I had a quick question about quotients of varieties. I am still not very good at them. Let $T$ be a torus, $\alpha$ a nontrivial character of $T$, and $S = (\textrm{Ker } \alpha)^0$. Since $T$ is commutative, $S$ is automatically normal in $T$, so I know that the quotient variety $T/S$ is an affine algebraic group. It is also connected, as a continuous image of $T$.

Let $G_m$ be the variety $k \setminus \{0\}$. Since $\alpha$ is a morphism of algebraic groups $T \rightarrow G_m$, we have that $G_m$ is a $T$-space, by the action $(a,x) \mapsto \alpha(a)x$. Since our field is algebraically closed, and $\alpha$ is nontrivial, we can argue further that $G_m$ is an equivariant $T$-space. Now $1 \in G_m$ has the property that $S$ is contained in its isotropy group (that is, $\textrm{Ker } \alpha$), so the universal property of quotient varieties implies that there is a unique equivariant morphism $\phi: T/S \rightarrow G_m$ such that $\phi(S) = 1$.

$\phi$ is given by the formula $\phi(aS) = \alpha(a)$, and also $\phi$ is surjective. Now I know that the dimension of $T$ is equal to the dimension of the kernel of $\alpha$ plus the dimension of the image of $\alpha$. And, the dimension of the kernel of $\alpha$ is the same as the dimension of $S$. Moreover, you can calculate the dimension of a quotient of varieties by subtracting the dimensions. So it follows that $\textrm{Dim } T/S = 1$.

But what I really want to conclude is that $T/S \cong G_m$ as algebraic groups. Should this be the case? This is what Springer says in 7.1.4.

  • $\begingroup$ Why are you taking the connected component? What is true is that $T/\ker \alpha\cong \mathbf{G}_m$, not modding out by the connected component. Did you consider the case when $T=\mathbf{G}_m$ (let's say in characteristic $0$)? Then for all non-trivial characters $\alpha$ one has that $(\ker \alpha)^\circ$ is trivial! $\endgroup$ Commented May 7, 2016 at 0:42
  • $\begingroup$ I am not sure why, that is just what Springer does in 7.1.4. He says "Fix $\alpha \in P'$. The group $G_{\alpha}$ contains the torus $S = (\textrm{Ker } \alpha)^0$...since $T/S$ is isomorphic to $G_m$ one has that $W_{\alpha}$ has order $\leq 2$." $\endgroup$
    – D_S
    Commented May 7, 2016 at 2:20

1 Answer 1


Yes, since the only connected 1-dim algebraic groups are $G_a$ and $G_m$, and $G_a$ is unipotent, $G_m$ "semisimple".

  • $\begingroup$ Thanks. But why does $T/S$ consist of semisimple elements (e.g. why is $(T/S)_s = T/S$)? $\endgroup$
    – D_S
    Commented May 17, 2016 at 0:04
  • 1
    $\begingroup$ canonical morphism $T \to T/S$ maps semisimple elements to semisimple elements. $\endgroup$
    – yisishoujo
    Commented May 17, 2016 at 2:42
  • $\begingroup$ That makes sense, thanks very much for answering. $\endgroup$
    – D_S
    Commented May 17, 2016 at 3:18

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