Given $n$, can $n$-dimensional linear algebraic groups over $\mathbb{C}$ be embedded into $\mathrm{GL}(N,\mathbb{C})$ for a uniformly bounded $N$?
Thanks so much for your reply!
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12$\begingroup$ Finite groups (i.e., 0-dimensional algebraic groups) cannot all be embedded into the same $\mathrm{GL}(N,\mathbb{C})$. So the problem should consider only algebraic groups that are connected. $\endgroup$– Dave Witte MorrisJun 8, 2015 at 19:01
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As Dave points out, it's very risky to include finite (dimension 0) affine algebraic groups (in the traditional sense) in your question. So I'm assuming $G$ is connected, in which case the answer to the question is YES.
Even more than in the case of Cayley's theorem for embedding finite groups into symmetric groups, the old theorem of Chevalley (in arbitrary characteristic) embeds an arbitrary affine algebraic group into a general linear group without any control over the size of that linear group. But if you appeal to the classication of semisimple groups along with the Levi decomposition in characteristic 0, you can control the size better. To start with, there are only finitely many (connected) semisimple groups of a given dimension $n$, so that already bounds $N$ for them. On the other hand, all $n$-dimensional algebraic tori are isomorphic to the direct product of $n$ copies of the multiplicative group, so they are also well-behaved with respect to embedding. At least in characteristic 0, all $n$-dimensional unipotent groups are isomorphic to vector groups and have bounded embeddings. Combined with the Levi decomposition over $\mathbb{C}$ (in the algebraic group setting), these individual results and the general Borel-Tits structure theory for reductive groups will provide $N$ indirectly.
ADDED: My initial thinking about the question was not precise enough to handle unipotent radicals or semidirect product decompositions here. So I would stick to reductive (connected) groups, where the structure and classification theorems easy to apply. Aside from that it's worth asking why people don't work more often with affine algebraic groups in linear embeddings: probably it's because such embeddings (though possible by Chevalley's theorem) are too loosely structured and seldom of much help in applications. Intrinsic properties such as Jordan-Chevalley decomposition and Bruhat decomposition are typically more useful in practice.
answered Jun 8, 2015 at 22:35
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2$\begingroup$ That $n$-dimensional unipotent groups have bounded embeddings in char. zero is part of Ado's theorem, and the best upper bound on $n$ is exponential. I don't understand your argument suggesting that it's just because they are "isomorphic to vector groups" (as what? as varieties?) $\endgroup$– YCorJun 9, 2015 at 9:22
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2$\begingroup$ Second, the Levi decomposition certainly helps, but you should say how: the group is $G=U\rtimes R$ with both the unipotent $U$ and the connected reductive $R$ having a faithful rep in some bounded dimension; why should $G$ also have a faithful rep in bounded dimension? $\endgroup$– YCorJun 9, 2015 at 10:22
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$\begingroup$ @YCor: Yes, my answer went too far. I'm comfortable with the case of connected reductive groups but less confident about handling either unipotent groups or semidirect products. As your first comment indicates, unipotent groups are especially tricky to handle. (See my added remarks.) $\endgroup$ Jun 9, 2015 at 14:26
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1$\begingroup$ Here's how to complete the proof in the reductive case: first it's OK (by the finiteness of cases) in the semisimple case. Let $G=TS$ be connected reductive and $Z=T\cap S$. Find a faithful rep of $S$ of bounded dimension: it splits as a sum of block on each of which $Z$ acts as a scalar. On each block extend the character from $Z$ to $T$ to get a representation of $T$ commuting with the $S$-rep and coinciding on $Z$, thus defining the $G$-rep, faithful on $S$. Sum with a faithful rep of $G/S$ of dimension $\dim(T)$ to get a faithful action of $G$. $\endgroup$– YCorJun 9, 2015 at 15:57
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$\begingroup$ @YCor: This approach looks reasonable, but I was thinking of using the classification of connected reductive groups more directly (in the style of SGA3, adapted to linear algebraic groups by T.A. Springer). In any case, finiteness of the number of such groups up to isomorphism for each dimension $n$ is probably not written down explicitly anywhere but is implicit in the classification by root data. $\endgroup$ Jun 10, 2015 at 18:19