# Continuous semigroup homomorphism of composition to additive structure

Let $$G$$ be the topological semigroup whose underlying space is $$C(\mathbb{R}^d,\mathbb{R}^d)$$ equipped with composition as semigroup operation and let $$H$$ be the topological group whose underlying space is $$C(\mathbb{R}^d,\mathbb{R}^d)$$ equipped with pointwise addition as group law; here $$C(\mathbb{R}^d,\mathbb{R}^d)$$ is equipped with the compact-open topology.

Is there a continuous (non-constant) semigroup homomorphism from $$G$$ to $$H$$?

• Nitpick: presumably you wish to exclude the trivial homomorphism $G \to \{0\}$ – Yemon Choi Nov 16 at 20:07
• Of course. I'll add the point. – AIM_BLB Nov 16 at 20:10
• You probably meant to take $G$ as the invertible functions on $\mathbb R^d$? – Christian Remling Nov 16 at 20:14
• @Christian Remling I'm looking for semi-group homomorphisms so there is no need for them to be invertible – AIM_BLB Nov 16 at 22:01

The new version, for semigroups, is much easier: there is no such homomorphism $$\varphi$$, for purely algebraic reasons. Consider just the constant functions $$c$$. Since $$cd=c$$ in the semigroup, you must map $$\varphi(d)=0$$. But for any $$f$$, $$fc$$ is also constant ($$=f(c)$$), so $$\varphi(f)+\varphi(c)=0$$ and thus $$\varphi(f)=0$$ as well.
• True, this is really obvious now. Though I feel that the group-theoretic version is not so simple.. where $G\subseteq C(\mathbb{R}^d;\mathbb{R}^d)$ comprised of invertible functions. – AIM_BLB Nov 16 at 23:33
• I don't see the need for the last sentence of this answer. You've already proved $\varphi(d)=0$ for all $d$ (not only constant $d$, because $cd=c$ for all $d$). – Andreas Blass Nov 17 at 0:12
• @AIM_BLB For invertible functions, consider $f(x) = -x \in C(\mathbb{R}^d, \mathbb{R}^d)$. Then $f \circ f = Id$, but there are no torsion points in the additive group. – user44191 Nov 17 at 0:19
• @user44191: I don't think it's that easy. This just shows that $f(x)=-x$ will have to be sent to $id=0$ in $H$. (Of course, there are lots of other elements of finite order for which this conclusion holds if $d>1$, as well as a large commutator subgroup that will have to get mapped to $0$, but I still don't think things are very obvious.) – Christian Remling Nov 17 at 0:55