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Let $k$ be a field of characteristic 0. The definition of a linear algebraic $k$-group of multiplicative type (m.t.) I've seen the most in the literature is that $G$ is of m.t. if it is a $\bar{k}/k$-twist of a closed subgroup of a torus. Some authors however define $G$ as a commutative group which is an extension of a finite group by a torus.

Why are these two definitions equivalent?

Also, from the first definition it's clear that a closed subgroup of a group of m.t. is again a group of m.t. From the second definition this is not that clear. Let me expand this a bit. Suppose $G$ is of m.t. Then $G$ is commutative and fits a s.e.s. $1 \to H \to G \to F \to 1$ with $H$ a torus and $F$ finite. Take a closed (commutative) subgroup $G'$ of $G$.

How do we construct a s.e.s. $ 1 \to H' \to G' \to F' \to 1$ with $H'$ a torus and $F'$ finite such that $G'$ fits it?

What I'm having trouble is seeing why $H'$ should be a torus again. My idea was to take the connected component of the identity of $G'$ and try to build a sequence with that, but I'm not too sure.

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2 Answers 2

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I don't know the answer to the first question, but I'll take a stab at the second. Suppose $G' \leq G$ is a closed subgroup of the commutative algebraic group $G$ of multiplicative type, fitting a short exact sequence $T \hookrightarrow G \twoheadrightarrow F$ with $T \cong (k^*)^n$ a torus and $F$ finite.

Let $T' = G'^0$ be the connected component of the identity (as suggested in the OQ). $T'$ is a connected subgroup of $G$, and $T = G^0$ is the maximal connected subgroup of $G$, so $T'$ is a subgroup of $T$. Thus $T'$ is a connected closed subgroup of a torus, hence also a torus.

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Let me try to complement Joshua Grochow's answer by answering the first question.

Begin by observing that $G$ being an extension of a finite group by a torus is equivalent to the same claim over $\overline{k}$ -- the forward implication is clear, and the reverse implication is clear as $G$ being an extension of a torus by a finite group is equivalent to $G^\circ$ being a torus, but this is equivalent (essentially by definition) to $(G^\circ)_{\overline{k}}$ being a torus, but basic group theory dictates that $(G^\circ)_{\overline{k}})=(G_{\overline{k}})^\circ$ (see [Milne, Proposition 1.34]).

So then, we see that we have quicly reduced ourselves to the following claim: if $k$ is algebraically closed and characteristic $0$, then $G$ embeds into a torus if and only if $G$ is an extension of a finite group by a torus.

The forward implication is clear as, by the above, it suffices to show that if $G$ embeds into a torus $T$ then $G^\circ$ is a torus. This has already been implicitly discussed in Josuha Grochow's answer, but this implies that $G^\circ$ is a torus (let me know if you don't understand this).

Conversely, suppose that $G$ is an extension of a finite group by a torus. Then, the key observe is that $G(k)$ consists entirely of semi-simple elements: this is clear for elements of $G^\circ(k)$ (as $G^\circ$ is a torus) and for elements of $\pi_0(G)(k)$ this is true by virtue of the fact that $k$ is characteristic 0 -- in characteristic $0$ every finite group consists of semi-simple elements (Hint: choosing a faithful embedding this is reduced to the well-known fact that if $k$ is of characteristic $0$ and $M$ in $\mathrm{GL}_n(k)$ is finite-order, then $M$ is semi-simple).

So then, take a faithful representation of $G\hookrightarrow \mathrm{GL}_{n,k}$. As the image of $G(k)$ consists entirely of commuting semi-simple elements, one knows from classical theory that, up to conjugation, we may assume that $G(k)$ lands inside of $D_n(k)$ (where $D_n$ is the group of diagonal matrices in $\mathrm{GL}_{n,k}$). As $G(k)$ is Zariski dense this implies that $G$ lands into $D_n$ and so $G$ embeds into a torus as desired.

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  • $\begingroup$ This is very helpful, thanks! What does the "twist" part of the question refer to? I don't see that explicitly in your answer, but I also don't know what it is, so I'm not sure I'd know it if it were there! $\endgroup$ May 17, 2022 at 5:48
  • $\begingroup$ @JoshuaGrochow Hey Joshua. Two groups $G$ and $H$ (at least in characteristic $0$ -- in characteristic $p$ there can be some ambiguity) are twists of each other if $G_{\overline{k}}\cong H_{\overline{k}}$. For instance, by definition, a group $G$ is a torus if and only if it is a twist of $\mathbb{G}_{m,k}^n$ for some $n$. Another example is that unitary groups are twists of the general linear group. $\endgroup$ May 17, 2022 at 5:59
  • $\begingroup$ Also, just to answer your implicit question: the second and third paragraphs which, in essence, allow me to reduce to the case when $k$ is algebraically closed secretly encodes the twist -- things that only depend on things up to twist can (often) be checked over $\overline{k}$. $\endgroup$ May 17, 2022 at 6:08
  • $\begingroup$ Ah, great, thanks! Under these assumptions, I had heard the terminology that G,H are "two [non-isomorphic] $k$-forms of $G_{\overline{k}}$". I've seen "twist" used in many other contexts, incl twisting by a cocycle - is that what's happening here? e.g. two k-forms of a group are always twists of one another by some Galois cocycle? $\endgroup$ May 17, 2022 at 18:09
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    $\begingroup$ Yeah, to be 100% accurate, a twist of $G$ really means a pair $(H,\iota)$ where $H$ is a group over $k$ and $\iota$ is an isomorphism $G_{\overline{k}}\to H_{\overline{k}}$. Then it's true that twists (up to equivalence) are classified by the Galois cohomology set $H^1(k,\mathrm{Aut}(G))$. $\endgroup$ May 18, 2022 at 6:01

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