I am solving a list of problems related to smooth vector bundles over a smooth manifold considered as locally free modules over the sheaf of smooth functions, to Cech cohomologies with coefficients in the sheaf of non-vanishing smooth functions, and to Picard groups. One task is to compute the Picard group of real projective spaces for all dimensions, that is, $\mathrm{Pic}(\mathbb{R}\mathit{P}^n)$ with $n\geq 1$. And I am stuck on this.
I recall that the Picard group $\mathrm{Pic}(M)$ of a smooth manifold $M$ is the group of isomorphic classes of real smooth rank 1 vector (line) bundles over this manifold with respect to tensor product of bundles. I know that the notion of Picard group make sense for every ringed space, but my task is only about the smooth case and I am not really familiar with algebraic geometry, so I would like to receive purely differentiably-geometric solution.
Now I provide some information and tools that I have for now that can help to solve this problem. I know definitions of Cech cohomologies $\check{H}^k(M, \mathcal{A}^{\times}_M)$ (more precisely, I was given only the definition of $\check{H}^1(M, \mathcal{A}^{\times}_M)$ and $\check{H}^0(M, \mathcal{A}^{\times}_M)$, but I believe that higher groups are not needed here), where $\mathcal{A}_M$ is the sheaf of smooth functions on $M$ and $\mathcal{A}^{\times}_M$ is the sheaf of non-vanishing smooth functions on $M$. I know that $\mathrm{Pic}(M)$ is isomorphic to $\check{H}^1(M, \mathcal{A}^{\times}_M)$. I know the Mayer-Vietoris sequence for Cech cohomologies. Finally, I know that $\check{H}^1(M, \mathcal{A}^{\times}_M)$ is isomorphic to $\check{H}^1(\mathcal{U}, \mathcal{A}^{\times}_M)$, where $\mathcal{U}$ is an open cover of $M$ such that $\check{H}^1(U, \mathcal{A}^{\times}_U)=O$ for every $U\in \mathcal{U}$. Generally, it is the only facts about all this that are assumed for solving of my problem (as it seems to me).
The previous task was about computing of $\mathrm{Pic}(S^n)$ for every $n\geq 1$. I did it via computing of $\check{H}^1(S^n, \mathcal{A}_{S^n}^{\times})$ with use of good covers. I got $\mathrm{Pic}(S^1)=\mathbb{Z}_2$ and $\mathrm{Pic}(S^n)=O$ for $n\geq 2$. Probably, these results should help in computing of $\mathrm{Pic}(\mathbb{R}\mathit{P}^n)$, but I have no idea. Goods covers here seem too complicated, for example, the affine cover consists of $n+1$ charts, so it seems hard to compute $\check{H}^1(\mathbb{R}\mathit{P}^n)$ in this way. The Mayer-Vietoris sequence also seems useless because of computational difficulties. May be, there should be an important step, but I do not see it. Thus, it would be great to get some advice.
Best Answer
The way I would do this is to first observe that the sheaf $\mathcal{A}^\times_M$ is isomorphic to $\mathcal{A}^+_M\oplus\{\pm 1\}$, where $\mathcal{A}^+_M$ is the sheaf of positive smooth functions and $\{\pm 1\}$ is the constant sheaf corresponding to the group $\{\pm 1\}\cong\mathbb{Z}/(2)$. The sheaf $\mathcal{A}^+_M$ is acyclic (it is isomorphic to $\mathcal{A}_M$ via exponentiation) so you just need to compute $H^1(M,\mathbb{Z}/(2))$.
Now this is not too hard to compute for $M=\mathbb{RP}^n$ using its covering by $n+1$ coordinate charts. Explicitly, if $U_i$ is the chart where the $i$th homogeneous coordinate is nonzero, $U_i\cap U_j$ has two connected components, one (call it $U_{ij}^+$) where the signs of the $i$th and $j$th coordinates are the same and one (call it $U_{ij}^-$) where they are different. So, a Cech $1$-cochain consists of a list of elements $a_{ij}^+,a_{ij}^-\in \{\pm 1\}$ for each $i,j$. To get a cocycle, you need $a_{ij}^\alpha a_{jk}^\beta=a_{ik}^\gamma$ whenever $\alpha\beta=\gamma$. Similarly, a $0$-cochain is a list of elements $a_i\in\{\pm 1\}$, and its coboundary is the $1$-cochain with $a_{ij}^+=a_{ij}^-=a_ia_j$. Now note that the $a_{ij}^+$ values of any $1$-cocycle can be obtained as such a coboundary, by taking $a_i=a_{i1}^+$. So, modulo coboundaries, we may assume any $1$-cocycle satisfies $a_{ij}^+=1$ for all $i,j$, which then implies the $a_{ij}^-$ values are all equal. There are thus two cohomology classes of $1$-cocycles: one where $a_{ij}^+=a_{ij}^-=1$ for all $i,j$ and one where $a_{ij}^+=1$ and $a_{ij}^-=-1$ for all $i,j$. This gives $H^1(\mathbb{RP}^n,\mathbb{Z}/(2))\cong\mathbb{Z}/(2)$ (assuming $n\geq 1$).
(Alternatively, if you have more machinery available, there are much simpler ways to reach this conclusion. In particular, if you know that sheaf cohomology with constant coefficients coincides with singular cohomology, you can use various tools of algebraic topology. For instance, it follows from the Hurewicz theorem and the universal coefficients theorem that $H^1(X,A)\cong\operatorname{Hom}(\pi_1(X),A)$ for any path-connected space $X$ and any constant coefficients group $A$, so you would just need to compute the fundamental group of $\mathbb{RP}^n$ which is easy using its universal cover $S^n$ for $n\geq 2$. Or, you can use cellular cohomology to see that $H^1(\mathbb{RP}^n)\cong H^1(\mathbb{RP}^2)$ for all $n\geq 2$ so you just need to compute it for $n=2$.
A more geometric perspective is that you can identify $H^1(M,\mathbb{Z}/(2))$ with the group of isomorphism classes of principal $\mathbb{Z}/(2)$-bundles over $M$. A principal $\mathbb{Z}/(2)$-bundle is just the same thing as a $2$-sheeted covering space, and $\mathbb{RP}^n$ has two of them, the trivial one and the double cover $S^n\to\mathbb{RP}^n$.)