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It is well-known that the character space, (i.e. the set of multiplicative characters) of a commutative, unital Banach algebra is non-empty.

But is there a complete characterization of when exactly the character space of Banach algebra is empty? Or even an incomplete characterization (ie certain classes of Banach algebras which we know have empty character space)?

Edit: I’d appreciate any responses, especially those with links for further reading!

TuringTester69
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    The kernel of a character is an ideal, so certainly simple Banach algebras have empty character space. – Aweygan Dec 02 '20 at 00:26
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    Characters of non-commutative algebras are not much studied because often there are none. The main source of counter examples is simple algebras, such as matrix algebras. Instead of characters people are often interested in irreducible representations which, for commutative *-algebras over the complexes are all one-dimensional (hence characters). For literature, look for spectral theory of non-commutative algebras. – Ruy Dec 02 '20 at 00:30

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Given a Banach algebra $B$ let $[B, B]$ (the commutator ideal) be the intersection of all kernels of morphisms of Banach algebras $B \to A$ where $A$ is commutative. This is a closed two-sided ideal of $B$, and the quotient $B/[B, B]$ is the abelianization of $B$; the universal commutative Banach algebra to which $B$ maps.

(Probably $[B, B]$ is the closure of the ideal generated by commutators $[b_1, b_2]$ but this construction of it manifestly has the appropriate universal property so it doesn't matter too much either way.)

Any character of $B$ must be a character of its abelianization, so the problem for arbitrary Banach algebras reduces immediately to the commutative case. A nonzero commutative Banach algebra always admits a character (by the Gelfand-Mazur theorem), so we get:

Proposition: $B$ admits a character iff $B/[B, B]$ is nonzero.

It's worth noting that $B/[B, B]$ can in fact be zero, for example if $B$ is simple and noncommutative (e.g. $B = M_n(\mathbb{C})$) as noted in the comments.

Qiaochu Yuan
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  • Good point @Qiaochu! It might be interesting to add that some people view $[B,B]$ as the closed linear span of (as opposed to closed ideal generated by) the commutators $[b_1,b_2] = b_1b_2 - b_2b_1$. With this definition one may easily prove that $B$ has a trace iff $[B,B]$ is a proper subspace, which is the case of $M_n(\mathbb C)$. However certain algebras have no trace (e.g. the compacts) so $[B,B]$ is sometimes equal to $B$. – Ruy Dec 02 '20 at 02:50
  • Thank you for the details. Do you know of a source where I can read more about the abelianization of Banach algebras? I’m not familiar with the concept. – TuringTester69 Dec 02 '20 at 04:41
  • @Turing: I'm not aware of a source; it doesn't appear to be a term used in the Banach algebra literature (perhaps because the abelianization often just isn't very interesting). Formally the existence of such a thing follows more or less from general categorical nonsense; it's formally analogous to the abelianization of a group or an ordinary ring (which is also often zero). But "commutator ideal" is a term people use so that might be a useful search term: https://www.google.com/search?q=%22banach+algebra%22+%22commutator+ideal%22 – Qiaochu Yuan Dec 02 '20 at 05:43
  • @QiaochuYuan thanks for your help. Just one more question. It's not obvious to me why a character in B must be a character in the abelianization (and vice versa). Could you clarify why that is the case or point me in the right direction? – TuringTester69 Dec 04 '20 at 17:56
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    @Turing: a character is a morphism $B \to \mathbb{C}$ of Banach algebras, and $\mathbb{C}$ is commutative, so the kernel of a character always contains $[B, B]$. (This is a little more concrete to see if you define $[B, B]$ to be the closed ideal generated by commutators so maybe I should've done it that way.) – Qiaochu Yuan Dec 04 '20 at 20:16