I am a physicist so forgive me if this question doesn't make sense.

You can start off by defining the Witt algebra, which I'll call $\mathfrak{g}$, as the complexified Lie algebra of vector fields on $S^1$.

$$ \mathrm{Vect}(S^1) \otimes \mathbb{C}. $$

Via Fouruier analysis we can see that this Lie algebra is spanned by $$ L_n = -i e^{-i n \theta} \frac{d}{d \theta} $$ for $n \in \mathbb{Z}$. These have the commutation relations

$$ [L_n, L_m] = (n - m) L_{n+m}. $$ This can be thought of as the Lie algebra of diffeomorphisms on $S^1$.

As per this question, there is one non-zero element of $H^2(\mathfrak{g}, \mathbb{C})$, a non trivial cocycle. This allows for you to define the central extension of the Witt algebra called the Virasoro algebra.

This is the Gelfand Fuchs cocycle. For two vector fields $h_1 \frac{d}{d\theta}$ and $h_2 \frac{d}{d\theta}$ it is

$$ c(h_1 \frac{d}{d\theta}, h_2 \frac{d}{d\theta}) = \int_{S^1} h_1 h_2'''d \theta $$

The answer to the other question ends with

The relation to algebraic topology is most easily seen in the language of de-Rham cohomology: Skew symmetric multlinear maps from $\mathfrak g$ to $\mathbb C$ correspond to left invariant differential forms on any Lie group with Lie algebra $\mathfrak g$ and the Lie algebra cohomology differential $\partial$ from above is induced by the exterior derivative (which maps left invariant forms to left invariant forms).

Now I'm not sure how to actually do this, but does this mean that the Gelfand Fuchs cocycle can somehow be "integrated" over some closed two-surface in the group, and the result will be non-zero? I'm curious if this is somehow possible, and what the exact two-surface is.


I have figured what the answer is, and it is "no."

My procedure for turning a non-trivial cocycle into a closed but not exact 2-form was correct. However, it won't work in the case of $\mathrm{Vect}(S^1)$. The reason is that the Lie group of $\mathrm{Vect}(S^1)$ is $\mathrm{Diff}(S^1)$, and that is not a compact Lie group. Therefore, it's second cohomology class will not match up with the Lie algebra cocycle as in the compact Lie group case. (The main reason is that you cannot integrate a constant two form over a non-compact space and get a finite answer in the end.)

Furthermore, topologically $\mathrm{Diff}(S^1) \cong O(2)$, so there is no non-trivial two surface anyway.


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