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May 14, 2016 at 2:40 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 14, 2016 at 2:31 answer added Mohammad Farajzadeh-Tehrani timeline score: 2
Mar 24, 2016 at 17:23 comment added Jason Starr There is a holomorphic and fiberwise linear map from the complex line bundle $T_\Sigma(-\sum_i p_i)$ to the complex line bundle $T_\Sigma$. That induces an injective map of sheaves $A(T_\Sigma(-\sum_i p_i)) \to A(T_\Sigma)$. Now compose that map with $du$. By abuse of notation, I am also calling that composition $du$.
Mar 24, 2016 at 15:42 comment added Mohammad Farajzadeh-Tehrani Thanks Jason, I think this sheaf theoretic description of yours was the key to make sense of Q2. Now I see where I was struggling: to make sense of the map $du\colon \mathbb{C}^\infty(T\Sigma(-p)) \to \mathbb{C}^\infty (u^*TX)$. Originally, $du$ is a map from $T\Sigma$ to $u^*TX$, how do the marked points affect it then?
Mar 24, 2016 at 11:53 comment added Jason Starr One clarification: for a holomorphic vector bundle $E$, $E(-\sum_i p_i)$ is a holomorphic vector bundle. When I write $E(-\sum_i p_i)$, I really mean the fine, $C^\infty$ sheaf $A(E(-\sum_i p_i))$. As a subsheaf of $A(E)$, this is different from the subsheaf of $C^\infty$ sections that vanish at each $p_i$. For instance, on the complex plane, the real coordinate $x$ is a section of $A(\mathbb{C})$ that vanishes at $0$, but it does not equal a $C^\infty$, $\mathbb{C}$-valued function times the complex coordinate $z=x+iy$, which is a generator of $A(\mathbb{C}(-\underline{0}))$.
Mar 24, 2016 at 11:43 comment added Jason Starr Yes, all of the sheaves in those complexes are the fine, $C^\infty$ sheaves, not the holomorphic sheaves. Of course for a holomorphic vector bundle $E$ on $\Sigma$ with associated fine, $C^\infty$ sheaf $A(E)$ of $C^\infty$ sections, the complex in degrees $[0,1]$, $\overline{\partial}:A(E)\to \Omega^{1,0}\otimes A(E)$ is an acyclic resolution of the sheaf $\mathcal{O}(E)$ of holomorphic sections,. Hence the complex has hypercohomologies equal to $H^0(\Sigma,\mathcal{O}(E))$, resp. $H^1(\Sigma,\mathcal{O}(E))$.
Mar 24, 2016 at 11:25 comment added Mohammad Farajzadeh-Tehrani Thanks Jason. By $T{\Sigma}(-\sum p_i)$ you mean sheaf of smooth (and not the usual holomorphic) sections of $T\Sigma$ that vanish at those points, correct?
Mar 24, 2016 at 10:23 comment added Jason Starr Typo correction: Every $f^*T_X$ should be $u^*T_X$. Also, the visible chain homomorphism of complexes from the first complex to the second complex is injective, and the cokernel is the complex concentrated in degrees $[-1,0]$, $T_\Sigma(-\sum_i p_i) \to \Omega^{1,0}_\Sigma\otimes T_\Sigma(-\sum_i p_i)$. This short exact sequence of complexes gives the long exact sequence in your question. Also, this shows that in the case that $(\Sigma,p_1,\dots,p_k)$ is not stable, then your long exact sequence should begin with $\text{Lie}(\text{Auto}(\Sigma,p_1,\dots,p_k))$.
S Mar 24, 2016 at 1:56 history suggested Silvia Ghinassi CC BY-SA 3.0
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Mar 24, 2016 at 1:50 comment added Jason Starr Your first column is the hypercohomology of the complex of sheaves on $\Sigma$ in degrees $[0,1]$, $f^*T_X\to \Omega^{1,0}_\Sigma\otimes f^*T_X$. Assume $\Sigma$ is smooth. Form the new complex concentrated in degrees $[-1,1]$, $T_\Sigma(-\sum_i p_i)\to f^*T_X \oplus (\Omega^{1,0}_{\Sigma}\otimes T_\Sigma(-\sum_i p_i)) \to \Omega^{1,0}\otimes f^*T_X$. The first map is $(du,\overline{\partial})$, and the second map is $(Du\overline{\partial},-\text{Id}\otimes du)$. The hypercohomology is $\text{Def}(f)$ and $\text{Obs}(f)$. There is a map of these complexes giving your long exact sequence.
Mar 24, 2016 at 1:43 review Suggested edits
S Mar 24, 2016 at 1:56
Mar 24, 2016 at 1:38 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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Mar 24, 2016 at 1:02 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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Mar 24, 2016 at 0:38 comment added Mohammad Farajzadeh-Tehrani I am just trying to write down its analytical description explicitly. Since I don't well understand the short exact sequence from which such long exact sequence is constructed (in the analytical setting) I can't simply follow the algebraic definition of connecting map. $D_u\bar\partial$ is of the form $\bar\partial+A$ where $\bar\partial$ defines a holomorphic structure on $u^*TX$ and $A$ is some degree 0 map (which depends on Nijenhueis tensor). So $Def(u)$ and $Obs(u)$ are deformations of $H^0_{\bar\partial}(TX)$ and $H^1_{\bar\partial}(TX)$. That makes me a bit confused.
Mar 24, 2016 at 0:32 comment added Jason Starr Mohammad, is there any reason to expect that the analytic description of this connecting map is different from the algebraic description? In particular, just to check, are you deforming the Cauchy-Riemann equation?
Mar 24, 2016 at 0:27 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 3.0
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Mar 24, 2016 at 0:22 history asked Mohammad Farajzadeh-Tehrani CC BY-SA 3.0