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I have two questions on complex geometry.

First one is that why the existence of almost complex structure on tangent bundle on real 2n-dimensional manifold is a topological question?

Wikipedia describes it as a topological question. I think that mean there is some homology or cohomology group associated to topological property whose vanishing or nonvanishing pertains to the existence.

Would you explain why it should be a topological question intuitively and may I suggest exact topological invariant which captures the existence property?

Second, in Huybrechts's book 'Complex Geometry', when it comes to Euler exact sequence on $P^n$, he mentioned that there is natural inclusion map such that $O(-1)\rightarrow\oplus_{j=0}^{n}O$. I can hardly come up with any idea on what this inclusion map is.

If possible, would you let me know the exact expression for this map? (here, $O(-1)$ is the tautological line bundle sheaf on $P^n$ and $O$ is the holomorphic sheaf of the trivial line bundle)

Hoping to get some shedding light in your reply.

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3 Answers

up vote 10 down vote accepted

(1) Why is the existence of an almost-complex structure a topological question? Suppose $M$ is a $2n$-manifold. The tangent bundle is classified by some map $M\to BGL_{2n}(\mathbb{R})$; $M$ admits an almost-complex structure if and only if this map admits a lift to $BGL_{n}(\mathbb{C})$ (that is, an almost-complex structure is the same as endowing the tangent bundle with the structure of a complex vector bundle). The existence of such a lift depends only on the homotopy type of the map $M\to BGL_{2n}(\mathbb{R})$, and is thus a topological question. As such, it can be analyzed via standard methods in obstruction theory, which I will leave for you to google.

(2) Suppose we have a projective space $\mathbb{P}V$, where $V$ is some vector space of dimension $n+1$. Then a map $X\to\mathbb{P}V$ is the same as a line bundle $\mathcal{L}$ on $X$ and a surjective map $V\otimes \mathcal{O}_X\to \mathcal{L}$. The identity map on $\mathbb{P}V$ classifies such data---namely, a map $V\otimes \mathcal{O}_{\mathcal{P}V}\to \mathcal{O}(1)$. Explicitly, this map is given by identifying the global sections of $\mathcal{O}(1)$ with $V$, via the standard computation of the cohomology of line bundles on projective space.

Now your map is dual of this map. (Picking a basis $(x_i)$ of $V=\Gamma(\mathbb{P}V, \mathcal{O}(1))$, this map is given by multiplication by $(x_i)$.)

By the way, the Euler exact sequence admits a very geometric interpretation. Namely, there is a sequence of maps $\mathbb{A}^{n+1}\setminus\{0\}\to \mathbb{P}^n\to \operatorname{pt}$, where the first map is the usual quotient by $\mathbb{C}^*$. This induces a short exact sequence of cotangent bundles $$0\to \pi^*\Omega^1_{\mathbb{P}^n}\to \Omega^1_{\mathbb{A}^{n+1}\setminus\{0\}}\to \Omega^1_{(\mathbb{A}^{n+1}\setminus\{0\})/\mathbb{P}^n}\to 0.$$

Each of these sheaves admits an (equivariant) action of $\mathbb{C}^*$, and so descend to sheaves on $\mathbb{P}^n$, giving exactly the Euler exact sequence.

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Dear Daniel, Thanks for great reply penetrating the fundamentals. Due to my ignorance, I can't fully grasp your answer, but it stimulate me to study more. After studying related some concepts appearing your answer and a meditatin,and then I will look again your reply. Thanks in advance for hindsight I will get before long. Regards, –  Jude Feb 22 '13 at 5:03
    
I'm happy I could help, Jude--is there any specific point on which you'd like clarification? –  Daniel Litt Feb 22 '13 at 5:11
    
Dear Daniel, Thanks for your kindness. On your first reply, you mentioned there is some map classifying tangent bundles but I don't know what the map is such. If it is not so rude and the answer might be simple, would you explain some basic idea underlies the construction of the map? –  Jude Feb 22 '13 at 5:45
    
@Jude: Real vector bundles of rank $n$ over a finite CW complex $X$ are naturally in bijection with homotopy classes of maps from $X\to BGL_n(\mathbb{R})$, e.g. the Grassmannian of $n$-planes in $\mathbb{R}^\infty$. This is a well-known fact from algebraic topology; see e.g. en.wikipedia.org/wiki/Classifying_space –  Daniel Litt Feb 26 '13 at 21:45
    
@Daniel: Thanks again. Classifying spaces is the one I didn't know yet. I am sorry for asking some basic question and really grateful for your benovelence.(espescially making a kind link to wiki) –  Jude Mar 7 '13 at 10:11
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If $(x_0:x_1:\dots:x_n)$ are the homogeneous coordinates on $P^n$ then the map $O(-1) \to O^{n+1}$ is given by $s \mapsto (sx_0,sx_1,\dots,sx_n)$.

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Thanks Sasha. At first glimpse, it seems a little weird because sxi is not an element in O. However, in Serge's reply, I can see the point you are meaning. Thanks for suggesting explicit map –  Jude Feb 22 '13 at 5:12
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Answer to the second question. Represent $\mathbb P^n$ as the space of lines (1-dimensional linear subspaces) in the $(n+1)$-dimensional vector space $E$. Then $O_{\mathbf P^n}(-1)$ is the linear bundle on $\mathbb P^n$ s.t. its fiber over the point of $\mathbb P^n$ corresponding to the line $\ell\subset E$ is the line $\ell$ itself. This bundle is naturally embedded in the trivial bundle with the fiber $E$, which becomes isomorphic to $O^{n+1}$ once you have chosen a basis in $E$.

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Dear Serge, Thanks for your clear remark. You enlightened the point I am missing. –  Jude Feb 22 '13 at 4:55
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