How can I compute the cotangent sheaf for a projective variety?

Question

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.

Answer 1

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 $X$.