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I'm trying to solve an exercise from Humphreys' Linear algebraic groups. It asks to show that center Z of a nilpotent algebraic group G has positive dimension if G has positive dimension.

I'm trying to consider the lower central series of G, say $C_{0}G=G \supset C_{1}G \supset ....C_{n-1}G\supset C_n{G}=\{e\} $. By definition of being nilpotent, we get that $C_{n-1}G\subset Z$ and $C_{n-1}G\neq \{e\}$. I'm stuck on how to proceed further.

There are 2 things which might be useful: $C_{n-1}G$ is closed and normal in G and if G is connected then the center Z has positive dimension. Any help is appreciated.

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2 Answers 2

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Choose the biggest $i$, s.t. $C^{i}G$ is not discrete, then $C^{i}(G)^{0}$ has positive dimension, and $C^{i+1}(G)$ is discrete. Now take any element $g\in G$, consider the map $C^{i}(G)^{0}\longrightarrow C^{i+1}(G)$, $x\longrightarrow [x,g]$.

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Suppose that $C_{n-1}G$ is discrete. Let $x$ an element of $C_{n-2}G$ and $G_0$ the connected component of the identity of $G$. Consider the map $f_x:C_0\rightarrow G$ defined by $f_i(g)=[x,g]=xgx^{-1}g^{-1}$, the image of $f_x$ is contained in $C_{n-1}G$. Since $C_{n-1}G$ is discrete and $C_i$ connected, we deduce that $f_x$ is constant, and for every $g\in G, f_x(g)=f_x(x)=1_G$. This implies that $C_{n-2}G\subset Z(G)$.

If $C_{n-2}G$ is not discrete. done, if not we repeat the argument with $C_{n-3}G$, and we stop for the lowest $i$ such that $C_{n-i}G$ is not discrete but $C_{n-i+1}G$ is discrete. $C_{n-i}G$ is in the center of $G$.

This shows that the center of $G_0$ has dimension $>0$.

Consider the map $g:G\rightarrow Aut(Z(G_0))$ defined by $g(x)(y)=xyx^{-1}$, the kernel of $g$ contains $G_0$. $g$ induces a morphism $h:G/G_0$. Remark that $h$ induces a morphism $Lie(h):G/G_0\rightarrow Aut(Lie(Z(G_0))$ by $Lie(h)(A)={d\over{dt}}_{t=0}h(exp(tA))$. The image of $h$ is a nilpotent subgroup of the group of automorphism of $Z(G_0)$. The image of $Lie(h):G/G_0\rightarrow Aut(Lie(Z(G))$ is also nilpotent. We deduce that there exists a non zero vector $x$ of $Lie(Z)$ such that $Lie(h)(g)(x)=x$ for every $g\in G/G_0$. $x$ is in the center of $G$

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  • $\begingroup$ Thank you for your answer. I have one question: Its clear that when $C_i$ is the connected component containing $x$, then $f_x=id_G$. How do we show $f_x=id_G$ when the connected component doesn't contain $x$? $\endgroup$
    – user567863
    Jun 4, 2019 at 0:58
  • $\begingroup$ Thank you for your help. I'm confused with the notation. From $x\in Lie(Z)$, how do we get that $x\in Z(G)$? $\endgroup$
    – user567863
    Jun 4, 2019 at 14:40

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