4
$\begingroup$

Let $H$ be a closed subgroup of the algebraic group $G$, $C = C_G(H)$. Prove that $\mathfrak{c} = \mathscr{L}(C_G(H)) \subseteq \mathfrak{c}_{\mathfrak{g}}(\mathfrak{h}) = \{ \mathrm{x} \in \mathfrak{g} | [\mathrm{x}, \mathfrak{h}] = 0\}$.

As $\frak{c}$ and $\frak{h}$ are the sets of elements of $\mathfrak{g}$ vanishing on the definition polynomials of closed subgroups $C$ and $H$ respectively, I think of proving this using definition polynomials.

Suppose that $\phi: G \times G \rightarrow G$ is the multiplication map of $G$. Then $\phi^*:K[G] \rightarrow K[G] \otimes K[G]$ maps $f \in K[G]$ to $f \circ \phi$, i.e., to $ \sum_i g_i \otimes h_i$, such that for any $x,y \in G$, $f(xy) = \sum_i g_i(x) h_i (y)$. For $\mathrm{x}, \mathrm{y} \in \mathscr{L}(G)$, and $f \in K[G]$, $\mathrm{xy}(f) = \sum_i \mathrm{x}(g_i) \mathrm{y}(h_i)$, where $\sum_i g_i \otimes h_i = \phi^*(f)$.

So, it has to be proved that for any $\mathrm{x} \in \mathfrak{c}$, $\mathrm{y} \in \mathfrak{h}$ and $f \in K[G]$, $\mathrm{xy}(f) - \mathrm{yx}(f) = 0$. With $\phi^*(f) = \sum_i g_i \otimes h_i$, that is $\sum_i\mathrm{x}(g_i)\mathrm{y}(h_i) = \sum_i\mathrm{y}(g_i)\mathrm{x}(h_i)$.

But the map $\phi^*$ is not quite clear to me, and I don't know how to obtain the equality $\sum_i\mathrm{x}(g_i)\mathrm{y}(h_i) = \sum_i\mathrm{y}(g_i)\mathrm{x}(h_i)$ by the vanishing property of $\mathrm{x}$ and $\mathrm{y}$ on definition polynomials.

(Here, the Lie algebra of an algebraic group is considered the same with its tangent space at the identity. I hope this is not wrong.)

Would you please fix this or show another way?

Sincere thanks for any answer, hint, reference comment or viewing.

  • 0
    I [gave this answer recently](http://math.stackexchange.com/questions/717505/for-an-element-x-in-an-algebraic-group-g-why-do-we-have-mathscrlc-gx/720054#720054), it seems like the same question.2014-04-03

0 Answers 0