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Let $G$ be a linear algebraic group over an algebraically closed field $k$, and $T$ a maximal torus of $G$.

Suppose we have two cocharacter $\mu, \mu' : \mathbb{G}_m \to T$, which are conjugate under $G$ i.e. there exists $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$.

Question. Can we always choose $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$ and such that $g$ normalizes $T$?

If $G = \mathrm{GL}_n$ then this is true: suppose that $T$ is the diagonal torus. Then a cocharacter of $T$ is just determined by its weights on the basis vectors of the natural representation of $G$ (i.e. integers $k_i$ such that $\mu(z)e_i = z^{k_i} e_i$).

Since $\mu'$, $\mu$ are $G$-conjugate, the weights of $\mu$ are a permutation of the weights of $\mu'$. We can choose an element of $G$ which induces this permutation on the basis vectors.

For general $G$, we can embed $G$ in $\mathrm{GL}_n$ for some $G$, such that $T$ is the intersection of $G$ with the diagonal torus. The weights of $\mu$ on the basis vectors of the corresponding representation are a permutation of the weights of $\mu'$. The question is whether we can choose this permutation (since the weights need not be distinct, there may be several choices for the permutation) so that it is induced by $N_G(T)$.

For my application, to show that the number of $G$-conjugates of $\mu$ in $T$ is finite, it is enough to know this for $\mathrm{GL}_n$, but it would be nice to know if it is true in general.

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Is your algebraic group $G$ semisimple or linearly reductive? –  Victor Protsak Jun 22 '10 at 11:49
    
No, no conditions are needed on $G$. –  Martin Orr Jun 22 '10 at 15:33
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2 Answers

up vote 4 down vote accepted

Yes. That $\mu'(z) = g \mu(z) g^{-1}$ means that $g T g^{-1}$ centralizes the image of $\mu'$. Thus, $T$ and $g T g^{-1}$ are maximal tori in the centralizer of the image of $\mu'$, so there exists $h$ in this centralizer such that $g T g^{-1} = h T h^{-1}$. Now replace $g$ by $h^{-1} g$. It doesn't seem that you need G to be semisimple or reductive for this, but I may be missing something here?

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For the finiteness question which is mentioned by Martin we'd need that $N_G(T)/Z_G(T)$ is finite for general $G$, and over an algebraically closed field this is pretty clear. –  BCnrd Jun 22 '10 at 13:43
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In fact something better is true (properly formulated!) over any field $k$, using $k$-rational conjugacy and maximal $k$-split $k$-tori. I will give a precise statement and proof below, with $G$ any affine $k$-group of finite type (no reductivity or smoothness hypotheses, though lack of smoothness is a red herring, as will be explained below; lack of reductivity is more serious with general $k$). In the end it will come down to the same idea as in the argument which is given by an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta, but we need preparations to make it work over any field and for any $G$ (even with smoothness restrictions, there is work to do). To stick with a single reference, below I refer to things in "Pseudo-reductive groups".

Let $G$ be an affine group scheme of finite over a field $k$, and $S$ a maximal $k$-split $k$-torus in $G$ (these exist, for dimension reasons). I first claim that any two such $S$ are $G(k)$-conjugate, which is really the whole point. We can replace $G$ with the identity component of the smooth Galois descent of the Zariski-closure of $G(k_s)$ in $G_{k_s}$ (see Lemma C.4.1 in "Pseudo-reductive groups" for why this is a smooth $k$-subgroup of the original $G$) to reduce to the case when $G$ is smooth and connected, and then it is very well-known when $k$ is algebraically closed (the case of interest in the question), less widely-known but due to Grothendieck (in SGA3) by a nice short inference from the algebraically closed case when $k = k_s$, and not at all widely known but announced (without published proof) by Borel & Tits; a proof of the Borel-Tits result is given in Theorem C.2.3 of the book "Pseudo-reductive groups". The proof in the general case involves a lot of work (but not the main results of the book).

With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$. I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$ and without smoothness hypotheses on $G$ (a red herring, by the trick in the preceding paragraph). This can be proved in several ways. Here is one. The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is \'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof). Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$).

Remark: By Proposition C.2.10, if $G$ is smooth and connected then $W(G,S)$ is a constant $k$-group and $W \hookrightarrow W(G,S)(k)$ is an isomorphism. That's nice to know, but we make no use of it.

OK, so finally we can state the general result:

Theorem: Let $G$ be an affine group of finite type over a field $k$, and $S$ a maximal $k$-split $k$-torus. Define $W$ as above, let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise for ${\rm{X}}_ {\ast}(S)$.

Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map $$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$ is bijective. In particular, $G(k)$-orbits on ${\rm{X}}_ {\ast}(G)$ have finite non-empty intersection with ${\rm{X}}_ {\ast}(S)$.

Proof: Letting $G' \subseteq G$ denote the Galois descent of the Zariski closure of $G(k_s)$ in $G_ {k_s}$, so $G'$ is a smooth closed $k$-subgroup of $G$ (see Lemma C.4.1 once again), clearly ${\rm{X}}_ {\ast}(G') = {\rm{X}}_ {\ast}(G)$, $G'(k) = G(k)$, and $S$ is contained in $G'$. Thus, we can replace $G$ with $G'$ without affecting what is to be proved, and so we can assume $G$ is smooth (but maybe not connected).
We now adapt the argument from the proof of Lemma C.3.5 (which treats a variant for pseudo-reductive $k$-groups and maximal $k$-tori). Any $k$-homomorphism $\mathbf{G}_ m \rightarrow G$ has image which is a $k$-split $k$-torus, so it lands in a maximal $k$-split $k$-torus. But all such $k$-tori are $G(k)$-conjugate to our friend $S$, as explained above (even without smoothness), so we get the surjectivity.

Now consider injectivity (which was the original MO question!). This will go exactly as in an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta's answer, using what was done above, but for convenience of a reader who may be nervous about general $k$ we now give the argument. For cocharacters $\mu$ and $\mu'$ of $S$ over $k$, suppose $g \in G(k)$ satisfies $g.\mu' = \mu$ in ${\rm{X}}_ {\ast}(G)$ (where $g.\lambda: t \mapsto g \lambda(t) g^{-1}$ denotes the $G(k)$-action on $k$-rational cocharacters $\lambda$ of $G$). Thus, $\mu'$ is valued in the commutative $g^{-1} S g$, so $g^{-1} S g$ and $S$ are both contained in the centralizer $Z_G(\mu')$ of $\mu'$ (which is the same as the centralizer of the image of $\mu'$, so it is represented by a closed $k$-subgroup of $G$). Hence, $g^{-1} S g$ and $S$ are contained in $H := Z_G(\mu')$. Clearly $g^{-1} S g$ and $S$ are visibly maximal $k$-split $k$-tori in $H$, so there exists $h \in H(k)$ such that $h^{-1} (g^{-1} S g) h = S$. It is harmless to replace $g$ with $gh$, yet this brings us to the case $g \in N_ {G(k)}(S)$. QED

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