Why does a unitary in the Calkin algebra always lift to an (co-)isometry?

Question

Cf. the the title, consider a separable infinite-dimensional Hilbert space H, and the short exact sequence $$0 \to \mathcal{K}(H) \to \mathcal{B}(H) \to \mathcal{Q}(H) \to 0,$$where $\mathcal{K}(H)$ is the compact operators and $\mathcal{Q}(H)$ is the quotient, also known as the Calkin algebra.

Let $u$ be a unitary in $\mathcal{Q}(H)$. This means that $u = T + \mathcal{K}$, for a $T \in \mathcal{B}(H)$ where the differences $TT^* - I$ and $T^*T - I$ are compact. How can one show that $u$ lifts to an isometry, or a co-isometry? That is, why is there an isometry (or co-isometry) $S \in \mathcal{B}(H)$ such that $\pi (S) = u$, where $\pi$ denotes the quotient mapping? (Equivalently $S - T$ is compact)

This is an exercise in Rørdam's book on K-theory. In the previous part of the exercise one shows that whenever $E$ and $F$ are projections in $\mathcal{B}(H)$ with $Rank(E) \leq Rank(F)$ there exists a partial isometry $V$ with $V^*V = E$ and $VV^* \leq F$. I am assuming that one should use this somehow, possibly together with some results about the index map $\delta_1$, but I am really stuck on this one. Any help is appreciated!

Answer 1

Here is an argument that does not appeal to index theory and uses the suggestion in the book.

Let $T$ such that $\pi(T)=u$. Write $T=V|T|$ for the polar decomposition of $T$. Then, $V$ is a partial isometry, i.e. $V^*V$ and $VV^*$ are projections.

Since $\pi(T^*T)=\pi(I)$ and $\pi$ is a $*$-homomorphism, $\pi(|T|)=\pi ((T^*T)^{1/2})=\pi(I)$. It follows that $\pi(V)=\pi(T)$.

We have that $\pi (V) $ is a unitary and $V^*V$ is a projection, so $I-V^*V$ is a compact projection; thus, finite-rank. Similarly with $I-VV^*$.

Now, if $V^*V=I$ or $VV^*=I$, then $V$ is, respectively, an isometry or a co-isometry. If $I-V^*V$ and $I-VV^*$ are both nonzero, we have two cases:

• $\dim\text{Rank}(I-V^*V)\leq\dim\text{Rank}(I-VV^*)$. By the previous exercise in the book, let $W$ be a partial isometry with $W^*W=I-V^*V$ and $WW^*\leq I-VV^*$. By conjugating the inequality with $V^*$, we get $$ 0=V^*(I-VV^*)V=V^*WW^*V. $$ It follows that $W^*V=0$. Then $V+W$ is an isometry, since $$ (V+W)^*(V+W)=V^*V+W^*W+W^*V+V^*W=V^*V+W^*W=I. $$

• $\dim\text{Rank}(I-V^*V)\geq\dim\text{Rank}(I-VV^*)$. Similar to the previous case, now the isometry will be $V^*+W^*$.

Finally, note that since $W^*W$ is finite rank, $$0=\pi(W^*W)=\pi(W)^*\pi(W),$$ so $\pi(W)=0$. Then $$ \pi(V+W)=\pi(V)=\pi(T)=u. $$

Answer 2

If $u \in Q(H)$ is unitary, let $T_1 \in B(H)$ such that $\pi(T_1) = u$, then consider $$ T = T_1R^{-n} $$ where $R$ denotes the left or right shift (which exists on your separable infinite dimensional Hilbert space) so that $T$ has index zero. Now $\exists S_1 \in GL(B(H))$ such that $$ T-S_1 \in K(H) $$ Hence, $S:=S_1R^n$ satisfies $$ T_1 -S\in K(H) $$ and is an isometry or co-isometry.