Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In order to learn about vector bundles, I would like to draw the tautological vector bundle over the complex projective line

$$ E = \{(x,v) \in \mathbb{CP}^1 \times \mathbb{C}^2 : v \in x \} .$$

Identifying the complex projective line with the Riemann sphere, $\mathbb{CP}^1 \cong S^2$, I hope that it might be possible to visualize this bundle by attaching small planes to each point of the sphere, similar to how one can visualize the tangent bundle of sphere.

In other words, I'm looking for an embedding $E \hookrightarrow S^2 \times \mathbb{R}^3$ into a trivial bundle. (Obviously, $E$ has to be viewed as a 2-dimensional real vector bundle for this to make sense.) I am aware that such a thing might not exist, in which case I would like to learn why.

share|cite|improve this question
up vote 2 down vote accepted

I claim that there is no bundle embedding of the realification of the tautological bundle $\mathcal{O}(-1)$ into the trivial (real) bundle $S^2 \times \mathbb{R}^3$. Suppose there were; then we could take the orthogonal complement $N$, and we would obtain a decomposition $$ N \oplus \mathcal{O}(-1)_{\mathbb{R}} = S^2 \times \mathbb{R}^3$$ where $N$ is a real line bundle. But real line bundles on any compact CW complex $X$ are classified by the first Chern class, which lives in $H^1(X, \mathbb{Z}/2)$ (since the infinite 1-Grassmannian is a $K(\mathbb{Z}/2, 1)$). However, $H^1(S^2, \mathbb{Z}/2)=0$, and so $N$ is trivial.

It follows that if such an embedding existed, then $\mathcal{O}(-1)_{\mathbb{R}}$ would be stably trivial. This is, however, not the case. Stable trivialty would imply that the Stiefel-Whitney classes were trivial, as the product formula for them shows. However, we know that the top Chern class in $H^2(\mathbb{CP}^1, \mathbb{Z})$ generates the group, and also that (Proposition 3.8 in Hatcher's Vector Bundles and K-theory, available here) implies that the top Stiefel-Whitney class is the image of the top Chern class. But the image of a generator in $H^2(\mathbb{CP}^1, \mathbb{Z})$ in $H^2(S^2, \mathbb{Z}/2)$ is nonzero.

share|cite|improve this answer
Thank you for this nice proof, which I unfortunately do not understand at all. :D Is there a "pedestrian" proof relying on, say, explicit descriptions of some cohomology classes (I'm fine with assuming that everything is differentiable, so we can use de-Rham cohomology) or am I better off learning what Stiefel-Whitney classes are? – Greg Graviton Feb 19 '11 at 16:45
@Greg, there is little hope of turning Akhil's argument into something that can be explained to someone who's beginning with vector bundles. If you keep up the work, though, it should become accessible soonish :) – Mariano Suárez-Alvarez Feb 19 '11 at 17:03
@Greg: Dear Greg, I don't know how to explain this using de Rham cohomology. But there is a simple axiomatic approach one can take towards the Chern and Stiefel-Whitney classes; if one believes that such exist and satisfy the key properties (which is basically how I think of them myself), then nothing in the above proof will be mysterious. (See ch. 4 of Milnor's Characteristic classes for the S-W half developed in this way.) Note that the actual construction can be done quickly via the Leray-Hirsch theorem; cf. Hatcher's book referred to above. – Akhil Mathew Feb 19 '11 at 18:27

I think you can't do much better than to visualize the Hopf fibration $S^{3} \to S^{2}$, which can be defined by restricting the tautological line bundle to $S^{3} \subset \mathbb{C}^{2}$ (in other words, the fiber over each point $x \in \mathbb{CP}^{1}$ is the circle of points of length $1$ in $x$). Most of the pictures are obtained by stereographically projecting $S^{3}\smallsetminus \{N\}$ to $\mathbb{R}^{3}$. Just google for Hopf fibration and you'll get lots of nice pictures.

share|cite|improve this answer

I do not think this is possible. But you can try to do this over $\mathbb{S}^{2}\times \mathbb{R}^{4}$. The most basic reason for this to be impossible is the 'complex twist' involved in the tautological bundle made its rigid enough that become impossible to put into $\mathbb{S}^{3}$ (which is too tight). I doubt if one need to work with characteristic classes back and forth, this feels killing a mosquito with a big hammer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.