# $K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras

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?

• All star-HMs between Cstar algebras are automatically norm-decreasing (assuming you mean what I would call contractive) 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 )
• 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. 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 $$\mathbb{C}$$.)