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I am quite new to the subject, so I apologize in advance for anything sloppy which I may write.
The homotopy group $\pi_3(SU(2))= \mathbb{Z}$ classifies maps of the form $S^3 \simeq SU(2) \xrightarrow{} SU(2)$, where $\mathbb{Z}$, roughly speaking, measures how many times $SU(2)$ wraps onto itself.
These maps, and their homotopy classification, appear in the standard construction of instantons via bundles. For example, we have that the map $g_0: x \in S^3 \mapsto e \in SU(2)$, with $e$ being the identity, belongs to the class $0$ of $\pi_3(SU(2))$.
The following map belongs to the class $1$: $g_1: (x_1,x_2,x_3,x_4)\in S^3 \mapsto \frac{1}{r} (x^4 \mathbb{1} + \sum_i x_i \sigma_i) \in SU(2)$, where $r^2=x_1^2 + x_2^2 + x_3^2 +x_4^2$, $\sigma_i$ are the Pauli matrices and $\mathbb{1}$ is the identity.
For $n>1$, the map $g_n: x \mapsto r^{-n}(x^4 \mathbb{1} + \sum_i x_i \sigma_i)^n$ belongs to the class $n$ of $\pi_3(SU(2))$.
Clearly, all these maps can be thought to be maps from $SU(2)$ to $SU(2)$. $g_0$ and $g_1$ are homomorphisms, i.e. $g(x y)=g(x)g(y)$. However, since $SU(2)$ is not abelian, it is evident that $g_n$ with $n>1$ is $not$ a homomorphism.
Now, finally to my question: are there homomorphisms $SU(2) \xrightarrow{} SU(2)$ that belong to the class $n$ of $\pi_3(SU(2))$ for $n>1$? If so, could you give me an example?
Thanks in advance!

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1 Answer 1

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No, there a no homomorphisms belonging to class $n$ for $|n| > 1$.

More generally, we have the following:

Proposition: Suppose $G$ is a connected simple Lie group (meaning all proper normal subgroups are finite). Then every homomorphism is either trivial or an isomorphism.

Proof: Suppose $f:G\rightarrow G$ is non-trivial. Moving to the Lie algebra level, the induced map $f_\ast$ must also be non-trivial. So $\ker f_\ast$ is a proper ideal, and hence is trivial. This implies $f_\ast$ is an isomorphism, which, in turn, implies $f$ is an isomorphism. $\square$

Of course, a trivial map induces the $0$ map on $\pi_k$ for any $k$.

On the other hand, an isomorphism must induce an isomorphism on $\pi_k$ for any $k$. Thus, in your particular case, only $n\in \{0, \pm 1\}$ is possible.

In fact, $n = -1$ is not possible when $G = SU(2)$. Probably the easiest way to see this is to note that $SU(2)$ has trivial outer automorphism group, so all automorphisms are isotopic to the identity map, which clearly is the $n=1$ case.

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  • $\begingroup$ Thank you! I guess that the opposite is true for $U(1)$, where there are homomorphisms belonging to class $n$ for any $|n|>1$. Is that the case? $\endgroup$
    – User175a23
    Nov 20, 2022 at 18:46
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    $\begingroup$ yes, that is the case $\endgroup$ Nov 21, 2022 at 4:59
  • $\begingroup$ Thanks again. As a final follow up, would things change if I were to look at $SO(3)$, which is not connected, instead of $SU(2)$? $\endgroup$
    – User175a23
    Nov 21, 2022 at 13:33
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    $\begingroup$ $SO(3)$ is connected, but not simply connected. It is still a simple group, so the above argument applies. For disconnected groups, one must worry about basepoints, etc, but the Proposition still holds with the obvious modification. E.g., if $G$ is disconnected with identity component $G^0$, any homomorphism must map $G^0$ to $G^0$, the induced map on $\pi_k$ will be an isomorphism. $\endgroup$ Nov 21, 2022 at 13:52

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