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$\mathit{Diff}(S^1)$ refers to the group of orientation preserving diffeomorphisms of the circle. The semigroup of annuli $\mathcal A$ is its "complexification": the elements of $\mathcal A$ are isomorphism classes of annulus-shaped Riemann surfaces, with parametrized boundary.

Both $\mathit{Diff}(S^1)$ and $\mathcal A$ have central extensions by $\mathbb R$, and my question is about their relationship.


♦ The group $\mathit{Diff}(S^1)$ carries the so-called Bott-Virasoro cocycle, which is given by $$ B(f,g) = \int_{S^1}\ln(f'\circ g)\;\; d\;\ln(g'). $$ The corresponding centrally extended group is $\widetilde{\mathit{Diff}(S^1)}:=\mathit{Diff}(S^1)\times \mathbb R$, with product given by $(f,a)\cdot(g,b):=(f\circ g,a+b+B(f,g))$.

♦ The elements of the central extension $\widetilde{\mathcal A}$ of $\mathcal A$ have a very different description.
An element $\widetilde{\Sigma}\in\widetilde{\mathcal A}$ sitting above $\Sigma\in\mathcal A$ is an equivalence class of pairs $(g,a)$, where $g$ is a Riemannian metric on $\Sigma$ compatible with the complex structure, and $a\in\mathbb R$. There's the extra requirement that the boundary circles of $\Sigma$ be constant speed geodesics for $g$.

The equivalence relation involves the Liouville functional: one declares $(g_1,a_1)\sim (g_2,a_2)$ if $g_2=e^{2\varphi}g_1$, and $$ a_2-a_1=\int_\Sigma {\textstyle\frac 1 2}(d\varphi\wedge \ast d\varphi+4\varphi R), $$ where $R$ is the curvature 2-form of the metric $g_1$.


It is reasonable to believe that the restriction of the central extension $\widetilde{\mathcal A}$ to the "subgroup" $\mathit{Diff}(S^1)\subset \mathcal A$ is $\widetilde{\mathit{Diff}(S^1)}$. But I really don't see why that's should be the case.

Any insight? How does one relate the Bott-Virasoro cocycle to the Liouville functional??

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One reason these look different is that the "subgroup" $\widetilde{\operatorname{Diff}}(S^1)$ looks a little odd in $\widetilde{\mathcal{A}}$. In particular, you'd like to take a thin annulus with boundaries parametrized in two different ways. But you require that the boundary be parametrized by constant-speed geodesics, which requires you to blow up the metric near the boundary by a conformal factor $\phi$ depending on the speed of the parametrization. –  Dylan Thurston Apr 14 '11 at 10:12
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3 Answers

There's a very geometric interpretation of the Virasoro-Bott cocycle in terms of projective structures on Riemann surfaces, which I hope should relate directly to the Liouville functional.

Namely to describe a 1-dimensional central extension of a Lie algebra it's equivalent to give an affine space over the dual of the Lie algebra with a compatible "coadjoint" action of the Lie algebra. (This models the hyperplane in the dual to the central extension, consisting of functionals taking value one on the central element.)

In the case of vector fields on the circle (I'm thinking of the Laurent series model, but shouldn't be hard to translate to smooth functions) the dual to the complexified Lie algebra is canonically the space of quadratic differentials (with Laurent coefficients) with its canonical Diff S^1 action by rotation. There's a canonical affine space over quadratic differentials, namely the space of projective structures (ie atlases into projective space with Mobius transitions). The cocycle describing this affine space (ie the transformation property of projective structures under general, not just Mobius, changes of coordinates) is the Schwarzian derivative, which translates directly into the Virasoro-Bott cocycle when you write the corresponding central extension (this is all explained in detail eg in my book with Frenkel, Vertex Algebras on Algebraic Curves).

In any case the Liouville functional relates to the conformal factor you need to change your given metric to one with constant curvature, so should have a natural geometric relation with projective structures, but I don't see it right now..

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One major issue in trying to compare the Bott-Virasoro story to the Liouville story is that the former doesn't make sense for annuli with non-zero thickness, while the latter doesn't make sense for annuli with zero thickness. Does your projective structure story bridge that gap? I have the feeling that your answer is entirely at the Lie algebra level... The last paragraph is interesting though: could you elaborate what you mean by it? –  André Henriques Apr 14 '11 at 20:10
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There's a more geometrically natural description of a $\mathbb{Z}$-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 $\widetilde{\mathcal{A}}$.

Presumably these two extensions embed in the $\mathbb{R}$-central extension you describe.

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I spent some time trying to find an explicit map to the $\mathbb{R}$ extension, without luck so far. It's a bit mysterious. –  Dylan Thurston Apr 14 '11 at 9:04
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The center of the the universal central extension of $\mathrm{Diff}(S^1)$ is $\mathbb Z\times\mathbb R$ (the $\mathbb Z$ has to do with "minimal energy", while the $\mathbb R$ has to do with "central charge"). So your answer is orthogonal to my question. –  André Henriques Apr 14 '11 at 10:23
    
@André: Thanks! I think I've been confused about this for years. –  Dylan Thurston Apr 14 '11 at 11:34
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Let me give a method for answering the problem. I haven't yet done the relevant integral to get an actual answer.

In the setting you originally laid out for $\mathcal{A}$, actual diffeomorphisms are not included; however, you can find an annulus close to a diffeomorphism $f$ by taking an annulus of the form $[0,\epsilon] \times S^1$ with $\epsilon$ very small, and parametrizing the two sides differently: parametrize the left boundary by $x \mapsto (0,f(x))$ and parametrize the right boundary by $x \mapsto (1,x)$. Call this parametrized annulus $A_0(f)$.

$A_0(f)$ is not of the type that you apply the Liouville functional to, since the left boundary is not constant-speed geodesic. So let's construct another metrized torus where both boundaries are constant-speed. Let $B_\epsilon:[0,\epsilon] \to [0,1]$ be a bump function which is $1$ in a neighborhood of $0$ and $0$ in a neighborhood of $\epsilon$. Let $A_1(f)$ be the annulus $A_0(f)$ with the metric rescaled as $$ g_1(s,t) = \exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)g_0(s,t) = \exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)(ds^2 + dt^2). $$ Then $A_1(f)$, with the same boundary parametrization as in $A_0(f)$, is now parametrized by constant-speed geodesics. Call $(A_1(f), 0) \in \widetilde{\mathcal{A}}$ the canonical annulus of the diffeomorphism $f$.

Now suppose we have two diffeomorphisms $f,g$. We can glue the associated cannonical annuli $A_1(f), A_1(g)$. The result will represent the diffeomorphism $g \circ f$; but the metric will not be the canonical metric on $A_1(g \circ f)$. We have to rescale the metric by an appropriate scaling factor $\phi$ to get to the canonical metric. (Actually, the conformal width of the annulus is also now $2\epsilon$ rather than $\epsilon$, but that shouldn't matter.) The integral in the problem statement for this rescaling presumably reduces to the Bott-Virasoro integral as $\epsilon$ gets sufficiently small.

I'd still like to understand this geometrically well enough to not have to do the integral.

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