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It is a well-known result that any affine algebraic group is a closed subgroup of some $\mathrm{Gl}_n(\Bbbk)$. However, I would like to see a proof for that, so I looked it up in various books, more precisely in

  • Procesi, Lie Groups - An Approach through Invariants and Representations, the theorem on page 172
  • Borel, Linear Algebraic Groups, 2nd Edition, Proposition 1.10
  • Alexander H.W. Schmitt, Geometric Invariant Theory and Decorated Principal Bundles, Theorem 1.1.3.3

The proof is always more or less the same and always lacks an important argument (in my opinion). Let me give you a quick overview:

If $\mu:G\times G\to G$ denotes the multiplication morphism and $\Bbbk[G]=\Bbbk[f_1,\ldots,f_n]$ then we may assume that the $f_i$ span a $G$-invariant subspace $V\subseteq\Bbbk[G]$. One can show that the comorphism $\mu^\sharp:\Bbbk[G]\to\Bbbk[G]\otimes\Bbbk[G]$ maps $V$ into $\Bbbk[G]\otimes V$ and thus, $$ \mu^\sharp(f_i) = \sum_{j=1}^n \psi_{ij} \otimes f_j $$ for certain $\psi_{ij}\in\Bbbk[G]$. One then maps $g$ to the matrix $\Psi_g$ with entries $\psi_{ij}(g)$.

Now my question is: Why is this a homomorphism of groups?

I can tell you what I have done so far. One may consider the triple multiplication morphism $\nu:G\times G\times G\to G$ which is just $\nu:=\mu\circ(\mathrm{id}\times\mu)$, so we get $$\begin{align*} \nu^\sharp(f_i) &= (\mathrm{id}\times\mu)^\sharp\left(\sum\nolimits_{j=1}^n \psi_{ij} \otimes f_j\right) = \sum_{j=1}^n \psi_{ij} \otimes \mu^\sharp(f_j) \\ &= \sum_{j=1}^n \psi_{ij} \otimes \sum_{k=1}^n \psi_{jk} \otimes f_k = \sum_{k=1}^n \left(\sum\nolimits_{j=1}^n \psi_{ij} \otimes \psi_{jk}\right) \otimes f_k \end{align*}$$ so $f_i(abc) = \sum_{k=1}^n (\Psi_a\Psi_b)_{ik} \cdot f_k(c)$. Conversely, we may also understand $abc$ as the product of $ab$ with $c$, so $$\begin{align*} \sum_{k=1}^n (\Psi_a\Psi_b)_{ik} \cdot f_k(c) = f_j(abc) &= \sum_{k=1}^n (\Psi_{ab})_{ik} \cdot f_k(c). \end{align*}$$ Now, what we need are elements $c_j\in Z(f_k\mid k\ne j)\setminus Z(f_j)$ to show that $\Psi_a\Psi_b=\Psi_{ab}$. I have a vague feeling that $f_j$ can not be contained in $(f_k\mid k\ne j)$ because $G$ acts linearly on $V$, but for some reason I am stuck. So my question is:

  • Can you finish my proof, i.e. show that $Z(f_k\mid k\ne j)\setminus Z(f_j)\ne \emptyset$ or equivalently, $f_j\notin(f_k\mid k\ne j)$?
  • If not, can you give an alternative proof?
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Have you consulted the original sources? Have you looked at Milne's script on algebraic groups? –  Martin Brandenburg Jul 12 '12 at 9:50
    
I don't know the original sources, but in my experience, a graduate text book rather works out more details, not less, than the original paper. Still, I'd give it a look if I knew what paper it was. I also looked through Milne's stuff and I couldn't really easily identify the statement in any of his notes. –  Jesko Hüttenhain Jul 12 '12 at 10:18
    
You should also look at Humphrey's book, which contains (if I recall) much the same argument as Giulio's answer. –  Ryan Reich Dec 29 '13 at 18:37
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1 Answer

up vote 4 down vote accepted

Well, I think the idea is this. What does it mean to find an injective homomorphism $G\to \operatorname{GL}(W)$? It means to find a faithful representation $W$ of $G$. Somehow, the intuition tells me that the $V$ you are talking about is the right $W$, because it is spanned by elements that describe well $G$.

In fact, this is right: if an element $g$ fixes $V$, it fixes all of $k[G]$, hence it must be the identity. So we have an injective homomorphism (of groups, not talking of algebraic structure yet)

$$\rho:G\to \operatorname{GL}(V)$$

Now, what coefficients do the matrices of $\rho$ have? Well, we may suppose that the $f_i$'s are a basis of $V$ (they generate, so I can extract a basis). Now,

$$(\rho(g)f_i)(h)=f_i(hg)=\sum_j\psi_{i,j}\otimes f_j(g,h)=\sum_j\psi_{i,j}(g)f_j(h)$$ hence $$\rho(g)f_i=\sum_j\psi_{i,j}(g)f_j$$ So $\rho(g)=(\psi_{i,j}(g))_{1\leq i,j\leq n}$, and $\rho$ is in fact an injective homomorphism of algebraic groups.

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This looks really good! Thanks for tending to this very old question of mine and giving a satisfying answer. –  Jesko Hüttenhain Dec 29 '13 at 18:19
    
You're welcome! –  Giulio Bresciani Dec 29 '13 at 20:56
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