TITLE: The geometry of the solution set of a symmetric equation in four symmetric matrices
QUESTION [15 upvotes]: 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 $A_1^{-1} + A_2^{-1} = A_3^{-1}+ A_4^{-1}$.

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

REPLY [3 votes]: 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.