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Consider the orthogonal group $O(2l+1)$ ($l \geq 1$), is there an matrix in $O(2l+1)$ with its minimal polynomial being $(x-1)^{2l+1}$?

A similar question has been asked for the $O(2l)$ (this link : Minimal polynomial of unipotents in orthogonal group) . So I am asking now for the $2l+1$ case.

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  • $\begingroup$ The matrix then has only eigenvalue 1 and therefore it must be the identity matrix which has minimal polynomial x-1. Hence only for l=0 there exists such a matrix. In the link you provided they consider the split-orthogonal group which makes the problem probably more difficult. $\endgroup$
    – user100927
    Aug 2, 2017 at 11:33
  • $\begingroup$ I've edited it to $l \geq 1$, my question is for $O(3)$, $O(5)$ etc. $\endgroup$ Aug 2, 2017 at 12:08
  • $\begingroup$ Then the answer is no by the explanation I gave. $\endgroup$
    – user100927
    Aug 2, 2017 at 12:14
  • $\begingroup$ The question remains if you consider the standard orthogonal group, i.e. all matrices with A^TA=I or some kind of split orthogonal group. $\endgroup$
    – user100927
    Aug 2, 2017 at 12:15
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    $\begingroup$ It will be easier to say something if you explain what you mean by $0(m)$. $\endgroup$
    – abx
    Aug 2, 2017 at 13:36

1 Answer 1

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Assuming you mean $\mathrm{O}_{2l+1}(K)$ ($K$ algebraically closed of char ≠ 2): any member of the regular unipotent class in $\mathrm{SO}_{2l+1}(K)$ has Jordan form $J_{2l+1}$, hence the desired minimal polynomial: see e.g. Liebeck–Seitz, p. 57.

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  • $\begingroup$ In the above, if we replace $K$ by a finite field $\mathbb{F}_q$ (of char $\neq$ 2). What happens then? $\endgroup$ Aug 3, 2017 at 6:20
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    $\begingroup$ @UdayBhaskar: Same result, actually all of the unipotent classes will have a representative that is a matrix with entries in $\mathbb{F}_q$. $\endgroup$ Aug 3, 2017 at 10:52
  • $\begingroup$ @MikkoKorhonen That sounds right, and seems to be in Huppert (1980, Satz 3.2d) for any field of char ≠ 2. (In the finite case, signature automatically $(l+1,l)$ or $(l,l+1)$.) $\endgroup$ Aug 3, 2017 at 12:56

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