User pavel safronov - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:11:18Z http://mathoverflow.net/feeds/user/18512 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119139/are-rational-sections-of-a-vector-bundle-useful/119177#119177 Answer by Pavel Safronov for Are rational sections of a vector bundle useful? Pavel Safronov 2013-01-17T15:40:12Z 2013-01-17T15:40:12Z <p>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. <code>$$\mathrm{Bun}_G(X)\cong G(\mathbf{C}(X))\backslash\prod_{x\in X}' G(\mathbf{C}((t_x)))/G(\mathbf{C}[[t_x]]),$$</code> where $G(\mathbf{C}(X))$ is the group of rational maps from $X$ to $G$.</p> <p>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.</p> http://mathoverflow.net/questions/116709/identifying-t-bun-g-with-higgs-bundles/116733#116733 Answer by Pavel Safronov for Identifying $T^* Bun_G$ with Higgs bundles Pavel Safronov 2012-12-18T19:53:37Z 2012-12-18T19:53:37Z <p>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.</p> <p>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.</p> <p>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 $\phi\in H^0(U\times C, \mathrm{ad} P\otimes \omega_C)$.</p> http://mathoverflow.net/questions/107761/canonical-bundle-on-the-stack-of-g-bundles-on-a-curve/107784#107784 Answer by Pavel Safronov for canonical bundle on the stack of G bundles on a curve Pavel Safronov 2012-09-21T16:57:06Z 2012-09-24T18:24:38Z <p>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.</p> <p>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 <code>$Hom(D\times U, Bun_G(X))\rightarrow Hom(\mathrm{pt}\times U, Bun_G(X))$</code>, where <code>$D=\mathrm{Spec}\ \mathbf{C}[\epsilon]/\epsilon^2$</code>. Since this is a <em>Picard</em> groupoid, $\pi_0$ and $\pi_1$ are abelian groups, which are $H^0$ and $H^{-1}$ of the tangent complex.</p> <p>One can explicitly describe this groupoid as follows: its objects are $G$-bundles $P\rightarrow X\times U\times D$ together with an isomorphism <code>$P|_{X\times U\times\mathrm{pt}}\cong P_0$</code> on $X\times U$.</p> <p>$\pi_1$ 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 <code>$\pi_1=H^0(X\times U, \mathrm{ad}\ P_0)$</code>.</p> <p>The computation of $\pi_0$ is trickier, let me just write down the answer: <code>$\pi_0=H^1(X\times U, \mathrm{ad}\ P_0)$</code>. 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 <a href="http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Sept22%28Dmodstack1%29.pdf" rel="nofollow">notes</a> for details).</p> <p>So, on each affine $U$ mapping to $Bun_G(X)$ the tangent complex has cohomology computed by <code>$\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$</code>, where $\pi: U\times X\rightarrow U$. With some work one can show that the actual tangent complex <code>$f^* T_{Bun_G(X)}$</code> is quasi-isomorphic to <code>$\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$</code>.</p> <p>Finally, let me explain why this implies that the tangent complex on $Bun_G(X)$ is <code>$T_{Bun_G(X)}=\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E}[1]$</code> 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 <a href="http://www.math.harvard.edu/~gaitsgde/GL/QCohtext.pdf" rel="nofollow">here</a>.</p> <p>To conclude, the canonical bundle is the inverse of the determinant of the tangent complex, i.e. <code>$K=det(\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E})$</code>.</p> http://mathoverflow.net/questions/105932/example-of-a-form-linear-in-infinitely-many-variables/105954#105954 Answer by Pavel Safronov for Example of a form linear in infinitely many variables ? Pavel Safronov 2012-08-30T15:09:08Z 2012-08-30T15:09:08Z <p>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 $V=\bigwedge(H_+)\hat{\otimes}\bigwedge(H_-)^*$.</p> <p>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 $V$.</p> http://mathoverflow.net/questions/105263/are-there-f-un-lie-algebras/105270#105270 Answer by Pavel Safronov for Are there F_un Lie algebras ? Pavel Safronov 2012-08-22T21:17:03Z 2012-08-22T21:17:03Z <p>I will only attempt to answer the first question.</p> <p>$n$-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:</p> <ol> <li>Plain morphisms of pointed sets. The monoid of endomorphisms has cardinality $(n+1)^n$.</li> <li>Maps of pointed sets which are injective if you throw away the basepoints (see <a href="http://arxiv.org/abs/1006.0912" rel="nofollow">http://arxiv.org/abs/1006.0912</a>). It is not too hard to see that the cardinality of $End(\mathbf{F}_1)$ is <code>$$\sum_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{n!}{(n-k)!}=\sum_{k=0}^n\frac{(n!)^2}{k!(n-k)!^2}.$$</code> Here $k$ is the number of elements that don't go to the basepoint.</li> </ol> <p>Note, that in both cases the group of automorphisms is the same ($S_n$).</p> http://mathoverflow.net/questions/99643/why-does-bosonic-string-theory-require-26-spacetime-dimensions/99667#99667 Answer by Pavel Safronov for Why does bosonic string theory require 26 spacetime dimensions? Pavel Safronov 2012-06-15T01:44:23Z 2012-06-15T01:44:23Z <p>In addition to Chris Gerig's operator-language approach, let me also show how this magical number appears in the path integral approach.</p> <p>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.</p> <p>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 $\mathcal{M}$.</p> <p>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.</p> <p>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".</p> http://mathoverflow.net/questions/93250/what-differential-operators-have-schur-polynoms-as-eigenfunctions-can-this-be-d/93270#93270 Answer by Pavel Safronov for What differential operators have Schur polynoms as eigenfunctions ? Can this be deformed to trig. Calogero and Jack polynoms ? Pavel Safronov 2012-04-06T01:22:14Z 2012-06-12T20:26:41Z <p>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.</p> <p>Clearly, the numerator is an eigenfunction of the Laplace operator on the Cartan $\mathfrak{h}$: $$L_1(\delta s_\lambda):=\Delta (\delta s_\lambda) = (\lambda+\rho, \lambda+\rho) \delta s_\lambda.$$</p> <p>So, the Schur polynomials themselves are eigenfunctions of $M_1=\delta^{-1} L_1\delta$.</p> <p>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.</p> <p>The degeneration Jack -> Schur corresponds to taking the coupling constant $k=1$ in which case the Sutherland Hamiltonians are all free.</p> <p>You can see formulas for a general root system in <a href="http://arxiv.org/abs/hep-th/9403168" rel="nofollow">hep-th/9403168</a>: the conjugated quadratic Hamiltonian is given by (2.4).</p> http://mathoverflow.net/questions/99076/symplectic-structures-from-lagrangians/99080#99080 Answer by Pavel Safronov for Symplectic structures from Lagrangians? Pavel Safronov 2012-06-08T01:41:28Z 2012-06-08T01:41:28Z <p>The reference is Deligne-Freed, Classical Field Theory, chapter 2. I will follow their notation.</p> <p>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 $\mathcal{F}$.</p> <p>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 <em>linear over functions</em>. 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.</p> <p>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 $d\delta +\delta d=0$.</p> <p>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 $$\omega=\int_\Sigma Tr(\delta A\wedge \delta A).$$</p> <p>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.</p> http://mathoverflow.net/questions/79840/sl2-c-chern-simons-theory-in-genus-1/91680#91680 Answer by Pavel Safronov for SL(2,C) Chern-Simons theory in genus 1 Pavel Safronov 2012-03-20T01:47:32Z 2012-03-20T01:47:32Z <p>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 $k\neq 0$.</p> <p>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).</p> http://mathoverflow.net/questions/81983/determinant-line-does-not-depend-on-the-differential Determinant line does not depend on the differential Pavel Safronov 2011-11-27T04:14:52Z 2011-11-27T04:34:08Z <p>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<br> $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},\nabla_{X/S})$ is isomorphic to the determinant line bundle of the Dolbeault (Hodge)<br> cohomology $\det\mathbf{R}^\bullet\pi_*(\Omega^\bullet_{X/S}\otimes\mathcal{E},0)$?</p> <p>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?</p> http://mathoverflow.net/questions/81977/if-lieg-is-semi-simple-then-the-moment-map-exists/81981#81981 Answer by Pavel Safronov for If Lie(G) is semi simple then the moment map exists! Pavel Safronov 2011-11-27T02:45:08Z 2011-11-27T02:45:08Z <p>Let $\mathfrak{g}=Lie(G)$. The action of $G$ on $M$ gives a morphism of Lie algebras $a:\mathfrak{g}\rightarrow Vect_{symp}(M)$.</p> <p>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 <em>not</em> a morphism of Lie algebras in general.</p> <p>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 $b([x,y])-\{b(x),b(y)\}=c(x,y)=f([x,y])$.</p> <p>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.</p> http://mathoverflow.net/questions/79554/what-is-the-general-statement-of-hilbert-90/79556#79556 Answer by Pavel Safronov for What is the general statement of Hilbert 90? Pavel Safronov 2011-10-30T23:06:24Z 2011-10-30T23:06:24Z <p>$H^1$ 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).</p> http://mathoverflow.net/questions/78082/asymptotics-of-the-tba-equation Asymptotics of the TBA equation Pavel Safronov 2011-10-13T23:37:42Z 2011-10-22T19:20:45Z <p>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 $$\epsilon(\beta)=R\cosh\beta-\int\frac{d\beta'}{\pi\cosh(\beta-\beta')}\log(1+\exp(-\epsilon(\beta'))).$$</p> <p>A reference for this is Al.B. Zamolodchikov, Thermodynamic Bethe Ansatz in relativistic models, Nucl Phys B342 (1990) 695-720.</p> <p>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.</p> <p>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. <a href="http://arxiv.org/abs/0807.4723" rel="nofollow">http://arxiv.org/abs/0807.4723</a>, appendix C).</p> <p>Was this equation studied anywhere in the mathematical literature?</p> http://mathoverflow.net/questions/121246/weight-decomposition-and-eigenspaces-euler-vector-field Comment by Pavel Safronov Pavel Safronov 2013-02-11T16:37:24Z 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 $\pi^{-1}\mathcal{O}_X$. http://mathoverflow.net/questions/121246/weight-decomposition-and-eigenspaces-euler-vector-field Comment by Pavel Safronov Pavel Safronov 2013-02-09T00:59:16Z 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? http://mathoverflow.net/questions/120697/hitchin-fibration-outside-of-type-a Comment by Pavel Safronov Pavel Safronov 2013-02-04T16:49:23Z 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 $c$. http://mathoverflow.net/questions/120697/hitchin-fibration-outside-of-type-a Comment by Pavel Safronov Pavel Safronov 2013-02-04T16:47:08Z 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... http://mathoverflow.net/questions/119139/are-rational-sections-of-a-vector-bundle-useful/119177#119177 Comment by Pavel Safronov Pavel Safronov 2013-01-17T23:42:56Z 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 &quot;Quantization of the Hitchin system&quot;, section 5.3.10 and Frenkel-Ben-Zvi &quot;Vertex algebras and algebraic curves&quot;, section 20.3.5. http://mathoverflow.net/questions/118658/cohomology-of-the-gauss-manin-connection Comment by Pavel Safronov Pavel Safronov 2013-01-11T19:44:42Z 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. http://mathoverflow.net/questions/118395/is-there-a-relationship-between-tensor-or-form-bundles-and-iterated-tangent-cot Comment by Pavel Safronov Pavel Safronov 2013-01-09T06:10:52Z 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 $TM\otimes TM$. http://mathoverflow.net/questions/117075/a-question-on-isomp-1e-p-2e-rightrightarrows-x Comment by Pavel Safronov Pavel Safronov 2012-12-30T08:36:27Z 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 $T_X\rightarrow A_E$. http://mathoverflow.net/questions/117075/a-question-on-isomp-1e-p-2e-rightrightarrows-x Comment by Pavel Safronov Pavel Safronov 2012-12-24T05:44:44Z 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 $E$. http://mathoverflow.net/questions/114786/approximation-in-lie-algebras Comment by Pavel Safronov Pavel Safronov 2012-11-29T01:57:41Z 2012-11-29T01:57:41Z It seems you can still take $u = -l$. http://mathoverflow.net/questions/114786/approximation-in-lie-algebras Comment by Pavel Safronov Pavel Safronov 2012-11-29T00:24:39Z 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 $l(x_2)$. http://mathoverflow.net/questions/108649/de-rham-representatives-of-the-cohomology-classes-in-hsu3/108656#108656 Comment by Pavel Safronov Pavel Safronov 2012-10-02T23:44:32Z 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 $S^3\times S^5$. http://mathoverflow.net/questions/108201/inclusion-of-logarithmic-de-rham-complex-into-differentials Comment by Pavel Safronov Pavel Safronov 2012-09-27T01:40:43Z 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 $X$. http://mathoverflow.net/questions/108201/inclusion-of-logarithmic-de-rham-complex-into-differentials Comment by Pavel Safronov Pavel Safronov 2012-09-27T01:11:57Z 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 $D$. http://mathoverflow.net/questions/107761/canonical-bundle-on-the-stack-of-g-bundles-on-a-curve/107784#107784 Comment by Pavel Safronov Pavel Safronov 2012-09-24T18:27:08Z 2012-09-24T18:27:08Z Basically, you can compute the tangent &quot;space&quot; in families right from the beginning. I modified the answer accordingly.