Induced map on $K_{1}$-group

Question

Let $A$ be a unital $C^{\ast}$-algebra. Any automorphism $\alpha$ of $A$ induces a map on $K_{1}(A)$ by $\alpha_{\ast}[u]=[\alpha(u)]$.

Let the automorphism $\alpha$ be inner, does it follow that $\alpha_{\ast}$ is the identity map?

I guess the answer is negative because unitary equivalence is far from homotopy of unitaries. The answer may also depend on the type of the $C^{\ast}$-algebra. Any suggestions?

Answer 1

In short: this is essentially because $K_1(A)$ is Abelian, and it is true for every inner automorphism of every $C^*$-algebra. But if one wants to be careful, there are a few things to check and recall. I will assume like you that $A$ is unitary, but that's true in general with the same proof, replacing $A$ by $\widetilde{A}$.

1) Preliminary facts: let us denote $(K_1(A),+)$. This is an Abelian group for any $C^*$-algebra $A$ and by the standard picture, we have $K_1(A)=\{[u]\,;\,u\in U_\infty(A)\}$.

For every $v,w$ in the same $U_n(A)$, we have $$ [vw]=[wv]=[v]+[w]. $$

That's a direct consequence of Whitehead's lemma. Precisely $$ \pmatrix{u&0\\0&v}=\pmatrix{u&0\\0&1}\pmatrix{0&1\\ 1&0}\pmatrix{v&0\\0&1}\pmatrix{0&1\\ 1&0}\sim_h \pmatrix{uv&0\\0&1} $$ where the rhs homotopy follows from the continuous $2\times 2$ unitary path $$ \pmatrix{\cos t&\sin t\\ \sin t&-\cos t }. $$ And conjugation by the same path on the lhs allows to exchange $u$ and $v$, whence the result.

In particular, for every unitary $u_n$ in $U_n(A)$, we have $$ 0=[1_n]=[u_nu_n^*]=[u_n]+[u_n^*]. $$

Note: with the fact above, it is easy to deduce that $K_1(A)$ is indeed Abelian. With $u\in U_n(A)$ and $v\in U_m(A)$, we can assume $m=n$ up to adding some $1$'s down the diagonal. Then $[u]+[v]=[u\oplus v]=[(uv)\oplus 1]=[(vu)\oplus 1]=[v\oplus u]=[v]+[u]$.

2) Proof: now assume $\alpha$ is an inner automorphism of $A$, that is there exists $u$ unitary in $A$ such that $\alpha(x)=uxu^*$ for every $x\in A$. Let us compute $\alpha_\ast=K_1(\alpha)$ on $K_1(A)$, that is on $[v]$ for every $v\in U_n(A)$, for every $n\geq 1$.

The key point is that for $v=(v_{ij})\in U_n(A)$, still denoting by $\alpha$ the $*$-homomorphism induced by $\alpha$ on $M_n(A)$, we have $$ \alpha(v)=\pmatrix{\alpha(v_{ij})}=\pmatrix{uv_{ij}u^*}=\mbox{diag}(u,\ldots,u)\,v\,\mbox{diag}(u^*,\ldots,u^*)=u_nvu_n^* $$ where $u_n$ is the unitary of $U_n(A)$ made of a constant diagonal of $u$, and zero elsewhere.

Therefore $$ \alpha_\ast([v])=[\alpha(v)]=[u_nvu_n^*]=[u_n]+[v]+[u_n^*]=[v]+[u_n]+[u_n^*]=[v]. $$ Thus $\alpha_\ast=K_1(\alpha)$ is the identity on $K_1(A)$.

Of course, it suffices to use the property $[vw]=[wv]$ for $v,w\in U_n(A)$ to get $[u_nvu_n^*]=[u_n^*u_nv]=[v]$. But it was a good opportunity to recall that for a unitary $u_n\in U_n(A)$, $-[u_n]=[u_n^*]$.

3) Extension: more generally, for every approximately inner automorphism of $A$, we have $K_0(\alpha)=\mbox{id}$ and $K_1(\alpha)=\mbox{id}$.

Answer 2

If I may add a little to the above answer, $K_1(A)$ is by definition the group of homotopy classes of unitaries in the unitization $(K \otimes A)^{\sim}$ of its stabilization $K \otimes A$ (replace $A$ by $A^{\sim}$ in the non-unital case as julien stated above.)

Polar decomposition shows that every unitary in $(K \otimes A)^{\sim}$ is homotopic to one of the form $u \oplus (1 - 1_{M_n(A)} ) $ where $u \in M_n(A)$.

Also, for any unitary $u$ in a corner $p (K \otimes A)^{\sim} p$ and for any partial isometry $s$ with $s^*s = p$, $u \oplus (1-p)$ is homotopic to $sus^* \oplus (1-ss^*)$.

So every $K_1$ class can be represented by some unitary $u \in M_n(A)$ after adding $1$'s down the diagonal and you can move a unitary $u \in M_n(A)$ up and down the diagonal without changing its $K_1$-class. It is immediate that $K_1$ is abelian and inner-automorphisms descend to the identity map on $K_1$.