Representation of finitely generated groups If $\Gamma$ is a finitely generated group.  Consider the representation $\mathrm{Rep}(\Gamma, \mathrm{SL}_2(\mathbb{C} )):=\mathrm{Hom}(\Gamma, \mathrm{SL}_2(\mathbb{C} ))$. 
How can we show $\mathrm{Rep}(\Gamma, \mathrm{SL}_2(\mathbb{C} ))$ is an affine algebraic set?  I know we can get the polynomials from the relations between the generators, but I do not know how to write them down. 
Also if $\Gamma$ is not finitely generated, Is the representation space an affine algebraic set?  If not, what makes it work in the finitely generated case?
Could someone help me, please?
 A: Let us write $G = SL_{2}(\mathbb{C})$ and consider the simplest case when $\Gamma = F_{n}$ the free group on $n$ generators. 
By universal property $Hom(\Gamma, G) = G^{n}$ ( n-fold cartesian product). Since our $G$ happens to be an affine algebraic variety so is $G^{n}$. 
This shows that if you were to take $\Gamma$ as infinitely generated then the hom-set would not be finite dimensional variety and (depending on your point of view) varieties are supposed to be finite dimensional over the base field. It is not even an ind-variety!
Now suppose  in the general case $\Gamma$ is presented as $F_{m} \rightarrow F_{n} \rightarrow \Gamma_{1} \rightarrow 1$ where $F_{i},\; i =m,n$ is finitely generated free group with some generators. 
Then $Hom(\Gamma, G) = im(  \phi: G^{n} \rightarrow G^{m})$ where $\phi$ comes canonically from the presentation. We are reduced to showing that the image is an algebraic set. Since $G$ is an affine algebraic group this is equivalent to showing that the map induced on the respective rings  is an algebraic map. But, relations are nothing but equality of certain words and this translates in the rings to angebraic equality of some products of variables.
