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Let $G$ be a compact connected Lie group with semisimple Lie algebra ${\frak g}$. With the following reasoning, I show that there is no non-trivial Lie group homomorphism $$\chi:G\to S^1.$$ Is that true?


Proof. Let $\chi:G\to S^1\subseteq\Bbb C$ be a character. Then, $d\chi:{\frak g}\to i\Bbb R$ so $id\chi:{\frak g}\to \Bbb R$, i.e. $id\chi\in{\frak g}^*$. But since $\chi$ is a homomorphism, and $S^1$ is abelian, we have for all $g\in G$ and $X\in{\frak g}$ that $$({\operatorname{Ad}}_g^*id\chi)(X)=i\frac{d}{dt}\Big|_{t=0}\chi(\exp(t\operatorname{Ad}_{g^{-1}}(X)))=i\frac{d}{dt}\Big|_{t=0}\chi(g^{-1}\exp(tX)g)=i\frac{d}{dt}\chi(\exp(tX))=id\chi(X).$$ Thus, $\operatorname{Ad}_g^*id\chi=id\chi$ for all $g\in G$. Using the Killing form to identify $id\chi\in{\frak g}^*$ with an element $X_{\chi}\in{\frak g}$, we find $\operatorname{Ad}_g(X_{\chi})=X_{\chi}$ for all $g\in G$. Thus, $X_{\chi}\in\operatorname{Lie}(Z_G)=Z_{{\frak g}}$ (where $Z$ means "center"). But a semisimple Lie algebra has zero center and hence $X_{\chi}=0$. Thus, $id\chi=0$ and hence $\chi=1$. $$\tag{Q.E.D.}$$

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    $\begingroup$ The group has to be connected, for otherwise this is false. Take any compact $G$ and consider the Lie group $G\times\mathbb Z/7\mathbb Z$. It admits a non-trivial homomorphism to $S^1$. $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '16 at 8:21
  • $\begingroup$ @MarianoSuárez-Alvarez Right! I was using this assumption to deduce $d\chi=0$ $\implies$ $\chi =1$. Thanks! $\endgroup$ – Ado Apr 23 '16 at 8:23
  • $\begingroup$ You should have a question on Ado's theorem, right ? $\endgroup$ – Dietrich Burde Apr 23 '16 at 17:46
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The character $\chi$ induces a morphism $f:g\rightarrow R$, since $g$ is semi-simple, $[g,g]=g$, but the restriction of $f$ to $[g,g]$ is zero since $f([x,y])=[f(x),f(y)]=0$, thus $f=0$.

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If $\chi$ is a group homomorphism, then its differential at the identity is a Lie algebra homomorphism $f:\mathfrak g\to\mathbb R$. If $f$ is not zero, then its kernel $K$ is an ideal in $\mathfrak g$ which, as the algebra is semisimple, has a complement, that is, there is another ideal $L$ in $\mathfrak g$ such that $\mathfrak g=K\oplus L$. The ideal $L$ is one dimensional, because $f$ induces an isomorhism $\mathfrak g/K\to\mathbb R$. Now an ideal of a semisimple Lie algebra is itself a semisimple Lie algebra, and therefore it is impossible for $L$ be one-dimensional.

What happens if $f$ is zero (and $G$ connected, as it should be)?

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  • $\begingroup$ Just to be clear: The answer is yes and you are giving a second proof different from mine? $\endgroup$ – Ado Apr 23 '16 at 8:21

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