Line bundles on $S^2$ and $pi_2(mathbb{R}P^2)$

Real line bundles on $$S^2$$ are all trivial, but what about the following way to think about a line bundle: we view a line bundle on $$S^2$$ (thought of as living in $$3$$d space) as providing a map $$S^2 to mathbb{R}P^2$$ the space of lines; and since $$pi_2(mathbb{R}P^2) simeq mathbb{Z}$$ we should have a $$mathbb{Z}$$ worth of different line-bundles.

Clearly I’ve cheated very badly in the latter line of thought; but I perhaps am still a bit confused at what step I have cheated so badly; is it that the following intuition just makes no sense?

• if I have a manifold $$M hookrightarrow mathbb{R}^n$$ and a line bundle $$L mapsto M$$; can I always find a vector field $$V$$ on $$mathbb{R}^n$$ that is not zero anywhere on $$M$$, such that the line bundle $$L$$ looks like what you get from ‘integrating $$V$$ around $$M$$‘? I think this is another way of saying my inuition that line bundles come from maps $$M to mathbb{R}P^{n-1}$$ which might be totally wrong; and which I’m not too sure how to formalise

EDIT : I just realised since $$pi_2(mathbb{R}P^3)$$ is trivial, even if my intuition above was correct, you’d still have just the trivial line bundle on $$S^2$$ since you can always "view the lines in a higher dimension" and rotate them away; if that makes any sense. Sorry if this is all just gibberish: just trying to learn

Mathematics Asked by questions on December 29, 2020

1 Answers

One Answer

The classifying space of real line bundles is $$mathbb{RP}^{infty}$$, not $$mathbb{RP}^2$$; $$mathbb{RP}^2$$ instead classifies line subbundles of the trivial $$3$$-dimensional real vector bundle $$mathbb{R}^3$$ (and similarly $$mathbb{RP}^n$$ classifies line subbundles of the trivial $$n+1$$-dimensional real vector bundle $$mathbb{R}^{n+1}$$).

So what the calculation of $$pi_2(mathbb{RP}^2)$$ vs. $$pi_2(mathbb{RP}^n), n ge 3$$ reveals is that there are a $$mathbb{Z}$$'s worth of real line subbundles of $$mathbb{R}^3$$ on $$S^2$$ but that these bundles all become isomorphic after adding an additional copy of $$mathbb{R}$$. With a little effort it should be possible to write down these line subbundles and the resulting isomorphisms explicitly. Probably the normal bundle of the embedding $$S^2 to mathbb{R}^3$$ is a generator.

If $$X$$ is any compact Hausdorff space then every vector bundle on it is a direct summand of a trivial bundle, so for line bundles what this tells us is that every line bundle is represented by a map $$X to mathbb{RP}^n$$ for some $$n$$ (but isomorphisms of line bundles may require passing to a larger value of $$n$$ to define). This connects up nicely with the picture where $$mathbb{RP}^{infty}$$ is the filtered colimit of the $$mathbb{RP}^n$$'s, because a map $$X to mathbb{RP}^{infty}$$ has image contained in some $$mathbb{RP}^n$$ by compactness.

Correct answer by Qiaochu Yuan on December 29, 2020

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