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For a commutative Banach Algebra $A$ we can define $\Phi_A$, the character space of $A$, to be the set of all characters of $A$. (A character is a non-zero ring homomorphism $\phi:A\to \Bbb C$.)

What happens if we allow $A$ to be non-commutative?

An issue that I can think of is that the Gelfand Representation Theorem might not hold (although I am not sure if its proof relies on commutativity in any non-trivial way). Nevertheless, is there a merit in talking about, says, the character space of $\mathcal B(H)$, the algebra of bounded linear operators on $H$?

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  • $\begingroup$ As far as I can recall, a non-commutative Banach algebra there may be no (multiplicative) characters at all. I think you could take a group without any one-dimensional representations (I think any compact simple Lie group of dimension at least $2$ should work) and look at its group algebra. My representation theory is rusty at the very least, but that should give you an example. $\endgroup$
    – tomasz
    Dec 22, 2016 at 9:04
  • $\begingroup$ Unfortunately, not all Banach algebras have characters. For example, take a separable infinite-dimensional Hilbert space $H$. If $B(H)$ had a character, then its kernel would be a 1-codimensional norm-closed ideal in $B(H)$. But the only nontrivial such ideal in $B(H)$ is $K(H)$, the compact operators. So this is impossible. Thus $B(H)$ has no characters. For similar reasons, one can show that actually no $B(H)$ (no matter the size of $H$) has any characters unless $H = \mathbb{C}$. So we won't get a good Gelfand analogue. But I've outlined some options below. $\endgroup$
    – Josh Keneda
    Dec 22, 2016 at 9:22

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I'm not sure about how much of this goes through for general Banach algebras, but I can talk about the $C^*$ case.

For C$^*$-algebras (including $B(H)$), we can try to generalize the character space by using representation theory. If $A$ is an abelian C$^*$-algebra, then any character gives an irreducible representation of $A$. In fact, the character space is precisely the collection of all irreducible representations of $A$. So to generalize the character space (also sometimes called the spectrum of $A$), we gather up all of the irreducible representations of our algebra ($B(H)$ for example), and study that (after suitably topologizing it).

We can also study a related space, called the primitive spectrum of $A$, where we collect all of the kernels of irreducible representations instead of the representations themselves. Obviously there's a surjection from the spectrum of $A$ to the primitive spectrum, since we can just associate to each irreducible representation its kernel.

For commutative C$^*$-algebras, the character space, spectrum, and primitive spectrum all agree. So we really are generalizing the character space, at least in the C$^*$ setting.

One nice thing that we learn from Gelfand duality is that we can fully recover a commutative C$^*$-algebra $A$ from its spectrum, so the spectrum plays the role of a sort of dual object to $A$. Unfortunately, we can't recover C$^*$-algebras in general from their spectra: The spectrum of $K(H)$ is just a single point. In particular, we can't distinguish $M_n(\mathbb{C})$ from $M_m(\mathbb{C})$ by using their spectra.

For more, check out the wikipedia page here.

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  • $\begingroup$ Beautiful. Thank you! $\endgroup$
    – BigbearZzz
    Dec 22, 2016 at 13:38

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