Let $LG=\operatorname{Maps}(S^1,G)$ be the loop group of a compact Lie group $G$. I should add some adjectives to $G$, but for sake of simplicity let's just take $G=SU(2)$.

There is a central extension $$1\to S^1\to\widetilde{LG}\to LG\to 1$$ (these are classified by "level" in $H^3(G)$, but we may as well restrict attention to the "universal" such extension corresponding to a generator of this group). The constructions of this central extension that I have found so far (e.g. the one in Pressley--Segal) all go via first defining a closed $2$-form on $LG$, arguing it defines a unique $S^1$-bundle, and then putting a group structure on this bundle.

Is there a more intrinsic definition of $\widetilde{LG}$?

In other words, given a loop $\gamma:S^1\to G$, I would like to have an intrinsically defined principal $S^1$ homogeneous space (or, equivalently, a $1$-dimensional complex vector space).

For example, here is an answer "up to homotopy". Since $\pi_1(G)=\pi_2(G)=0$ and $\pi_3(G)=\mathbb Z$, given any loop $\gamma:S^1\to G$, the space of extensions $\bar\gamma:D^2\to G$ (i.e. $\bar\gamma|_{\partial D^2=S^1}=\gamma$) is homotopy equivalent to $\Omega^2G$ which is connected with fundamental group $\mathbb Z$. If we take the $1$-truncation of this space (add cells to kill all higher homotopy groups), we get $S^1$ (up to homotopy).

This gives an "intrinsically defined" space homotopy equivalent to $S^1$ defined in terms of a given loop $\gamma:S^1\to G$ (although it's not very explicit, and has questionable meaning/use). What about an honest one-dimensional complex vector space? (with a natural meaning). Even better, can we define intrinsically a holomorphic line bundle over the complexified loop group $LG_{\mathbb C}$?

  • 2
    $\begingroup$ It sounds like you are looking for the determinant line bundle on the affine Grassmannian $LG/LG_+$ or its pullback to the loop group? A related question is mathoverflow.net/questions/24845/… , and the latest word (maybe) is arxiv.org/abs/1804.02567 $\endgroup$ – David Ben-Zvi Oct 9 '18 at 19:04
  • $\begingroup$ or if you prefer a less algebraic version, the determinant bundle on the Grassmannian is the pullback from the line bundle defining the Plucker embedding of the Segal-Wilson Grassmannian, i.e. the construction of semi-infinite wedge powers $\endgroup$ – David Ben-Zvi Oct 9 '18 at 19:09

Fix a cocycle $\omega\in C^3(G, \mathbb R)$ such that $\omega(H_3(G, \mathbb Z)) = \mathbb Z$. (For $G = SU(2)$, we can take $\omega$ to be the standard volume form.) Fix a loop $L$ in $G$, and let $\partial^{-1}(L) \subset C_2(G, \mathbb Z)$ denote the set of 2-chains whose boundary is $L$. For $x,y \in \partial^{-1}(L)$, define an equivalence relation $x \sim y$ if $\omega(\partial^{-1}(x - y)) \in \mathbb Z$. Then $\partial^{-1}(L)/\sim$ is an $S^1$ torsor.

The above paragraph defines an $S^1$-bundle over $LG$, but it doesn't tell you how to multiply two elements of this bundle. For this, we use the fact that $\pi_2(G)$ is trivial and redefine $\partial^{-1}(L)$ to be maps of $D^2$ into $G$ which restrict to $L$ on the boundary. We can multiply the maps $D^2 \to G$ pointwise. This group structure will extend to the quotient if we choose $\omega$ to be invariant under left and right multiplication by elements of $G$. (I have not thought about this last claim as carefully as I should, so be skeptical here.)

Assuming I did not make a stupid error in the previous paragraph (time constraints!), we can summarize as follows: Elements of $\widetilde{LG}$ are represented by maps to $D^2$ to $G$, and two such maps are considered equivalent if the $\omega$-volume they cobound is an integer.

  • $\begingroup$ Indeed, that's a better way of saying what I said in the second to last paragraph of my question. $\endgroup$ – John Pardon Oct 9 '18 at 18:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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