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For any ring $A$, I wish to calculate the sheaf of differentials $\Omega_{\mathbb{P}_A^n/A}$. Let $S=A[x_0,\ldots,x_n]$, and let $S_{(x_i)}=A[x_{0/i},\ldots,\widehat{x_{i/i}},\ldots,x_{n/i}]$. I know that I have that $$\Omega_{\mathbb{P}_A^n/A}\vert_{D_+(x_i)}\cong\left(\Omega_{S_{(x_i)}/A}\right)^\tilde{}\cong\left(\bigoplus_{j\ne i}S_{(x_i)}\mathrm{d}x_{j/i}\right)^\tilde{}$$ but I don't see how to glue.

Dominic Wynter
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    This is the cotangent sheaf, for $\mathbb P^1$ you should have $\Omega_{\mathbb P^1} = \mathscr O(-2)$. The transition functions for the tangent sheaf of $\mathbb P^n$ are simply the derivatives of the usual transitions functions, and the matrix of transition functions for the cotangent sheaf is just the inverse of the tangent sheaf. –  Apr 22 '17 at 01:53
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    For the projective space $P^n$ it is much more convenient to use the Euler sequence $0 \to \Omega \to O(-1)^{n+1} \to O \to 0$, then transition functions. – Sasha Apr 22 '17 at 04:41

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Suppose $R= \mathbb{C}[x_1,\dots, x_n]$. Then $\Omega_{R/\mathbb{C}} \cong <dx_1, \dots, dx_n>$ as a free $R-$module. On $\mathbb{P}^n_\mathbb{C}$, this means that on the open set $U_0 = Spec(\mathbb{C}[x_1,\dots, x_n])$, we have that the sections of $\Omega_{\mathbb{P}^n/\mathbb{C}}$ over $U_0$ are generated by $<dx_1, \dots, dx_n>$. Now the transition functions for $\Omega_{\mathbb{P}^n/\mathbb{C}}$ are given by the Jacobian of the transition functions for the coordinates, as stated above.

For $\mathbb{P}^1$, the transition function from $U_0 = Spec(\mathbb{C}[x])$ to $U_1 = Spec(\mathbb{C}[x^{-1}])$ is $x\mapsto x^{-1}$. The Jacobian is $(-x^{-2})$. This says that on $\Omega_{\mathbb{P}^1_\mathbb{C}}$, if we call $a_1$ the coordinate on $U_1$, $da_1 = \frac{-1}{x^2}dx$ on $U_1$ in terms of the coordinate $x$ on $U_0$. This is the isomorphism to $\mathcal{O}_{\mathbb{P}^1}(-2)$ referred to above.

For $\mathbb{P}^2$, let's consider $U_0 = Spec(\mathbb{C}[x,y])$ and $U_2 = Spec(\mathbb{C}[xy^{-1},y^{-1}])$. I'll think of the transition functions as $(x,y) \mapsto (xy^{-1},y^{-1})$ so the Jacobian is given by $$ \left( \begin{matrix} y^{-1} & \frac{-x}{y^2} \\ 0 & -y^{-2} \end{matrix}\right) $$

If we call the coordinates $(a_1,a_2)$ on $U_2$, this says that $$da_1 = \frac{1}{y}dx - \frac{x}{y^2}dy $$ and $$da_2 = \frac{-1}{y^2}dy$$ on $U_2$, in terms of the $x,y$ coordinates on $U_0$.

locally trivial
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