Tautological bundle on $\mathbb P^n$ is topologically trivial over the complement of a hypersurface.

Question

I'm asking this because I am confused about the first paragraph of section 1.1 of this paper: https://arxiv.org/pdf/1407.7229.pdf. In particular, I don't see why the $\mathbb C^*$-bundle $\Pi_{d,n}\setminus\Sigma_{d,n}\to N_{d,n}$ is trivial. Is this obvious? What am I missing?

Answer 1

$\Bbb P^n$ is the quotient of $\Bbb A^{n+1}\setminus \{0\}$ by the $\Bbb G_m$ action given by rescaling the coordinates. Restricting to the complement of a hypersurface $V(f)=H\subset \Bbb P^n$, we see that the pre-image under the quotient map is also affine (as it is $D(f)$ where $f$ is considered as a polynomial on $\Bbb A^{n+1}$). Since everything in sight is affine, one may inspect coordinate rings to see that $k[D(f)]\cong k[\Bbb P^n\setminus H]\otimes k[x,x^{-1}]$ and then the product description is clear.

To see this description of the coordinate rings, one may reduce to the case of $f=x_0$ by the embedding of $\Bbb P^n \subset \Bbb P^{N}$ given by $\mathcal{O}(d)$ and a linear change of coordinates, and then it ought to be relatively clear.

Answer 2

The restriction of the tautological bundle ${\mathcal O}(-1)$ to the complement of a hypersurface $X \subset {\mathbb P}^n$ of degree $d > 1$ is NOT trivial. In fact, it is a generator of ${\mathrm{Pic}}({\mathbb P}^n \setminus X) \cong {\mathbb Z}/d{\mathbb Z}$.