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Let us consider affine Cremona group $GA_{n}$ ($n\gt 1$).

Question: do exist finite-dimensional representation of $GA_{n}$ ?

Thank you for the answer!

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Silly comment: I assume you are asking about faithful representations. Every group has the trivial one dimensional representation. (I don't have any more substantive comments, but I expect someone will.) – David Speyer Sep 14 '11 at 12:45
Yes, I would like to know about existence of finite-dimensional (of dimension bigger then 1) irreducible representation. – Andriy Sep 16 '11 at 14:32

I mentioned here (esp. Remark 5.3) that $GA_2$ has no faithful linear representation (finite-dimensional, over any field), but the proof can easily be modified to show that any linear representation of the kernel $SGA_2$ of the determinant of the Jacobian has no nontrivial linear representation (and thus any linear representation of $GA_2$ has an abelian image).

To summarize the proof: by the (elementary) computation done in the above link, if $M(x,y)=(x,x+y)$, then for every $n$ there exists a nilpotent subgroup $N$ of $SGA_2$ such that $M$ belongs to the $n$-th term of the descending central series of $N$. Nilpotency lengths of nilpotent subgroups in a given dimension are bounded. It easily follows that $M$ is killed by every linear representation of $SGA_2$. Also the normal subgroup generated by $M$ contains the linear, and thus the affine transformations of the plane and using generation of $GA_2$ by affine and elementary transformations it's easy to conclude. (I might include this further remark since the paper is not yet published.)

For $n\ge 3$ this still proves that every linear representation of $GA_n$ kills the affine transformations with determinant one, but I don't know if it's enough to conclude that all linear representations have an abelian image.

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Yes it would be interesting to add this nice remark. – Jérémy Blanc Jun 26 '12 at 8:32

There is one obvious representation which takes an automorphism of $\mathbb A^n$ to the determinant of its Jacobian. Let $G_n$ be its kernel. It carries a natural structure of infinite-dimensional algebraic group. Shafarevich proved that $G_n$ is simple as an algebraic group in this paper. Perhaps this suffices to answer your question but I am not sure.

Edit. According to Jérémy's comment below the proof of Shafarevich does not work. Apparently, the simplicity of $G_n$ as an algebraic group is an open problem for $n \ge 3$.

However it is known that $G_2$, as an abstract group, is not simple. This was proved by Danilov in this other paper. According to Furter and Lamy, Danilov shows that the normal subgroup generated by $(ea)^{13}$, where $a = (y,−x)$ and $e = (x,y+ 3x^5 − 5x^4)$, is a strict subgroup of $G_2$. For a more general statement see the paper By Furter and Lamy.

It is also known that planar Cremona group ${\mathrm{Bir}}(\mathbb P^2)$ is not simple. This was proved recently by Cantat and Lamy.

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Thanks a lot for your answer! But I also would like to put again my question and do it more precisely: Do exist finite-dimensional (of dimension bigger then 1) irreducible representation of G_{n} ? – Andriy Sep 16 '11 at 14:30
Thanks for the interesting information about non-simplicity of Bir(P^{2})! – Andriy Sep 16 '11 at 14:30
Pay attention, the proof of Shafarevich is known to be completely false. The fact that $G_2$ is simple as an infinite-dimensional algebraic group is not so hard to find, but it is still open if $G_n$ is simple or not, as an infinite-dimensional algebraic group, for $n\ge 3$. – Jérémy Blanc Jun 24 '12 at 20:43

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