# Relationship between the $2$-plane bundles over $S^2$ and $\mathbb{Z}$

I want to follow up on this answer by asking a few more questions (posting directly on the question didn't seem to "bump" the thread). I was trying to read the referenced text (Husemoller's Fiber Bundles), but I couldn't understand that much of it since the "Clutching Constructions" chapter was a bit terse.

Given that you know there's a one-to-one correspondence between the isomorphism classes of $2$-dimensional bundles on $S^2$ and $\mathbb{Z}$, how do you know which bundle corresponds to which integer? For example, consider something like the trivial, tangent, or cotangent bundles.

Or to expand my question a bit more generally, given a map $f:S^1\rightarrow GL(n,\mathbb{R})$, what does the bundle represented by $f$, $E_f$, look like?

I think once I see an example, I'll be able to understand clutching constructions much better. I appreciate any help!

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The bundle corresponding to $0\in \mathbb{Z}$ is, not surprisingly, the trivial bundle.

For $z\in \mathbb{Z}$, the bundle corresponding to $-z$ is the same bundle with opposite orientation on the fibers, so we may as well just try to understand $z = 1, 2, 3,...$. Let $E_z$ be the rank 2 oriented vector bundle corresponding to the positive integer $z$.

Now, stick an inner product on $E_z$ and let $M_z\subseteq E_z$ be the set of all unit length vectors. Note that $M_z = f^{-1}(1)$ where $f:E_z\rightarrow\mathbb{R}$ is $f(p) = |p|$ and that $1$ is clearly a regular value. Hence, $M_z$ is actually an embedded submanifold. $M_z$ is called the unit sphere bundle over $S^2$.

The projection map, when restricted to $M_z$, gives a bundle $S^1\rightarrow M_z\rightarrow S^2$.

Via some abstract nonsense, the number $z$ can be identified with the Euler class of this bundle. The Euler class is technically an element of $H^2(S^2;\mathbb{Z})$, but since we're bluring the distinction between positive and negative numbers, we can take it to be a well defined integer.

There is something called the Gysin sequence which is an exact sequence involving $S^2$ and $M_z$ which can be computed with knowledge of the Euler class. Suffice it to say that we can show, using the Gysin sequence, that $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$. This is how we tell the bundles apart. In particular, if we can construct a bundle where $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$, then we must have found them all.

1. The Euler class of the tangent bundle of an oriented manifold can be identified with the Euler characteristic times the fundamental class. Since the Euler characteristic of $S^2$ is $2$, the tangent bundle is $E_2$ (or $E_{-2}$). The cotangent bundle of a manifold is always bundle isomorhpic to the tangent bundle via the flat/sharp maps (assuming a background Riemannian metric).

2. As to what the bundle "looks like", here's how I think about it.

Start with the Hopf bundle $S^1\rightarrow S^3\rightarrow S^2$. This is a principal bundle, so there is a "preferred" direction on the $S^1$s. Now, think of each $S^1$ as sitting inside of an $\mathbb{R}^2$, with each $\mathbb{R}^2$ disjoint from every other $\mathbb{R}^2$. Give each $\mathbb{R}^2$ the orientation which makes the "preferred" direction on the $S^1$, say, counter clockwise. This is $E_1$. (Notice that in this case, $M_1 = S^3$ and we have $H^2(S^3) = 0$ as claimed.)

Now, there is a canonical $\mathbb{Z}/z\mathbb{Z}$ inside of $S^1$, the $z$th roots of unity. Dividing by this action, we get a bundle $S^1/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^3/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^2$.

The fiber is still a circle, but the total space becomes a lens space (or $\mathbb{R}P^3$ when $z=2$). Do the same trick of filling in each circle fiber with an oriented $\mathbb{R}^2$ to make the bundle $E_z$. In this case, the sphere bundle is the lens space.

Incidentally, this provides one of the more complicated proofs I know of that the unit tangent bundle of $S^2$ (i.e., $M_2$) is diffeomorphic to $\mathbb{R}P^3$.

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This doesn't directly answer your question, but the computations become simpler I think if you restrict to oriented 2-plane bundles. Now these are classified by maps (up to homotopy) $S^1 \to GL_+(2)$. But by putting a metric on any bundle and making sure our transition functions respect the metric, such bundles are actually classified by maps $S^1 \to SO(2) \simeq S^1$ (another way to see this is that $GL_+(2)$ deformation retracts onto $SO(2)$). But any map to a sphere to itself is determined up to homotopy by its degree. For maps $S^1 \to S^1$ this means they all look like $e^{i\theta}\mapsto e^{in\theta}$, with $n$ being the degree of the map.

Further since $SO(2) \simeq U(1) \subset GL(1,\mathbb C)$, we actually see that any oriented 2-plane bundle is a complex line bundle over $S^2 \simeq \mathbb C P^1$. These are classified by their first Chern classes, which lie in $H^2(S^2;\mathbb Z)$. Since the Chern class of the tautological line bundle $H \to \mathbb CP^1$ generates $H^2(S^2)$ and since the Chern class is a group isomorphism from line bundles (with tensor product as the operation) to $H^2(S^2;\mathbb Z)$, it follows that any complex line bundle is a tensor product of copies of $H$ and its dual $H^*$.

Under these correspondences the trivial bundle corresponds to the integer 0 since its transition functions can be made constant. $TS^2$ and $T^* S^2$ are isomorphic (as is always the case for tangent and cotangent bundles, which can be seen by putting a metric on your manifold) and it can be shown that $T \mathbb CP^1$ direct sum with the trivial complex line bundle is $H^* \oplus H^*$. From properties of the total chern class it follows that the first chern class of $T\mathbb CP^1$ is twice that of $H^*$.

I find Husemoller to be a little difficult in many places. I would recommend Hatcher's book on Vecotor bundles and K-theory (available freely on his website) and Milnor and Stasheff's Characteristic classes.

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For simply connected bases (more generally, things with $H^1(X;\mathbb{Z}/2\mathbb{Z}) = 0$), all vector bundles over them are automatically orientable, so your "restriction" at the beginning isn't really a restriction. – Jason DeVito Oct 28 '11 at 2:51
Ah, good point Jason! – Eric O. Korman Oct 28 '11 at 3:04