3
$\begingroup$

Consider split orthogonal group $O(2l)$ over a field of characteristic zero. We may assume the matrix of the bilinear form to be $\begin{pmatrix} O&I\\ I&O\end{pmatrix}$.

Let $u$ be a unipotent element in $O(2l)$.

Computations with $l=2$ show that the possible minimal polynomials of unipotents are $X-1, (X-1)^2, (X-1)^3$ but not $(X-1)^4$ as it would have been in $GL(4)$.

So my question is: what is the largest $d$ such that $(X-1)^d$ is a minimal polynomial of a unipotent in $O(2l)$?

In fact I would like to know what $d$ occur. In the case of the group $GL(n)$ the answer comes form Jordan canonical forms and unipotents correspond to a partition of $n$ and hence $d$ could take any value between $1$ and $n$.

Thanks in advance!

$\endgroup$
3
  • $\begingroup$ You'd be better of using the matrix with $1$'s on the anti-diagonal. Then you can take your unipotents to be upper-triangular, and this question should yield to direct calculation - my first thought is that the minimal polynomial of a unipotent $g$ will just depend on the dimension of the largest Levi subgroup that contains $g$ $\endgroup$
    – Nick Gill
    Jun 17, 2015 at 10:01
  • $\begingroup$ BTW, I've voted for this question to be reopened. In its current form it seems fine to me. $\endgroup$
    – Nick Gill
    Jun 17, 2015 at 10:03
  • $\begingroup$ @NickGill Thanks for the input. I will try to enlist all Parabolics thus Levi. Hope that gives good info. $\endgroup$ Jun 17, 2015 at 10:22

1 Answer 1

5
$\begingroup$

This has to do with SL(2) representation theory. The answer for $O(2l)$ is $d=2l-1$. First of all, let $W$ be the irreducible representation of $SL(2)$ of dimension $2l-1$. This preserves a non-degenerate quadratic form and hence the image of $SL(2)$ lies in $SO(2l-1)\subset SO(2l)$. The image of the nontrivial unipotent of $SL(2)$ is "regular" in $SO(2l-1$ and therefore it has minimal polynomial $(X-1)^{2l-1}$. Hence $d\geq 2l-1$.

I claim that $d$ cannot be $2l$: if so, by Jacobson-Morozov, there is an $SL(2)\subset SO(2l)$ passing through the unipotent whose minimal polynomial is $(X-1)^{2l}$. If the $SL(2)$ rep is irreducible, then the even dimensional rep has an invariant symplectic form and hence the $SL(2)$ cannot preserve a quadratic form. Hence the rep splits into a direct sum, which means that the minimal polynomial has degree strictly less than $2l$.

$\endgroup$
3
  • $\begingroup$ Thank you for the answer. Sounds good! However I would also like to know which d appears. Since GL(l) can be embedded inside O(2l), I can see all d from 1 up to l appear. How about d between l+1 up to 2l-1? $\endgroup$ Jun 18, 2015 at 4:49
  • $\begingroup$ this can again be read off from SL(2) representation theory. If a rep of SL(2) is to preserve a non-degenerate quadratic form, then the multiplicity of an even (dimensional) irreducible representation must be even. This is the only constraint. Now take the principal SL(2) in a product of these symplectic and orthogonal groups and the unipotent of SL(2) in the product is your general unipotent. The $d$ is the largest dimensional irrep of SL(2) occurring in this even dimensional rep. $\endgroup$ Jun 18, 2015 at 5:09
  • $\begingroup$ I think (I made a hasty calculation) all odd $d$ less than $2l$ and all even $d$ less than or equal to $l$ will appear $\endgroup$ Jun 18, 2015 at 5:52

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.