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Let $G$ be a Lie group and $C_r^*(G)$ and $C^*(G)$ be its reduced and maximal group $C^*$-algebras respectively. The left-regular representation of a group $G$ induces a surjective map

$$\lambda_G:C^*(G)\rightarrow C^*_r(G),$$

so $C_r^*(G)$ is a quotient of $C^*(G)$. I am wondering whether $C_r^*(G)$ is in fact a direct summand of $C^*(G)$?

Thanks for your help.

Edit: As YCor commented, more precisely what I am looking for is a $*$-homomorphism $f:C_r^*(G)\rightarrow C^*(G)$ such that $\lambda_G\circ f=\text{id}_{C_r^*(G)}$ and $f(x)y=0$ for all $x\in C_r^*(G)$ and $y\in\ker\lambda_G$. That is, a direct summand in the sense of $C^*$-algebra direct sums.

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    $\begingroup$ The statement of the question is false as such: the left regular rep induces a surjective map $\lambda_G: C^*(G)\rightarrow C^*_r(G)$. $\endgroup$
    – Alain Valette
    May 23, 2018 at 6:29
  • $\begingroup$ Thanks, you're right. I got it confused with the injection $\lambda_G:L^1(G)\hookrightarrow B(L^2(G))$ whose image is $C^*_r(G).$ I will edit the question accordingly. $\endgroup$
    – geometricK
    May 23, 2018 at 11:15
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    $\begingroup$ "is a direct summand": more precisely, are you asking whether the is a splitting of the surjection $C^*(G)\to C^*_r(G)$ by a $*$-homomorphism? $\endgroup$
    – YCor
    May 23, 2018 at 12:23
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    $\begingroup$ I don't know if it's the question, since the question is not clearly formulated (direct summand in which sense?). From your last comment I guess that you're asking about a splitting $f$ (a $*$-homomorphism $f:C^*_r(G)\to C^*(G)$ such that $\lambda_G\circ f=\mathrm{id}_{C^*_r(G)}$), such that, in addition, $f(x)y=0$ for every $x\in C^*_r(G)$ and every $y$ in the kernel of $\lambda_G$? $\endgroup$
    – YCor
    May 23, 2018 at 14:57
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    $\begingroup$ If you want to relate the K-theory of these algebras, why insist on a map that "goes the wrong way"? (There are non-amenable groups for which the canonical maps from K_*(full) to K_*(reduced) are isomorphisms, if this is the kind of result you want) $\endgroup$
    – Yemon Choi
    May 23, 2018 at 19:21

1 Answer 1

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The answer is negative. For a counter-example, consider the free group on two generators $\mathbb F_2$. It is well known that the full C*-algebra $C^*(\mathbb F_2)$ is residually finite dimensional, RFD for short (meaning that its finite dimensional representations separate points). This property is evidently hereditary in the sense that every subalgebra of a RFD C*-algebra is also RFD.

On the other hand the reduced C*-algebra $C^*_r(\mathbb F_2)$ is known to be simple, and hence it has no finite dimensional representation whatsoever.

Consequently $C^*_r(\mathbb F_2)$ is not isomorphic to a subalgebra of $C^*(\mathbb F_2)$, much less a direct summand of it.

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