Timeline for Bound on the number of connected components of a linear algebraic group $G<\mathrm{SL}_n$?
Current License: CC BY-SA 4.0
10 events
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| Aug 15, 2023 at 19:11 | comment | added | H A Helfgott | Right, thanks. If all examples split, then one could deal with the matter by Larsen-Pink. I now wonder whether there are simple examples where the exact sequence does not split, i.e., where we don't really have a finite group that we can deal with separately. | |
| Aug 15, 2023 at 18:38 | comment | added | Nate | It might be worth noting that $GL_n$ embeds into $SL_{n+1}$ via $g \to \begin{pmatrix}g&0\\0&det(g)^{-1}\end{pmatrix}$, so your $GL$ example is also an $SL$ example. | |
| Aug 15, 2023 at 13:58 | comment | added | H A Helfgott | In all of these examples, the exact sequence $1\to G_0\to G\to \Gamma\to 1$ splits, i.e., we can really see the quotient $G/G_0$ as a finite subgroup of $G$. What would be an example of a sequence (with $\Gamma$ big and nasty) in which this does not happen? | |
| Aug 15, 2023 at 13:52 | comment | added | H A Helfgott | If one allows a dependence on the degree of $G$ as a variety, then bounding the number of connected components of $G$ becomes trivial (it is bounded by the degree of $G$). | |
| Aug 15, 2023 at 13:50 | comment | added | H A Helfgott | I suppose any finite subgroup $\Gamma$ of $\mathrm{SL}_n(K)$ can be seen as "linear algebraic" in so far as it is a zero-dimensional (non-irreducible) variety - and the finite subgroups of $\mathrm{SL}_n(K)$ for $K$ of characteristic $\ne 0$ are harder to classify than for $K\ne 0$. Then one could try yo define more complicated examples by considering the product $\Gamma\cdot G$ for $G$ a linear algebraic group normalized by $\Gamma$. So the answer to my question has to be "no". | |
| Aug 15, 2023 at 13:44 | comment | added | H A Helfgott | @DaveBenson Well, Jordan's theorem is not true in characteristic $\ne 0$, at least not without additional conditions. Larsen-Pink amounts to a version that works for arbitrary characteristic. I suppose one cannot really do better than that? | |
| Aug 15, 2023 at 13:37 | comment | added | H A Helfgott | @RobertBryant Right, that's the general sort of thing I wanted to avoid (roots of unity) and your example makes it clear that requiring $G<\mathrm{SL}_n$ is not enough. Thanks. | |
| Aug 15, 2023 at 13:04 | comment | added | Dave Benson | Maybe you want to work modulo abelian normal subgroups, as in Jordan's Theorem and its variants. | |
| Aug 15, 2023 at 12:29 | comment | added | Robert Bryant | Maybe I don't understand the definition of 'linear algebraic group'. It seems to me that the subgroup $G_k\subset\mathrm{SL}(2,\mathbb{C})$ defined by the equations $g\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}g$ and $g^k=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has at least $k$ components. Is this $G_k$ not 'linear algebraic'? | |
| Aug 15, 2023 at 12:19 | history | asked | H A Helfgott | CC BY-SA 4.0 |