Mathematics Asked by questions on December 29, 2020

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

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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