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Is there a non trivial homomorphism $f: SU(2) \to \operatorname{Diff}(S^1)$?

(From the comments) By a previous question, we know that there is no nontrivial homomorphism $SU(2) \to O(2)$. Since $O(2)$ is a subgroup of $\operatorname{Diff}(S^1)$ (rotations are diffeomorphisms), maybe there is a nontrivial map in the larger group.

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  • $\begingroup$ Related to this question: math.stackexchange.com/questions/1302830/… $\endgroup$ May 28, 2015 at 18:36
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    $\begingroup$ if $f$ is non-trivial then the map $g:S^1\to S^1$, $g(z)=f(diag(z,z^{-1}))(1)$ ($S^1$ is seen here as the unit circle in $\mathbb C$), must have non-zero degree ($\pm1$ if $f$ is injective, or $\pm2$ if $f(-{\bf 1})=id$) [this has to be proven; sorry for being lazy]. So $g$ is not null-homotopic. However, $G:SU(2)\to S^1$, $G(X)=f(X)(1)$ is null-homotopic, as $SU(2)$ is 1-connected and thus $G$ lifts to a map $SU(2)\to\mathbb R$. So we get a contradiction. $\endgroup$
    – user8268
    May 28, 2015 at 18:56
  • $\begingroup$ It seems to be correct. But, how to compute the degree of $g$? $\endgroup$ May 28, 2015 at 19:56

1 Answer 1

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Theorem. There are no nontrivial homomorphisms $f$ (continuous or not) from $SU(2)$ to $Homeo(S^1)$.

Proof. First of all, it is known that $SO(3)=SU(2)/(\pm 1)$ is a simple group (as an abstract group); a proof of this is a very nice exercise in elementary group theory and geometry. If you cannot prove it, see for instance Berger’s book “Geometry-I”, section 8.5. Thus, it suffices to show that each homomorphism $f$ has kernel consisting more than just of the center of $SU(2)$. Observe that $SU(2)$ contains the binary Icosahedral group $G$, a nontrivial $Z_2$-central extension of the alternating group $A_5$. Thus, $f$ defines a homomorphism $f: G\to Homeo(S^1)$. I claim that each such homomorphism has to be trivial. Since $G$ is perfect, the image of $f$ is contained in the orientation-preserving subgroup of $Homeo(S^1)$. Now, if $g: S^1\to S^1$ is an orientation-preserving non-identity homeomorphism of finite order, it cannot have fixed points (prove first that each finite order homeomorphism of the interval has to be the identity map). Thus, each faithful topological action of $G$ (or its alternating quotient) on $S^1$ has to be fixed-point free, hence, defines a covering map $S^1\to S^1/f(G)$. However, the quotient $Y=S^1/f(G)$ has to be a topological circle, hence, by the basic covering theory, $G$ is isomorphic to the quotient $\pi_1(Y)/(i_*\pi_1(S^1))$, where $i_*$ is the monomorphism of fundamental groups induced by the covering map $S^1\to Y$. But such a quotient has to be cyclic. A contradiction. qed

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