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.



Relation between maximal and reduced group $C^*$-algebras
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.


