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

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.

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The easiest way to see this (working over $\mathbb{C}$) is: a vector bundle over a variety is never compact. –  GS Jan 27 '10 at 15:06
Yes, here's one: take $\mathbb{P}^2$ and remove any point. Every global section of the structure sheaf extends past this point to a global section over $\mathbb{P}^2$, and thus is constant. –  Charles Siegel Jan 28 '10 at 0:11

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!

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Actually, this is my first thought. But why vector bundle is never proper? Even more, can I say that the total space of an algebraic vector bundle is an affine varieties? –  Fei YE Jan 27 '10 at 16:17
The total space of an algebraic vector bundle over a projective variety can't be affine, as the zero section gives a nontrivial map from a projective variety to it! –  Kevin McGerty Jan 27 '10 at 16:20
Consider the map $(A^1 \setminus 0) \to A^1$ given by $f(x) = 1/x$. The map doesn't extend over zero, so $A^1$ is not proper by the valuative criterion. By a similar argument, $A^n$ is not proper. The total space of a vector bundle is locally $U \times A^n = A^n_U$, where $U$ is affine... –  Kevin H. Lin Jan 27 '10 at 16:28
Thanks Kevin. I understand that $\mathbb{A}^n$ is not proper and the total space of a vector bundle is locally affine and non-proper, but why or how this implies that the total space is non-proper. This argument seems not applicable to $\mathbb{P}^n$ which is locally isomorphic to $\mathbb{A^n}$. –  Fei YE Jan 27 '10 at 16:37
How about this: 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. –  Kevin H. Lin Jan 27 '10 at 16:43

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 :).

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Frankly speaking, I don't see why the stability under pullback of proper morphisms alone should imply properness of closed fibers in this case. In particular, I don't understand the "Thus" in "Thus a closed fibre of a locally trivial [...]": the fiber is a pullback along a map from SpecK, not to SpecK. What am I missing? –  Qfwfq Apr 17 '10 at 13:03
...Or are you assuming that, if $E\rightarrow X$ is a fiber bundle with proper total space $E$ (i.e. $E\rightarrow \operatorname{Spec}(\mathbb{k})$ is a proper morphism), than $E \rightarrow X$ is a proper morphism? –  Qfwfq Apr 17 '10 at 13:19
(note that $E \rightarrow X$ is not the pullback of $E \rightarrow \operatorname{Spec}(\mathbb{k})$ along $X \rightarrow \operatorname{Spec}(\mathbb{k})$) –  Qfwfq Apr 17 '10 at 13:25
Ohmygod! I just wanted a "K" in "SpecK", not that strange symbol that I see now! –  Qfwfq Apr 17 '10 at 15:05
It seemes to me that the question of unknown is why $E \to X$ is proper if $E \to \operatorname{Spec}(k)$ is proper. This is a conseguence of the following well known fact, that can be proven using the valutative criterion: If $f \colon X \to Y$ and $g \colon Y \to Z$ are such that $g \circ f$ is proper and $g$ is separated, then $f$ is proper. –  Ricky Aug 3 '10 at 0:29
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Heh, actually it is a rank zero vector bundle over itself. Don't forget this possibility, folks!

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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.