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Let $T$ be a torus of an algebraic group $G$, then $T$ acts on $\mathfrak{g}$, and $\mathfrak{g}$ has a decomposition:

$\mathfrak{g} = \mathfrak{c}_{\mathfrak{g}}(T) \oplus \coprod\limits_{\alpha \in \Phi} \mathfrak{g}_{\alpha}$.

Let $I(T)$ be the identity component in the intersection of all Borel subgroups of $G$ containing $T$, then for a subset $\Psi$ of $\Phi$, $\mathfrak{g}$ can also be decomposed as:

$\mathfrak{g} = \mathscr{L}(I(T)) \oplus \coprod\limits_{\alpha \in \Psi} \mathfrak{g}'_{\alpha}$.

For any $\alpha \in \Phi$, define $T_{\alpha} = (\mathrm{Ker} \alpha)^{\circ}$. Then there is a conclusion on singular tori (a torus is singular when it is contained in infinitely many Borel subgroups):

Let $S$ be a torus in $G$, $T$ a maximal torus containing it. Then $S$ is singular if and only if $S \subseteq T_{\alpha}$ for some $\alpha \in \Psi$.

Now consider the case when $K= \mathbb{C}$, $G=SL(n,K)$, $T = \{ \mathrm{diag} (d_1, \cdots d_n) | d_i \in K^*, i=1, \cdots, n \}$ and $S=\{ \mathrm{diag}(d, \cdots, d) | d \in K^* \}$. $S$ is in the center of $G$, so it is contained in every Borel subgroup of $G$. In addition, for any Borel subgroup $B$ of $G$, since $G$ is not solvable, $B \neq G$. $B$ doesn't contain $G^{\circ} = G$, thus $G/B$ is infinite and there are infinitely many Borel subgroups contained in $G$. All of them contain $S$, so, $S$ is singular. But for which $\alpha \in \Psi$, $S \subseteq T_{\alpha}$?

As $G$ is semisimple (thus reductive), $\Psi = \Phi$. Let $\alpha_i: T \rightarrow \mathbb{G}_m, \mathrm{diag} (d_1, \cdots, d_n) \mapsto d_i$. Is it true that $\Phi =\{ \alpha_1, \cdots, \alpha_n \}$? $T_{\alpha_i}= (\mathrm{Ker} \alpha_i)^{\circ} = \mathrm{Ker} \alpha_i = \{ \mathrm{diag} (d_1, \cdots, d_n) | d_1, \cdots, d_n \in K^*, d_i =1 \}$? If these are true, then no $T_{\alpha_i}$ would contain $S$.

Where is wrong?

Thank you very much.

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Now I know where I was wrong... When $G =SL(n,K)$, the root system is of type $A_n$, and a base consists of $ \{ \alpha_1, \cdots, \alpha_{n-1} \}$, and $\alpha_i: T \rightarrow \mathbb{G}_m$, $\mathrm{diag}(d_1, \cdots, d_n) \mapsto d_i/d_{i+1}$ (not $d_i$). (It seems that there are not many people interested in algebraic groups.) –  ShinyaSakai Aug 20 '11 at 18:33
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