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I found this bundle to be trivial. However I am having trouble to persuade one of my fellow graduate students that this is true. He reasoned as follows:

1) The inclusion map $i: \mathbb{CP}^{n}\rightarrow \mathbb{CP}^{n+1}$ give us a canonical splitting $$i^{*}(\mathbb{TCP}^{n+1})=\mathbb{TCP}^{n}\oplus N\mathbb{CP}^{n}$$

2) Calculating the Chern class using Whitney sum formula give us $$c_{1}i^{*}(\mathbb{TCP}^{n+1})=c_{1}(\mathbb{TCP}^{n})+c_{1}(N\mathbb{CP}^{n})$$

3) We know $c_{1}(\mathbb{TCP}^{n})=(n+1)a$ by Milnor-Stasheff. On the other hand by naturality we have $i^{*}[(n+2)a]=(n+2)a$. So the normal bundle is not trivial as it has class $a=(n+2)a-(n+1)a$.

We agree that $i^{*}(\mathbb{TCP}^{n+1})=\mathbb{TCP}^{n+1}|\mathbb{CP}^{n}$ by the nature of the pull back bundle. However I disagree that the normal bundle of $\mathbb{CP}^{n}$ is included in $\mathbb{TCP}^{n+1}|\mathbb{CP}^{n}$. He suggested the following exact sequence:

$$0\rightarrow \mathbb{TCP}^{n}\rightarrow \mathbb{TCP}^{n+1}|\mathbb{CP}^{n}\rightarrow N\mathbb{CP}^{n}\rightarrow 0$$which I think should not hold thinking about the situation of the sphere because there is no reason the normal vectors should be included in the tangent vectors in a higher dimension.

I also could not understand why $i^{*}$ is injective given its nature is a restriction map. He told me this is a "general phenomenon of $CW$-complexes that the $n$-th cohomology is determined by the $n$-th skeleton". I do not think this is entirely self-evident in this case, for there may be $H^{2}$ classes that is $0$ in $\mathbb{TCP}^{n+1}$ but not in its restriction to $\mathbb{TCP}^{n+1}|\mathbb{CP}^{n}$. Here we are working with $\mathbb{CP}^{2}$ and $\mathbb{CP}^{1}$. I suspect I might be wrong and map is injective. But I am not sure. So I decided to ask.

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Bombyx mori
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  • Dear user, To begin with, are you thinking of $\mathbb C P^n$ and $\mathbb C P^{n+1}$ as complex manifolds (and so discussing the holomorphic normal bundle, which is a complex line bundle on $\mathbb C P^n$) or as real manifolds? Since you are computing with Chern classes I am guessing the latter, but it seems worthwhile to check first. Regards, – Matt E Apr 22 '13 at 03:39
  • @MattE: We compute it as complex line bundles because he asked me the question in the context of $\mathbb{CP}^{1}$ sits in $\mathbb{CP}^{2}$ with a normal bundle which is a line bundle. – Bombyx mori Apr 22 '13 at 03:45
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    For your link on the triviality of the normal bundle of $\mathbb{C}P^n$, I don't think that it is the consequence of the triviality of normal bundle of $S^{2n+1}$. Here $S^1$ not only acts on base space, but also acts on fibers, which makes the quotient of this trivial bundle nontrivial. – Yuchen Liu Apr 22 '13 at 04:01
  • @jerrysciencemath: If a bundle is a complex line bundle and the underlying space is a complex manifold, then I believe you can view it as a principal $\mathbb{S}^{1}$ bundle in some sense. In other words you see $U(1)$ acts on the fiber over $\mathbb{CP}^{1}$. – Bombyx mori Apr 22 '13 at 04:10

1 Answers1

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No, the normal bundle is not trivial. It generates the group of all complex line bundles on $\mathbb CP^n$. Its Chern class is the positive generator of $H^2(\mathbb CP^n)$.

In general, the normal bundle is a quotient of the tangent bundle of the ambient manifold. You need a Riemannian (or Hermitian) metric to get a subbundle.

Ted Shifrin
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  • I doubt if this is true. – Bombyx mori Apr 22 '13 at 03:49
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    LOL ... I made five statements, and, unless my mathematical knowledge of many years has evaporated, I believe they're all correct. What precisely are you doubting? :) – Ted Shifrin Apr 22 '13 at 04:01
  • With all respect of your knowledge I do not think the normal bundle is such a quotient, if you mean the exact sequence in above. Thus the rest also does not make sense to me. – Bombyx mori Apr 22 '13 at 04:07
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    Without an inner product on the tangent space, one defines the fiber of the normal bundle, indeed, to be the quotient, modding out by the tangent space of the submanifold. How would you propose to do it? – Ted Shifrin Apr 22 '13 at 04:11
  • The normal bundle should not be defined by a quotient. But I agree I do not know how to define it canonically without a metric. I do not think in general the normal bundle sits inside the tangent bundle of the ambient space. – Bombyx mori Apr 22 '13 at 04:15
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    @user32240: Dear user, You seem to have basic confusions about the concepts involved. The normal bundle is (by definition) the quotient of the pull-back of the tangent bundle of the ambient variety by the tangent bundle of the variety. (This tangent bundle is a subbundle of the previously mentioned pull-back.) And, as Ted indicates, the normal bundle of $\mathbb CP^n$ in $\mathbb CP^{n+1}$ is non-trivial. This is related to the fact that two generic hyperplanes in $\mathbb CP^n$ have non-trivial intersection in a linear subspace of dimension $n-1$. Regards, – Matt E Apr 22 '13 at 04:31
  • @MattE: Dear Matt: I can not recall the definition of the normal bundle in Milnor&Stasheff or other books, but it seems absurd to me that the normal vectors would be included in the tangent vectors of the ambient space, as I wrote in the question. I need to think about the second part you told me. – Bombyx mori Apr 22 '13 at 04:36
  • @user32240: Dear user, As Ted and I wrote, they are canonically a quotient of the restriction of the tangent bundle of the ambient space to the space in question. You can also make them a sub, if you have a metric (but this depends on the choice of metric, and is not needed for your question; in Milnor and Stasheff, where they don't really consider complex manifolds, it is probably considered as a sub via a metric as a matter of course, since in the real manifold situation, metrics are easily obtained). I don't know what you find absurd; in the intuitive picture of a normal vector ... – Matt E Apr 22 '13 at 05:15
  • ... to a curve in the plane, the normal vector is pretty clearly tangent to the plane (what else could it be?). Regards, – Matt E Apr 22 '13 at 05:16
  • @user32240: P.S. Do you understand the difference between an inclusion and a quotient? Even in your question, you write down an exact sequence (a correct one) which exhibits the normal bundle as a quotient, and then write "I think should not hold" (but you are mistaken) and go on to speak about an inclusion of the normal vectors, although there is no such inclusion being asserted in the exact sequence in question. – Matt E Apr 22 '13 at 05:19
  • @user32240: P.P.S. Even more generally, if you don't understand why the exact sequence in your question is correct, what do you think a normal vector, or the normal bundle, actually is? – Matt E Apr 22 '13 at 05:20
  • @MattE: The exact sequence is provided to me by the fellow graduate student, who obviously got exacerbated by my ignorance and left that as the final explanation. It seems to me by thinking $\mathbb{S}^{1}$ in $\mathbb{R}^{2}$, the normal vectors of $\mathbb{S}^{1}$ may not really be in the tangent space at that point in $\mathbb{R}^{2}$. But this is wrong as you pointed out. – Bombyx mori Apr 22 '13 at 05:22
  • @MattE: I see the difference between the quotient and the inclusion, but I thought for the exact sequence to hold every vector in $\mathbb{TCP}^{n+1}|\mathbb{CP}^{n}$ has to split into $a+b$, with $a\in \mathbb{TCP}^{n}$ and $b\in N\mathbb{CP}^{n}$, one should be able to intuitively place the normal vector into the tangent bundle of the ambient space, which I thought is impossible. – Bombyx mori Apr 22 '13 at 05:25
  • @MattE: I think the normal bundle is similar to the base space by the tabular neighborhood theorem, using exponential maps in local coordinates. So it was not intuitively clear to me why the normal directions would be fit into the whole tangent bundle. – Bombyx mori Apr 22 '13 at 05:26
  • @MattE: So I will accept the answer. I have checked $i^{*}$ is injective and all steps of his proof is correct, though I am still unclear why Ryan's argument fails in this case, maybe it is because he regard the manifold as the quotient of the real manifold instead of the complex one. – Bombyx mori Apr 22 '13 at 05:28
  • @user32240: Dear user, Regarding your last comment, you should reread jerrysciencemath's comment above. You responded to it, but didn't seem to understand the actual point. (Your comment about a complex line bundle being a $\mathbb C^{\times}$-bundle, or --- what is more-or-less the same --- a $U(1)$-bundle, is correct, but not relevant to jerrysciencemath's point.) Regards, – Matt E Apr 22 '13 at 05:42
  • @MattE: Thanks for answering such a low level question at such a late hour. I need to think. – Bombyx mori Apr 22 '13 at 05:50