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 $LG_{\mathbb C}$?

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    $\begingroup$ 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.

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  • $\begingroup$ 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

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