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 $GL(4)$. 



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



Minimal polynomial of unipotents in orthogonal group
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 $2l$. 


