2
$\begingroup$

Given a surjective * - homomorphism $\phi: A \rightarrow B$ of C*-algebras and a projection $p\in B$, does $\phi^{-1}(p)$ always contain a projection? If untrue in general, does it hold for $A=B(H)$, $B=B(H)/K(H)$ and $\phi$ equal to the quotient map?

$\endgroup$
4
$\begingroup$

As Martin answered, its true with $B(H)$ and the Calkin algebra. Its not true in general - there is an example in 2.2.10 of Rørdam's K-theory book. The example is to is to take $A = C([0,1])$ and $B = \mathbb{C} \oplus \mathbb{C}$ and take the map $A \to B$ given by $f \mapsto (f(0),f(1))$.

Edit: An interesting case where this is true is the following (which also gives the $B(H)$ and Calkin case). If $A$ is real rank zero, then any projection in a quotient lifts to a projection in the algebra. This is lemma 3.1.13 in Farah's book. The proof is actually quite short and involves the characterization of real rank zero in terms of an approximate unit of projections.

Edited more: its true of (norm) ultraproducts $\prod_{\mathcal{U}}A_n$ and of $\prod_nA_n/\oplus_n A_n$ (the ability to lift projections to $\prod_nA_n$, that is), which can be shown directly (for any C*-algebras). I initially said one can do this alluding to the fact that $\ell^{\infty}$ has real rank zero, but I don't think thats true - although it this still gives it for $\ell^{\infty}/c_0$ (a direct argument can be done like in the ultrafilter case, or like how Martin did for $B(H)$).

$\endgroup$
1
  • $\begingroup$ Nice. $ \ \ \ $ $\endgroup$ – Martin Argerami May 8 at 3:58
4
$\begingroup$

Not sure about the general case.

In your particular case, yes, it's true. Suppose that $\phi(T)$ is a projection. Since $\phi(T^*)=\phi(T)^*=\phi(T)$, you get that $$ \phi\Big(\frac{T+T^*}2\Big)=\phi(T), $$ so you may assume without loss of generality that $T$ is selfadjoint. From $\phi(T-T^2)=0$, we get that $T-T^2$ is compact. The spectrum of a compact selfadjoint operator is either finite or a sequence that converges to zero. And $$ \sigma(T-T^2)=\{\lambda-\lambda^2:\ \lambda\in\sigma(T)\}. $$ If $t-t^2=s-s^2$, we rewrite this as $t-s=t^2-s^2=(t-s)(t+s)$. So either $s=t$ or $s=1-t$. Thus, since $\sigma(T-T^2)$ is countable, so is $\sigma(T)$. If $\lambda$ is an accumulation point for $\sigma(T)$, there exists a sequence $\{\lambda_n\}\in\sigma(T)$ with $\lambda_n\ne\lambda_m$ and $\lambda_n\to\lambda$. Then $\lambda_n-\lambda_n^2\to\lambda-\lambda^2$, so $\lambda-\lambda^2$ is an accumulation point for $\sigma(T-T^2)$ (note that infinitely many points in the sequence $\{\lambda_n-\lambda_n^2\}$ are distinct). So $\lambda-\lambda^2=0$, which forces $\lambda=0$ or $\lambda=1$. We have shown that the only two possible accumulation points of $\sigma(T)$ are $0$ and $1$.

Let $r\in (0,1)\setminus\sigma(T)$. Then $1_{(-\infty,r)}$ and $1_{(r,\infty)}$ are continuous on $\sigma(T)$. So by continuous functional calculus we write $$ T=R+I-S, $$ where $R=1_{(-\infty,r)}(T)$ and $S=I-1_{(r,\infty)}(T)$ are selfadjoint operators with $R(I-S)=0$ and with countable spectrum with only zero as a possible accumulation point. This allows us to write $$ R=\sum_k\alpha_kP_k,\qquad S=\sum_j\beta_jQ_j $$ where $\{\alpha_k\}$ and $\{\beta_j\}$ are sequences that converge to zero, and the $\{P_k\}$ and $\{Q_j\}$ are families of pairwise orthogonal projections. With some relabelling, we may write $$ R=\sum_k\alpha_kP_k+\alpha_k'P_k',\qquad S=\sum_j\beta_jQ_j+\beta_j'Q_j', $$ where the projections $P_k$ and $Q_j$ have infinite rank, and the projections $P_k'$ and $Q_j'$ have finite rank. Now $$\tag1 \phi(R)=\sum_k\alpha_k\phi(P_k),\qquad\phi(S)=\sum_j\beta_j\phi(Q_j). $$ We have $\phi(T^2)=\phi(T)$, which is $$ \phi(R^2)+\phi((I-S)^2)=\phi(R)+\phi(I-S). $$ Multiplying by $\phi(R)$ we get $\phi(R)^3=\phi(R)^2$; being a selfadjoint operator, this equality tells us that the spectrum is $\{0,1\}$ and so $\phi(R)$ is a projection. Similarly, $\phi(I-S)$ is a projection, and so is $\phi(S)$. Going back to $(1)$ we get that $\alpha_k,\beta_j\in\{0,1\}$, for all $k$ and $j$. If we now form $$ P=\sum_k\alpha_kP_k+\sum_j(1-\beta_j)Q_j, $$ this is a projection (note that $P_kQ_j=0$ for all $k,j$, this comes from $R(I-S)=0$). And $$ \phi(P)=\phi(R)+\phi(I-S)=\phi(T). $$

$\endgroup$

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.