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Consider the following topological groups

  • $\operatorname{Homeo}(\mathbb{R}^d)$ be the topological group of all homeomorphism from $\mathbb{R}^d$ onto itself; equipped with the compact-open topology (and composition as operation).
  • $G\triangleq C(\mathbb{R}^d,\mathbb{R}^d)$ equipped with pointwise addition as group law; here $C(\mathbb{R}^d,\mathbb{R}^d)$ is equipped with the compact-open topology,
  • $L^p\triangleq L^p(\mathbb{R}^d;\mathbb{R}^d)$ is the space of Lebesgue-Bochner Integrable Functions on $\mathbb{R}^d$ with additive group-structure.

Is there a surjective continuous group homomorphism from $\operatorname{Homeo}(\mathbb{R}^d)$ onto any $L^p$ (for some $1\leq p< \infty$) or onto $G$?

This is a follow-up/variant of this question, which only requires the map to be a semi-group homomorphism.

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    $\begingroup$ This is widely referred to as the (self-)homeomorphism group of $\mathbf{R}^d$. $\endgroup$
    – YCor
    Commented Nov 20, 2019 at 10:01
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    $\begingroup$ You still need to edit... Homeo(R^d) is a subset of C(R^d,R^d), not in (natural) bijection. Also I don't see what $p$ means in $G^p$, and why you're using the same letter $G$. $\endgroup$
    – YCor
    Commented Nov 20, 2019 at 10:14
  • $\begingroup$ Sounds completely in an opposite direction to the previous question (and to the title)... yes you have continuous linear forms $L^p\to\mathbf{R}$, and then composing with 1-parameter subgroups you get nontrivial homomorphism to the given groups. I guess the question will be modified soon; don't mind checking carefully what you're asking. $\endgroup$
    – YCor
    Commented Nov 20, 2019 at 10:27
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    $\begingroup$ I guess that the group $\mathrm{Homeo}^+(\mathbf{R}^d)$ of index 2 is perfect and so the answer would be no. But it's hard to find references (except for $d=1,2$). $\endgroup$
    – YCor
    Commented Nov 20, 2019 at 10:59
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    $\begingroup$ Yes, for $d=1$ the answer is no. Indeed J. Schreier and Ulam classified the normal subgroups of $\mathrm{Homeo}(\mathbf{R})$: there are very few (the whole group, the oriented one of index 2, the compactly supported one, and the trivial subgroup) and in particular the abelianization is reduced to $C_2$. $\endgroup$
    – YCor
    Commented Nov 20, 2019 at 11:10

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