Timeline for Why is a convex variety called convex?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 8, 2020 at 4:11 | comment | added | Tabes Bridges | @YuhangChen this is a trick you should get familiar with. The only way I can remember the full calculation of cohomology for line bundles on $\mathbb P^n$ is via the slogan "line bundles of non-negative degree have global sections. the only other cohomology is $H^n$ forced by those global sections and Serre duality." | |
| Jan 8, 2020 at 3:54 | vote | accept | Yuhang Chen | ||
| Jan 8, 2020 at 3:54 | comment | added | Yuhang Chen | @TabesBridges Thanks for the detailed explanations. The application of Serre duality to compute $h^1$ is pretty neat! | |
| Jan 8, 2020 at 2:18 | comment | added | Tabes Bridges | @YuhangChen Sorry, I suppose a constant map to a variety $X$ will actually have $f^* T_X$ trivial of rank $\dim X$ (too late to edit previous comment). So instead the RR calculation reads $\dim X - h^1(\Sigma,f^* T_X) = (\dim X)(1-g)$, which reduces again to $h^1(\Sigma, f^* T_X) = 0$. | |
| Jan 8, 2020 at 2:11 | comment | added | Tabes Bridges | @YuhangChen... in this case is $h^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(-2)\otimes \mathcal O_{\mathbb P^1}(-2)) = h^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(-4)) = 0$, where $\omega_C$ is the canonical bundle (here the same as the holomorphic cotangent bundle), and we use that negative degree line bundles have no sections. | |
| Jan 8, 2020 at 2:09 | comment | added | Tabes Bridges | @YuhangChen First, the last term on the RHS of Riemann-Roch is $+1$, not $-1$. If you plug $h^0(\Sigma, L) = 1$ and $\deg L = 0$ (which uniquely determine that $L$ is trivial), you get $1 - h^1(\Sigma,L) = 0 - g + 1$, which does indeed simplify to $h^1 = g$, but in general $h^1(C,\mathcal O_C)$ is one way of computing the genus of a smooth, projective curve. In case 2 I'm referring to the computation of the cohomology of line bundles on projective space. This particular fact can also be recovered from Serre duality, which identifies $h^1(C,L) = h^0(C, \omega_C\otimes L^\vee)$ which.... | |
| Jan 7, 2020 at 23:49 | comment | added | Yuhang Chen | @TabesBridges In case one, did you get the result from the Riemann-Roch theorem, i.e., $h^0(\Sigma,L) -h^1(\Sigma,L) = \deg L -g-1$ where the line bundle $L = f^*T_X \cong \Sigma \times \mathbb{C}$? In case 2, excuse my ignorance but what did you mean by standard calculations? | |
| Jan 7, 2020 at 20:37 | comment | added | Tabes Bridges | In case one, $f^*T_X$ is trivial, so $H^1 = g(\Sigma) = 0$. In case 2, $f^* T_X = \mathcal O_{\mathbb P^1}(2)$ so $H^1 = 0$ by standard calculations. So yes, both of these are unobstructed. | |
| Jan 7, 2020 at 16:40 | comment | added | Yuhang Chen | So both cases have zero obstructions, right? | |
| Jan 7, 2020 at 15:37 | history | edited | Ben McKay | CC BY-SA 4.0 |
added two examples
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| Jan 7, 2020 at 15:24 | comment | added | Yuhang Chen | Thanks for the explanation. It is very intuitive and helpful. Also thanks for the reference. Can you say something about $H^1(\Sigma, f^*T_X)$ in the following two cases? Case 1: The map $f$ is a constant, i.e., it maps the source curve to a point. Case 2: The map $f$ is an identity from the line $\mathbb{CP}^1$ to itself. | |
| Jan 7, 2020 at 14:58 | history | edited | Ben McKay | CC BY-SA 4.0 |
added notion of tangent space
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| Jan 7, 2020 at 10:45 | history | answered | Ben McKay | CC BY-SA 4.0 |