The n-Category Café

The prospect of a wolf under the bed is also nasty, but I can live with the prospect as long as nobody actually exhibits such a wolf!

If anyone gives up and wants to know the answer to my puzzle, they can click here.

The always infallible Wikipedia has a footnote claiming, without proof or reference, that the axiom of choice is needed to show that the dual of an infinite-dimensional vector space is nonzero.

However, let it be known that all my previous comments on this blog entry concerned Banach spaces, and when I said ‘dual’ I meant the space of continuous linear functionals. Is the statement that every infinite-dimensional Banach space has a nontrivial dual in this sense equivalent to the axiom of choice? This is a different question than your question about vector spaces.

I’d bet a doughnut the answer is ‘yes’. But I don’t know a reference, not even a reference to a footnote in a Wikipedia article.

Incidentally, in an appropriate sense, your two suggestions are almost the same. In (some parts of) Banach space theory, people would say that your first example is just a noncommutative version of the second, in the same sense that the Schatten classes are noncommutative versions of $\ell^p$ spaces. This is because the space of bounded operators on a countable-dimensional Hilbert space is the space of operators whose sequence of singular values lies in $\ell^\infty$, and the compact operators are those whose singular values converge to $0$.

I hadn’t heard of Banach limits.

For anyone too lazy to click: a Banach limit is a recipe for taking the limit of any bounded sequence of reals. (I’ll say this for the reals rather than the complex numbers, because I only know the definition in the real case.) Formally, a Banach limit is a continuous linear map $\phi: \ell^\infty \to \mathbf{R}$ such that

• if $x$ converges then $\phi(x) = \lim_{n\to\infty} x_n$
• $\phi(x_2, x_3, \ldots) = \phi(x_1, x_2, \ldots)$
• $\phi(x) \geq 0$ if $x_n \geq 0$ for all $n$.

Elsewhere on this thread it’s been mentioned that without some form of the axiom of choice, it’s impossible to construct any continuous linear map $\ell^\infty \to \mathbf{R}$ apart from those of the form $x \mapsto \sum_n c_n x_n$ where $c \in \ell^1$. It’s easy to see that no element of $\ell^1$ gives a Banach limit. Thus, the existence of Banach limits depends on the axiom of choice.

It struck me that this was very similar to the situation in a post of mine a few months ago: means. For example, there’s no nontrivial recipe for taking the “mean” of binary digits without using the axiom of choice. (In standard terminology: the existence of a nonprincipal ultrafilter on an infinite set depends on the axiom of choice.) Similarly, to prove the amenability of any infinite abelian group, you need the axiom of choice.

A Banach limit is a kind of mean. For example, if $x$ is a genuinely convergent sequence, you might think of its limit as a kind of “average of $x_1, x_2, \ldots$, ignoring any finite segment at the beginning”. In the means post, I also talked about Arrow’s Theorem; and you might also think of a Banach limit as a voting system, whereby $x_1, x_2, \ldots$ get to vote for what their limit should be.

Is that also true for $(L^\infty(\mathbb{R}))^* - L^1(\mathbb{R}))$?

I think so. I believe that the story is the same for $L^p(X,\mu)$ whenever $L^1(X,\mu)$ is infinite dimensional and separable (which can of course easily be characterized by simple properties of the measure space $(X,\mu)$, but I’m too busy/lazy to remember/reproduce those right now). The details are probably in Schechter’s book, which is cited in the MathOverflow post I referred to, and which I don’t have a copy of to check.

“Always” and “never” are overstatements, as evidenced by the fact that the space of convergent sequences also has a standard name: $c$. But $c_0$ certainly is discussed more often, partly because it shows up more in the wild (as for example here), partly because it fits more simply into the family of other classical spaces ($c_0^* \cong \ell^1$, with the duality given in the familiar way), and partly because from a Banach space geometric point of view $c$ and $c_0$ are essentially the same thing anyway ($c$ and $c_0$ are isomorphic, and $c_0$ is a 1-codimensional 1-complemented subspace of $c$).

For what it’s worth, $c_0$ is a commutative C*-algebra without unit, and $c$ — I’d never heard that name, but it’s nice! — is the corresponding C*-algebra with unit.

If someone hands you a C*-algebra without unit, say $A_0$, you can give it a unit by forming

$$A = A_0 \oplus \mathbb{C}$$

and giving $A$ a multiplication in the more or less obvious way, where $A_0$ has the same multiplication as before, but $1 \in \mathbb{C}$ is decreed to be the multiplicative unit.

In the case at hand,

$$c = c_0 \oplus \mathbb{C}$$

means that every convergent sequence can be uniquely written as a sequence that converges to zero, plus a constant sequence.

Now, a commutative $C^*$-algebra with unit is always isomorphic to the algebra of all continuous functions on some compact Hausdorff space.

Puzzle: what is this space for $c$?

Puzzle: what is this space for $\ell^\infty$?

These puzzles should be familiar to people who have studied this stuff a bunch, but they might be fun for people who haven’t.