Given a projective variety $X \subset \mathbb{P}^n_\mathbb{C}$, how can I compute the cotangent sheaf? Is it just the sheafification of the kahler differentials? For example, if I take the K3 surface $$ X = \textbf{Proj}(R) = \textbf{Proj}\left( \frac{\mathbb{C}[x,y,z,w]}{(x^4 + y^4 + z^4 + w^4)} \right) $$ then the kahler differentials of the underlying graded ring is the graded module $$ \frac{Rdx \oplus Rdy \oplus Rdz \oplus Rdw}{x^3dx + y^3dy + z^3dz + w^3dw} $$ Given a finite presentation of a smooth projective variety $S/(f_1,\ldots, f_k)$, the diffrerential of the $f_i$'s should always be homogeneous.

  • $\begingroup$ Does your description work locally? $\endgroup$ – Mariano Suárez-Álvarez Dec 23 '16 at 23:29
  • $\begingroup$ Do you mean in an affine chart? $\endgroup$ – 54321user Dec 23 '16 at 23:38
  • $\begingroup$ For example. ${}{}$ $\endgroup$ – Mariano Suárez-Álvarez Dec 23 '16 at 23:56

The answer is yes. The sheaf $\Omega_X^1$ is the sheafification of the module $\Omega_{S/\mathbb C}^1$, where $S$ is the homogeneous coordinate ring of $X$.

If you want to compute the cohomology of $\mathbb \Omega_X^1$, you can use standard exact sequences:

The Euler sequence: $$ 0 \to \Omega_{\mathbb P}^1|X \to \mathscr O_X(-1)^{N+1} \to \mathscr O_X \to 0. $$ The cotangent sequence: $$ 0 \to \mathscr I/\mathscr I^2 \to \Omega_{\mathbb P}^1|X \to \Omega_X^1 \to 0. $$ Note that if $X$ is a hypersurface of degree $d$, then $\mathscr I/\mathscr I^2 = \mathscr O_X(-d)$.

Tensoring the ideal sequence we also get the sequence (in the case of a hypersurface; the general case is similar): $$ 0 \to \Omega_{\mathbb P}^1(-d) \to \Omega_{\mathbb P}^1 \to \Omega_{\mathbb P}^1|X \to 0. $$

Now playing around with long exact sequences should give you all the information needed to compute the cohomology of $X$.

  • $\begingroup$ Oh, so if I evaluate $\Omega^1_X$ on one of the affine opens $U_i$ by inverting one of the coordinates, this will be isomorphism to the localization of $\Omega^1_{S/\mathbb{C}}$ $\endgroup$ – 54321user Dec 24 '16 at 20:03
  • $\begingroup$ @user251222 Yes, I believe that's true. $\endgroup$ – Fredrik Meyer Dec 25 '16 at 11:14

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