Let $A$ and $B$ be two $C^*$-algebras, and let $p:A \to B$ be a surjective norm-decreasing $*$-homomorphism which is injective on a dense $*$-sub-algebra of $A$. Can such a map have non-trivial kernel, and if so, is it possible that the $K$-theory groups of $A$ and $B$ can be non-isomorphic?

  • 2
    $\begingroup$ All star-HMs between Cstar algebras are automatically norm-decreasing (assuming you mean what I would call contractive) $\endgroup$
    – Yemon Choi
    Mar 4 '20 at 23:54

Yes to both.$\newcommand{\Cst}{{\rm C}^*}$ The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any non-amenable discrete group will provides examples for your first question.

There are non-amenable groups $G$ for which $q$ induces an isomorphism on $K$-theory — I think the standard name for such groups is $K$-amenable. Lance proved that free groups have this property. But infinite groups with Kazhdan's Property (T) do not have this property because the so-called Kazhdan projection in the full group $\Cst$-algebra lies in the kernel of $q$. (Thanks to Jamie Gabe in comments for clarifying/sharpening my original statement.)

(See also this MO question $*$-algebras, completions, and $K$-theory )

  • 4
    $\begingroup$ It might be worth adding that the Kazhdan projection $p$ has non-trivial $K_0$-class in the kernel of $q$ since $p$ is mapped to a generating projection in $K_0(\mathbb C) = \mathbb Z$ by the trivial representation. $\endgroup$
    – Jamie Gabe
    Mar 5 '20 at 0:14

There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map.

(My first example took $A = C[0,3]$ and let $A_0$ be the set of all polynomials in $x$ with rational coefficients and $p: A \mapsto \mathbb{C}$ the point evaluation at $x = e$. Since $e$ is transcendental, $p$ is injective on $A_0$. But $A_0$ is not a $*$-subalgebra over $\mathbb{C}$.)


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.