The n-Category Café

so if $X$ has a String structure then the Spin bundle does not lift to a String 2-bundle??

It does. The obstructions are the same. And “String structure” means that the obstruction vanishes.

$$\Omega^\bullet( -- , \Omega^\bullet(S^1))$$

That notation is a bit inpenetrable

In fact it is a typo, too. What I meant to write was

$$\Omega^\bullet(maps( -- , \Omega^\bullet(S^1))) \,.$$

If it were exactly the internal hom, I could simply write

$$hom(--, \Omega^\bullet(S^1)) \,.$$

But we are still not sure what the best abstract-nonsense name of $$\Omega^\bullet(maps( -- , \Omega^\bullet(S^1)))$$ is, are we?

We’ve talked about that privately, but for completeness and in case any of our readers is wondering, let me summarize the situation in more detail, as reviewed and explained nicely in

Katsuhiko Kuribayashi, On the vanishing problem of String classes (pdf)

Originally (I had once collected some list of literature here) at a time when higher bundles hadn’t been much around yet, a String structure on a Spin manifold $X$

$$\array{ Spin(n) &\hookrightarrow& Q \\ && \downarrow \\ && X }$$

had been conceived as the vanishing of the obstruction of lifting the corresponding loop bundle over loop space $\Omega X$

$$\array{ \Omega Spin(n) &\hookrightarrow& \Omega Q \\ && \downarrow \\ && \Omega X }$$

through the Kac-Moody central extension

$$1 \to U(1) \to \hat \Omega Spin(n) \to \Omega Spin(n) \to 1$$

of the loop group.

The obstructing line 2-bundle (“lifting gerbe”) to this lift is classified by a degree 3-class called $\mu(Q)$ in the above text:

$$\mu(Q) \in H^3(\Omega X, \mathbb{Z}) \,.$$

One would expect that this 3-class is the transgression of a 4-class down on $X$. And that’s essentially true:

On any Spin bundle $Q \to X$ we have a “fractional Pontrjagin” four class denoteded $$\frac{1}{2}p_1(Q) \in H^4(X,\mathbb{Z})$$

and it is true that

$$(\frac{1}{2}p_1(Q) = 0) \Rightarrow (\mu(Q) = 0) \,.$$

Kuribayashi in his article addresses the question whether the converse holds: does $\mu(Q)$ always come from transgressing a four-class down on $X$ to loop space?

The answer is: yes, if $H^4(X,\mathbb{Z})$ is torsion free.

$$(H^4(X,\mathbb{Z}) torsion free and H^2(X,\mathbb{R}) \leq 1) \Rightarrow ( (\frac{1}{2}p_1(Q) = 0) \Leftarrow (\mu(Q) = 0)) \,.$$

So the vanishing of $\frac{1}{2}p_1(Q)$ is a slightly stronger condition, in general, than the vanishing of $\mu(Q)$. Except when torsion vanishes, then both are equivalent.

It’s an issue familiar throughout the study of $n$-bundles: transgressing an $n$-bundle to a space of maps with a $d$-dimensional source yields an $(n-d)$-bundle on the space of maps, but not every $(n-d)$-bundle on the space of maps will, in general, arise this way. Only those which have sufficiently high “multiplicative structure” remembering that they came from a higher structure downstairs will.

(If this sounds mysterious: the underlying mechanism is really most simple when you conceive all $n$-bundles in terms of their fiber-assigning $n$-functors and realize that transgression of these $n$-functors is just hitting them with an internal hom. I had once discussed that, with Bruce Bartlett, in a bit more detail for a simple class of cases in the entry Multiplicative structure of transgressed $n$-bundles (pdf, blog)

I think what should really be addressed as “the string class” is $\frac{1}{2}p_1(Q)$ instead of just $\mu(Q)$, even though historically it was other way round.

A discussion of the relation:

Pontrjagin class / CS 3-bundle down on $X$

$$\leftrightarrow$$

String 3-class /Kac-Moody lift obstruction up on $L X$

now appears in the list of example in section 9.3.1.

What is currently equation 488 highlights the source of the derivative terms in the loop group cocycles, i.e. the fact that a Lie algebra cocycle

$$\mu = \langle \cdot, [\cdot, \cdot]\rangle$$

turns, after transgression to loops, into the Kac-Moody 2-cocycle

$$(f,g) \mapsto \int_{S^1} \langle f, g'\rangle \,.$$