Let $G$ be an algebraic group with closed normal subgroup $N$. Suppose that $N$ and $G/N$ are both unipotent. Does it imply that $G$ is also unipotent?
1 Answer
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By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, $V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent.
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$\begingroup$ Your "definition" is not the usual one, since the notion of "unipotent" arises at first in the treatment of Jordan decomposition in a linear algebraic group (over an algebraically closed field of arbitrary characteristic). $\endgroup$ Commented Feb 15, 2013 at 1:07
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1$\begingroup$ In Jantzen's RAGS, he asserts that for arbitrary k-group schemes (k a commutative ring), one takes this as the defining characteristic of a unipotent k-group scheme. $\endgroup$ Commented Feb 15, 2013 at 1:19
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1$\begingroup$ There is some variation in the definition in the literature, but the definition I give essentially agrees with that in SGA 3, Demazure Gabriel, Conrad-Gabber-Prasad, and Springer 1998. See XV 4.2 of the notes mentioned in an earlier comment. $\endgroup$– anonCommented Feb 15, 2013 at 1:48
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$\begingroup$ Dear anon: The definition of unipotence in SGA3 seems to "essentially agree" with yours only in the sense that they're logically equivalent; the proof of the equivalence isn't trivial. To be precise, the definition in C-G-P is taken to be the one in SGA3, which in turn is given in XVII 1.1 (and 1.2): for some (equivalently, any) algebraically closed extension $K$ of $k$, $G_K$ has a composition series whose successive quotients are $K$-subgroups of $\mathbf{G}_{\rm{a}}$ (also see 3.2, 3.5, and 4.1.1(i),(ii) in Expose XVII). $\endgroup$ Commented Feb 15, 2013 at 2:50
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3$\begingroup$ @pranavk In the old days, people had a lot of trouble defining things like "unipotent algebraic group". See Jim's comment, where he requires Borel-Chevalley classification theory to prove that an extension of unipotent algebraic groups is unipotent. It's hard to imagine a more natural definition than that an algebraic group G is unipotent if, for every finite-dimensional representation V, there exists a basis of V such that G acts through $U_n$ (upper triangular matrices with one's on the diagonal). The chapter where unipotent groups are studied, and the defns shown equivalent is 10 pages long $\endgroup$– anonCommented Feb 15, 2013 at 3:12
$G$is connected though it's not an issue in characteristic 0. Anyway, the answer is definitely yes. If you already have the full Borel-Chevalley structure theory of (connected) linear algebraic groups in hand, it's an easy consequence: modulo the unipotent radical of$G$, you get a reductive group (which can't be unipotent if nontrivial). Naturally you might prefer short-cuts. $\endgroup$