\dim U'\cap U+\dim\pi_{U}(W')\geq 2n'-r'+1$.
Applying Lemma <ref> part <ref>, we have that this last inequality is an equality and that $U'\cap U\subset \pi_{U}(W')$ is the unique flag making $P$ a nested $s$-compression space.
In particular $\dim U'\cap U=s+1+n'-r'$ and $\dim \pi_U(W')=n'-s$.
Lemma <ref> part <ref> now gives $\dim \pi_W(U')+\dim W'\cap W=n-n'$.

We compare some of the above dimensions and use $n-r=n'-r'$ and $r'=2s+1$.
Comparing the dimensions of $U'\cap U$ and $U'$ gives $s\leq t$.
Next, note that $U'\cap U$ is contained in $\pi_U(W')$ and they have the same dimension which means $\pi_U(W')=U'\cap U$.
We conclude that $\pi_U(U')=U'\cap U$ implying $\pi_W(U')=U'\cap W$.
Therefore $\pi_W(U')\subset W'\cap W$ is a flag in $W$ and by Lemma <ref> part <ref>, $\pi_W(U')\perp_{P'} W'\cap W$.

The exact sequences $0\to U'\cap U\to U' \xrightarrow{\pi_W} W$ and $0\to W'\cap W\to W' \xrightarrow{\pi_U} U$  give the dimensions of the flag $\pi_W(U')\subset W'\cap W$ as

\begin{equation}\label{eq:difference-in-dims}
\begin{split}
\dim \pi_W(U')&=\dim U' - \dim U'\cap U = t-s,\\
\dim W'\cap W &= \dim W' - \dim \pi_U(W') = (n-n')-(t-s).
\end{split}
\end{equation}

Since $P'$ is general and $\pi_W(U')\perp_{P'} W'\cap W$, by Lemma <ref> we must have $t-s=0$.

To complete the proof of $\compk{1}{s}\backslash\bigcup_{s'\neq s}\compk{1}{s'}\neq\emptyset$, we need to produce an $n'\times n'$ matrix $P$ such that $[P]$, as a point on $\fano{1}{\mathrm{SD}_{n'}^{r'}}$ is a nested $t$-compression space only for $t=s$.
Take the nested $s$-compression space

\[
0   & B \\
B^t & 0
\end{pmatrix},\quad\text{with}\quad
\begin{array}{cccc|c}
z_0 & z_1    &        &     &                 \\
& \ddots & \ddots &     & \text{\large 0} \\
&        & z_0    & z_1 &
\end{array}
\right)_{s\times (s+1+n'-r')},
\]

where the entries not shown in $B$ are all zero.
In <cit.>, the authors prove this point is not a $t$-compression space for $t<s$ and by Corollary <ref>, it is also not a $t$-compression space for $t>s$ since $s$ is the middle integer in $0,\ldots,r'-1$.
Therefore, $[P]$ is a nested $t$-compression space only for $t=s$.
The last statement in <ref> holds because $\bigcap_s \compk{1}{s}$ contains the point

\[
\begin{pmatrix}
z_0 & z_1 &                 & \\
z_1 &     &                 & \\
&     & \text{\Large 0} & \\
&     &                 &
\end{pmatrix}.
\]

Part <ref>: the map in (<ref>) is proper with image $\compk{1}{s}$.
For a proper map, the dimension of fibers is upper semi-continuous on the target, therefore it is enough to show the fiber over $[Q]$ is a single point.
Since $\Gr(k+1,\mathcal{U}(s))$ is a bundle over $\comp{s}$ and by Corollary <ref>$, \comp{s}$ is isomorphic in its reduced structure to $\flag{s+1+n-r,n-s}{\KK^n}$ , it is enough show a flag $[U',W']\in\flag{s+1+n-r,n-s}{\KK^n}$ satisfying $U'\perp_Q W'$ is uniquely determined by $[Q]$.

Using (<ref>) and $t-s=0$, we have that $U'=U'\cap U$ and we saw in the proof of part <ref> that $U'\cap U$ is uniquely determined by $[P]$ (the uniqueness is due to Lemma <ref> and does not depend on the representative matrix $P$).
We claim $W'=U'\oplus W$, which shows that $W'$ is also uniquely determined by $[P]$.
By (<ref>), $\dim W'\cap W=n-n'=\dim W$, i.e. $W\subset W'$.
We have subspaces $U',W\subset W'$ with $U'\cap W=0$ (since $U'\subset U$ and $U\cap W=0$) and

\[
\dim U'+\dim W=(s+1+n'-r')+ (n-n')=n-s=\dim W'.
\]

Therefore, $W=U'\oplus W$.

The closed subscheme $\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$ has dimension

\[
\dim \compk{k}{s} = (s+1+n-r)(r-2s-1)+s(n-s)+(\kappa(s)-k)(k+1).
\]

In particular for $k=1$, each irreducible component $\compk{1}{s}\subset\fano{1}{\SD[r]{n}}$ has dimension $nr+(s-1)(n-r)-5$ and when $r=n$, the components are equidimensional and of the expected dimension.

The second statement follows from the first.
By Theorem <ref> part <ref>, there exists a line $[L]\in\compk{1}{s}\subset\fano{1}{\SD[r]{n}}$ for which there are only finitely many flags $U\subset W$ in $\flag{s+1+n-r,n-s}{\KK^n}$ such that $U\perp_L W$.
After acting by $\gl{n}$, we may assume the standard $s$-compression space $[Q]\in\compk{\kappa(s)}{s}$ contains this line.
Take a $k$-plane $[P]\in\compk{k}{s}$ with $L\subset P\subset Q$.
Then there are only finitely many flags $U\subset W$ with $U\perp_P W$.
The map in (<ref>) is proper with image $\compk{k}{s}$ and the fiber over $[P]$ is finite.
By upper semi-continuity of fiber dimensions, the fiber over a general point of $\compk{k}{s}$ is finite.

\begin{align*}
\dim \compk{k}{s} &= \dim \Gr\left(k+1,\mathcal{U}(s)\right)\\
&=\dim \comp{s}+(\kappa(s)-k)(k+1)\\
&=\dim \flag{s+1+n-r,n-s}{\KK^n}+(\kappa(s)-k)(k+1)\\
\end{align*}

In Section <ref> we will prove Proposition <ref> which gives that all irreducible components of $\fano{1}{\SD[r]{n}}$  (except $\compk{1}{\frac{r-1}{2}}$ when $r$ is odd) are generically non-reduced.

Let $0\leq s,s'\leq\lfloor\frac{n-1}{2}\rfloor$ with $s\neq s'$.
Assume that $k\leq\min\{\kappa(s),\kappa(s')\}$ (or equivalently the closed subschemes $\compk{k}{s}$ and $\compk{k}{s'}$ of $\fano{k}{\SD[r]{n}}$ are non-empty).
Then $\compk{k}{s}$ and $\compk{k}{s'}$ are distinct and one does not contain the other.

Let $Q$ be the standard $s$-compression space.
By Theorem <ref> part <ref>, there is a line $L$ inside $Q$ such that $L$ is not a nested $s'$-compression space.
Any $k$-plane inside $Q$ containing $L$ is also not a nested $s'$-compression space, i.e. $\compk{k}{s}$ is not contained in $\compk{k}{s'}$.
The statement is now proved since the argument is symmetric in $s,s'$.

Next, we prove Theorem <ref> characterizing the irreducibility of $\fano{k}{\SD[r]{n}}$.

If $\fano{k}{\SD[r]{n}}$ is irreducible, then there is only one $0\leq s\leq \left\lfloor\frac{n-1}{2}\right\rfloor$ such that $k\leq \kappa(s)$.
This is because if $s\neq s'$ and $k\leq \min\{\kappa(s),\kappa(s')\}$, then $\compk{k}{s},\compk{k}{s'}$ would lie on one irreducible component, namely $\fano{k}{\SD[r]{n}}$.
Let $\mathcal{U}$ be the universal bundle on $\fano{k}{\SD[r]{n}}$.
The natural map from the Grassmann bundle $\Gr(2,\mathcal{U})$ to $\fano{1}{\SD[r]{n}}$ then has an irreducible image containing both $\compk{1}{s},\compk{1}{s'}$ which contradicts Theorem <ref> part <ref>.
Using the convexity of $\kappa(s)$ in $s$, we see that $k\leq \kappa(s)$ for only one integer $s$ if and only if the conditions on $k$ hold.

Conversely, if the conditions on $k$ hold, then by <cit.>, every point of $\fano{k}{\SD[r]{n}}$ is a compression space, i.e. $\fano{k}{\SD[r]{n}}=\compk{k}{0}$ or $\compk{k}{s_{\max}}$, depending on whether the first or second condition on $k$ holds.
We conclude that the scheme is irreducible.

§ SMOOTHNESS

§.§ Dimensions of tangent spaces

Deformation theory provides a tool to compute the dimensions of Zariski tangent spaces at points of $\fano{k}{\SD[r]{n}}$.
We can use these dimensions to detect the smoothness of points.

We describe the general setup.
Let $S=\KK[x_{1,1},x_{1,2},\ldots,x_{n,n}]$ be the homogeneous coordinate ring of the projective space $\PP^{\binom{n+1}{2}-1}$ of all symmetric matrices and let $I$ be the ideal of $\SD[r]{n}$, generated by the $r\times r$ minors of the generic symmetric matrix in (<ref>).
Let $J$ be the homogeneous ideal of a $k$-plane $Q$ inside $\SD[r]{n}$ and choose an isomorphism $S/J=\KK[z_0,\ldots,z_k]$.
Then we can write $Q$ as an $n\times n$ matrix whose entries are linear forms in $z_0,\ldots, z_k$.
The tangent space to the Fano scheme $\fano{k}{\SD[r]{n}}$ at $[Q]$
is isomorphic to the space of first-order deformations of $Q$ inside $\SD[r]{n}$ which is known to be isomorphic to $\Hom_{\mathcal{O}_{\SD[r]{n}}}(\mathcal{J},\mathcal{O}_Q)$, where $\mathcal{J}$ is the ideal sheaf of $Q$ in $\SD[r]{n}$ (see <cit.>).

Since $J$ is generated by homogeneous linear polynomials, $J/I$ and $S/J$ are saturated $S/I$-modules and by the equivalence of the categories of quasi-coherent sheaves on $\SD[r]{n}$ and saturated $S/I$-modules we have

\[
\Hom_{\mathcal{O}_{\SD[r]{n}}}(\mathcal{J},\mathcal{O}_Q)\simeq \Hom_{S/I}(J/I,S/J)_0,
\]

where the right-hand side is the space of degree-preserving maps of graded $S/I$-modules.
In this section we will compute dimensions of Zariski tangent spaces at select points $[Q]\in\fano{k}{\SD[r]{n}}$ by analyzing the corresponding spaces $\Hom_{S/I}(J/I,S/J)_0$.
We say a minor of a matrix is $\ell$-anchored if it involves the first $\ell$ rows and columns.

Let $[Q]\in\fano{k}{\SD[r]{n}}$ be a nested $s$-compression space of the form

\begin{equation}
\label{eq:s-comp-space-block-form-r}
\begin{pNiceArray}{ccccc}[first-row,first-col]
&&&&& \\
s&&B&&C&D \\
& &  &&& \\
& &&  && \\
\end{pNiceArray},
\end{equation}

where the blocks $B,\ldots,E$ have entries in $\KK[z_0,\ldots,z_k]_1$.
If $r-2s-1>0$ and $\det E=0$, then

\[
\dim T_{[Q]}\fano{k}{\SD[r]{n}}=\left( \binom{n+1}{2}-1-k \right)(k+1).
\]

On the other hand if $r-2s-1=0$ or $\det E\neq 0$, then

\[
\dim T_{[Q]}\fano{k}{\SD[r]{n}}=a_{\det} + (s+1+n-r)(r-2s-1)(k+1)+(\kappa(s)-k)(k+1),
\]

where $a_{\det}$ is the dimension of the $\KK$-vector space of all symmetric matrices $A$ of size $s+1+n-r$ over $\KK[z_0,\ldots,z_k]_1$ such that the $s$-anchored $(2s+1)\times(2s+1)$ minors of

\begin{equation}
\label{eq:permissible-A}
\begin{pNiceArray}{cccc}[first-row,first-col]
&&&& \\
s&&&0&D \\
&&&& \\
\end{pNiceArray}
\end{equation}

As explained above, $\dim T_{[Q]}\fano{k}{\SD[r]{n}}=\dim\Hom_{S/I}(J/I,S/J)_0$.
A set of generators for $J/I$ consists of $x_{i,j}$ corresponding to the entries in the zero blocks in (<ref>) together with $\kappa(s)-k$ linearly independent linear forms $f_i$.
An element $\phi\in\Hom_{S/I}(J/I,S/J)_0$ is determined by specifying the images $\phi(x_{i,j})$ and $\phi(f_i)$ in $\KK[z_0,\ldots,z_k]_1$.
These images must be chosen in a way that every relation satisfied by the generators $x_{i,j}, f_i$ is also satisfied by their images.

The Koszul relations $a_1(a_2)-a_2(a_1)$ for $a_i\in J$ are obviously mapped to zero.
The only other relations come from writing the $r\times r$ minors of the generic symmetric matrix (generators of $I$) as $S/I$-linear combinations of the generators of $J$.
A minor that is not $(r-s-1)$-anchored has at least $s+2$ of the last $s+1+n-r$ columns and is automatically sent to zero in $S/J$ since every term in the minor involves two $x_{i,j}$ corresponding to positions in the zero blocks, hence two generators of $J/I$.

We inspect each term of an $(r-s-1)$-anchored minor up to Koszul relations.
Such a minor involves $s+1$ of the last $s+1+n-r$ columns in (<ref>).
If a term involves a position in block $B$ or $C$, then it  involves fewer than $s$ positions in block $D$, hence involves two positions in the zero blocks.
Similarly, if a term has a position in the $(r-2s-1)\times (s+1+n-r)$ zero block, then it will have fewer than $r-2s-1$ positions in block $E$, thus involves a position in the $(s+1+n-r) \times (r-2s-1)$ block, i.e. two positions in the zero blocks.
To remove the terms that are mapped to zero from consideration, we may replace blocks $B, C, C^t$, and the $(r-2s-1)\times (s+1+n-r)$ and $(s+1+n-r) \times (r-2s-1)$ blocks with zeros in the generic symmetric matrix.

Let $A$ be the matrix filled with $\phi(x_{i,j})$ for those $(i,j)$ in the lower right block of size $s+1+n-r$ in (<ref>).
The entries of $A$ must be such that the $(r-s-1)$-anchored $r\times r$ minors of

\begin{equation}
\label{eq:s-comp-blocks-zero-substituted}
\begin{pNiceArray}{ccccc}[first-row,first-col]
&&&&& \\
s&&0&&0&D \\
& &  &&& \\
& &&  && \\
\end{pNiceArray},
\end{equation}

vanish, since these minors are the images in $S/J$ of the $r\times r$ minors of the generic symmetric matrix under the map $\phi$.
Note that each anchored minor involves exactly one position in $A$.
Therefore, the fact that $\phi$ sends generators of $I$ to zero imposes conditions only on $\phi(x_{i,j})$ in the $A$ block.
This means there are no restrictions on the images of $x_{i,j}$ in the $(r-2s-1)\times (s+1+n-r)$ block and the images of $f_i$s.
Let $a_{\det}$ be the dimension of the space of all symmetric matrices $A$ such that the $(r-s-1)$-anchored $r\times r$ minors of (<ref>) vanish.

An $(r-s-1)$-anchored $r\times r$ minor of the matrix in (<ref>) is up to a sign the product of $\det E$ and an $s$-anchored $(2s+1)\times(2s+1)$ minor of (<ref>).
If $\det E=0$, then there are no restrictions on what the entries of $A$ are.
Therefore we can freely map generators of $J/I$ to elements in $\KK[z_0,\ldots,z_k]_1$ and we get

\[
\dim \left(\Hom_{S/I}(J/I,S/J)_0\right)=\left( \binom{n+1}{2}-1-k \right)(k+1).
\]

If $r-2s-1=0$ or $\det E\neq 0$, then $a_{\det}$ is also the dimension of the space of all symmetric matrices such that the $s$-anchored $(2s+1)\times (2s+1)$ minors in (<ref>) vanish.
Since there are no restrictions on the images of $x_{i,j}$ in the $(r-2s-1)\times (s+1+n-r)$ block and $f_i$s we have

\[
\dim \left(\Hom_{S/I}(J/I,S/J)_0\right) = a_{\det} + (s+1+n-r)(r-2s-1)(k+1)+(\kappa(s)-k)(k+1).
\]

To compute the dimension of the tangent space at a general point of $\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$, we first do the case when $n$ is odd, $r=n$, and $s=\frac{r-1}{2}$.
But before that, we need two preliminary lemmas.

Let $L\geq 0$ be an integer and $\alpha,\beta,\gamma\in \CC$.

\[
\sum_{\ell=0}^L \binom{\alpha+\beta \ell}{\ell} \binom{\gamma-\beta \ell}{L-\ell} =\sum_{\ell=0}^L \binom{\alpha+\gamma-\ell}{L-\ell} \beta^{\ell}.
\]

In particular the sum $\sum_{\ell=0}^L \binom{\alpha-\ell}{\ell} \binom{\gamma+\ell}{L-\ell}$ depends only on $\alpha+\gamma$ and $L$.

Let $n\geq3$ be odd and set $s=\frac{n-1}{2}$.
Suppose $A=[a_{i,j}]$ and $D$ are $(s+1)\times(s+1)$ and $s\times (s+1)$ matrices respectively.

\[
\det\left(
\begin{array}{c|ccc}
0   & & D & \\
\hline
& &   & \\
D^t & & A & \\
& &   & \\
\end{array}
\right)_{n\times n}=-\sum_{1\leq i,j\leq s+1}(-1)^{i+j}a_{i,j}p_i p_j,
\]

where $p_i$ is the minor of $D$ after deleting its $i$-th column.

Suppose $a_{i,j}$ are independent forms.
Note that we are not assuming $A$ is symmetric.
In the expansion of the above determinant, due to the upper left zero block, every term involves exactly one entry in $A$.
To get the coefficient of $a_{i,j}$, we expand the determinant along row $i$ and obtain $-(-1)^{i+j}p_i p_j$.

For an integer $s\geq 2$, define the matrix $D_s(z_0, z_1,z_2)$ as

\begin{equation}\label{eq:block-D}
\begin{pmatrix}
z_0 & z_1    & z_2    &        &     \\
& \ddots & \ddots & \ddots &     \\
&        & z_0    & z_1    & z_2 \\
&        &        & z_0    & z_1
\end{pmatrix}_{s\times (s+1)},
\end{equation}

where the entries not shown are all zero.

We are now equipped to prove the following.

Let $r=n$ be odd and suppose $1\leq k\leq \kappa(s)$ where $s=\frac{r-1}{2}$.
There exists a point $[Q]\in\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$ such that

\[
\dim T_{[Q]}\fano{k}{\SD[r]{n}}=s(s+1)+\left(\kappa(s)-k\right)(k+1).
\]

Consider a nested $s$-compression space $[Q]$ of the form

\begin{equation}
\label{eq:middle-s-compression}
\begin{array}{c|ccc}
B   & & D & \\
\hline
& &   & \\
D^t & & 0 & \\
& &   & \\
\end{array}
\right),
\end{equation}

where $B$ and $D$ are  $s\times s$ and $s\times (s+1)$ matrices respectively, with entries in $\KK[z_0,\ldots,z_k]_1$ such that the linear span of entries in $Q$ has $\KK$-dimension $k+1$.
We claim that $[Q]$ is the desired point for suitable $D$ and arbitrary $B$.

By Lemma <ref>, $\dim T_{[Q]}\fano{k}{\SD[r]{n}}=a_{\det}+\left(\kappa(s)-k\right)(k+1)$,
where $a_{\det}$ is the $\KK$-dimension of the space of $(s+1)\times(s+1)$ matrices $A$ over $\KK[z_0\ldots,z_k]_1$ with

\begin{equation}
\label{eq:det-0}
\det\left(
\begin{array}{c|ccc}
0   & & D & \\
\hline
& &   & \\
D^t & & A & \\
& &   & \\
\end{array}
\right)=0.
\end{equation}

It remains to exhibit for $1\leq k\leq \kappa(s)$ a matrix $D$ for which $a_{\det}=s(s+1)$.
The matrix $D$ that we will choose is such that its rows are $\KK$-linearly independent and the space of $(s+1)\times(s+1)$ matrices in the $\KK$-row span of $D$ has trivial intersection with the space of $(s+1)\times(s+1)$ matrices in the $\KK$-column span of $D^t$.
For such $D$, we can readily identify a space of matrices $A$ of dimension $s(s+1)$ satisfying  (<ref>), namely those matrices $A=\tilde{D}+\tilde{D}^t$ where $\tilde{D}$ is an $(s+1)\times(s+1)$ matrix in the $\KK$-row span of $D$.
Then, it remains to show that if $A$ is a matrix satisfying (<ref>), it can be written as $A=\tilde{D}+\tilde{D}^t$ with $\tilde{D}$ as described above.

Case $k=1$:
Take $D=D_s(z_0,z_1,0)$  (see (<ref>)).
Let $A=[a_{i,j}]$ be a symmetric matrix satisfying (<ref>) with $a_{i,j}\in\KK[z_0,z_1]_1$.
We claim $A=\tilde{D}+\tilde{D}^t$ with $\tilde{D}$ as above.
Equivalently, we show that we can zero out the entries of $A$ in (<ref>) through suitable  operations using the blocks $D, D^t$.
We may assume (after  operations) that no entries of $A$ except possibly $a_{s+1,s+1}$ involves $z_0$.
But $a_{s+1,s+1}$ cannot involve $z_0$ because otherwise by Lemma <ref>, the determinant in (<ref>) would have a $z_0^{2s+1}$ term.
Therefore, we can write $a_{i,j}=\alpha_{i,j}z_1$ for suitable scalars $\alpha_{i,j}$.

Furthermore, by using successive  operations, we can push $z_1$-terms in the matrix $A$ to the boundary of this matrix without producing $z_0$ in the entries of $A$ (see Figure <ref>), i.e. we may assume $a_{i,j}=0$ for $1<i,j<s+1$.
By Lemma <ref>, expanding the determinant in (<ref>) gives

\[
\sum_{i,j}(-1)^{i+j}a_{i,j}p_i p_j=0,
\]

where $p_i=z_0^{i-1} z_1^{s+1-i}$, $1\leq i\leq s+1$.

\[
\sum_{i,j}(-1)^{i+j}\alpha_{i,j} z_0^{i+j-2} z_1^{2s+3-(i+j)}=0.
\]

Inspecting the $(z_0,z_1)$-degree, we see that for a fixed $\ell$, the sum $\sum_{i+j=\ell}(-1)^{i+j}\alpha_{i,j}$ vanishes.
Since $A$ has entries only at its boundary, we get $a_{i,j}=\alpha_{i,j}z_1=0$.

\left(
\begin{array}{cc|ccc}
& & z_0 & z_1 & \\
& & & z_0 & z_1 \\
\hline
z_0 & & & & \\
z_1 & z_0 & & \alpha z_1 & \\
& z_1 & & &
\end{array}
\right)
original matrix

\left(
\begin{array}{cc|ccc}
& & z_0 & z_1 & \\
& & & z_0 & z_1 \\
\hline
z_0 & & &  &\frac{\alpha}{2}z_1 \\
z_1 & z_0 &  & 0 & \\
& z_1 & \frac{\alpha}{2}z_1& &
\end{array}
\right)
after two operations

pushing $z_1$ to the boundary by row-column operations

Case $k=2$:
Take $D=D_s(z_0,z_1,z_2)$.
Let $A=[a_{i,j}]$ be a symmetric matrix satisfying (<ref>) where $a_{i,j}\in\KK[z_0,z_1,z_2]_1$.
Similar to the previous case we show that we can zero out the block $A$ through  operations by using the blocks $D, D^t$.
Using what we proved for $k=1$, we may assume that after suitable  operations on the matrix in (<ref>), the entries of $A$ are scalar multiples of $z_2$.
We claim this reduces $A$ to zero.
In Lemma <ref>, we will prove that the polynomials $p_i p_j$ for $1\leq i\leq j\leq s+1$ are $\KK$-linearly independent where $p_i$ is the minor of $D$ after deleting the $i$-th column.
Expanding the determinant in (<ref>) gives

\[
\sum_{1\leq i,j\leq s+1} (-1)^{i+j} a_{i,j} p_i p_j =0,
\]

which by linear independence of $p_i p_j$ yields $a_{i,j}=0$.

Case $k>2$:
Take $D=D_s(z_0,z_1,z_2)+D'$ where $D'$ is any $s\times (s+1)$ matrix with entries in $\KK[z_3,\ldots,z_k]_1$ such that the $\KK$-dimension of entries in $Q$ is $k+1$.
Assume $A$ has entries in $\KK[z_0,\ldots,z_k]_1$ and satisfies (<ref>).
Using the case $k=2$, we can assume that after  operations, $A$ does not involve $z_0,z_1,z_2$.
We prove this reduces $A$ to zero.
To see that for a fixed $2<\ell\leq k$, the entries of $A$ do not involve $z_{\ell}$, set all the $z_i$ except $z_0,z_1,z_2,z_{\ell}$ equal to zero.
The entries of $A$ are now scalar multiples of $z_{\ell}$.
By the same argument as in the case $k=2$, all entries of $A$ are zero.
Since $\ell$ was arbitrary, we are done.

We used the following lemma in the proof of the previous proposition.

Let $s\geq 2$ and consider $D_s(z_0,z_1,z_2)$ in (<ref>).
For $1\leq i \leq s+1$, let $p_i$ be the minor of $D_s(z_0,z_1,z_2)$ after deleting the $i$-th column.

\begin{equation}
\label{eq:expr-p_i}
p_i=\sum_{\ell=0}^{\frac{s-i+1}{2}}(-1)^{\ell} \binom{s-i-\ell+1}{\ell} z_0^{i+\ell-1} z_1^{s-i+1-2\ell} z_2^{\ell}.
\end{equation}

* for $1\leq i,j,i',j'\leq s+1$ with $i+j\neq i'+j'$, the monomials appearing in $p_i p_j$ are distinct from those in $p_{i'} p_{j'}$ ,

* for any fixed integer $2\leq d \leq 2(s+1)$, the polynomials $p_i p_j$ for $1\leq i\leq j\leq s+1$ and $i+j=d$ are $\KK$-linearly independent.

In particular, the polynomials $p_i p_j$ for $1\leq i\leq j\leq s+1$ are $\KK$-linearly independent.

Let us denote by $p_{i,s}$ the determinant of the matrix obtained by deleting column $i$ in $D_s(z_0,z_1,z_2)$.
We then have

\begin{equation}
\label{eq:initial-conditions}
p_{s+1,s}=z_0^s,\quad p_{s,s}=z_0^{s-1} z_1,\quad p_{s-1,s}=z_0^{s-2}z_1^2-z_0^{s-1}z_2.
\end{equation}

Moreover, by expanding the determinant along the last row we obtain the recursion

\begin{equation}
\label{eq:recursion}
p_{i,s}=z_1 p_{i,s-1}-z_0 z_2 p_{i,s-2}\qquad(1\leq i\leq s-2).
\end{equation}

The equations in (<ref>) and (<ref>) uniquely determine $p_{i,s}$ for all $s\geq 2$ and $1\leq i\leq s+1$.
To prove the expression for $p_i$ in (<ref>), it is enough to show that this expression satisfies both (<ref>) and (<ref>).
Let $q_{i,s}$ be the right-hand side of (<ref>).
It is easy to verify $q_{i,s}$ satisfies (<ref>).
To show it satisfies the recursion, we compute

\begin{equation*}
\begin{split}
z_1 & q_{i,s-1}-z_0 z_2 q_{i,s-2}\\
& = z_1\sum_{\ell=0}^{\frac{s-i}{2}}(-1)^{\ell}\binom{s-i-\ell}{\ell}z_0^{i+\ell-1} z_1^{s-i-2\ell} z_2^{\ell} - z_0 z_2\sum_{\ell=0}^{\frac{s-i-1}{2}}(-1)^{\ell}\binom{s-i-1-\ell}{\ell} z_0^{i+\ell-1} z_1^{s-i-1-2\ell} z_2^{\ell}\\
& = \sum_{\ell=0}^{\frac{s-i}{2}}(-1)^{\ell}\binom{s-i-\ell}{\ell}z_0^{i+\ell-1} z_1^{s-i+1-2\ell} z_2^{\ell} + \sum_{\ell=1}^{\frac{s-i+1}{2}} (-1)^{\ell}\binom{s-i-\ell}{\ell-1} z_0^{i+\ell-1} z_1^{s-i+1-2\ell} z_2^{\ell}\\
& = \sum_{\ell=0}^{\frac{s-i+1}{2}}(-1)^{\ell}\binom{s-i+1-\ell}{\ell} z_0^{i+\ell-1} z_1^{s-i+1-2\ell} z_2^{\ell}\\
& = q_{i,s}.
\end{split}
\end{equation*}

Proof of part <ref>: Using the expression for $p_i$, we have

\[
p_i p_j =\sum_{\ell=0}^{\frac{s-i+1}{2}} \sum_{\ell'=0}^{\frac{s-j+1}{2}} (-1)^{\ell+\ell'} \binom{s-i-\ell+1}{\ell}\binom{s-j-\ell'+1}{\ell'}z_0^{i+j+(\ell+\ell')-2} z_1^{2s-(i+j)+2-2(\ell+\ell')} z_2^{\ell+\ell'},
\]

and a similar expression for $p_{i'}p_{j'}$.
If $p_i p_j$ and $p_{i'}p_{j'}$ share a monomial, then by inspecting the $(z_0,z_2)$-degree, we see that $i+j=i'+j'$.

Proof of part <ref>: Let $\tilde{p}_{i,j}=(p_i p_j)|_{z_0=z_1=1}$.
It is enough to show that the polynomials $\tilde{p}_{i,j}$ for $1\leq i\leq j\leq s+1$, $i+j=d$ are $\KK$-linearly independent or equivalently the polynomial $\tilde{p}_{i_0,j_0}$ together with $\tilde{p}_{i,j}-\tilde{p}_{i+1,j-1}$ for $i_0\leq i\leq j-2, i+j=d$ are linearly independent.
Here $i_0$ is the smallest integer for which there is an integer $j_0$ with $i_0+j_0=d$ and $1\leq i_0\leq j_0\leq s+1$.
We prove the latter statement by showing that if $i\leq j-2$, then the term with the smallest degree in $\tilde{p}_{i,j}-\tilde{p}_{i+1,j-1}$ has degree $s-j+2$ (note that $\tilde{p}_{i_0,j_0}$ has a degree zero term).
This would complete the proof of part <ref>.

To show the smallest degree in $\tilde{p}_{i,j}-\tilde{p}_{i+1,j-1}$ is $s-j+2$, in the expression

\[
\tilde{p}_{i,j}=\sum_{\ell=0}^{\frac{s-i+1}{2}} \sum_{\ell'=0}^{\frac{s-j+1}{2}} (-1)^{\ell+\ell'} \binom{s-i-\ell+1}{\ell}\binom{s-j-\ell'+1}{\ell'} z_2^{\ell+\ell'},
\]

we make the change of variable $L=\ell+\ell'$ to get

p̃_i,j    =        ∑_L=0^s-i+j/2+1∑_0≤ℓ≤L
0≤L-ℓ≤s-j+1(-1)^L s-i-ℓ+1ℓs-j-(L-ℓ)+1L-ℓz_2^L

=        ∑_L=0^s-j+2∑_0≤ℓ≤L
0≤L-ℓ≤s-j+1(-1)^L s-i-ℓ+1ℓs-j-(L-ℓ)+1L-ℓz_2^L

+ (terms with degree higher than $s-j+2$)

=        ∑_L=0^s-j+1∑_ℓ=0^L(-1)^L s-i-ℓ+1ℓs-j-(L-ℓ)+1L-ℓz_2^L

+∑_ℓ=1^s-j+2(-1)^s-j+2s-i-ℓ+1ℓℓ-1s-j+2-ℓ z_2^s-j+2

+ (terms with degree higher than $s-j+2$)

=        ∑_L=0^s-j+2∑_ℓ=0^L(-1)^L s-i-ℓ+1ℓs-j-(L-ℓ)+1L-ℓz_2^L

-(-1)^s-j+2s-i+10-1s-j+2 z_2^s-j+2

+ (terms with degree higher than $s-j+2$).

Similar to (<ref>),

\begin{equation}
\label{eq:ptilde2}
\begin{split}
\tilde{p}_{i+1,j-1} = & \sum_{L=0}^{s-j+2}\sum_{\ell=0}^{L}(-1)^L \binom{s-i-\ell}{\ell}\binom{s-j-(L-\ell)+2}{L-\ell}z_2^L\\
&+\text{(terms with degree higher than $s-j+2$)}.
\end{split}
\end{equation}

By Jensen's formula (Lemma <ref>), the double summations in (<ref>) and (<ref>) are equal, hence

\[
\tilde{p}_{i,j}-\tilde{p}_{i+1,j-1}=-z_2^{s-j+2} +\, \text{(terms with higher degree)}.
\]

This completes the proof of part <ref>.
The last statement of the lemma follows from parts <ref> and <ref>.

Using Fano schemes, we can prove the following result on symmetric matrices.

Let $k,s\geq 1$ and assume $D$ is a general $s\times (s+1)$ matrix with entries in $\KK[z_0,\ldots,z_k]_1$.
Then a symmetric matrix $A$ of size $s+1$ over $\KK[z_0,\ldots,z_k]_1$ satisfies

\begin{equation}
\label{eq:general-D}
\det
\begin{pmatrix}
0   & D \\
D^t & A
\end{pmatrix}
\end{equation}

if and only if $A=\tilde{D}+\tilde{D}^t$, where $\tilde{D}$ is an $(s+1)\times (s+1)$ matrix whose rows are in the row span of $D$.

One direction is clear.
For the other, let $n=r=2s+1$ and consider the Fano scheme $\fano{k}{\SD[r]{n}}$.
Without loss of generality we may assume $k\leq \kappa(s)$.
Up to $\gl{n}$ action, a general point $[Q]\in\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$ is of the form (<ref>) where $B$ and $D$ are matrices filled with general linear forms.

The dimension of $\compk{k}{s}$, by Corollary <ref> is $s(s+1)+\left(\kappa(s)-k\right)(k+1)$ and by Proposition <ref>, the Fano scheme contains a point in $\compk{k}{s}$ with tangent space dimension equal to $\dim\compk{k}{s}$.

This means the dimension of the tangent space to $\fano{k}{\SD[n]{n}}$ at $[Q]$ equals $\dim\compk{k}{s}$.
This dimension together with Lemma <ref> give that for a general $D$, the dimension of all matrices $A$ satisfying (<ref>) is $a_{\det}=s(s+1)=s(n-s)$.
Since $D$ is general, the space of all matrices $A$ satisfying (<ref>) contains the subspace of those matrices of the form $\tilde{D}+\tilde{D}^t$ where $\tilde{D}$ is an $(s+1)\times (s+1)$ matrix in the row span of $D$ and this subspace has dimension $s(n-s)$ too.
Therefore every matrix $A$ satisfying (<ref>) is of the required form.

Next, we compute the dimension of the tangent space at a general point of $\compk{k}{s}$.

Let $[Q]$ be a general point of $\compk{k}{s}\subset\fano{k}{\SD[r]{n}}$.

\[
\dim T_{[Q]}\left( \fano{k}{\SD[r]{n}} \right)=s(s+1+n-r)+(s+1+n-r)(r-2s-1)(k+1)+(\kappa(s)-k)(k+1).
\]

Up to $\gl{n}$ action, $Q$ is the matrix in (<ref>) in which the entries in blocks $B,\ldots,E$ are general linear forms in $z_0,\ldots,z_k$.
Since $E$ is general, $\det E\neq 0$ and by Lemma <ref>,

\[
\dim T_{[Q]}\left( \fano{k}{\SD[r]{n}} \right)=a_{\det}+(s+1+n-r)(r-2s-1)(k+1)+(\kappa(s)-k)(k+1).
\]

Our task is to show $a_{\det}=s(s+1+n-r)$.
Call a matrix $A$ of size $s+1+n-r$ over $\KK[z_0,\ldots,z_k]_1$ permissible if all $s$-anchored $(2s+1)\times(2s+1)$ minors of (<ref>) vanish.
The space of all permissible $A$ contains a subspace of dimension $s(s+1+n-r)$, namely the set of all matrices $\tilde{D}+\tilde{D}^t$ where $\tilde{D}$ is a square matrix of size $s+1+n-r$ whose rows are in the row span of the general matrix $D$.

Conversely, we show if $A$ is a permissible matrix, then it is of the form $\tilde{D}+\tilde{D}^t$.
Consider the principal $s$-anchored $(2s+1)\times(2s+1)$ minors of (<ref>).
By assumption these minors vanish and are of the form

\[
\det\begin{pNiceArray}{cc}[first-row]
s & s+1\\
0 & D' \\
D'^t & A'\\
\end{pNiceArray},
\]

for submatrices $D', A'$ of $D$ and $A$ respectively.
By Corollary <ref>, the vanishing of the above determinant implies $A'$ is the sum of a unique matrix in the row span of $D'$ and its transpose.
Since this holds for $A'$ in every principal $s$-anchored $(2s+1)\times(2s+1)$ minor in (<ref>), it also holds for $A$, i.e. $A=\tilde{D}+\tilde{D}^t$ where rows of $\tilde{D}$ are in the row span of $D$.
This completes the proof.

§.§ Smoothness

In this section we use the dimensions of tangent spaces of $\fano{k}{\SD[r]{n}}$ to detect components and give results on the smoothness of our Fano schemes.

Let $r$ be odd, $s=\frac{r-1}{2}$ and $k\leq \kappa(s)$.
The subscheme $\compk{k}{s}$ of $\fano{k}{\SD[r]{n}}$ is an irreducible component that contains a smooth point of $\fano{k}{\SD[r]{n}}$.
Moreover, when $k=\kappa(s)$, every point of $\compk{k}{s}$ is a smooth point of $\fano{k}{\SD[r]{n}}$.

By Proposition <ref> and Corollary <ref>, when $r-2s-1=0$,
a general point of $\compk{k}{s}$ has tangent space dimension equal to $\dim\compk{k}{s}$.
Therefore $\compk{k}{s}$ is a component containing a smooth point of $\fano{k}{\SD[r]{n}}$.
When $k=\kappa(s)$, by Corollary <ref>, $\compk{\kappa(s)}{s}$ is a single $\gl{n}$ orbit, proving the last statement.

Next, we prove Proposition <ref> which states that our Fano schemes can have generically non-reduced components.

Let $[Q]$ be a general point of $\compk{k}{s}$.
The irreducible component $\compk{k}{s}$ is generically reduced if and only if $\dim T_{[Q]}\left( \fano{k}{\SD[r]{n}} \right)=\dim \compk{k}{s}$.
By Proposition <ref> and Corollary <ref>

\[
\dim T_{[Q]}\left( \fano{k}{\SD[r]{n}} \right)-\dim \compk{k}{s} = (r-2s-1)\left((s+1+n-r)k-s\right).
\]

We now have the desired result since $\left((s+1+n-r)k-s\right)>0$.

§ FURTHER QUESTIONS

In our investigation of the Fano schemes $\fano{k}{\SD[r]{n}}$, we were able to prove that the closed subschemes $\compk{k}{s}$ consisting of nested $s$-compression spaces are irreducible components when $k=1$ or $\kappa(s)$ or when $r=2s+1$.
When $r=2s+1$, the proof used the dimensions of tangent spaces since in this case $\compk{k}{s}$ is generically reduced.
By Proposition <ref> when $r\neq 2s+1$, such a proof will not work.
The proof of the case $k=1$ (Theorem <ref>) relied on the Kronecker canonical form for a pencil of matrices.

Moreover, we resolved the case $k=\kappa(s)$ (Corollary <ref>) by looking at affine charts corresponding to Borel fixed points but when $k<\kappa(s)$, the affine chart of a Borel fixed nested $s$-compression space could contain a point that is not a nested $s$-compression space.

Is $\compk{k}{s}$ an irreducible component of $\fano{k}{\SD[r]{n}}$ for arbitrary $n,r,k$ and $s$?

In Section <ref>, we gave partial results on the smoothness of $\fano{k}{\SD[r]{n}}$.
We believe by using the results of this section together with those in Section <ref>, one can fully characterize the smoothness of these schemes and prove the following.

The scheme $\fano{k}{\SD[r]{n}}$ is smooth if and only if $r$ is odd and

\[
\max\left\{ \kappa(0),\kappa\left(\frac{r-3}{2}\right) \right\} < k \leq \kappa\left(\frac{r-1}{2}\right).
\]

Our next question involves de Seguins Pazzis' theorem <cit.> which states that when $k>\max\{\kappa(1),\kappa(s_{\max}-1)\}$, the Fano scheme consists only of compression spaces.
Here $s_{\max}=\lfloor\frac{r-1}{2}\rfloor$.
It is interesting to know if this bound can be improved.
One reason this is important is that when we know the closed points of $\fano{k}{\SD[r]{n}}$ are all compression spaces, then we can classify the irreducible components of the scheme.

Given $n$ and $r$, what is the largest value $k$ for which $\fano{k}{\SD[r]{n}}$ has a non-compression point?

Looking at the above question in a different way, it would be interesting if there is an invariant for a subspace of symmetric matrices of rank less than $r$ that can detect being a compression space.
This invariant could lead to a characterization of when we have only compression spaces.

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