1
$\begingroup$

Let $A, B \in \mathbb{R}^{d \times d}$ denote two symmetric positive definite matrices. I am interested in solutions $V_r \in \mathbb{R}^{d \times r}, 1 < r < d$ to the system of quartic polynomials $$ V_r^\top [A, B] V_r = [V_r^\top A V_r, V_r^\top B V_r] $$ and $$ V_r^\top V_r = I_r, $$ where $[A,B] = AB-BA$ denotes the matrix commutator. In other words, I wish to find orthonormal matrices $V_r$ whose quadratic form distributes over the commutator of $A$ and $B$.

By inspection we know this system is satisfied if $V_r$ corresponds to any $r$ eigenvectors of $A$, or if $V_r$ corresponds to any $r$ eigenvectors of $B$. However, you cannot combine subsets of eigenvectors of $A$ with $B$, except in the special case where they commute.

My conjecture is that these are the only solutions. Since my constraints are ``(real) algebraic varieties''---they are at most quartic degree polynomials in $d$ dimensions, and there are $r(r-1)$ such constraints in total---I suspect there must be tools which allow us to count the number of points where all of these curves intersect. In particular, if we can show that the number of intersections are less than or equal to $2 {d \choose r}$, then my conjecture follows.

I would appreciate if anybody could provide references to such theorems, or perhaps disprove my original conjecture!

$\endgroup$
1
  • $\begingroup$ The notion of degree is slightly misleading, since the degree is in terms of the matrices, not in terms of the entries. That probably helps you if anything, since high degrees tend to force the variety to be general type. However, such finiteness results in arithmetic geometry are very hard to obtain, and you're likely better off thinking about the specific shape of your equations than trying to apply general out-of-the-box results. $\endgroup$ Commented Dec 4, 2023 at 19:49

1 Answer 1

0
$\begingroup$

When $r = 1$, any unit vector $V_r$ satisfies the equations, not just eigenvectors of $A$ and $B$. This disproves the conjecture.

$\endgroup$
2
  • $\begingroup$ Yes, sorry, I have edited the comment to specify that $r > 1$. $\endgroup$
    – mtcli
    Commented Dec 6, 2023 at 1:00
  • $\begingroup$ But in fact the conjecture is false anyways. Taking $A$ to be block diagonal, we can choose $V_r$ to be the block projector, and the equality is satisfied for any $B$. $\endgroup$
    – mtcli
    Commented Dec 6, 2023 at 12:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .