4
$\begingroup$

Let $F$ be a field and write $$\mathfrak{s}\mathfrak{p}_4(F)=\left\{\left(\begin{array}{cc} A & B \\ C & -A^T \\ \end{array}\right)\mid A,B,C\in M_2(F), B=B^T, C=C^T\right\}$$ for the symplectic Lie algebra over $F$ of dimension $10$. It is well known that the symplectic group $\text{Sp}(4,F)=\{Q\in M_{2n}(F)\mid J^TQJ=Q\}$, with $J=\left(\begin{array}{cc} O & I \\ -I & O \\ \end{array}\right)$, acts by conjugation on $\mathfrak{s}\mathfrak{p}_4(F)$ (the adjoint action), so $Q\mathfrak{s}\mathfrak{p}_4(F)Q^{-1}\subseteq\mathfrak{s}\mathfrak{p}_4(F)$ for any $Q\in\text{Sp}(4,F)$.

Question: Does there exists for every $M\in\mathfrak{s}\mathfrak{p}_4(F)$ some element $Q\in\text{Sp}(4,F)$ such that $$QMQ^{-1}=\left(\begin{array}{cc} A & B \\ O & -A^T \\ \end{array}\right)?$$ In other words, is every $M\in\mathfrak{s}\mathfrak{p}_4(F)$ conjugate (by an element of $\text{Sp}(4,F)$) to an element in "upper triangular form"?

Furthermore, I am interested in possible normal/standard forms for the orbits of $\mathfrak{s}\mathfrak{p}_4(F)$ under the action of $\text{Sp}(4,F)$. Any references are very welcome!

Remark. If it makes life easier, you can assume that $F$ is algebraically closed and also that $\mathfrak{s}\mathfrak{p}_4(F)$ is semisimple. I am interested in the cases where $F$ has prime characteristic.

$\endgroup$
3
  • 1
    $\begingroup$ The answers for algebraically closed fields and general fields should be very different (algebraically closed field yes, general field no). Are you interested in only the algebraically closed field case or the general case also? $\endgroup$
    – Will Sawin
    Jan 16, 2021 at 0:12
  • 1
    $\begingroup$ Just the algebraically closed case would be enough :), including both the zero and prime characteristic case if possible. $\endgroup$
    – user299843
    Jan 16, 2021 at 1:54
  • $\begingroup$ This is equivalent to asking whether every element in $\mathfrak{sp}_4$ preserves a Lagrangian plane, if I'm correct. $\endgroup$
    – YCor
    Jan 16, 2021 at 18:56

1 Answer 1

1
$\begingroup$

In the case $F=\mathbb{R}$, this is known as the Hamiltonian Schur form, and it is known that there are some obstructions: for instance, it is simple to see that the eigenvalue 0 has even multiplicity in $$ \left(\begin{array}{cc} A & B \\ O & -A^T \\ \end{array}\right) $$ because it appears the same number of times in the spectrum of $A$ and of $-A^T$. So every structured matrix in which the 0 eigenvalue has odd multiplicity cannot be reduced to that form. Similarly, with a Jordan form of $A$, one sees that blocks of odd size must come in pairs. Examples of structured matrices with 'forbidden' eigenstructures can be constructed. The same kind of reasoning applies to generic fields, I think. A more thorough reference: https://link.springer.com/article/10.1007%2Fs10013-020-00394-3

$\endgroup$
2
  • $\begingroup$ If I am not mistaken, the characteristic polynomial of a matrix in $\mathfrak{s}\mathfrak{p}_4(F)$ is of the form $T^4+aT^2+b$ for some $a,b$. In particular, $0$ always has even multiplicity as an eigenvalue of a matrix in $\mathfrak{s}\mathfrak{p}_4(F)$. Do you mean with "coming in pairs" that if $\lambda$ is an eigenvalue so is $-\lambda$? I will have a look later at the reference you gave. $\endgroup$
    – user299843
    Jan 16, 2021 at 19:17
  • $\begingroup$ @user299843 The first version was wrong. Jordan blocks must come in pairs, I think, e.g., a $\lambda=0$ block of size 3 and a block of size 1 doesn't work. (I am more familiar with the version with complex conjugates, to be honest, so I am not sure everything works properly. I wrote the first part just not to make this a link-only answer.) :) $\endgroup$ Jan 16, 2021 at 20:46

Your Answer

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

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