Charles Rezk had highlighted in MO:q/123624 that "Nearby homomorphisms from compact Lie groups are conjugate", and in consequence -- further highlighted in Remark 2.2.1 of his Global Homotopy Theory and Cohesion -- that for $G$ and $\Gamma$ Lie groups with $G$ compact, the homomorphism space $\mathrm{Hom}(G,\Gamma) \subset \mathrm{Maps}(G,\,\Gamma)$ decomposes under the adjoint action of $\Gamma$ as:

$$ \mathrm{Hom}(G,\,\Gamma) \;\;\;\;\simeq\; \underset{ { [\phi] \in } \atop { H^1_{\mathrm{Grp}}(G,\,\Gamma) } }{\coprod} \; \Gamma/\mathrm{Stab}_\Gamma(\phi) \;\;\;\;\; \in \Gamma \mathrm{Act}(\mathrm{TopSp}) \,. $$

I am wondering: Is the generalization of this statement to crossed homomorphisms still true?

To recall, for $G$ and $\Gamma$ two groups, and in addition for

$$ \alpha \;\colon\; G \longrightarrow \mathrm{Aut}_{\mathrm{Grp}}(\Gamma) $$

an action of $G$ on $\Gamma$ by group automorphisms, a function $\phi \;\colon\; G \longrightarrow \Gamma$ is a crossed homomorphism if

$$ \underset{g_1, g_2 \in G}{\forall} \;\;\; \phi(g_1 \cdot g_2) \;=\; \phi(g_1) \cdot \alpha(g_1) \big( \phi(g_2) \big) \,. $$

Moreover, a pair $\phi$, $\phi'$ of such crossed homomorphisms is "crossed conjugate" (this terminology I seem to making up, but the definition is standard) if there exists $\gamma \in \Gamma$ such that

$$ \underset{g \in G}{\forall} \;\;\; \phi'(g) \;=\; \gamma^{-1} \cdot \phi(g) \cdot \alpha(g)(\gamma) \,. $$

So I am wondering: Are nearby crossed homomorphisms crossed conjugate?

I like to think about this as follows (by this Prop.):

The crossed-conjugation groupoid of crossed homomorphisms is isomorphic to the groupoid of (strict, i.e. $\mathrm{Grpd}$-enriched) sections of the groupoid bundle $\mathbf{B} (\Gamma \rtimes_\alpha G) \xrightarrow{ \mathbf{B}\mathrm{pr}_2 } \mathbf{B} \Gamma$ which is the groupoidal delooping of the semidirect product group projection.

In other words:

A crossed homomorphism $\phi \,\colon\, G \xrightarrow{\;} \Gamma$ is the same as an actual homomorphism of the form $\big( \phi(-),\, (-)\big) \;\colon\; G \longrightarrow \Gamma \rtimes G$; and under this identification, a crossed conjugation is just a plain conjugation with elements of the form $(\gamma, \mathrm{e})$ in $\Gamma \hookrightarrow \Gamma \rtimes G$.

In this equivalent reformulation, the question becomes:

For $G$ and $\Gamma$ Lie groups, with $G$ compact, are nearby homomorphisms of the form $\big( \phi(-),\, (-) \big) \;\colon\; G \xrightarrow{\;} \Gamma \rtimes G$ conjugate not just by any element in $\Gamma \rtimes G$, but by an element of the form $(\gamma, \mathrm{e})$ in $\Gamma \hookrightarrow \Gamma \rtimes G$?

Using just the fact that nearby plain homomorphisms out of compact Lie groups are plain conjugate, one can immediately deduce that this crossed generalization is true in the special case that the action $\alpha \,\colon\, G \to \Gamma \rtimes G$ restricts to the trivial action on the center of $G$. However, this special case does not seem to be too interesting. For example in the archetypical application to twistings of Real K-theory, $G$ is abelian.

So it looks like to make progress here one needs to go into any one of the proofs that construct the conjugation of nearby homomorphisms and see if they can be enhanced to incorporate the above "section constraint" in the case that the codomain group is a semidirect product with the domain group. I admit that I haven't really looked into these proofs yet, firing off this question first to see if anyone has thought about it.


Yes, this is true.

The space of crossed homomorphisms from $G$ to $\Gamma$ could be identified, by taking graphs, with the space $Y$ consisting of subgroups of $\Gamma\rtimes G$ which intersect $\Gamma$ trivially and project onto $G$. You ask for discreteness of $Y$ modulo the $\Gamma$-conjugation action. Every such subgroup is the image of an homomoprhism $G\to \Gamma\rtimes G$, thus may be identified as an element of the space $\text{Hom}(G,\Gamma\rtimes G)$ modulo the action of $\text{Aut}(G)$ by pre-composition. Let me write $X$ for the subspace of $\text{Hom}(G,\Gamma\rtimes G)$ which is the preimage of $Y$. We know, by local rigidity, that $\text{Hom}(G,\Gamma\rtimes G)$ is discrete mod the action of $\Gamma\rtimes G$ by post-composition via the inner action. Thus also $X$, which is a subspace $\text{Hom}(G,\Gamma\rtimes G)$, and $Y$, which is a qoutient of $X$, are discrete mod the action of $\Gamma\rtimes G$. We are thus left to show that $\Gamma$ acts transitively on each $\Gamma\rtimes G$-orbit in $Y$. Note that every element in $y\in Y$ is a subgroup of $\Gamma\rtimes G$ satisfying $\Gamma\cdot y=\Gamma\rtimes G$ and the subgroup $y$ clearly stabilizes the element $y$ by the conjugation action. Thus indeed $\Gamma$ acts transitively on the orbit of $y$.

  • $\begingroup$ Thanks for replying! Allow me to see if I am following the argument you propose: I was thinking of identifying the space of crossed homomorphisms with the subspace of actual homomorphisms $G \to \Gamma \rtimes G$ which are the identity on G. Given a pair of nearby such, the existing proof produces an element $(\gamma,g)$ which conjugates them into each other, and the problem seems to be to show that this element can be chosen to be of the form $(\gamma, e)$. Now, your point is, if I understand well, that (continued in next comment)... $\endgroup$ Sep 14 '21 at 7:55
  • $\begingroup$ Your point seems to be (correct me if I am misunderstanding) that we can just as well identify the space of crossed homomorphisms $G \to \Gamma$ with that of actual homomorphisms $G \to \Gamma \rtimes G$ which are an inner automorphism on the $G$ factor, modulo that inner automorphism action. I certainly see that. But I admit to not yet following how that solves the problem: It seems to me that this goes to show that every conjugation in $\Gamma \rtimes G$ may be understood as taking a crossed hom. to a crossed hom. -- but we need this statement for any prescribed pair of nearbys. No? $\endgroup$ Sep 14 '21 at 8:00
  • $\begingroup$ @UrsSchreiber I am sorry, but I don't understand the last sentence of your last comment. $\endgroup$
    – Uri Bader
    Sep 14 '21 at 12:45
  • $\begingroup$ If I hand you a pair of nearby homomorphisms $G \to \Gamma \rtimes G$ which are both the identity on $G$, how do you propose to construct an element $\gamma \in \Gamma$ such that conjugation with $(\gamma, e) \,\in\, \Gamma \rtimes G$ conjugates these two homomorphisms into each other? $\endgroup$ Sep 14 '21 at 12:48
  • $\begingroup$ So you want a construction, not a proof of exitance? I can give a somewhat more constructive proof, I think. I will add it as another answer when I get the time. $\endgroup$
    – Uri Bader
    Sep 14 '21 at 16:39

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