Does every finitely generated free solvable group embed into the group of polynomial automorphisms of some C^n?
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1$\begingroup$ This is related to the Jacobian conjecture, and so I've retagged as algebraic-geometry and commutative-algebra. I don't have an answer, but have you looked for stuff in van den Essen's book "Polynomial automorphisms and the Jacobian conjecture"? It seems like the first place to start looking. $\endgroup$– Charles SiegelCommented Oct 25, 2009 at 19:42
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$\begingroup$ sorry, what does "free solvable" mean? The free group I know are not very solvable. $\endgroup$– Ben Webster ♦Commented Oct 25, 2009 at 20:17
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$\begingroup$ That's a good point. I was just ignoring the word free as meaningless there. $\endgroup$– Charles SiegelCommented Oct 25, 2009 at 21:28
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$\begingroup$ "Free solvable" could mean e.g. a free k-step solvable group. Explicitly, this is F/D^k(F), where F is a free group and D^*(F) its derived series. I don't know the context well enough to know if this is what was meant. $\endgroup$– Tom ChurchCommented Oct 25, 2009 at 21:50
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1 Answer
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Wilhelm Magnus (W. Magnus, Über $n$-dimensionale Gittertransformationen. Acta Math. 64 (1935), no. 1, 353--367.) seems to have shown that the free metabelian group on $n$ generators has a faithful representation of degree $2$. It follows that there is a copy of the free solvable group on $n$ generators and of derived length $2$ inside $GL_2(\mathbb{C})$.
That answers your question affirmatively in a special case.
answered Nov 23, 2009 at 4:28