User pavel safronov - MathOverflow

Answer by Pavel Safronov for Are rational sections of a vector bundle useful?

2013-01-17T15:40:12Z

One can recover a bundle on a Riemann surface completely given a rational frame. This goes under the name of Weil uniformization of the moduli of bundles. More precisely, there is a map from the moduli of $G$-bundles on $X$ with a rational frame to $$\prod_{x\in X}' G(\mathbf{C}((t_x)))/G(\mathbf{C}[[t_x]])$$ by sending a section to its Taylor expansion around the zeros and poles. This map is an isomorphism and the moduli space of bundles is obtained by forgetting the frame, i.e. $$\mathrm{Bun}_G(X)\cong G(\mathbf{C}(X))\backslash\prod_{x\in X}' G(\mathbf{C}((t_x)))/G(\mathbf{C}[[t_x]]),$$ where $G(\mathbf{C}(X))$ is the group of rational maps from $X$ to $G$.

I don't think you can get much information from a single section: let's say $E = L_1\oplus L_2$ is a sum of two line bundles. Suppose the section $t$ of $E$ comes from a section of $L_1$, which allows you to compute $c_1(L_1)$. Then neither $c_1(E)=c_1(L_1)+c_2(L_2)$ nor $c_2(E)=c_1(L_1)c_1(L_2)$ are determined by it.

Answer by Pavel Safronov for Identifying $T^* Bun_G$ with Higgs bundles

2012-12-18T19:53:37Z

The tangent complex to $Bun_G(C)$ can be identified with $T_{Bun_G(C)}=\mathbf{R}\pi_*\mathrm{ad}\ P[1]$, where $\pi:Bun_G(C)\times C\rightarrow Bun_G(C)$ is the natural projection and $P$ is the universal bundle.

Then the cotangent stack is $T^* Bun_G(C) = Spec Sym (T_{Bun_G(C)})$. Maps from $U$ into the total space of the bundle $T^* Bun_G(C)\rightarrow Bun_G(C)$ are the same as maps $U\rightarrow Bun_G(C)$ together with a section of the dual sheaf of $T_{Bun_G(C)}$. Relative Serre duality identifies $\mathcal{Hom}(T_{Bun_G(C)}, \mathcal{O})$ with $\mathbf{R}\pi_*\mathcal{Hom}(\mathrm{ad}\ P, \omega_C)\cong\mathbf{R}\pi_*(\mathrm{ad}\ P\otimes\omega_C)$ using the Killing form.

So, maps $U\rightarrow H^0(T^* Bun_G(C))$ to the underlying ordinary stack are identified with $G$-bundles over $U\times C$ and a section $\phi\in H^0(U\times C, \mathrm{ad} P\otimes \omega_C)$.

Answer by Pavel Safronov for canonical bundle on the stack of G bundles on a curve

2012-09-21T16:57:06Z

The tangent complex is a complex of quasi-coherent sheaves on $Bun_G(X)$, i.e. a compatible family of complexes of quasi-coherent sheaves on every affine scheme $f:U\rightarrow Bun_G(X)$ (i.e. a $G$-bundle $P_0\rightarrow X\times U$) mapping smoothly.

The global sections $\Gamma(U, f^* T_{Bun_G(X)})$ of the tangent complex over $U$ is the complex associated to the Picard groupoid given by the fiber of $Hom(D\times U, Bun_G(X))\rightarrow Hom(\mathrm{pt}\times U, Bun_G(X))$ , where $D=\mathrm{Spec}\ \mathbf{C}[\epsilon]/\epsilon^2$ . Since this is a Picard groupoid, $\pi_0$ and $\pi_1$ are abelian groups, which are $H^0$ and $H^{-1}$ of the tangent complex.

One can explicitly describe this groupoid as follows: its objects are $G$-bundles $P\rightarrow X\times U\times D$ together with an isomorphism $P|_{X\times U\times\mathrm{pt}}\cong P_0$ on $X\times U$.

$\pi_1$ computed at the trivial bundle $P=P_0\times D$ is the group of $G$-equivariant automorphisms $P\rightarrow P$ which commute with the projection to $X\times U\times D$. This is precisely the space of vertical vector fields $\pi_1=H^0(X\times U, \mathrm{ad}\ P_0)$ .

The computation of $\pi_0$ is trickier, let me just write down the answer: $\pi_0=H^1(X\times U, \mathrm{ad}\ P_0)$ . The derivation relies on the fact that there are no deformations over an affine scheme, so you just have to pass to an affine cover of $X\times U$ (see Sam Raskin's notes for details).

So, on each affine $U$ mapping to $Bun_G(X)$ the tangent complex has cohomology computed by $\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$ , where $\pi: U\times X\rightarrow U$. With some work one can show that the actual tangent complex $f^* T_{Bun_G(X)}$ is quasi-isomorphic to $\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$ .

Finally, let me explain why this implies that the tangent complex on $Bun_G(X)$ is $T_{Bun_G(X)}=\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E}[1]$ for $p:Bun_G(X)\times X\rightarrow Bun_G(X)$. The point is that for nice morphisms of stacks $p$, $\mathbf{R} p_*$ can be defined by the usual pushforward for every base change to an affine scheme. See the proof of proposition 2.1.1 here.

To conclude, the canonical bundle is the inverse of the determinant of the tangent complex, i.e. $K=det(\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E})$ .

Answer by Pavel Safronov for Example of a form linear in infinitely many variables ?

2012-08-30T15:09:08Z

Consider the Hilbert space $H=L^2(S^1, \mathbf{C}^n)$ and the subspaces $H_+, H_-$ of positive and negative Fourier modes. One can construct the Hilbert space of infinite wedge products $V=\bigwedge(H_+)\hat{\otimes}\bigwedge(H_-)^*$.

Just as $H$ carries an action of $LSU(n)=Maps(S^1, SU(n))$ by pointwise multiplication, $V$ carries an action of the central extension $\widetilde{LSU(n)}$. It turns out that all irreducible positive-energy level 1 representations occur as summands in $V$.

Answer by Pavel Safronov for Are there F_un Lie algebras ?

2012-08-22T21:17:03Z

I will only attempt to answer the first question.

$n$-dimensional vector space over $\mathbf{F}_1$ is the same as a pointed set with $n+1$ elements. It is natural to call $GL_n(\mathbf{F}_1)$ the group of automorphisms of $\mathbf{F}_1^n$ and $\mathfrak{gl}_n(\mathbf{F}_1)$ the monoid of endomorphisms. There are at least two notions of morphisms:

Plain morphisms of pointed sets. The monoid of endomorphisms has cardinality $(n+1)^n$.
Maps of pointed sets which are injective if you throw away the basepoints (see http://arxiv.org/abs/1006.0912). It is not too hard to see that the cardinality of $End(\mathbf{F}_1)$ is $$\sum_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{n!}{(n-k)!}=\sum_{k=0}^n\frac{(n!)^2}{k!(n-k)!^2}.$$ Here $k$ is the number of elements that don't go to the basepoint.

Note, that in both cases the group of automorphisms is the same ($S_n$).

Answer by Pavel Safronov for Why does bosonic string theory require 26 spacetime dimensions?

2012-06-15T01:44:23Z

In addition to Chris Gerig's operator-language approach, let me also show how this magical number appears in the path integral approach.

Let $\Sigma$ be a compact surface (worldsheet) and $M$ a Riemannian manifold (spacetime). The string partition function looks like $$Z_{string}=\int_{g\in Met(\Sigma)}dg\int_{\sigma\in Map(\Sigma,M)}d\sigma\exp(iS(g,\sigma)).$$ Here $Met(\Sigma)$ is the space of Riemannian metrics on $\Sigma$ and $S(g,\sigma)$ is the standard $\sigma$-model action $S(g,\sigma)=\int_{\Sigma} dvol_\Sigma \langle d\sigma,d\sigma\rangle$. In particular, $S$ is quadratic in $\sigma$, so the second integral $Z_{matter}$ does not pose any difficulty and one can write it in terms of the determinant of the Laplace operator on $\Sigma$. Note that the determinant of the Laplace operator is a section of the determinant line bundle $L_{det}\rightarrow Met(\Sigma)$. The measure $dg$ is a 'section' of the bundle of top forms $L_g\rightarrow Met(\Sigma)$. Both line bundles carry natural connections.

However, the space $Met(\Sigma)$ is enormous: for example, it has a free action by the group of rescalings $Weyl(\Sigma)$ ($g\mapsto \phi g$ for $\phi\in Weyl(\Sigma)$ a positive function). It also carries an action of the diffeomorphism group. The quotient $\mathcal{M}$ of $Met(\Sigma)$ by the action of both groups is finite-dimensional, it is the moduli space of conformal (or complex) structures, so you would like to rewrite $Z_{string}$ as an integral over $\mathcal{M}$.

Everything in sight is diffeomorphism-invariant, so the only question is how does the integrand change under $Weyl(\Sigma)$. To descend the integral from $Met(\Sigma)$ to $Met(\Sigma)/Weyl(\Sigma)$ you need to trivialize the bundle $L_{det}\otimes L_g$ along the orbits of $Weyl(\Sigma)$. This is where the critical dimension comes in: the curvature of the natural connection on $L_{det}\otimes L_g$ (local anomaly) vanishes precisely when $d=26$. After that one also needs to check that the connection is actually flat along the orbits, so that you can indeed trivialize it.

Two references for this approach are D'Hoker's lectures on string theory in "Quantum Fields and Strings" and Freed's "Determinants, Torsion, and Strings".

Answer by Pavel Safronov for What differential operators have Schur polynoms as eigenfunctions ? Can this be deformed to trig. Calogero and Jack polynoms ?

2012-04-06T01:22:14Z

Schur polynomials are defined to be the characters of irreducible representations of $G=GL_n$. By the Weyl character formula, $$s_\lambda=\frac{1}{\delta}\sum_{w\in W}(-1)^{l(w)}e^{w(\lambda+\rho)},$$ where $\delta$ is the Vandermonde determinant.

Clearly, the numerator is an eigenfunction of the Laplace operator on the Cartan $\mathfrak{h}$: $$L_1(\delta s_\lambda):=\Delta (\delta s_\lambda) = (\lambda+\rho, \lambda+\rho) \delta s_\lambda.$$

So, the Schur polynomials themselves are eigenfunctions of $M_1=\delta^{-1} L_1\delta$.

The operators $L_1$ and $M_1$ admit one-parameter deformations $L_k$ and $M_k=\delta^{-k} L_k\delta^k$ to the quadratic Sutherland (trigonometric Calogero-Moser) and the Sekiguchi operators. Eigenfunctions of $M_k$ are the Jack polynomials.

The degeneration Jack -> Schur corresponds to taking the coupling constant $k=1$ in which case the Sutherland Hamiltonians are all free.

You can see formulas for a general root system in hep-th/9403168: the conjugated quadratic Hamiltonian is given by (2.4).

Answer by Pavel Safronov for Symplectic structures from Lagrangians?

2012-06-08T01:41:28Z

The reference is Deligne-Freed, Classical Field Theory, chapter 2. I will follow their notation.

Let $M$ be a spacetime manifold (for simplicity assume oriented) and $\mathcal{F}$ the space of fields. $d$ is the de Rham differential along $M$ and $\delta$ the differential along $\mathcal{F}$.

If the action $S$ is local, in the sense that $S=\int_M L$ for $L\in\Omega^{0,n}(\mathcal{F}\times M)$, then the procedure is the following. Find a form (variational one-form) $\gamma\in\Omega^{1,n-1}(\mathcal{F}\times M)$, such that $\alpha=\delta L+d\gamma\in\Omega^{1,n}(\mathcal{F}\times M)$ is linear over functions. What this means is that $\alpha(f\xi)=f\alpha(\xi)$ for every vector field $\xi\in T\mathcal{F}$ and a function $f\in\mathcal{O}(M)$. Then the symplectic form is defined to be $\omega=\int_H\delta\gamma\in\Omega^2(\mathcal{F})$ for $H$ a hypersurface in $M$. It is closed on the space of classical solutions, but may be degenerate.

In the case of Chern-Simons, this is literally true if $\mathcal{F}$ is the affine space of connections before modding out by the gauge transformations. I will consider the compact group Chern-Simons, the complex group version is similar. Let $M=\Sigma\times\mathbf{R}$ and $H=\Sigma\times\{0\}$. $$S=\int_M Tr(A\wedge dA+\frac{2}{3}A\wedge A\wedge A).$$ Then $$\delta L=Tr(\delta A\wedge dA-A\wedge \delta dA+2\delta A\wedge A\wedge A).$$ In this formula only the second term is not linear over functions. It can be killed off if one takes $$\gamma=Tr(A\wedge \delta A).$$ Its derivative is $$d\gamma = Tr(dA\wedge \delta A-A\wedge d\delta A),$$ so the nonlinear term in $\delta L+d\gamma$ disappears, since $d\delta +\delta d=0$.

Here the choice of $\gamma$ is unique if one assumes it itself is linear over functions. In the end one gets the standard symplectic form on the space of flat connections $$\omega=\int_\Sigma Tr(\delta A\wedge \delta A).$$

One should note that the action is not local in the naive sense on the space of connections mod gauge: the Lagrangian changes by a closed form under gauge transformations. However, the symplectic form does descend to the space of classical solutions mod gauge.

Answer by Pavel Safronov for SL(2,C) Chern-Simons theory in genus 1

2012-03-20T01:47:32Z

Let me call your $\omega$ as $\omega_I$. The symplectic form you get from the Chern-Simons action is $k\omega_I+s\omega_K$, where $\omega_K$ is one of the Kähler forms on the Hitchin space, which, in particular, is exact. If you choose a real polarization as Witten does, the Hilbert space is $\Gamma(Bun_GX,Det^{\otimes k})$, where $Det$ is the determinant bundle whose first Chern class $[\omega_I]$. One should note that the polarization is not the naive vertical polarization on $T^*Bun_GX$ since the fibers are not Lagrangian for $k\neq 0$.

Narasimhan-Seshadri identifies $Bun_GX$ with the character variety for the compact group, which in genus 1 is $T\times T\ /W$. So, the Hilbert space is $\Gamma(T\times T\ /W, Det^{\otimes k})$, precisely what Witten claims after eq. (5.11).

Determinant line does not depend on the differential

2011-11-27T04:14:52Z

Let $\pi:X\rightarrow S$ be a proper morphism of schemes over $\mathbf{C}$ and $\mathcal{E}$ a vector bundle on $X$ with a relative flat connection $\nabla_{X/S}$. Is it true that the determinant line bundle of the de Rham cohomology
$\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},\nabla_{X/S})$ is isomorphic to the determinant line bundle of the Dolbeault (Hodge)
cohomology $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},0)$?

Heuristically, the determinant line bundle behaves like the Euler characteristic, which does not depend on the differential. What is a reference for such a statement?

Answer by Pavel Safronov for If Lie(G) is semi simple then the moment map exists!

2011-11-27T02:45:08Z

Let $\mathfrak{g}=Lie(G)$. The action of $G$ on $M$ gives a morphism of Lie algebras $a:\mathfrak{g}\rightarrow Vect_{symp}(M)$.

Since $\mathfrak{g}$ has trivial abelianization, $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, i.e. any element can be decomposed into commutators. An easy computation shows that a commutator of two symplectic vector fields is a Hamiltonian vector field, so you can define a linear map $b:\mathfrak{g}\rightarrow C^\infty(M)$, which is not a morphism of Lie algebras in general.

Pick any two elements $x,y\in\mathfrak{g}$ and observe, that $b([x,y])-\{b(x),b(y)\}$ is a constant, since $\{b(x),b(y)\}$ is a Hamiltonian function for $[x,y]$. Call it $c(x,y)$: it defines a two-cocycle on $\mathfrak{g}$ which is furthermore trivial (by semisimplicity $H^2(\mathfrak{g})=0$). Therefore, there is an element $f\in\mathfrak{g}^*$, such that $b([x,y])-\{b(x),b(y)\}=c(x,y)=f([x,y])$.

Finally, define the map $\mathfrak{g}\rightarrow C^\infty(M)$ by $x\mapsto b(x)-f(x)$, you can easily check that it is a morphism of Lie algebras.

Answer by Pavel Safronov for What is the general statement of Hilbert 90?

2011-10-30T23:06:24Z

$H^1$ computed via sheaf cohomology coincides with the Cech $H^1$, which can be interpreted as giving transition functions. In particular, $H^1(X, GL_n)$ is in bijection with the set of rank $n$ vector bundles on $X$ in the Zariski topology (Theorem 11.4 in Milne's notes on etale cohomology).

Asymptotics of the TBA equation

2011-10-13T23:37:42Z

The Thermodynamic Bethe Ansatz equation is an integral equation that was derived by Yang and Yang to study some interacting systems. In the simplest case, it is $$\epsilon(\beta)=R\cosh\beta-\int\frac{d\beta'}{\pi\cosh(\beta-\beta')}\log(1+\exp(-\epsilon(\beta'))).$$

A reference for this is Al.B. Zamolodchikov, Thermodynamic Bethe Ansatz in relativistic models, Nucl Phys B342 (1990) 695-720.

I am interested in the asymptotics of the solutions as $R\rightarrow 0$. Zamolodchikov gives a heuristic argument that for small $R\cosh\beta$ we can neglect the first term. Therefore, the $\beta\rightarrow\beta+const$ invariance is restored and the solution $\epsilon(\beta)$ becomes independent of $\beta$ for small $R\cosh\beta$. Can one deduce a more precise asymptotic behavior? In particular, I would like to write down the small $R$ corrections.

One can easily show that this integral operator acting on $\exp(-\epsilon)$ maps the ball of radius $(e^R-1)^{-1}$ in $C^0(\mathbf{R})$ to itself. Furthermore, it is a contraction for large $R$ (see e.g. http://arxiv.org/abs/0807.4723, appendix C).

Was this equation studied anywhere in the mathematical literature?

Comment by Pavel Safronov

2013-02-11T16:37:24Z

The pullback locally looks like $\pi^{-1}M\otimes\mathcal{O}_{\mathbf{C}^\times}$. The decomposition $\mathcal{O}_{\mathbf{C}^\times}=\oplus\mathbf{C}\cdot z^l$ is precisely the eigenspace decomposition for $z\partial_z$. There is also an invariant way to write the pullback as $\pi^{-1}M\otimes_{\pi^{-1}\mathcal{O}_X}\mathcal{O}_L$; observe that $E$ acts trivially on $\pi^{-1}\mathcal{O}_X$.

Comment by Pavel Safronov

2013-02-09T00:59:16Z

If you have a $\mathbf{C}^\times$-bundle $L\rightarrow X$, you get an action map $\mathrm{Lie}\mathbf{C}^\times\rightarrow H^0(L, T_L)$. The image of $z\partial_z$ under this map is the required vector field. Isn't that what you're asking?

Comment by Pavel Safronov

2013-02-04T16:49:23Z

... In general, however, $c$ does not carry an algebra structure and there is no analogous $\mathrm{rk} G:1$ cover, only a (ramified) $|W|:1$ cover parametrizing Borels containing $c$.

Comment by Pavel Safronov

2013-02-04T16:47:08Z

Just to illustrate the necessity of cameral covers for general groups. If you have a regular Higgs field $\phi\in H^0(X, \mathrm{ad} P\otimes K)$, its centralizer $c\subset \mathrm{ad} P$ will be a subbundle of abelian Lie algebras (its rank is the rank of the group). When $G=GL_n$, the multiplication on $\mathrm{ad} P$ descends to a commutative algebra structure on $c$, so you can take the relative Spec of $c$ to obtain the spectral cover...

Comment by Pavel Safronov

2013-01-17T23:42:56Z

Unfortunately, I am not aware of a reference. I would guess the proof essentially follows from the factorization property of the space of $G$-bundles with a rational section known as the Beilinson-Drinfeld Grassmannian. The factorization property of the latter follows from the Beauville-Laszlo theorem. There are many places where it's mentioned, e.g. Beilinson-Drinfeld's "Quantization of the Hitchin system", section 5.3.10 and Frenkel-Ben-Zvi "Vertex algebras and algebraic curves", section 20.3.5.

Comment by Pavel Safronov

2013-01-11T19:44:42Z

It's probably obvious, but since the Gauss-Manin connection on $\mathcal{H}^i(Y/S)$ is obtained as the D-module pushforward of $\mathcal{O}_Y$ to $S$ and the de Rham cohomology is the pushforward to a point, there is a spectral sequence relating $H^j_{dR}(S, (\mathcal{H}^i(Y/S), \nabla^i_{GM}))$ and the de Rham cohomology of $Y$ itself.

Comment by Pavel Safronov

2013-01-09T06:10:52Z

The relative tangent bundle to $p: TM\rightarrow M$ is $p^* TM$, so you have an exact sequence $0\rightarrow p^* TM\rightarrow TTM\rightarrow p^*TM\rightarrow 0$. For smooth manifolds it is split, hence $TTM\cong p^*TM\oplus p^*TM$. This is different from, say, (the pullback of) $T^{(2,0)}M$, which is $TM\otimes TM$.

Comment by Pavel Safronov

2012-12-30T08:36:27Z

The Atiyah bundle $A_E$ is the bundle of $G$-invariant vector fields on $E$. These integrate to isomorphisms of torsors between different fibers, i.e. your $Isom$ groupoid. The Atiyah bundle has an anchor map $A_E\rightarrow T_X$ given by the differential of the projection $E\rightarrow X$. It is a map of Lie algebras, which is an infinitesimal version of the groupoid multiplication. The analogy is Lie algebroids are to groupoids, as Lie algebras are to groups. Finally, a connection is a splitting $T_X\rightarrow A_E$.

Comment by Pavel Safronov

2012-12-24T05:44:44Z

Perhaps, to see why such a groupoid might arise, one should look at the infinitesimal version. It is simply $TE / GL(n)$ and is known as the Atiyah bundle of $E$ (it is a Lie algebroid as opposed to the finite version being a groupoid). It appears, for example, when one defines connections on $E$.

Comment by Pavel Safronov

2012-11-29T01:57:41Z

It seems you can still take $u = -l$.

Comment by Pavel Safronov

2012-11-29T00:24:39Z

I may be missing something, but let $u=0$ and $B_{x_2}$ be a Borel whose Lie algebra contains $l(x_2)$.

Comment by Pavel Safronov

2012-10-02T23:44:32Z

A quick way to compute the cohomology is to note that the Leray-Serre spectral sequence for $SU(3)\rightarrow SU(3)/SU(2)\cong S^5$ is degenerate, so the cohomology coincides with the cohomology of the product $S^3\times S^5$.

Comment by Pavel Safronov

2012-09-27T01:40:43Z

Sorry, I didn't understand your question at first. Forms in $j_*\Omega^k_U$ are definitely not $C^\infty$ on $X$, since they are not even defined at $D$. Let me try to be more precise. The kernel of $\Omega^k_X(*D)\rightarrow j_*\Omega^k_U$ consists of meromorphic forms on $X$ which vanish on $U$. Since they are zero on an open set, they are zero on the whole $X$.

Comment by Pavel Safronov

2012-09-27T01:11:57Z

$j_*\Omega^k_U$ is the sheaf of differential forms which are holomorphic on $U$. You have an inclusion $\Omega^k_X(*D)\subset j_*\Omega^k_U$ of forms meromorphic along $D$. Furthermore, $\Omega^k_X(log D)\subset \Omega^k_X(*D)$ as forms having a first-order pole along $D$.

Comment by Pavel Safronov

2012-09-24T18:27:08Z

Basically, you can compute the tangent "space" in families right from the beginning. I modified the answer accordingly.