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The following question appears in an example in page 116, Representation theory of semisimple groups, A. W. Knapp.

Consider $G=\mathrm{Ad}(SL(3,\mathbb R))$. It is a subgroup of $GL(8,\mathbb C)$. Let $U$ be the compact real form of $G$. Then $U$ is isomorphic to $SU(3)/\mathbb Z_3$.

Here $\mathbb Z_3$ is the foundamental group of $U$. Is $Z_3=\{0,1,2\}$?

Here $U$ is the analytic subgroup of $\mathfrak{u}=\mathfrak{k}\oplus i\mathfrak{p}$. Can we obtain $U$ via $SU(3)$ and the map $\mathbb{Ad}$?

Qinghua Pi
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1 Answers1

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One identifies $\mathbb{Z}_3$ with the center of $SU(3)$, namely $$Z(SU(3))=\{\zeta I_3:\zeta^3=1\}.$$

I assume that you mean $G=Ad(SL_3(\mathbb{C}))$. Assuming this, we have the adjoint representation $$Ad:SL_3(\mathbb{C})\rightarrow GL(\mathfrak{sl}_3(\mathbb{C})).$$ Its kernel is $Z(SL_3(\mathbb{C}))=Z(SU(3))\cong\mathbb{Z}_3$. Therefore, the image of $SU(3)$ under $Ad$ is a closed Lie subgroup of $GL(\mathfrak{sl}_3(\mathbb{C}))$ isomorphic to $SU(3)/\mathbb{Z}_3$.

Peter Crooks
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