The geometry of the solution set of a symmetric equation in four symmetric matrices

Question

Let $n$ be a natural number. We can view the space of invertible symmetric matrices over a field as an open subset of$\mathbb A^{(n^2+n)/2}$. Inside the fourth power of this space, we have the closed subscheme consisting of tuples satisfying $A_1 + A_2 = A_3 + A_4$ and $A_1^{-1} + A_2^{-1} = A_3^{-1}+ A_4^{-1}$.

1. Is this subscheme a complete intersection of dimension $n^2+n$?

2. How many irreducible components does this subscheme have?

The motivation is that this would evaluate the fourth moment of symplectic Kloosterman sums, in the same way that Kloosterman's classical argument evaluates the fourth moment of the usual Kloosterman sums. However, I don't expect techniques from number theory to be helpful here.

The $n=1$ case has three irreducible components of dimension $2$, given by equations as follows $(x_3=x_1,x_4=x_2),(x_3=x_2,x_4=x_1),(x_2=-x_1,x_4=-x_3)$. Using these, we can make $3^n$ $2n$-dimensional families of diagonal examples. All of these are contained in an irreducible component of dimension at least $n^2+n$, as the scheme is defined by only $n^2+n$ equations in $2n^2 +2n$-space. However, only for a few obvious ones can I locate an $n^2+n$-dimensional family containing them.

---

Let me express what I think remains to be done after the two answers already given. Using Alex Gavrilov's algebraic formulation, I think we can classify all the irreducible components that arise when $A_1+A_2$ and $A_1^{-1}+A_2^{-1}$ are invertible. One might conjecture that all irreducible components arise on invertible irreducible component on one subspace and an irreducible component where $A_1+A_2=0$ on another subspace, as in David Speyer's answer. So the main thing to do that I don't know how to do is verify or disprove this conjecture.

Answer 1

I'll describe a number of $n^2+n$ dimensional subvarieties. I have not yet found a point which isn't on one of them (in particular, they subsume all of your $2n$ dimensional families). If we are very optimistic, perhaps this is your intersection.

Before starting, it seems nicer to change signs, so the equations are $W+X+Y+Z=0$ and $W^{-1}+X^{-1}+Y^{-1}+Z^{-1}=0$.

Let $n=d_1+d_2+d_3$. Choose any decomposition of $\mathbb{C}^n$ as $V_1 \oplus V_2 \oplus V_3$ with $\dim V_i = d_i$; the space of ways to do this is $n^2-d_1^2-d_2^2-d_3^2$ dimensional. Then take our quadratic forms $(W,X,Y,Z)$ so that $V_1$, $V_2$ and $V_3$ are mutually orthogonal, with $W+X=Y+Z=0$ on $V_1$, $W+Y=X+Z=0$ on $V_2$ and $W+Z=X+Y=0$ on $V_3$. This adds $(d_1^2+d_1)+(d_2^2+d_2)+(d_3^2+d_3)$ dimensions, so $n^2+n$ in all.

---

I found another $6$ dimensional solution for $n=2$ which is not contained in any of the above. I suspect that we can also build that solution into direct sums to get more $n^2+n$ dimensional solutions as I did above, but I haven't checked the dimension counts carefully.

Recall that $$\begin{bmatrix} a&b \\ c&d \end{bmatrix}^{-1} = \frac{1}{\det \left[ \begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \right]} \begin{bmatrix} d&-b \\-c&a \end{bmatrix}.$$ Therefore, if we have a $2 \times 2$ solution to $$W+X+Y+Z=0, \det W = \det X = \det Y = \det Z \quad (\ast)$$ then it will also obey $W^{-1}+X^{-1}+Y^{-1}+Z^{-1}=0$.

Now, on the space of $2 \times 2$ symmatric matrices, $\det$ is a quadratic form. Let $A$, $B$ and $C$ be orthogonal with the respect to the associate inner product ($6$ dimensions). Then $\det(\pm A \pm B \pm C) = \det(A) + det(B)+\det(C)$. Choosing the $4$ sign options with an even number of minuses, we get a $6$-dimensional space of solutions to $(\ast)$.

---

I have an $4 \times 4$ solution which I don't think is on any of the components I've found above. Haven't figured out how to embed it it in a $4^2+4$ dimensional family yet but, of course, it must do so. (The intersection of $n^2+n$ equations on a $2(n^2+n)$-dimensional variety must have every component of dimension $\geq n^2+n$.)

I start with a $2\times 2$ solution which is not symmetric. Let $$I = \begin{bmatrix} i&0 \\ 0&-i \end{bmatrix} \qquad J = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} \qquad K = \begin{bmatrix} 0&i \\ i&0 \end{bmatrix}.$$ (This is the standard embedding of the quaternions in $\mathrm{Mat}_{2 \times 2} (\mathbb{C})$.) Then take the $4$ matrices of the form $\pm p I \pm q J \pm r K$ with an even number of $-1$'s. They clearly sum to $0$, and $$(\pm p I \pm q J \pm r K)^{-1} = \frac{1}{p^2+q^2+r^2} (\mp p I \mp q J \mp r K)$$ so their inverses do as well.

Now we need some trickery to make this solution symmetric. We can replace $I$, $J$ and $K$ by $iI$, $iJ$ and $iK$ and this still works, with the benefit that $iI$, $iJ$ and $iK$ are now Hermitian. If we embed $2\times 2$ complex matrices into $4 \times 4$ real matrices in the standard way, then Hermitian becomes symmetric. In other words, use $(\pm p I' \pm q J' \pm r K')$ where $$I' = \begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{bmatrix} \ J' = \begin{bmatrix} 0&0&0&1\\0&0&-1&0\\0&-1&0&0\\1&0&0&0 \end{bmatrix} \ K' = \begin{bmatrix} 0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0 \end{bmatrix}$$

Answer 2

This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.

Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)

Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.

[EDIT] (The above is almost true except there are other components.) Consider first the case when $Y$ and $Z$ are general. We may chose a square root $B^{1/2}$ and consider matrices $$x=B^{1/2}XB^{1/2},\,\,a=B^{1/2}AB^{1/2}.$$ In this terms, the above equation is $$xa=ax,\,x^2+ax+a=0.$$ In the general case, it has $2^n$ solutions (consider $a$ in the diagonal form.) I presumed that all these solutions are Galois conjugate, but on second thought, no. At least, there are two obvious solutions $X=-Y$ and $X=-Z$ which are not conjugate to anything. For $n>1$ there are more solutions.

So, there are four components, $X+Y=0, X+Z=0, Y+Z=0$ and the fourth one (probably reducible) comes from extra solutions of the above equation (for $n>1$). So, it is not much. Still, the above argument has a point: it should be possible to count the components once you figure out how to divide these $2^n$ solutions between them.

Answer 3

This non-answer grew too long for a comment.

Let's change coordinates, $B_i = \check\rho(t)A_i$, where $\check\rho(z)$ is the diagonal matrix $diag(t,t^2,t^3,\ldots,t^n)$. Then the original equations $A_i = A_i^T, A_1+A_2 = A_3+A_4, A_1^{-1}+A_2^{-1} = A_3^{-1}+A_4^{-1}$ become $B_i = \check\rho(t) B_i^T \check\rho(t^{-1})$ and the other two unchanged. If you take $t\to\infty$ then instead of considering symmetric matrices you're considering upper triangular. There's a serious issue that these may not be enough equations to define the limit (may not be "Gröbner enough"), but if they define a complete intersection then (they are Gröbner enough and) the original was a complete intersection.

The natural action of $GL(n)$ on the original four matrices, $g\cdot A_i = g A_i g^T$, degenerates/extends to an action of $B\times B$, left and right independent multiplication by upper triangular matrices. Using that we can assume $B_1 = \bf 1$, and still have a conjugation action on the other three $B_i$. So we've reduced to the question, is the space of quadruples $(1,B_2,B_3,B_4)$ of upper triangular matrices, satisfying your original two matrix equations, only $n+1\choose 2$-dimensional. Or even stronger, is (this space)/(the residual $B$-action) only $1$-dimensional.

For $n=2$, the resulting space is indeed still a complete intersection, now with 12 components, of degrees {3,3,3,3,3,3, 2,2,2, 1,1,1}. I didn't look into how the original components break into these.