Invariance under change of cohordinates cohomology projective space There is this statement in Hartshorne remark 7.1.1 which I don't understand: he says that $\displaystyle \frac{x_0^n}{x_1\cdots x_n}d\left(\frac{x_1}{x_0}\right)\wedge \cdots \wedge d\left(\frac{x_n}{x_0}\right)$ determines a generator of $H^n(\mathbb{P}^n,\omega)$ which is invariant under change of variables. I don't understand why it is true.
If for example as change of coordinates I take a transposition, say the map
sending $(x_0,\ldots,x_n)$ to $(x_0,x_2,x_1,x_3,\ldots,x_n)$, then the pull back of the differential should become
$$
\begin{align}
& \frac{x_0^n}{x_1\cdots x_n} d\left(\frac{x_2}{x_0}\right) \wedge d\left(\frac{x_1}{x_0}\right) \wedge d\left(\frac{x_3}{x_0}\right) \wedge \cdots \wedge d\left(\frac{x_n}{x_0}\right) \\[10pt]
= & -\frac{x_0^n}{x_1\cdots x_n} d\left(\frac{x_1}{x_0}\right) \wedge\cdots \wedge d\left(\frac{x_n}{x_0}\right).
\end{align}
$$
 A: Note that you don't really have defined an element of $H^n(\mathbb{P}^n,\omega)$. Rather, if you denote by $U_0,\dots,U_n$ the standard covering of $\mathbb{P}^n$, you defined an element of $H^0(U_0\cap\dots\cap U_n,\omega)$. (Think about a change of coordinates such that $x_0'=x_0+x_1$).
Now there exists a map $\partial:H^0(U_0\cap\dots\cap U_n,\omega)\rightarrow H^n(\mathbb{P}^n,\omega)$, however this map depends on the $U_i$'s and in particular of the order ! In fact, if you change the order of the $U_i$ by a permutation $\sigma$, the corresponding map $\partial'$ will satisfy $\partial'=\epsilon(\sigma)\partial$ where $\epsilon(\sigma)$ is the signature of the permutation.
Note that your transposition does not preserve all the $U_i$. Rather, it permutes two of them, so you need to put a sign to get your class in $H^n(\mathbb{P}^n,\omega)$, and now you see it is invariant by this transposition.
Think about the case where there is only two open set $U,V$. Then the map $\partial$ is the boundary map of the Mayer-Vietoris long exact sequence. It does change of sign if you switch $U,V$.
The same is true with a covering consisting of more open subsets. In this case, $\partial$ comes from $\check{C}^n({U_0,\dots,U_n},\omega)\rightarrow\check{H}^n(\mathbb{P}^n,\omega)$. Depending on your definition of the Cech complex, the sign comes up more or less easily. See here for a comparison of the methods (using unordered cochain, ordered cochain and alternating cochain) : math.stanford.edu/~conrad/papers/cech.pdf.
In general, I think that signs in homological algebra are sometimes really painful to deal with. If a sign comes up and I am not happy about it, I wonder if it does not come from a careless computation of a boundary morphism.
