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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.

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    $\begingroup$ 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. $\endgroup$
    – Connor Malin
    Commented Jul 6, 2020 at 20:17
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    $\begingroup$ A non connected example of this is given by $H= \mathbb{Z}/2 \times \mathbb{Z}/2$ and $G= \mathbb{Z}/4$. $\endgroup$
    – Connor Malin
    Commented Jul 6, 2020 at 20:28
  • $\begingroup$ @ConnorMalin Don't all such examples need to be disconnected? $\endgroup$
    – Najib Idrissi
    Commented Jul 6, 2020 at 20:33
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    $\begingroup$ @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. $\endgroup$
    – Tyrone
    Commented Jul 6, 2020 at 20:40

3 Answers 3

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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.

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    $\begingroup$ 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''. $\endgroup$
    – Igor Belegradek
    Commented Jul 6, 2020 at 23:45
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    $\begingroup$ 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. $\endgroup$
    – Hang
    Commented Jul 7, 2020 at 0:29
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    $\begingroup$ @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. $\endgroup$
    – Igor Belegradek
    Commented Jul 7, 2020 at 1:48
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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.

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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.

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  • $\begingroup$ Thanks! This is a good point, but it could be more interesting to assume the connectedness. $\endgroup$
    – Hang
    Commented Jul 6, 2020 at 19:46

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