# Are there homomorphisms $SU(2) \xrightarrow{} SU(2)$ with higher homotopy?

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?

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.

• 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? Nov 20, 2022 at 18:46
• yes, that is the case Nov 21, 2022 at 4:59
• Thanks again. As a final follow up, would things change if I were to look at $SO(3)$, which is not connected, instead of $SU(2)$? Nov 21, 2022 at 13:33
• $SO(3)$ 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. Nov 21, 2022 at 13:52