In the previous post, we saw that if $ G\times X\to X$ is an algebraic group acting on a variety $ X$ and $ F\subseteq k[X]$ is a finite-dimensional subspace then there exists a finite dimensional subspace $ E\subseteq k[X]$ with $ E\supseteq F$ such that $ E$ is invariant under translations.

Recall that if $ g\in G$ and $ f\in k[X]$ then the translation $ \tau_g(f)(y) = f(g^{-1}y)$ so that $ E$ being invariant under translations means that $ \tau_g(E) = E$ for all $ g\in G$. Now, let’s use the method outlined in the previous post to actually construct a three-dimensional representation of $ G = \mathrm{SL}_2(k)$.

The Example

In this setting we specialise to the case where $ G$ is an algebraic group acting on itself via multiplication: $ m:G\times G\to G$ is given by a Hopf algebra homomorphism $ \Delta:k[G]\to k[G]\otimes_kk[G]$. Of course, in this case we will need to actually choose some finitely generated Hopf $ k$-algebra as our ring of functions of $ \mathrm{SL}_2(k)$. Let’s use $ k[\mathrm{SL}_2] = k[T_1,T_2,T_3,T_4]/(T_1T_4 – T_2T_3 – 1)$. Thus we think of elements of $ \rm{SL}_2(k)$ as homomorphisms $ k[\mathrm{SL}_2]\to k$ corresponding to the matrix:

$ \begin{pmatrix}T_1 & T_2 \\ T_3 & T_4\end{pmatrix}$

The comultiplication map is then easily checked to be given by

$ T_1\mapsto T_1\otimes T_1 + T_2\otimes T_3$
$ T_2\mapsto T_1\otimes T_2 + T_2\otimes T_4$
$ T_3\mapsto T_3\otimes T_1 + T_4\otimes T_3$
$ T_4\mapsto T_3\otimes T_2 + T_4\otimes T_4$

Of course, this is nothing other than the rule for multiplying two matrices!

Now, we remarked previously that if we have a function $ f\in k[\rm{SL}_2]$ we can send it through the comultiplication map to get $ \Delta(f) = \sum h_i\otimes k_i$ and then the subspace generated by the $ k_i$ will be stable under all left translations, because it is the subspace given by all the translates of $ f$, and the translate of a translate of $ f$ is itself clearly a translate of $ f$.

As an example, take $ f = T_1T_3$. Then

$ \Delta(f) = T_1T_3\otimes T_1^2 + T_1T_4\otimes T_1T_3 + T_2T_3\otimes T_3T_1 + T_2T_4\otimes T_3^2$.

Hence there is a representation given by $ \mathrm{SL}_2$ acting on the subspace $ V = \mathrm{span}(T_1^2 , T_1T_3 , T_3^2)$. It is given on $ k$-points by the group homomorphism $ \mathrm{SL}_2(k)\to \mathrm{Aut}(V)$ by

$ \begin{pmatrix} a & b \\ c & d\end{pmatrix}\mapsto\begin{pmatrix}ad + bc & -2bd & -2ac\\ -cd & d^2 & c^2 \\ -ab & b^2 & a^2\end{pmatrix}$.

This can be checked by direct computation. Note that I chose $ \mathrm{SL}_2$ here to avoid writing out details involving the determinant, but the same technique will work for any algebraic group.

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