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.


3 Answers 3


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 $$I' = \begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{bmatrix} \ J' = \begin{bmatrix} 0&0&0&1\\0&0&-1&0\\0&-1&0&0\\1&0&0&0 \end{bmatrix} \ K' = \begin{bmatrix} 0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0 \end{bmatrix}$$

  • $\begingroup$ Where does that last equality come from? $\endgroup$
    – MTyson
    Aug 23, 2017 at 2:01
  • $\begingroup$ Oh, grr. I'm wrong. $XY^{-1}ZX^{-1}$ does not equal $ZY^{-1}$. I'll delete that part shortly. $\endgroup$ Aug 23, 2017 at 2:03
  • $\begingroup$ Interesting! I think I can show this gives all the two by two quadruples with fixed determinant. I also agree with your suspicion that this extends to higher dimension. $\endgroup$
    – Will Sawin
    Aug 23, 2017 at 20:38
  • $\begingroup$ @WillSawin Here is a strategy for proving this is a complete intersection: Rewrite the equations using $8$ symmetric matrices: $W+X+Y+Z=A+B+C+D=0$, $AW=BX=CY=DZ=1$; this gives us a closed subvariety of $4n^2+4n$ dimensional space. Taking the limit as we "zoom out" turns these equations into $W+X+Y+Z=A+B+C+D=0$, $AW=BX=CY=DZ=0$. (continued) $\endgroup$ Aug 23, 2017 at 21:35
  • $\begingroup$ Surprisingly, the equations $AW=0$ with $A$ and $W$ symmetric still have a $\binom{n+1}{2}$ dimensional space of solutions -- it breaks up into $n+1$ components according to the ranks of $A$ and $W$. It would be enough to show that, for each of the $(n+1)^4$ components of $AW=BX=CY=DZ=0$, the intersection with the linear space $A+B+C+D=W+X+Y+Z=0$ has dimension $n^2+n$. $\endgroup$ Aug 23, 2017 at 21:35

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.

  • $\begingroup$ It's not as bad as you think in $n=1$, as for one of the components $Y$ and $Z$ are not general. $\endgroup$
    – Will Sawin
    Aug 23, 2017 at 15:35
  • $\begingroup$ Do you have a reference for the claim about solutions of an algebraic Ricatti equation being Galois conjugate? $\endgroup$
    – Will Sawin
    Aug 23, 2017 at 15:38
  • $\begingroup$ Actually, I do not know any reference related to the actual algebra of this equations. (They are used almost exclusively in applied math.) The thing is, this particular equation is actually a system: a Riccati equation together with $ABX=XBA$. This system has a more or less explicit solution. Denote $x=B^{1/2}XB^{1/2}$ and $a=B^{1/2}AB^{1/2}$. The system may then be written as $$xa=ax,\,x^2+ax+a=0.$$ The solution of it is obvious once $a$ is written in the diagonal form. For a general case, there are $2^n$ solutions depending on the choice of square roots. $\endgroup$ Aug 24, 2017 at 10:43
  • $\begingroup$ I believe it is possible to prove that all these solutions are Galois conjugate, though it may require some effort. $\endgroup$ Aug 24, 2017 at 10:46
  • $\begingroup$ I mean, they are conjugate for a general matrix $a$. Just now I spotted that in the case $A=B=0$ the solution is an arbitrary $X$, and this is again a component of dimension $n^2+n$, so I was wrong! $\endgroup$ Aug 24, 2017 at 11:11

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.

  • $\begingroup$ Cool idea! I don't think your stronger claim about a one-dimensional quotient can be true. The diagonal elements are invariant under conjugation, so the dimension of the quotient is at least the dimension of the space of diagonal solutions, which is $n$. Of course, these solutions are fixed under conjugation by diagonal matrices so it doesn't contradict the claim that the full space is ${n+1 \choose 2}$-dimensional. $\endgroup$
    – Will Sawin
    Sep 26, 2017 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.