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Hello!

Let $E$ and $F$ be two vector bundles and let $f:E\rightarrow F$ be a bundle map. Then the kernel of $f$ is not always a subbundle of $E$. Does somebody have a simple example? Does there exist any simple conditions for which the kernel is a subbundle?

Thank you!

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If, for example, the dimension of the kernel is locally constant, then it should be a subbundle. –  Dylan Wilson May 29 '11 at 4:52
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-1 since everything here can be found in every introduction to vector bundles, which one should probably read a little bit before asking a question here (see the FAQ). –  Martin Brandenburg May 29 '11 at 9:14
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@Martin Brandenburg: Well, I am a graduate student and I have already read books about Differential Geometry, which is the subject I like the most. That's true that I never read a textbook on vector bundles yet although I am planning to read the Husemoller. Moreover I guessed that this question would perhaps be too simple. However, if you feel like this question is too easy for you, why don't you just give some answer or a reference to a textbook that you find important/interesting (like other people below)? It would be more respectful and constructive than your somehow unpleasant comment. –  Benjamin May 29 '11 at 10:07
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-1 @Benjamin: It may be a little rough but basically what Martin is saying is that this question, while very reasonable for someone learning about vector bundles, has nothing to do with math research. The FAQ states that all question should be about research problems or close. Giving an answer to the question encourages them. There is another site,mathstackexchange, which is perfect for these questions...go have a look! –  Daniel Pomerleano May 29 '11 at 15:29
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Dear Benjamin: Please, Martin and Daniel's remarks, however objectionable, are directed exclusively to the question, and not at all to the questioner. –  Giuseppe Tortorella May 30 '11 at 7:43

3 Answers 3

up vote 3 down vote accepted

I think this may be an example:

Consider the map from $S^1 \times \mathbb{R}^2 \rightarrow S^1 \times \mathbb{R}^2$ given by the linear map $f_\theta: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that takes the basis vectors $e_1, e_2$ to $e_1, e_2^\theta$, where $e_2^\theta$ is $e_2$ rotated by $\theta$. This should be a map of bundles over $S^1$, but the kernel isn't a subbundle as it has no trivializing neighborhood around, say, $\theta =\pi/2$.

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@Dylan Wilson: Thank you again! –  Benjamin May 29 '11 at 5:54

Another example. Let $$E = [0,1] \times \mathbb{R} $$ be the trivial line bundle over the unit interval. We have a bundle homomorphism $\phi: E \to E $ defined by $\phi(x,t)=(x,xt)$ which has kernel $$ \left( \{0\} \times \mathbb{R} \right) \cup \left( (0,1] \times {0} \right).$$ This is not a sub-bundle because it does not have locally-constant rank.

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@long trail: Thank you! –  Benjamin May 29 '11 at 5:54
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There's a theorem to the effect that this is "the only thing that can go wrong" which states that a bundle map has subbundle kernel if and only if it has subbundle image if and only if it has locally constant rank. I think that there's another equivalent condition, too. I would check Allen Hatcher's Vector Bundles and K-Theory book. –  long trail May 29 '11 at 6:05
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See also Atiyah's K-Theory book and the various remarks on "strict bundle maps" therein. –  Dylan Wilson May 29 '11 at 7:29

Here is a standard example (which you can find also in the big book of Demailly) in the complex category with the language of sheaves which perhaps may be of some interest for you.

Take $X=\mathbb C^3$, $\mathcal F=\mathcal O^{\oplus 3}$, $\mathcal G=\mathcal O$ and $$ \varphi\colon\mathcal O^{\oplus 3}\to\mathcal O,\quad (u_1,u_2,u_3)\mapsto\sum_{j=1}^3z_j\,u_j(z_1,z_2,z_3). $$ Since $\varphi$ yelds a surjective bundle morphism on $\mathbb C^3\setminus\{0\}$, it is easy to see taht $\ker\varphi$ is locally free of rank $2$ over $\mathbb C^3\setminus\{0\}$. However, by looking at the Taylor expansion of the $u_j$'s at $0$, you can check that $\ker\varphi$ is the $\mathcal O$-submodule of $\mathcal O^{\oplus 3}$ generated by the three sections $(-z_2,z_1,0)$, $(-z_3,0,z_1)$ and $(0,z_3,-z_2)$, and that any two of these three sections cannot generate the $0$-stalk $(\ker\varphi)_0$.

Thus, $\ker\varphi$ is not locally free.

On the other hand, if you have a surjective bundle morphism, the kernel is always locally free.

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