Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can a representation of a Banach algebra be unbounded?

To clarify, I'm not asking about a representation as unbounded operators, but rather a homomorphism $\pi: A \to B(H)$ for some Hilbert space $H$, with the property that $\sup_{a \neq 0} \frac{\|\pi(a)\|}{\|a\|} = \infty$.

This question is inspired by chapter 2.5 of Arveson's Spectral Theory, which proves that every representation of a Banach *-algebra is contractive. This raises the question of what happens when a Banach algebra does not have an (isometric) involution, or when a homomorphism does not respect it.

I know of examples of unbounded homomorphisms of Banach algebras (the simplest: if $E$ is an infinite-dimensional Banach space, turn both $E$ and $\mathbb{C}$ into Banach algebras by defining all products to be zero, and any unbounded functional on $E$ will become an unbounded homomorphism), but none in which the codomain is $B(H)$.

share|improve this question
    
Perhaps I am confused: isn't your example of an unbounded homomorphism from $E$ to $\mathbb{C}$ also an example of the kind you want? Can't you compose your homomorphism with the isometric embedding $\mathbb{C}\to B(H)$ obtained by sending $1$ to the identity operator? –  Dima Shlyakhtenko Aug 10 '11 at 1:03
    
This imbedding won't be a homomorphism because $\mathbb C$ was endowed with an unusual product rule in this construction. –  fedja Aug 10 '11 at 3:15
4  
@fedja: thanks, you a right -- but then if you pick a nonzero operator $T\in B(H)$ with $T^2=0$, the map from $\mathbb{C}$ with zero multiplication sending $1$ to $T$ is a homomorphism? Then what I wrote works fine. –  Dima Shlyakhtenko Aug 10 '11 at 4:45
1  
Interesting - I wonder what can be done in the semisimple case? –  Ollie Margetts Aug 10 '11 at 13:36
1  
This kind of question is sometimes referred to as an "automatic continuity" problem, since one is asking for conditions on a domain or codomain algebra ensuring that a homomorphism between Banach algebras is automatically continuous. Off the top of my head, I don't recall results where the codomain is $B(H)$, but will think about this some more –  Yemon Choi Aug 10 '11 at 22:42

1 Answer 1

up vote 2 down vote accepted

As Dima Shlyakhtenko indicated, there are examples, obtained for example by modifying your zero product example to $x\mapsto\begin{bmatrix}0&f(x)\\\ 0&0\end{bmatrix}$ for some unbounded linear functional $f$.

A generalization of the result you mentioned is found in Theorem 4.1.20 of Rickart's General theory of Banach algebras. A special case of that theorem says that if the range of the representation is a $*$-subalgebra of $B(H)$, then the representation is automatically continuous.

What looks like a good reference is Dales's Banach algebras and automatic continuity, but I don't currently have a copy. It is referenced in an Encyclopaedia of mathematics article by the same author called "Automatic continuity for Banach algebras," which also gives some other references that might help. According to the article, there are some Banach spaces $E$ such that every homomorphism from $B(E)$ to another Banach algebra is continuous. (It doesn't say which ones.)

share|improve this answer
    
Thanks for the reference! Dales's book doesn't have many results for representations, but here are some items of interest: --Thm 5.1.4: Every simple representation (no invariant subspaces) of a Banach algebra on a normed space is continuous. --Cor 5.4.12: Homomorphisms from $B(E)$ are continuous if $E \simeq E \oplus E$. --Many basic questions are still open, like automatic continuity of homomorphisms from $C^*$-algebras to Banach algebras if the codomain is another $C^*$-algebra or the map is surjective. --Automatic continuity of maps from $C(\Omega)$ depends on the continuum hypothesis. –  Dave Gaebler Sep 20 '11 at 18:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.