Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety? - MathOverflow

Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?

2010-01-27T14:59:11Z

Is it possible that $\mathbb{P}^n$ is an algebraic vector bundle over some algebraic variety? This is an interesting question that my friend asked in a student seminar. I believe that the answer is NOT. Because the only global sections of $\mathcal{O}_{\mathbb{P}^n}$ are constants. However as a total space of vector bundles $E$ over $X$, the global sections are all in $\Gamma(X, \oplus Sym^n E)=\oplus \Gamma(X,Sym^n E)$ which may not be $\mathbb{C}$ in general. The trouble case is when $\Gamma(X, E)=0$? Am I correct? Is there any other explanation?

Edit: Thanks to everyone for your comments and answer. It seems that compactness is the correct way to follow.

However, I still wondering that why the argument on sections doesn't work. In other words, is there an example of a non-proper algebraic variety whose structure sheaf only has constant global sections.

Answer by Kevin Lin for Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?

2010-01-27T16:10:44Z

Stephen Griffeth's argument works over any field. The total space of a vector bundle is never proper (follows by, e.g., valuative criterion for properness). On the other hand, $P^n$ is always proper.

Here is an argument that the total space of a vector bundle is not proper: A fiber of a vector bundle is isomorphic to $A^n$ . Moreover, the fiber considered as a subscheme is a closed subscheme. You can map $(A^1 \setminus 0)$ into the fiber by a map like $f(x)=1/x$. This map can't extend over zero, because if it did, then zero would be sent to something in the closure of the fiber. But the fiber is already closed, so zero would be sent to the fiber, contradiction.

Edit: Of course, amaanush's answer is a much better answer than mine!

Answer by amaanush for Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?

2010-01-27T18:40:13Z

That's right. I think the one sentence rephrasing of Kevin's argument would be proper morphisms are stable under pullback (this is obvious by the very definition of properness which is stated through the pullbacks).

Of course a variety is proper if and only if the structure map to the point scheme (spec k) is proper. Thus a closed fibre of a locally trivial fibre-bundle, with total-space proper, is proper (complete)-- which cannot be the case for a vector bundle.

On the other hand that does not apply to $\mathbb{A}^n \hookrightarrow \mathbb{P}^n$, as the pullback will just give $\mathbb{A}^n \stackrel{id}{\longrightarrow} \mathbb{A}^n$ is proper, which is fine :).

Answer by Michael Thaddeus for Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?

2010-08-02T23:28:32Z

Heh, actually it is a rank zero vector bundle over itself. Don't forget this possibility, folks!

Answer by Charles Staats for Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety?

2010-08-03T01:49:47Z

In answer to your last question, on why the "sections" argument does not work: Let $X = \mathbb{P}^n_k$, and let $E$ be the line bundle over $X$ with sheaf of sections given by $\mathcal{O}(-1)$. Then the global sections of $E$ are given by $$\Gamma(E, \mathcal{O}_E) \cong \Gamma(X, Sym(\mathcal{O}(-1))) \cong \bigoplus_{n=0}^{\infty} \Gamma(X, \mathcal{O}(-n)) \cong k.$$ Thus, the global sections of $E$ are all constant. However, as the arguments given in the other answers show, $E \to \text{Spec} k$ is not proper.