# Total space of the line bundle $\mathcal{O}(1)$ over $\mathbb{P}^n$

It is well known that total space of the tautological line bundle $\mathcal{O}(-1)$ over projective space $\mathbb{P}^n$ is closed subvariety of $\mathbb{P}^n\times\mathbb{A}^{n+1}$. My question is how to realize total space of $\mathcal{O}(1)$ over $\mathbb{P}^n$ in such manner, i.e. I need an embedding of $Tot(\mathcal{O}(1))$ in simple variety and defining equations. Thanks.

-

It is the complement $\mathbb{P}^{n+1} - \{x\}$ of a point in a projective space.

-
Shouldn't $\mathbb{P}^{n+1}\setminus\{x\}\to \mathbb{P}^n$ be the tautological line bundle? – Fei YE Nov 7 '10 at 1:59
The tautological line bundle contains a divisor, isomorphic to $\mathbb{P}^n$, with normal bundle of degree $-1$ (namely, the zero section). On the other hand every effective compact divisor in $\mathbb{P}^{n+1}-\{x\}$ has a positive normal bundle. In particular, a hyperplane avoiding $x$ has normal bundle of degree $1$. The projection from the point identifies the complement of $x$ with the total space of this normal bundle. – Tony Pantev Nov 7 '10 at 2:20
Thank you, Tony. Maybe you know such simple description of $Tot(O(n))$ for $n>1$ also? – Klim Puhov Nov 8 '10 at 14:23
You can get $Tot(\mathcal{O}(n))$ in a similar manner by deleting a point from a weighted projective space. But this is more contrived. And is not really better than thinking of $Tot(\mathcal{O}(n))$ as a toric variety. So this probably is not what you want. – Tony Pantev Nov 9 '10 at 1:22
$Tot(\mathcal{O}_{\mathbb{P}^{n}}(k))$ is $\mathbb{P}^{n+1}(1,1,\ldots,1,k) - \{x\}$, where $x = (0:0:\ldots:0:1)$. – Tony Pantev Apr 7 at 12:13

Dear Luther King, since you ask for equations, let me add them to Tony's beautifully geometric answer.

Consider $\mathbb P^{n+1}$ with homogeneous coordinates $(z_0:z_1:\ldots:z_{n+1})$ and $\mathbb P^{n}$ enbedded as the hyperplane $z_0=0$. If $x\in\mathbb P^{n+1}$ is the point $x=(1:0:\ldots:0)$, the required total space $T=Tot \mathcal O_{\mathbb P^{n}}(1)$ is the complement $T=\mathbb P^{n+1} \setminus \{x\}$ of $x$ in $\mathbb P^{n+1}$. The fiber of our bundle $\mathcal O_{\mathbb P^{n}}(1)$ at the arbitrary point $(0:z_1:\ldots:z_{n+1}) \in \mathbb P^n$ is the set of all $(\lambda:z_1:\ldots:z_{n+1})$ with $\lambda \in k$ (base field) .

The one-dimensional vector space structure on the fiber is given by $(\lambda:z_1:\ldots:z_{n+1})+(\mu:z_1:\ldots:z_{n+1})=(\lambda +\mu:z_1:\ldots:z_{n+1})$ and similarly for products by scalars. Beware that we definitely do not have an isomorphism from our fiber to $k$ defined by $(\lambda:z_1:\ldots:z_{n+1})\mapsto \lambda$: this is not even a well-defined map. This is not surprising: after all $\mathcal O_{\mathbb P^{n}}(1)$ is not a trivial bundle!

A more canonical approach (edited) Readers of the canonical faith may suppress coordinates as follows.

Consider a $k$ - vector space $V$ of dimension $n+1$, its projectivization $\mathbb P (V)$ and its embedding $\mathbb P (V) \to \mathbb P (k \oplus V)$ sending the point $\mathbb P (l) \in \mathbb P (V)$ to the point $\mathbb P (0\oplus l) \in \mathbb P (k \oplus V)$. The vector bundle $\mathcal O _{\mathbb P (V)} (1)$ then has as total space the open subset $T\subset \mathbb P (k \oplus V)$ obtained by deleting $x=\mathbb P (k \oplus 0)$ from $\mathbb P (k \oplus V)$ i.e. $T= \mathbb P (k \oplus V)\setminus x$. The fiber over $\mathbb P (l)$ [identified with $\mathbb P (0\oplus l)$] is $\mathbb P (k \oplus l)\setminus x$, a vector space of dimension one with origin the point $\mathbb P (0 \oplus l)$ but with no prefered isomorphism to $k$.

Elementary geometry It may clarify the above to recall that given a one dimensional projective space $\mathbb P$ over $k$, if you delete a point $x$ from it you get a one dimensional affine space $\mathbb P \setminus x$ and if you choose in it an origin, you get a one dimensional vector space, but that vector space has no prefered isomorphism with the vector space $k$.

-
Thanks for detailed answer, Georges! – Klim Puhov Nov 7 '10 at 20:22

For me the best description of $Tot(O(1))$ is the tautological --- as the relative spectrum of the sheaf of algebras $A = O \oplus O(-1) \oplus O(-2) \oplus \dots$ on $P^n$: $$Tot(O(1)) = Spec_{P^n}(A).$$ This allows to work with $Tot(O(1))$ more effectively than any other description. For example, a coherent sheaf on $Tot(O(1))$ can be represented by a quasicoherent sheaf $F$ on $P^n$ together with a morphism $F(-1) \to F$ inducing on $F$ a structure of an $A$-module.

-
Yes, I know about such description, but for my purposes the above one is preferable. – Klim Puhov Nov 7 '10 at 20:35