0
$\begingroup$

If $U$ is a closed subgroup of $GL(V)$ consisting of unipotent elements, show that $\log x$ $(x \in U)$ belongs to $\mathfrak{u}$.

Here, for an unipotent element $x \in GL(V)$, $\log x = \sum_{i=1}^{\infty}(-1)^{i+1}(x-1)^i/ i$. $\mathfrak{u}$ is the Lie algebra of $U$. All can be represented in matrix form.

If I am not mistaken, $x$ can be transformed into a triangular matrix with diagonal elements all equal $1$. So $T_{ij}$ $(1 \leq j < i \leq n)$, $T_{kk}$ $(1 \leq k \leq n)$ are all in $\mathscr{I}(U)$, the defining ideal of $U$. Suppose that $\mathrm{dim}V = n$, then every element of $y \in U$, satisfies $(y -1)^n =1$. Then how can I find a polynomial $f(T_{11}, \cdots, T_{nn}) \in \mathscr{I}(U)$, according to the equality $(y -1)^n =1$? I think this might be helpful for the proof.

Then, how can I prove it?

1 Answers 1

1

Since $x \in U$ is unipotent, by its very definition it means that $x-1$ is nilpotent, i.e. $\exists n \in \mathbb{N}$, such that $(x-1)^k = 0$ $\forall k \ge n$.

The defining series for $\log(x)$ then becomes a polynomial of order $\le n$.

What you need to show is that $(x-1)^k \in \mathfrak{u}$ for $1 \le k < n$, then $\log(x) \in \mathfrak{u}$, because $\mathfrak{u}$ is a linear space.

It is easy to see that $\exp( (x-1)^k )$ is unipotent, and $(x-1)^k \in \mathfrak{u}$. Indeed: $ \exp( (x-1)^k ) - 1 = (x-1)^k \left( 1 + \frac{1}{2} (x-1)^k + \ldots + \frac{1}{n!} (x-1)^{(n-1) k} \right) $ Hence $\left(\exp( (x-1)^k ) - 1\right)^n = (x-1)^{n k} \left( 1 + \ldots \right)^n = 0$.

  • 0
    Thank you so much. I think I still don't know some elementary facts. As you have said "the closedness of $U$ somehow implies that any element, whose exponential is in $U$ must be from $\mathfrak{u}$". I see that $\exp ((x-1)^k)$ is unipotent, but why must it be in $U$? (Then we can infer that $(x-1)^k \in \mathfrak{u}$.) Moreover, when $G$ is an algebraic group of unipotent elements, is it true that $\exp: \mathfrak{g} \rightarrow G$ and $\log: G \rightarrow \mathfrak{g}$ are both bijections? I sincerely wish to get your answer if it won't take you too much trouble.2011-10-24