Affine Cremona group 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!
 A: 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.
A: 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.
