# Representations of $\mathbb{G}_m$

I know that the multiplicative affine group scheme $$\mathbb{G}_m$$ is diagonalizable, since the algebra that represents it is $$k[X,X^{-1}]$$, which is isomorphic to the group algebra $$k[\mathbb{Z}]$$. Note that I am using the definition of Waterhouse, "Introduction to Affine Group Schemes": an affine group scheme is diagonalizable iff it is represented by $$k[M]$$ for some abelian group $$M$$.

Does the fact that $$\mathbb{G}_m$$ is diagonalizable imply that for every representation $$\mathbb{G}_m \to GL_n$$ there are (compatible? (in what sense?)) bases for each $$k-$$algebra $$R$$ such that the image of $$R^\times$$ in $$GL_n(R)$$ consists of diagonal matrices wrt those bases?

Does this imply that every representation of $$\mathbb{G}_m$$ splits into $$1-$$dimensional representations? If not, how can I prove that it does split?

You can assume that $$k$$ is a field, if you need it, but I wouldn't mind a more general answer.

This is not an answer but it's too long to be a comment, I'm just hoping it will shed a tiny bit of light.

I'm not an expert on affine group schemes at all, but here's what I can recollect about this thing.

Suppose $$G=Spec(k[M])$$ where $$M$$ is some abelian group and $$X= Spec(R)$$ is some affine scheme, $$R$$ a $$k$$-algebra, then a $$G$$-action on $$X$$ is essentially the same thing as an $$M$$-grading on $$R$$.

Indeed it corresponds to a map $$R\to R\otimes k[M] = \displaystyle\bigoplus_{m\in M} R$$, and the axioms for the group action tell you exactly that this map is a decomposition onto an $$M$$-grading (you have to get your hands into it to see what's happening - if you want I can add some details at this level, where I still understand things).

Now a representation $$G\to GL_n$$ is the same as an action of $$G$$ on $$\mathbb{G}_a^n = Spec(k[X_1,...,X_n])$$; so it corresponds to an $$M$$-grading of $$k[X_1,...,X_n]$$, and if you look at points on a specific $$k$$-algebra $$A$$ I think I remember a teacher of mine saying that this grading corresponds to an eigen-decomposition of the action of $$G(A)$$ on $$\mathbb{G}_a^n(A) = A^n$$, here I don't remember the details though (that's why this is not an answer, just a rather long comment).

Let me add that if $$k$$ is a field, a group-like element of $$k[M]$$ is an element of $$M$$ : indeed $$\Delta (\sum_m a_m m) = \sum_m a_m m\otimes m$$, so if it is group-like, then for all $$m,n$$, $$a_ma_n = \delta_{m,n}a_m$$ , therefore it must be $$0$$ or some element of $$M$$ : characters of $$G$$ correspond exactly to $$M$$, and again, if I recall correctly (but I'm fuzzy on the details) the eigenspaces of above correspond precisely to these characters of $$G$$.

When $$k$$ is a field, the answer to your questions is yes, but I don't know how to prove it without using the general structure theory of linear algebraic groups. Let $$\rho: \mathbb G_m \rightarrow \operatorname{GL}_n$$ be the given representation. It suffices to replace $$\mathbb G_m$$ by $$\operatorname{Im} \rho = \mathbb G_m/\operatorname{Ker} \phi$$ and assume $$\rho$$ is a faithful representation: if $$\rho$$ is not the trivial homomorphism, then $$\operatorname{Dim Ker} \rho = 0$$, and so the image of $$\rho$$ is isomorphic to $$\mathbb G_m$$ (even though $$\rho$$ is not the identity on $$\mathbb G_m$$). This is because the $$k$$-rational image of a split torus is a split torus, and split tori are determined by their dimension.

So the question reduces to considering a $$k$$-closed subgroup $$H$$ of $$\operatorname{GL}_n$$ which is isomorphic to $$\mathbb G_m$$. In particular, $$H$$ is a split torus, and every split torus is contained in a maximal split torus. Every maximal split torus in $$\operatorname{GL}_n$$ is conjugate to the group of diagonal matrices by some matrix $$g \in \operatorname{GL}_n(k)$$.

So $$\operatorname{Int} g$$ defines an isomorphism of group schemes of $$H$$ onto a closed subgroup scheme of the group $$T$$ of diagonal matrices in $$\operatorname{GL}_n$$. This isomorphism will give you the splitting of $$\rho$$ into one-dimensional representations. Also, this isomorphism of group schemes is what will transfer each group $$H(R) \cong R^{\ast}$$ of $$R$$-rational points to a subgroup of diagonal matrices in $$T(R) = (R^{\ast})^n$$.