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!

  • $\begingroup$ 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 '15 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 '15 at 10:03
  • $\begingroup$ @NickGill Thanks for the input. I will try to enlist all Parabolics thus Levi. Hope that gives good info. $\endgroup$ – Anupam Singh Jun 17 '15 at 10:22

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 $2l$.

  • $\begingroup$ 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? $\endgroup$ – Anupam Singh Jun 18 '15 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. $\endgroup$ – Venkataramana Jun 18 '15 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 $\endgroup$ – Venkataramana Jun 18 '15 at 5:52

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