Dear all,

The next Banach spaces webinar is on Friday September 11 9AM CDT (e.g., Dallas, TX time). Please join us at

https://unt.zoom.us/j/512907580

Speaker: Bence Horváth, Czech Academy of Sciences

Title: When are surjective algebra homomorphisms of $\mathcal{B}(X)$ automatically injective?

Abstract. A classical result of Eidelheit asserts that if $X$ and $Y$ are Banach spaces then they are isomorphic if and only if their algebras of operators $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are isomorphic as Banach algebras, in the sense that there is a continuous bijective algebra homomorphism $\psi: , \mathcal{B}(X) \rightarrow \mathcal{B}(Y)$. It is natural to ask whether for some class of Banach spaces $X$ this theorem can be strengthened in the following sense: If $Y$ is a non-zero Banach space and $\psi: \mathcal{B}(X) \rightarrow \mathcal{B}(Y)$ is a surjective algebra homomorphism, is $\psi$ automatically injective?

It is easy to see that for a ``very nice'' class Banach spaces, such as $c_0$ and $\ell_p$, where $1 \leq p < \infty$, the answer is positive. Further examples include $\ell_{\infty}$ and $( \oplus_{n=1}^{\infty} \ell_2^n )_{c_0}$ and its dual space $\left( \oplus_{n=1}^{\infty} \ell_2^n \right)_{\ell_1}$, and the arbitrarily distortable Banach space $\mathbf{S}$ constructed by Schlumprecht. In recent joint work with Tomasz Kania it was shown that ``long'' sequence spaces of the form $c_0(\lambda)$, $\ell_{\infty}^c(\lambda)$ and $\ell_p(\lambda)$ (where $1 \leq p < \infty$) also enjoy this property.

In the other direction, with the aid of a result of Kania--Koszmider--Laustsen we will show that for any separable, reflexive Banach space $X$ there is a Banach space $Y_X$ and a surjective algebra homomorphism $ \psi: , \mathcal{B}(Y_X) \rightarrow \mathcal{B}(X)$ which is not injective.

* For more information about the past and future talks, and videos please visit the webinar website http://www.math.unt.edu/~bunyamin/banach

Upcoming schedule

September 18: Chris Phillips, University of Oregon

Thank you, and best regards,

Bunyamin Sari