# Smooth map homotopic to Lie group homomorphism

Let $$G$$ and $$H$$ be connected Lie groups. A Lie group homomorphism $$\rho:G\to H$$ is a smooth map of manifolds which is also a group homomorphism.

Question: Can we find a smooth (or real-analytic) map $$f:G\to H$$ which is not homotopic to any Lie group homomorphism?

For example, if $$G=H=S^1$$, it seems the answer is no. For simplicity, we may begin with the same question but assuming some extra conditions, such like (i) $$G,H$$ are torus, (ii) $$G,H$$ are compact, etc.

• If you construct two Lie groups such that $H \simeq G$, but $BH \not\simeq BG$ then any homotopy equivalence $H \simeq G$ cannot be homotopic to a homomorphism, since applying $B$ to it would deloop it to a weak equivalence. Jul 6 '20 at 20:17
• A non connected example of this is given by $H= \mathbb{Z}/2 \times \mathbb{Z}/2$ and $G= \mathbb{Z}/4$. Jul 6 '20 at 20:28
• @ConnorMalin Don't all such examples need to be disconnected? Jul 6 '20 at 20:33
• @NajibIdrissi $SO(4)$ and $S^3\times SO(3)$. It is true however that if $G,G'$ are connected simple lie groups, then $G,G'$ are isomorphic if and only if they are homotopy equivalent. Jul 6 '20 at 20:40

If $$G$$ is a compact simply-connected simple Lie group, then any nontrivial homomorphism $$G\to G$$ is an automorphism (it is injective because $$G$$ is simple, and any immersion of closed connected manifolds of the same dimension is covering map), and in particular it has degree $$\pm 1$$. For example, if $$f: S^3\to S^3$$ is a map of degree $$d$$ with $$|d|>1$$, then $$f$$ is not homotopic to a homomorphism.

On the other hand, by obstruction theory any self-map of an $$n$$-torus is homotopic to a map induced by multiplication by an $$n\times n$$ matrix with integer entries, which is a homomorphism.

• In fact, compact connected simple Lie groups have (see en.wikipedia.org/wiki/List_of_simple_Lie_groups) finite fundamental group and hence their self-coverings are diffeomorphisms. Thus in the above answer one can drop the assumption $G$ is simply-connected''. Jul 6 '20 at 23:45
• Thank you for the great answer! Could you explain more about the self-map of torus? I'm not familiar with obstruction theory. Or, any reference is also good enough.
– Hang
Jul 7 '20 at 0:29
• @Hang If $X$, $K$ are connected CW complexes and $K$ has contractible universal cover, then the set of homotopy classes $[X,K]$ of maps from $X$ to $K$ is bijective to the set of conjugacy classes of the induced $\pi_1$-homomorphism (e.g. Spanier, "Algebraic Topology", Ch.11, Section1, Theorem 11). Thus $[T^n, T^n]$ is bijective to the set of endomorphisms of $\mathbb Z^n$. Those are $n\times n$ matrices over $\mathbb Z$. Any such matrix (as a map of $\mathbb R^n$) descends to a self-map of an $n$-torus. Jul 7 '20 at 1:48

As Igor shows, every endomorphism of a simple Lie group $$G$$ has degree $$\in\{0,\pm 1\}$$.

On the other hand, every compact Lie group admits self maps of other degrees. Namely, the $$k$$-th power map $$g\mapsto g^k$$ has degree $$k^r$$, where $$r$$ is the rank of the group. So, each $$k$$ with $$|k|\geq 2$$ gives an example of a smooth map which is not homotopy equivalent to a homomorphism.

One way to compute the degree of the $$k$$-th power map is as follows. First, we can find an element $$g\in G$$ which lies in a unique maximal torus $$T^r$$ and which is also a regular value of the $$k$$-th power map. The uniqueness of the maximal torus implies that all $$k$$-th roots of $$g$$ lie in $$T^r$$, so this reduces the degree calculation to $$T^r$$, where it is obvious.

I’m not sure if this is what you’re looking for, but the map $$\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$ given by sending $$0\mapsto 1\,, 1\mapsto 0$$ isn't homotopic to a homomorphism. More generally if the codomain is disconnected the answer to your question seems to be positive.

• Thanks! This is a good point, but it could be more interesting to assume the connectedness.
– Hang
Jul 6 '20 at 19:46