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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?

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It doesn't make sense to ask whether something which is a priori a set is an affine algebraic set, in the same way that it doesn't make sense to ask whether something which is a priori a set is a group. (It does make sense to ask whether a subset of an affine algebraic set is an algebraic subset, but $\text{Hom}(\Gamma, \text{SL}_2(\mathbb{C}))$ doesn't come with a canonical embedding into an affine algebraic set.) The best you can ask for is that there exists a canonical affine algebraic structure, but this is extra structure, not a property. –  Qiaochu Yuan Sep 20 '12 at 0:29
Anyway, I think this is a good exercise to do for yourself. Try the case $\Gamma = \mathbb{Z}$ first, then perhaps $\Gamma = F_n$, then perhaps $\Gamma = \mathbb{Z}^n$. If you get stuck then there is something more basic that you don't understand and you should be asking questions about that instead. –  Qiaochu Yuan Sep 20 '12 at 0:32
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1 Answer 1

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.

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