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In a matrix Lie group $G$, we say that $g\in G$ is unipotent if $$(g-I)^n=0 $$ for some $n\in \mathbb{N}.$

I read in a Tao's article, that

More generally, we say that an element $g$ of a Lie group $G$ is unipotent if its adjoint action $x \mapsto gxg^{-1}$ on the Lie algebra $\mathfrak{g}$ is unipotent.

How can I show that in the matrix Lie group case these definitions coincide, i.e., if $(g-I)^n=0$, then $(Ad_g(x)-x)^k =0$ for some $k\in \mathbb{N}$ and $x$ in the Lie algebra of $G$?

Is this even true?

calculatormathematical
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1 Answers1

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Suppose $g$ is unipotent. Then $g=1+n$ for nilpotent $n$. Then $$\mathrm{Ad}_g(x)-x =gxg^{-1}-x =(gx-xg)g^{-1} =(nx-xn)g^{-1}.$$ Because $n$ is nilpotent, there is a filtration $$0=V_0\subset V_1 \subset\dots \subset V_k=V$$ such that $n(V_i)\subset V_{i-1}$. Since $$g^{-1}=(1+n)^{-1}=1-n+n^2-\dots,$$ $g^{-1}$ preserves the filtration. Define $\phi(x)=(nx-xn)g^{-1}$. You can check for yourself that $\phi^m(x)$ takes $V_i$ into $V_{i+k-m}.$ Then $\phi^{2k}=0,$ so $\phi$ is nilpotent and $\mathrm{Ad}_g$ is unipotent.

subrosar
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