User peter dalakov - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:52:03Z http://mathoverflow.net/feeds/user/6278 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118930/algebraic-definition-of-the-kuranishi-map/118958#118958 Answer by Peter Dalakov for Algebraic definition of the Kuranishi map Peter Dalakov 2013-01-15T10:23:47Z 2013-01-15T10:56:53Z <p>In characteristic zero the answer is positive.</p> <p>If $L^\bullet$ is the dgla (differential graded Lie algebra) governing your deformation problem (the Kodaira-Spencer dgla $\oplus A^{0,p}(X,T_X)$ in your example), the formal Kuranishi theorem states that for every splitting $\delta$ of $L^\bullet$, there exists a hull $Kur^\delta_{L^\bullet}\to Def_{L^\bullet}$, the <em>formal Kuranishi space</em>. Here $Def_{L^\bullet}:Art_{k}\to Sets$ is the deformation functor associated to $L^\bullet$. A <em>splitting of $L^\bullet$</em> is a degree $-1$ linear map, $\delta:L^\bullet\to L^\bullet[-1]$, such that $\delta^2=0$, $d\delta d=d$, $\delta d\delta=\delta$. It plays the role of $d^\ast G$ in Hodge theory, where $G$ is Green's operator and $d^\ast$ is the adjoint of $d$.</p> <p>Manetti's article that Francesco mentions is a great introduction. Another reference that I am very fond of is Goldman and Millson's paper <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.ijm/1255988270" rel="nofollow">The Homotopy invariance of Kuranishi space</a>, where they compare the algebraic and analytic description of the hull.</p> http://mathoverflow.net/questions/109272/whats-bad-about-unstable-sheaves/109302#109302 Answer by Peter Dalakov for What's "bad" about unstable sheaves? Peter Dalakov 2012-10-10T15:09:24Z 2012-12-25T15:59:21Z <p>Stable sheaves are <em>simple</em>, i.e., $\textrm{End}E\simeq \mathbb{C}$. One thing that you want to avoid is the jumping of the automorphism group in a family.</p> <p>A classical example is to consider a hyperelliptic curve $X$, and $[L]\in\textrm{Pic}^{g-1}X$. If $\pi:X\to \mathbb{P}^1$ is the $g^1_2$, then Grothendieck-Riemann-Roch plus Riemann-Hurwitz tell you that $\pi_\ast L\simeq \mathcal{O}(a-1)\oplus \mathcal{O}(-a-1)$, where $a=h^0(L)$. So you can take a take a family of line bundles over the unit disk $\{L_t\}_{t\in\Delta}$, with $h^0(L_0)=1$, $h^0(L_t)=0$ for $t\in\mathbb{C}^\ast$. Then the generic element will be semistable, $\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus 2}$, with automorphism group $GL_2$, and over zero you have $\mathcal{O}\oplus \mathcal{O}(-2)$, unstable, with 5-dimensional automorphism group.</p> <p>And of course you need boundedness, see Donu's comment.</p> <p><b>ADDENDUM</b></p> <p>Here is an example of how allowing unstable bundles messes up uniqueness of limits (and hence separatedness). Let $X$ be a curve of genus $g\geq 2$, and let $E$ be a semi-stable rank two bundle with $\det E\simeq \mathcal{O}_X$. Let $[L]\in \textrm{Pic}^d X$, $d\geq 2g$. Then $E\otimes L$ is semi-stable of determinant $L^2$. It is globally generated and surjects onto $L^2$, and so $E$ fits in an extension $$0\longrightarrow L^{-1}\longrightarrow E\longrightarrow L\longrightarrow 0.$$ Now, take a DVR $R$, $\textrm{Spec }R=\{p,0\}$, where $p$ is the generic point and $0$ the closed point, and consider a family of bundles $\mathcal{F}$ over $\textrm{Spec }R$, for which $\mathcal{F}_0\simeq E$. One can show that if $\mathcal{F}'$ is the elementary transformation of $\mathcal{F}$ along $L$, then $\mathcal{F}_p'\simeq \mathcal{F}_p$, but $\mathcal{F}'_0$ fits in an extension $$0\longrightarrow L\longrightarrow \mathcal{F}'_0\longrightarrow L^{-1}\longrightarrow 0.$$ However, by the choice of $L$, $H^1(X,L^2)=0$, so $\mathcal{F}'_0\simeq L\oplus L^{-1}$, an unstable bundle.</p> http://mathoverflow.net/questions/111694/what-is-geometric-intuition-of-special-lagrangian-manifolds/111932#111932 Answer by Peter Dalakov for What is geometric intuition of special Lagrangian manifolds? Peter Dalakov 2012-11-09T19:51:41Z 2012-11-09T21:58:00Z <p>Let $(M,g,J,\Omega)$ be a Calabi-Yau $n$-fold. Then $\textrm{Re }\Omega$ is a calibration on $(M,g)$. Let $L\subset M$ be a real submanifold with $\dim_{\mathbb{R}}L=n$. You have the following</p> <p><b>Proposition</b> </p> <p>$L$ is a special Lagrangian if and only if it admits an orientation making it into a calibrated (for $\textrm{Re }\Omega$) submanifold of $(M,g)$. <i>In that case, it is volume-minimising in its homology class.</i></p> <p>See Propositions 10.1 and 7.1 in the "Calabi-Yau manifolds..." book by Gross, Huybrechts and Joyce.</p> <p>Here $\textrm{Re }\Omega$ being a calibration means that at each point $p\in M$, and for every oriented tangent $n$-plane $V\subset T_{M,p}$, one has $\left.\textrm{Re }\Omega\right|_V\leq vol_V$, where $vol$ is the volume form of $g$. Then $L$ being a calibrated submanifold for $\textrm{Re }\Omega$ means that on the tangent spaces $T_{L,p}$ of $L$ the above inequality becomes equality.</p> http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau/110246#110246 Answer by Peter Dalakov for Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold Peter Dalakov 2012-10-21T16:39:37Z 2012-10-21T16:50:45Z <p><strong>Disclaimer:</strong> I am not sure what kind of "explanation" you are looking for. I would guess that you are after the observation (due to Hitchin), that complex Lagrangian submanifolds become special Lagrangian after rotating the complex structure.</p> <p><strong>Observation:</strong> Let $X$ be a hyperkaehler manifold. Let $\{I,J,K\}$ be a triple of complex structures, satisfying the quaternionic identities, and let $\{\omega_I,\omega_J,\omega_K\}$ be the respective Kaehler forms. Let $M\subset (X,I,\omega_I)$ be a complex-lagrangian submanifold for the complex-symplectic form $\omega^c= \omega_J+i\omega_K$. Then $M$ is a special lagrangian submanifod of $(X,J, \omega_J,\Omega = (\omega_K+i\omega_I)^{\dim_{\mathbb{C}} M})$.</p> <p>(Actually, if $\dim_{\mathbb{C}} M$ is odd you must either take $i\Omega$ as your holomorphic volume form, or use the more relaxed definition of special Lagrangian. ) </p> <p>Here "complex-Lagrangian" means that $M\subset (X,I)$ is a complex submanifold and $\left. \omega^c\right|_M=0$.</p> <p>So given a real-analytic Kaehler manifold, you embed it as the zero-section of the cotangent bundle, take the Kaledin-Feix metric on a (formal) tubular neighbourhood, and rotate the complex structure.</p> http://mathoverflow.net/questions/109430/projectives-in-the-category-of-coherent-sheaves-on-a-projective-variety/109597#109597 Answer by Peter Dalakov for Projectives in the category of coherent sheaves on a projective variety Peter Dalakov 2012-10-14T10:25:54Z 2012-10-14T10:33:32Z <p>If you are trying to lift a specific homomorphism, or to analyse extensions of vector bundles on curves, notice the following (almost obvious) observation:</p> <p>Given a short exact sequence of vector bundles $$0\to E\to W\to F\to 0,$$ a vector bundle homomorphism $f:V\to F$ lifts to a homomorphism $V\to W$ if and only if the extension class $\delta(W)$ satisfies $$\delta(W)\in \ker\left(H^1(X,\underline{Hom}(F,E))\to H^1(X,\underline{Hom}(V,E)) \right).$$ This is Lemma 3.1 in <a href="http://www.jstor.org/discover/10.2307/1970807?uid=3737608&amp;uid=2&amp;uid=4&amp;sid=21101315942057" rel="nofollow">Narasimhan - Ramanan</a>.</p> http://mathoverflow.net/questions/69096/why-is-the-degreerank-ratio-of-a-vector-bundle-called-its-slope/69131#69131 Answer by Peter Dalakov for Why is the degree:rank ratio of a vector bundle called its "slope"? Peter Dalakov 2011-06-29T17:26:45Z 2012-10-03T13:34:16Z <p>This is an addition to Angelo's comment. Given a vector bundle $E$, you can consider its Harder-Narasimhan filtration and assign to each element of the filtration a point in the degree-rank plane. The HN-polygon is the polygon obtained by connecting the dots. S.S.Shatz discussed the behaviour of the HN-polygon under specialisation in <a href="http://archive.numdam.org/article/CM_1977__35_2_163_0.pdf" rel="nofollow">The decomposition and specialisation of algebraic families of vector bundles</a> I beleive this is where the term orginated, see also <a href="http://www.jstor.org/stable/37156" rel="nofollow">Atiyah-Bott</a>, section 7 (p.565). </p> <p><strong>Addendum:</strong></p> <p>You can gain minor visual gratification from looking at the degree-rank plane as follows. If $F\subset E$ is a subbundle, then </p> <p>$$\deg \underline{Hom}(F,E)= \textrm{rk}F\deg E-\textrm{rk}E \deg F= \left| \begin{array}{cc} \deg E &amp; \deg F\\ \textrm{rk} E &amp; \textrm{rk} F\\ \end{array} \right|.$$ Also, $F$ destabilises $E$ exactly when the above determinant has negative sign.</p> http://mathoverflow.net/questions/107168/weitzenbock-identities/107174#107174 Answer by Peter Dalakov for Weitzenböck Identities Peter Dalakov 2012-09-14T12:53:26Z 2012-09-14T12:53:26Z <p>As far as references are concerned, you can read about the Bochner technique in H.Wu, "The Bochner technique in differential geometry". For some applications of the Lichnerowicz formula you can check Ch.3 of Berline, Getzler, Vergne, "Heat Kernels and Dirac operators", and Ch.3, 5 of T.Friedrich's "Dirac Operators in Riemannian geometry".</p> http://mathoverflow.net/questions/103705/global-definition-of-the-dolbeault-complex-of-a-vector-bundle/103714#103714 Answer by Peter Dalakov for Global Definition of the Dolbeault Complex of a Vector Bundle Peter Dalakov 2012-08-01T18:13:52Z 2012-08-01T20:40:27Z <p>To get the Dolbeault complex, you need a choice of <em>holomorphic</em> structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty_M\otimes_{\mathcal{O}_M}\mathcal{E}$ and $\overline{\partial}_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$ is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying $$\overline{\partial}_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}_E(\sigma),$$ for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that $\left. \overline{\partial}_E\right| _{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$. You can then extend the Dolbeault operator to $\overline{\partial}_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$, $\overline{\partial}_E^2=0$, by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution $$0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$ The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.</p> <p>You cannot do any of this without the holomorphic structure. Differently put, you need the total space of $E$ to be a complex manifold and the projection $E\to M$ to be holomorphic. You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.</p> <p><strong>ADDENDUM</strong></p> <p>The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure $\overline{\partial}_E$ is non-trivial. It can be phrased as follows:</p> <p>$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that $D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.</p> http://mathoverflow.net/questions/102116/riemannian-metrics-as-sections-of-a-vector-bundle/102125#102125 Answer by Peter Dalakov for Riemannian metrics as sections of a vector bundle Peter Dalakov 2012-07-13T10:03:11Z 2012-07-13T10:21:51Z <p>An orthogonal or hermitian structure on $E$ is a section of <i>fibre bundle</i> (which is not a vector bundle). I will deal with the complex case. </p> <p>The Lie algebra $\mathfrak{gl}(n)$ decomposes into $$\mathfrak{gl}(n)\simeq \mathfrak{u}(n)\oplus \textrm{Herm}_n,$$ where $\textrm{Herm}_n$ is the vector space of Hermitian $n\times n$ matrices. The exponential map sends it to the positive-definite hermitian matrices: $$\exp: \textrm{Herm}_n\to \textrm{Herm}_n^+,$$ which form a convex domain ("cone") . More importantly, this cone is actually $$\textrm{Herm}_n^+ = GL(n,\mathbb{C})/U(n).$$ To see this, consider the action of $GL(n,\mathbb{C})$ on the hermitian matrices by $(T,h)\mapsto \overline{T}^t h T$. Now, do all of this "fibrewise": if $P$ is the frame bundle of $E$, an hermitian metric is a section of the associated bundle with fibre $GL(n,\mathbb{C})/U(n)$.</p> <p>More coneptually, a choice of reduction of the structure group of a principal $G$-bundle $P$ to a subgroup $K$ is eqivalent to a choice of section of the associated $G/K$ bundle. The hermitian metric is a reduction of the structure group from $GL(n,\mathbb{C})$ to $U(n)$.</p> <p><b>Aside:</b></p> <p>Since this question may be a related to your other question</p> <p><a href="http://mathoverflow.net/questions/100906/hermitian-christoffel-symbols" rel="nofollow">http://mathoverflow.net/questions/100906/hermitian-christoffel-symbols</a></p> <p>let me say that if you have a complex manifold $X=(M,I)$ with Riemannian metric $g$ on $M$ (compatible with $I$) you can extend $g$ to the complexified tangent bundle $T_{X,\mathbb{C}}$ either as a $\mathbb{C}$-<i>bilinear</i> pairing (as in Kobayashi &amp; Nomizu), or as a <i>sesquilinear</i> pairing $g_{\mathbb{C}}$: see section 1.2 of Huybrechts, "Complex geometry". Now, $T_{X,\mathbb{C}}\simeq T^{1,0}\oplus T^{0,1}$. With the former choice $T^{p,q}$ are isotropic sub-bundles, and only the off-diagonal pairing is non-trivial. With the latter choice (Huybrechts, Griffiths &amp; Harris), the two subbundles are <i>orthogonal</i> and $\left. g_{\mathbb{C}}\right|_{T^{1,0}}$ is $\frac{1}{2}h$, $h= g-i\omega$ (and the conjugate of that on $T^{0,1}$). This turns $E = T^{1,0}$ into an hermitian vector bundle. </p> <p>In indices, in the first case you have $h_{ab}=0= h_{\overline{a}\overline{b}}$, $h_{a\overline{b}}\neq 0$, while in the second the other way around.</p> http://mathoverflow.net/questions/99912/almost-complex-structure-approach-to-deformation-of-compact-complex-manifolds/99926#99926 Answer by Peter Dalakov for Almost Complex Structure approach to Deformation of Compact Complex Manifolds Peter Dalakov 2012-06-18T17:44:34Z 2012-06-19T15:35:19Z <p>This approach to deformations is taken, for instance, in all of the original papers of Kodaira-Spencer and Nirenberg. You can have a look at <i> On the existence of deformations of complex analytic structures</i>, Annals, Vol.68, No.2, 1958</p> <p><a href="http://www.jstor.org/discover/10.2307/1970256?uid=3737608&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=47699092130607" rel="nofollow">http://www.jstor.org/discover/10.2307/1970256?uid=3737608&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=47699092130607</a></p> <p>but there are many other papers by the same authors. </p> <p>For a nice and compact exposition, you can look at these class notes of Christian Schnell: <a href="http://homepages.math.uic.edu/~cschnell/pdf/notes/kodaira.pdf" rel="nofollow">http://homepages.math.uic.edu/~cschnell/pdf/notes/kodaira.pdf</a></p> <p>Of course, the Maurer-Cartan equation and deformations (of various structures) via dgla's have been used by many other people since the late 1950-ies: Goldman &amp; Millson, Gerstenhaber, Stasheff, Deligne, Quillen, Kontsevich.</p> <p>Regarding the formula: that's a typo, indeed. You have two eigen-bundle decompositions, for $I$ and $I_t$: </p> <p>$$T_{M, \mathbb{C}} = T^{1,0}\oplus T^{0,1}\simeq T^{1,0}_t\oplus T^{0,1}_t$$</p> <p>and you write $T^{0,1}_{t}=\textrm{graph }\phi$, where $\phi: T^{0,1}_M\to T^{1,0}_M$. So actually</p> <p>$$\phi = \textrm{pr}^{1,0}\circ \left.\left(\textrm{pr}^{0,1}\right)\right|_{T^{1,0}_t}^{-1}.$$</p> <p>In local coordinates, $$\phi = \sum_{j,k=1}^{\dim_{\mathbb{C}} M}h_{jk}(t,z)d\overline{z}_j\otimes \frac{\partial}{\partial z_k},$$ and $T^{0,1}_t$ is generated (over the smooth functions) by</p> <p>$$\frac{\partial}{\partial \overline{z_j}} + \sum_{k=1}^{\dim_{\mathbb{C}}M}h_{jk}\frac{\partial}{\partial z_k}.$$</p> <p>Regarding the question "where does $t$ come from?", the answer is "From Ehresmann's Theorem": given a proper holomorphic submersion $\pi:\mathcal{X}\to \Delta$, you can choose a holomorphically transverse trivialisation $\mathcal{X}\simeq X\times \Delta$, $X=\pi^{-1}(0)= (M,I)$. In this way you get yourself two (almost) complex structures on $X\times \Delta$, which you can compare.</p> <p><strong>ADDENDUM</strong> I second YangMills' suggestion to have a look at Chapter 2 of Gross-Huybrechts-Joyce. You can also try Chapter 1 of K. Fukaya's book <i>"Deformation Theory, Homological algebra, and Mirror Symmetry"</i>, as well as the Appendix to <i>Homotopy invariance of the Kuranishi Space</i> by Goldman and Millson (Illinois J. of Math, vol.34, No.2, 1990). In particular, you'll see how one uses formal Kuranishi theory to avoid dealing with the convergence of the power series for $\phi(t)$. For deformations of compact complex manifolds, the convergence was proved by Kodaira-Nirenberg-Spencer. Fukaya says a little bit about the convergence of this series in general, i.e., for other deformation problems.</p> http://mathoverflow.net/questions/94485/connections-on-line-bundles-over-the-torus/94496#94496 Answer by Peter Dalakov for Connections on line bundles over the torus Peter Dalakov 2012-04-19T05:19:32Z 2012-04-26T06:17:22Z <p>The Kaehler polarisation is a choice of <em>complex structure</em>, $I$, on your torus $M\simeq S^1\times S^1$, which is <em>compatible</em> with the symplectic form. In other words, your torus becomes a <em>complex manifold</em> $X=(M,I)$, and $I$ satisfies $I^t\omega I=\omega$, $g=\omega I>0$. In particular, this tells you that $\omega$ is a positive $(1,1)$-form, i.e., locally $\omega = \frac{i}{2}h dz \wedge d\overline{z}$, $h>0$. Of course, since $\dim X=1$, $\omega$ is a $(1,1)$-form for any choice of $I$, but this is not so in higher dimensions.</p> <p>The connection $D$ can now be decomposed as $D= D^{1,0} + D^{0,1}$. Explicitly, $D^{0,1} = \frac{1}{2}(1+iI)D$ and $D^{1,0} = \frac{1}{2}(1-iI)D$. Or, if you prefer, choose a local frame $s$ of $L$, and write $$D = d + A_1 dq + A_2 dp = (\partial + B_1 dz) + (\overline{\partial} + B_2 d\overline{z})$$</p> <p>Since $\omega$ is of type $(1,1)$, the condition $D^2 = -i\omega$ translates to $$(D^{0,1})^2=0,\quad (D^{1,0})^2=0,\quad D^{0,1}D^{1,0} +D^{1,0} D^{0,1}=-i\omega .$$ Again, the first two equations hold automatically if $\dim X=1$, but not in general.</p> <p>This means that on $L$ you get the structure of a <em>holomorphic</em> line bundle $\mathcal{L}$ by taking the Dolbeault operator to be $\overline{\partial}_L= D^{0,1}$. Locally, if you choose a (smooth) frame $s$, this is the above $\overline{\partial}_L = \overline{\partial} + B_2 d\overline{z}$.</p> <p>Then the "Hilbert space" is $H^0(X,\mathcal{L})= \ker\overline{\partial}_L \subset A^0(L)$, where $A^0(L)$ is the (infinite-dimensional) vector space of global smooth sections of $L$. With respect to a local trivialisation, a section $\sigma =fs$ is holomorphic if $$\frac{\partial f}{\partial\overline{z}} + B_2 f=0.$$</p> <p>The space $H^0(X,\mathcal{L})$ is finite-dimensional for very general reasons. If you actually want to know how big it is, you look at the (Hirzebruch)-Riemann-Roch formula, which tells you that $$\dim H^0(X,\mathcal{L})- \dim H^1(X,\mathcal{L}) = \deg \mathcal{L} + 1-g = \deg \mathcal{L}.$$ Then $\dim H^0(X,\mathcal{L})=\deg \mathcal{L}$ if $\deg\mathcal{L}>0$, $\dim H^0(X,\mathcal{L})= 0$ if $\deg\mathcal{L}&lt;0$ (or if $\deg\mathcal{L}=0$ and $\mathcal{L}$ is nontrivial), and $\dim H^0(X,\mathcal{L})=1$ if $\mathcal{L}=\mathcal{O}_X$. You can describe the elements of $H^0(X,\mathcal{L})$ quite explicitly using theta-functions.</p> <p>Notice that everything - in particular $H^0(X,\mathcal{L})$ - depends on the choice of complex structure $I$, and even the dimension may jump as you vary the complex structure (if $\deg L=0$). But if you choose your prequantum line bundle to be sufficiently positive, the vector spaces $H^0(X,\mathcal{L})$ "glue" into a vector bundle over the upper half-plane.</p> <p><strong>ADDENDUM:</strong></p> <p>Let me spell out the relation with theta-functions a bit more explicitly. This is a huge classical topic (see the references at the end), so I'll be a bit sketchy.</p> <p>Write $V=\mathbb{C}$ and $\Lambda = \mathbb{Z}\oplus \tau \mathbb{Z}$, $\textrm{Im}\tau>0$. Then $\pi: V\to X=V/\Lambda$ is the universal covering map, and it is a local biholomorphism. </p> <p>If your symplectic structure descends from the form $\omega= c\ dq\wedge dp$ on $V=\mathbb{C}$, the "Bohr-Sommerfeld condition" tells you that $c=\frac{2\pi}{\textrm{Im}\tau}n$, $n\in \mathbb{Z}$. Notice that $\frac{1}{2\pi}\omega$ is the imaginary part of the hermitian form $H(z,w)=\frac{n}{\textrm{Im}\tau}z\overline{w}$ on $V$ and takes integer values on the lattice.</p> <p>Suppose $\mathcal{L}$ is as above, with $c_1(\mathcal{L}) =\frac{1}{2\pi}\left[\omega\right]$, $\deg \mathcal{L}=n$.</p> <p>As any holomorphic line bundle on $V$ is trivial, we can choose a global holomorphic trivialisation $\pi^\ast \mathcal{L}\simeq V\times \mathbb{C}$, and identify $\mathcal{L}= \pi^\ast \mathcal{L}/\Lambda$ with $V\times\mathbb{C}/\sim$. The equivalence relation is $(z,t)\sim (z+\lambda, a(\lambda,z)t)$, $\lambda\in\Lambda=\pi_1(X)$. Here $a: \Lambda\times V\to \mathbb{C}^\times$ are the <em>multipliers</em> or <em>factors of automorphy</em>. They are 1-cocyles of $\Lambda=\pi_1(X)$ with values in the entire nonvanishing functions $H^0(V,\mathcal{O}^\times_V)$, i.e., they are holomorphic in $z$ and satisfy $$a(\lambda+\mu,z)= a(\mu,\lambda+z)a(\lambda,z).$$ The multipliers do not, in general, take values in $U(1)$! Changing the trivialisation of $\pi^\ast \mathcal{L}$ replaces $a$ by a coboundary, and we have $H^1(\Lambda,H^0(V,\mathcal{O}^\times_V))\simeq H^1(X,\mathcal{O}^\times_X)$.</p> <p>Now, we can identify $H^0(X,\mathcal{L})\simeq H^0(V,\mathcal{O}_V)^{\pi_1(X)}$. I.e., we can identify $H^0(X,\mathcal{L})$ with the entire functions which satisfy the functional equation $$\theta(z+\lambda) = a(\lambda,z)\theta(z),$$ and these are by definition <em>theta-functions</em>. </p> <p>The explicit structure of the factors of automorphy is known. For instance, in each degree $n$ you have a canonical choice, $$a_0(\lambda,z) = \exp\left( n\pi i ab \right)\exp \frac{n\pi}{\textrm{Im}\tau}\left(z\overline{\lambda}+ \frac{1}{2}|\lambda|^2\right),\quad \lambda=a+b\tau.$$ You can get the other bundles of the same degree by multiplying $a_0$ with characters $\chi\in \textrm{Hom}(\Lambda,U(1))\simeq Pic^0(X)$.</p> <p>Finally, a word about the inner product. We can identify $A^0(\mathcal{L})$, the smooth sections of $\mathcal{L}$, with smooth functions on $V$, satisfying the same functional equation as the theta-functions. In terms of this identification the fibrewise hermitian product on $\mathcal{L}$ is $$\langle f,g \rangle(z) = f(z)\overline{g(z)}\exp \left(-\frac{n\pi}{\textrm{Im}\tau}|z|^2 \right).$$ The $L^2$-inner product on $H^0(X,\mathcal{L})$ is obtained by integrating this quantity over $X$. </p> <p>Line bundles on a complex torus are described in great detail in Birkenhake and Lange, <em>Complex Abelian Varieties</em> (especially Chapter 2), Griffiths and Harris, <em>Principles of AG</em>, Chapter 2, Section 6, and D.Mumford, <em>Abelian Varieties</em>, Chapter 1, Section 2.</p> <p>Here we have been looking at a 1-dimensional torus, but if you decide to consider the geometric quantisation of $U(1)$ Chern-Simons theory on a higher genus curve, you will need the case of tori of higher dimensions.</p> http://mathoverflow.net/questions/93642/de-rham-theorem-for-differentiable-spaces/93689#93689 Answer by Peter Dalakov for De Rham theorem for differentiable spaces Peter Dalakov 2012-04-10T17:32:03Z 2012-04-11T18:19:54Z <p>Let $X$ be a topological space, and $Symp_\bullet (X)$ the simplicial set with simplices $$Symp_m(X) = \{ \textrm{singular } m- \textrm{simplices } \sigma: \Delta^m\to X\}$$ (and usual faces and degeneracies). Let $A^\bullet(X)$ be the complex of rational polynomial differential forms on $Symp_\bullet(X)$. If $k$ is a $\mathbb{Q}$-algebra, denote $A^\bullet_k(X)=A^\bullet(X)\otimes_\mathbb{Q}k$. Dennis Sullivan proved that:</p> <p><strong>The integration map</strong> $$I : A^\bullet_k(X)\to S^\bullet(X,k)$$ <strong>induces a $k$-algebra isomorphism on cohomology.</strong></p> <p>Here $S^\bullet(X,k)$ is the complex of singular $k$-valued cochains.</p> <p>The theorem is one of the versions of a more powerful result, see D.Sullivan, <a href="http://www.numdam.org/item?id=PMIHES_1977__47__269_0" rel="nofollow">Infinitesimal Computations in Topology</a>, Theorem 7.1. In this form the theorem is stated in R.Hain's paper <em>The de Rham homotopy theory of complex algebraic varieties I</em>, section 2 (K-theory, 1987).</p> http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative/92656#92656 Answer by Peter Dalakov for Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? Peter Dalakov 2012-03-30T07:59:58Z 2012-04-01T08:10:34Z <p>This is actually a question of linear algebra. The apparently inappropriate choice of sign stems from the fact that if $W$ is a (complex) vector space, and $A$ an endomorphism of $W$, then $$\ker (A-\lambda 1)\subset W$$ is the eigenspace for eigenvalue $(+\lambda)$.</p> <p>In our case, we start with a real vector space, $V$, and an endomorphism $I$, satisfying $I^2=-1$. We extend it $\mathbb{C}$-linearly to an endomorphism $I_{\mathbb{C}}$ of $V\otimes \mathbb{C}$, and decompose the latter into $\pm i$ eigenspaces, usually called $V^{1,0}$ and $V^{0,1}$. The identity transformation decomposes into a sum of (eigenspace) projectors: $$1 = \frac{1}{2}\left( 1 - i I_{\mathbb{C}}\right) + \frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right),$$ and we have a $\mathbb{C}$-linear isomorphism $V\subset V\otimes\mathbb{C}\to V^{1,0} = \ker\frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right)$. Under this isomorphism $V\ni v\mapsto \frac{1}{2}(v-iIv)\in V^{1,0}$. </p> <p>The complex structure $I$ induces a complex structure on the dual space $V^\vee$: this is the dual (transpose) endomorphism: $I^\vee \alpha(v)=\alpha(Iv)$. Its complexification, $I^\vee_\mathbb{C}$, induces an eigenspace decomposition of $V^\vee\otimes\mathbb{C}$.</p> <p>For $\mathbb{R}^2 = (\mathbb{R}^2) ^\vee$ the "standard" complex structure $I$ is represented by the matrix $\left(\begin{array}{rr} 0&amp;-1\\ 1&amp;0\\ \end{array}\right)$, and $I^\vee$ by $\left(\begin{array}{rr} 0&amp;1\\ -1&amp;0\\ \end{array}\right)$. Notice that $I$ is skew-symmetric and the $\pm i$ eigenspaces of $I$ and $I^\vee$ are interchanged! </p> <p>If you apply the above isomorphism $V\simeq V^{1,0}$ ( resp. $V\simeq V^{0,1}$) to the first standard basis vector $e_1\in\mathbb{R}^2$, you will get $$\left\{ \frac{1}{2} \left( \begin{array}{r} 1\\ -i\\ \end{array} \right), \frac{1}{2} \left(\begin{array}{r} 1\\ i\\ \end{array}\right) \right\},$$ an eigenbasis (for $I_\mathbb{C}$) of $\mathbb{C}^2$, where the eigenvalues are ordered $\{+i,-i\}$. Its dual basis is $$\left\{ \left(\begin{array}{r} 1\\ i\\ \end{array} \right), \left(\begin{array}{r} 1\\ -i\\ \end{array} \right) \right\},$$ consisting of eigenvectors for $I^\vee_\mathbb{C}$, with eigenvalues $\{+i,-i\}$.</p> <p>Now rephrase all of the above in terms derivations. The $\mathbb{C}$-isomorphism $V\simeq V^{1,0}$ gives you $\frac{\partial}{\partial x}\mapsto \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)$, and $$\left\{ \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) \right\}$$ is a $\mathbb{C}$-basis of $\mathbb{R}^2\otimes \mathbb{C}$, dual to $\left\{dx + i dy, dx-idy \right\}$. As the latter is conventionally denoted by $\left\{dz,d\overline{z}\right\}$, it is natural to denote its dual basis by $\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$.</p> <p>Notice that with this convention $dz\left(\frac{\partial}{\partial \overline{z}} \right) =0$ and $d\overline{z}\left(\frac{\partial}{\partial \overline{z}} \right) =1$, which is highly desirable, as mentioned in the comments.</p> <p><strong>ADDENDUM</strong></p> <p>In sum, we have to make a compatible choice of:<p></p> <p>1) Eigenvectors of $I$<br> 2) Eigenvectors of $I^\vee$.<br></p> <p>We tend to put bars on the eigenvectors with eigenvalue $(-i)$. Eigenvectors are determined up to a nonzero scalar, so <em>a priori</em> we have to choose 4 of these (in $\mathbb{C}^2=\mathbb{R}^2\otimes\mathbb{C}$ ). However, the conditions:<p> a) $\frac{\partial}{\partial \overline{z}} = \overline{\frac{\partial}{\partial z}}$<br> b)$\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$ is a dual $\mathbb{C}$-basis to $\left\{dz,d\overline{z}\right\}$<br> leave us only with the freedom of replacing $\frac{\partial}{\partial z}$ by a scalar multiple. </p> <p>As an example, if you take the complex structure $\left(\begin{array}{rr} -1&amp;-2\\ 1&amp;1\\ \end{array}\right)$, then, up to that single scalar ambiguity, we have $$dz = dx+ (1+i)dy, \ d\overline{z} = dx + (1-i) dy$$ $$\frac{\partial}{\partial z}=\frac{1}{2}\left( (1+i)\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left( (1-i)\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)$$</p> http://mathoverflow.net/questions/91294/characterization-of-the-lie-derivative/91328#91328 Answer by Peter Dalakov for Characterization of the Lie derivative Peter Dalakov 2012-03-15T20:05:40Z 2012-03-15T20:05:40Z <p>Let me add my two cents to the great answers of Dick Palais and Tom Goodwillie. Let $A^\bullet(M)=\bigoplus_{k=0}^\infty A^k(M)$ be the graded algebra of differential forms on your manifold $M$. Then, for a vector field $X$, $\iota_X$ is the unique derivation of degree $(-1)$, satisfying $\iota_X(\alpha)=\alpha(X)$ for all $\alpha\in A^1(M)$, and $d$ is the unique derivation of degree $(+1)$, satisfying $df(X)=X(f)$ for all $f\in A^0(M)$.</p> <p>If $D_1$ and $D_2$ are graded derivations of degrees $p$ and $q$ respectively, their graded commutator $$D_1D_2 - (-1)^{pq}D_2D_1$$ is a derivation of degree $p+q$. In our case, this is $$L_X = d\iota_X + \iota_X d,$$ a degree zero derivation.</p> http://mathoverflow.net/questions/89505/reference-for-tensors-on-graded-spaces-needed/89524#89524 Answer by Peter Dalakov for Reference for Tensors on graded spaces needed Peter Dalakov 2012-02-25T21:45:22Z 2012-02-25T21:45:22Z <p>One pedagogical treatment of 1) is Chapter VIII from Marco Manetti's <i>Lectures on deformations of complex manifolds</i></p> <p><a href="http://lanl.arxiv.org/abs/math/0507286" rel="nofollow">http://lanl.arxiv.org/abs/math/0507286</a></p> <p>Another reference is Quillen's <i>Rational homotopy theory</i>, Appendix B. It is published in <i>The Annals of Math.</i>, Vol.90, No.2, 1969.</p> http://mathoverflow.net/questions/88740/good-book-on-riemann-surfaces-and-galois-theory/88745#88745 Answer by Peter Dalakov for Good book on Riemann surfaces and Galois theory? Peter Dalakov 2012-02-17T18:07:43Z 2012-02-17T18:07:43Z <p>I think V.I.Arnold's lectures <a href="http://www.amazon.com/Abels-Theorem-Problems-Solutions-International/dp/1402021860" rel="nofollow">"Abel's Theorem in Problems and Solutions" </a> may be a great supplementary reading. The book is basic but beautiful.</p> http://mathoverflow.net/questions/86782/metric-connections-on-a-lie-group/86822#86822 Answer by Peter Dalakov for Metric Connections on a Lie Group Peter Dalakov 2012-01-27T12:54:24Z 2012-01-27T13:20:03Z <p>Yes. </p> <p>Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z)$$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z)$$ $$+2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$</p> <p>where $$T(X,Y)=\nabla_X Y- \nabla_Y X -[X,Y]$$ is the torsion of $\nabla$ and $$N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y).$$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.</p> <p>Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$(\nabla_X Y)_e=\epsilon [X,Y],$$ $\epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0,$$ hence the names of the connections. But then you get $$0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.</p> http://mathoverflow.net/questions/85560/are-nilpotent-orbits-degenerations-of-semi-simple-orbits/85564#85564 Answer by Peter Dalakov for Are nilpotent orbits degenerations of semi-simple orbits ? Peter Dalakov 2012-01-13T09:23:36Z 2012-01-14T16:24:35Z <p>There was a conjecture of Grothendieck, proved by Brieskorn in 1970, which says that:</p> <p>Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$ of ADE type, and $x\in \mathfrak{g}$ a <em>subregular</em> nilpotent element. Let $S\subset \mathfrak{g}$ be a transversal (in $x$) slice to the $G$-orbit of $x$. Then the restriction $(S,x)\to (\mathfrak{t}/W,0)$ of the adjoint quotient is a semi-universal deformation of the corresponding simple ADE singularity.</p> <p>For other types the situation is a bit trickier, but is discussed in Slodowy, <em>Four lectures on simple groups and singularities</em>.</p> <p>Also, Lehn, Namikawa and Sorger</p> <p><a href="http://lanl.arxiv.org/abs/1002.4107" rel="nofollow">http://lanl.arxiv.org/abs/1002.4107</a></p> <p>give a classification of the nilpotent orbits for which the restriction of the adjoint quotient to the Slodowy (Kostant) slice gives a universal Poisson deformation of the central fibre.</p> <p>I may edit this later, as I am not sure if this is the kind of statement you are looking for.</p> http://mathoverflow.net/questions/85467/when-is-a-coadjoint-orbit-an-integrable-system-in-a-weak-sense-explained-below/85478#85478 Answer by Peter Dalakov for When is a coadjoint orbit an integrable system (in a weak sense explained below)? Peter Dalakov 2012-01-12T09:47:52Z 2012-01-12T09:47:52Z <p>Here is a related result of Azad, Ban and Biswas which you may find interesting.</p> <p>Let $G$ be a complex semisimple Lie group, $\mathfrak{g}=\textrm{Lie}(G)$. Let $c\in \mathfrak{g}$ be a semi-simple element and $\mathcal{O}:=\textrm{Ad}(G)\cdot c$. Let $\Omega$ be the (holomorphic) Kostant-Kirillov symplectic form. Suppose all eigenvalues of $\textrm{ad}(c)$ are real. Then $(\mathcal{O},\textrm{Re }\Omega)\simeq T^\vee (G/P)$, where $P\subset G$ is any parabolic with Levi component $Z(c)$, the centraliser of $c$ in $G$. Here $\simeq$ means ''an isomorphism of real symplectic manifolds'', and $T^\vee (G/P)$ is equipped with the Liouville symplectic form.</p> <p>A similar result holds if all eigenvalues of $\textrm{ad}(c)$ are purely imaginary, but then you have to take $\textrm{Im }\Omega$.</p> <p>This generalises a theorem of Arnold for $SL(n,\mathbb{C})$.</p> http://mathoverflow.net/questions/80413/analog-of-residue-for-meromorphic-quadratic-differentials/80426#80426 Answer by Peter Dalakov for Analog of residue for meromorphic quadratic differentials Peter Dalakov 2011-11-08T20:53:17Z 2011-11-11T13:46:18Z <p>Your definition definitely works <strong>when the quadratic differential has poles of order at most two.</strong> Let $X$ be a Riemann surface and $z$ a local coordinate centered at $x\in X$. The principal part of order at most 2 of a meromorphic differential $\omega$ is of the form $$R_2(\omega) = \left( \frac{a}{z^2} + \frac{b}{z}+ \ldots \right) dz^{\otimes 2}$$ and $Res^2_x (\omega):=a$. Principal parts of order at most $n$ form a rank $n$ vector bundle $R_n$, and there is a filtration $R_n\supset R_{n-1}\supset\ldots$ . In particular, we have a short exact sequence $$0 \to R_1\to R_2 \to \mathcal{O}_X\to 0,$$ and the last map is $Res^2:R_2\to \mathcal{O}_X$.</p> <p>Stalkwise, this can be identified with $$0\to \mu_x/\mu_x^2\to \mathcal{O}_{X,x}/\mu_x^2\to \mathcal{O}_{X,x}/\mu_x\to 0.$$ A more familiar form of the first sequence is $$0\to \Omega_X^1\to J\to \mathcal{O}_X\to 0$$ where $J$ is the unique non-trivial extension of $\mathcal{O}_X$ by $\Omega_X^1$ (corresponding to the Kaehler class). Or, if you want, $J= J^1(K^{1/2}_X)\otimes K^{-1/2}_X$.</p> <p>The theory of residues of quadratic differentials is quite classical. A great overview is A.N.Tyurin, "On periods of quadratic differentials", Russ.Math.Surveys, 33 (169), 1978.</p> http://mathoverflow.net/questions/79850/software-for-computing-baker-campbell-hausdorff/79851#79851 Answer by Peter Dalakov for Software for Computing Baker-Campbell-Hausdorff Peter Dalakov 2011-11-02T17:46:52Z 2011-11-02T17:46:52Z <p>I don't have a first-hand experience, but hope this is helpful anyway. K.Engo, A.Marthinsen and H.Munthe-Kaas have done a lot of work on numerical methods for solving ODE on manifolds (and Lie groups in particular). See for example their paper ''DiffMan: an object-oriented Matlab toolbox for solving differential equations on manifolds'', Appl. Numerical Mathematics, 39 (2001), p.323 where they discuss a particular package.</p> http://mathoverflow.net/questions/73307/rank-2-flat-bundles-on-an-elliptic-curve-via-extensions/73356#73356 Answer by Peter Dalakov for Rank 2 flat bundles on an elliptic curve, via extensions Peter Dalakov 2011-08-21T20:26:52Z 2011-08-21T20:26:52Z <p>Here is an answer which doesn't invoke non-Abelian Hodge theory.</p> <p>In his paper <em>"Complex Analytic Connections in Fibre Bundles"</em> (Trans.AMS, v.85, 1957), Atiyah showed the following: Let $V$ be a holomorphic vector bundle on a (smooth, compact) curve, and let $V=\oplus_i V_i$ be the decomposition of $V$ into indecomposable bundles. Then $V$ admits a holomorphic connection (necessarily flat, for dimension reasons) if and only if $\deg V_i=0, \forall i$. In particular, <em>any</em> indecomposable bundle of degree zero admits a holomorphic connection. See Propositions 17,19 and Theorem 10 from the paper. In particular, for any $L\in Pic^0$, the bundles $L\oplus L^{\vee}$ and $L\otimes I_2$ admit a holomorphic connection, whereas a rank 2 bundle of the form $F\oplus F^{\vee}$, $\deg F\neq 0$ never admits a holomorphic connection. (Here $I_2$ is the unique non-trivial extension of $\mathcal{O}$ by $\mathcal{O}$.) Now you can use the calculation that Michael Thaddeus mentions to see that your $V$ admits a holomorphic connection as long as the extension is non-split.</p> http://mathoverflow.net/questions/72498/what-is-the-role-of-contact-geometry-in-the-hamiltonian-mechanics/72533#72533 Answer by Peter Dalakov for What is the role of contact geometry in the hamiltonian mechanics? Peter Dalakov 2011-08-10T00:36:48Z 2011-08-10T00:36:48Z <p>In mechanics you often want to study systems whose Hamiltonian function depends on time (explicitly). For example, you can look at the motion of a charged particle in a <i>time-dependent</i> electric field. In such cases you are solving an ODE in the "extended phase space" ($\mathbb{R}^7$ in the above example), and not in $\mathbb{R}^6\simeq T^\vee \mathbb{R}^3$. Also, the translation between Hamiltonian and Lagrangian formulation of mechanics goes via Legendre transform, which fits very nicely in the framework of contact geometry. Contact geometry also enters mechanics through Hamilton-Jacobi theory and the "method of characteristics".</p> <p>So you can think of contact geometry as the odd-dimensional ("non-stationary") analogue of symplectic geometry. You can go from one to the other by "symplectisation". For example, if you start with a manifold, $M$, then you have a tautological contact structure on $X=\mathbb{P}T^\vee_M$. The symplectisation of $X$ is $T^\vee _M-{ 0 }$, with the canonical symplectic form. Symplectisation maps contact diffeomorphisms to symplectomorphisms, etc. etc. In the opposite direction, if you are given a symplectic manifold$(M,\omega)$ with $[\omega]=0\in H^2(M,\mathbb{R})$, you can build a line bundle $E\to M$ with a contact structure.</p> http://mathoverflow.net/questions/64873/newlander-nirenberg-theorem-for-general-vector-bundles/64899#64899 Answer by Peter Dalakov for Newlander-Nirenberg theorem for general vector bundles Peter Dalakov 2011-05-13T14:37:18Z 2011-05-13T14:37:18Z <p>Another great reference is S.Kobayashi, "Differential Geometry of Complex Vector Bundles", Chapter I, Proposition 3.7.</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/34054#34054 Answer by Peter Dalakov for Why is cotangent more canonical than tangent? Peter Dalakov 2010-07-31T20:01:18Z 2010-07-31T20:01:18Z <p>There is an asymmetry'' already at the linear algebra level: given a (fin.dim.) vector space $V$, you get a canonical indefinite pairing and a symplectic form on $V\oplus V^\vee$, while you do not get these on $V\oplus V$. </p> http://mathoverflow.net/questions/123942/how-many-flat-connections-has-a-line-bundle-in-algebraic-geometry/123945#123945 Comment by Peter Dalakov Peter Dalakov 2013-03-10T19:52:14Z 2013-03-10T19:52:14Z You do have though that all of the Chern classes vanish: the characteristic ring is generated by the Atiyah class, which is zero: see Thm 4 in Atiyah's paper. http://mathoverflow.net/questions/123942/how-many-flat-connections-has-a-line-bundle-in-algebraic-geometry/123945#123945 Comment by Peter Dalakov Peter Dalakov 2013-03-10T19:37:32Z 2013-03-10T19:37:32Z A priori, there is no reason for the connection $\nabla +\overline{partial}$ to have curvature of type (1,1) (and hence to be flat). It is the curvature of the Chern connection that satisfies a reality property (if you have fixed an hermitian structure). Also, you mean to write that $\nabla+\overline{\partial}$ is a connection on the complex line bundle corresponding to $\mathscr{L}$, not the holomorphic one. http://mathoverflow.net/questions/122729/why-is-it-important-that-partial-derivatives-commute Comment by Peter Dalakov Peter Dalakov 2013-02-25T11:38:34Z 2013-02-25T11:38:34Z If the torsion vanishes, horizontal 1-forms are closed: $\nabla\alpha =0\Rightarrow d\alpha=0$, hence locally exact. You can use the &quot;potentials&quot; as local coordinates, etc. http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate Comment by Peter Dalakov Peter Dalakov 2012-12-05T17:49:20Z 2012-12-05T17:49:20Z I totally agree with Sam, though I would tend to say that we're working with the manifold $\mathbb{R}$, together with a volume form $vol=dx$, so one could write this operator as $d/vol$, if so desired. http://mathoverflow.net/questions/115289/conformal-blocks-for-beginners Comment by Peter Dalakov Peter Dalakov 2012-12-05T17:39:17Z 2012-12-05T17:39:17Z Right, of course, you then have to work with bundles on singular curves. Though I think there has been work in that direction. (There was a paper of I.Kausz on this, I believe). But at least you will have a well-motivated (for AG-ers) description in the interior. http://mathoverflow.net/questions/115289/conformal-blocks-for-beginners Comment by Peter Dalakov Peter Dalakov 2012-12-04T11:19:18Z 2012-12-04T11:19:18Z Have you tried starting with the &quot;obvious&quot; introduction/motivation: the moduli of vector bundles of fixed det,coprime rank &amp; deg (and n=0) when the coarse moduli space is smooth, with $Pic\simeq \mathbb{Z}$. You can then discuss the Verlinde bundle over Teichmueller space. If you want to be really concrete, you can mention e.g. rank 2 bundles of degree zero and fixed determinant for $g=2$, when the coarse moduli space is $\mathbb{P}^3$. And then you can say that you want to upgrade this to a fancier version, living on $\overline{\mathcal{M}}_{g,n}$? Or is this too trivial for your audiences? http://mathoverflow.net/questions/114787/what-is-teichmuller-theory-and-its-history Comment by Peter Dalakov Peter Dalakov 2012-11-28T21:48:30Z 2012-11-28T21:48:30Z Sasha, I would suggest the (somewhat passionate) review of W.Abikoff for Nag's book &quot;The complex analytic theory of Teichmueller spaces&quot;: <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183555146" rel="nofollow">projecteuclid.org/&hellip;</a> http://mathoverflow.net/questions/114414/is-there-any-holomorphic-version-of-the-tubular-neighborhood-theorem Comment by Peter Dalakov Peter Dalakov 2012-11-26T11:21:08Z 2012-11-26T11:21:08Z The proofs of both the differential-geometric and symplectic versions use some sort of exponential map/flow/Poincare-lemma-type argument. Heuristically, this explains why you need a very special complex manifold if you want to have a holomorphic version: you must have some sort of holomorphic exponential mapping/flow. http://mathoverflow.net/questions/114414/is-there-any-holomorphic-version-of-the-tubular-neighborhood-theorem Comment by Peter Dalakov Peter Dalakov 2012-11-26T11:15:14Z 2012-11-26T11:15:14Z You can also look at P.Griffiths' paper &quot;The extension problem in complex analysis&quot;, Am.J.M., Vol.88, No.2 1996 for related questions. http://mathoverflow.net/questions/112256/what-are-the-applications-of-stokes-theorem-in-electrostatic-field Comment by Peter Dalakov Peter Dalakov 2012-11-13T08:47:56Z 2012-11-13T08:47:56Z Sir George's surname is actually Stokes, not Stoke. http://mathoverflow.net/questions/111430/tangent-bundle-and-normal-bundle-in-self-product Comment by Peter Dalakov Peter Dalakov 2012-11-04T16:32:10Z 2012-11-04T16:32:10Z @Will, @LMN: Exactly! A vector $PQ$ in $\mathbb{R}^n$ has a &quot;tail&quot; (P) and &quot;head&quot; (Q). A (smooth) vector field on $\mathbb{R}^n$ can be thought of as a map $\mathbb{R}^n\to\mathbb{R}^n\times\mathbb{R}^n$, $P\mapsto (P,Q)$. And $P\mapsto (P,P)$ is the zero section, $PQ=OQ-OP$. To make sense of this properly you need the tangent sequence of the diagonal, see Michael's or Sandor's answer. I am saying something really trivial, though. http://mathoverflow.net/questions/111430/tangent-bundle-and-normal-bundle-in-self-product Comment by Peter Dalakov Peter Dalakov 2012-11-04T09:20:06Z 2012-11-04T09:20:06Z Morally, this is actually the way we introduce vectors in (pre)calculus or in school, right? We talk of pairs of points $P$,$Q$ in $\mathbb{R}^2$ (up to equivalence), the &quot;diagonal pair&quot; $(P,P)$ being (equivalent to) zero. http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau Comment by Peter Dalakov Peter Dalakov 2012-10-21T15:46:41Z 2012-10-21T15:46:41Z Or maybe you are asking how the existence of the Kaledin-Feix metric gives you a special lagrangian embedding? http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau Comment by Peter Dalakov Peter Dalakov 2012-10-21T14:30:47Z 2012-10-21T14:30:47Z I am not exactly sure what you mean by &quot;explanation&quot;, but since you're asking for references, have a look at Birte Feix's thesis &quot;Hyperkaehler metrics on cotangent bundles&quot;. There she constructs the HK metric in a different way. See also <a href="http://mathoverflow.net/questions/46752/is-the-cotangent-bundle-to-a-kahler-manifold-hyperkahler" rel="nofollow" title="is the cotangent bundle to a kahler manifold hyperkahler">mathoverflow.net/questions/46752/&hellip;</a> http://mathoverflow.net/questions/107168/weitzenbock-identities Comment by Peter Dalakov Peter Dalakov 2012-09-14T13:30:26Z 2012-09-14T13:30:26Z The Weitzenboeck technique is mostly used to show that some (first order, self-adjoint) differential operator $D$ (acting on sections of an hermitian bundle over a compact manifold $M$) has trivial kernel. Since the $L^2$ norm of $Ds$ is $\int_M (D^2s,s)$, the goal is to write $D^2$ as some &quot;standard Laplacian&quot; plus a 0-order scalar term, $D^2=\Delta +A$. If $A$ is pointwise positive, by the maximum principle for $\Delta$ you get that $Ds=0$ implies $s=0$.