Why are characters of algebraic groups interesting?

Question

Are characters of an algebraic group the same thing as one-dimensional representations? Since a character of $G$ is $\rho:G\to \Bbb{G}_m\cong \text{GL}(1,\Bbb C)$ it seems these are the same thing? Am I right?

Why are characters interesting?

Answer 1

The character group $X(G)$ of a connected, reductive group $G$ is as interesting as the character group of its radical $A = \mathscr R(G)$. Here $A$ is the unique maximal torus lying in the center of $G$. It is the connected component of the center of $G$.

The result I'm getting at is that the restriction map $X(G) \rightarrow X(A)$ is an injection, and its image is a subgroup of finite index of $X(A)$.

Proof: Let $G_D$ be the derived group of $G$. Then $G_D \cap A$ is finite, and the product map $(x,y) \mapsto xy^{-1}$ induces an isomorphism of algebraic groups

$$(G_D \times A)/N \rightarrow G$$

where $N = \{ (x,x) : x \in G_D \cap A \}$. We can identify the character group of $(G_D \times A)/N$ with those characters of $G_D \times A$ which are trivial on $N$. Since $N$ is finite, this is a finite index subgroup of $X(G_D \times A)$. Now $X(G_D \times A) = X(G_D) \times X(A)$, with $X(G_D)$ the trivial group: any semisimple group such as $G_D$ coincides with its derived group, making any character on it trivial. The composition of all these maps

$$X(G) \xrightarrow{\cong} X(G_D \times A/N) \rightarrow X(G_D \times A) \xrightarrow{\cong} X(A)$$

is the restriction map. $\blacksquare$

So characters of $G$ can be identified with characters of a torus.

More generally, if $G$ is defined over a field $F$, then restriction induces an isomorphism of $X(G)_F$, the subgroup of $X(G)$ consisting of characters which are defined over $F$, with a subgroup of finite index of $X(A_G)$, where $A_G$ is the unique maximal $F$-split subtorus of the center of $G$.