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.

  • 5
    $\begingroup$ The statement of the question is false as such: the left regular rep induces a surjective map $\lambda_G: C^*(G)\rightarrow C^*_r(G)$. $\endgroup$ 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
  • 2
    $\begingroup$ "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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.