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2 corrected R vs Z central extension

There's a more geometrically natural description of a central $\mathbb{Z}$-central extension in both cases.

For $\operatorname{Diff}(S^1)$, the central extension are diffeomorphisms $f: \mathbb{R} \to \mathbb{R}$ which are equivariantly periodic, i.e. $f(x+1) = f(x) + 1$.

For $\mathcal{A}$, the central extension consists of holomorphic structures on the strip $I \times \mathbb{R}$ (where $I$ is an interval) which are invariant under translation by $1$ in the $\mathbb{R}$ direction, together with an equivariant parametrization of both boundaries by $\mathbb{R}$. These are considered up to equivariant isomorphism. (If the interval $I$ has length $> 0$, then you can assume the parametrization is by the identity map by shearing your holomorphic structures. But this way makes the compatibility with $\widetilde{\operatorname{Diff}}(S^1)$ more obvious.)

In both cases the composition is clear, and it's also obvious that $\widetilde{\operatorname{Diff}}(S^1)$ acts on $\widetilde{\mathcal{A}}$.

These

Presumably these two extensions are surely isomorphic to your explicit extensions, but I'll have to work out embed in the equivalence later$\mathbb{R}$-central extension you describe.

1

There's a more geometrically natural description of a central extension in both cases.

For $\operatorname{Diff}(S^1)$, the central extension are diffeomorphisms $f: \mathbb{R} \to \mathbb{R}$ which are equivariantly periodic, i.e. $f(x+1) = f(x) + 1$.

For $\mathcal{A}$, the central extension consists of holomorphic structures on the strip $I \times \mathbb{R}$ (where $I$ is an interval) which are invariant under translation by $1$ in the $\mathbb{R}$ direction, together with an equivariant parametrization of both boundaries by $\mathbb{R}$. These are considered up to equivariant isomorphism. (If the interval $I$ has length $> 0$, then you can assume the parametrization is by the identity map by shearing your holomorphic structures. But this way makes the compatibility with $\widetilde{\operatorname{Diff}}(S^1)$ more obvious.)

In both cases the composition is clear, and it's also obvious that $\widetilde{\operatorname{Diff}}(S^1)$ acts on $\widetilde{\mathcal{A}}$.

These two extensions are surely isomorphic to your explicit extensions, but I'll have to work out the equivalence later.