How can one compare these two 4-manifolds

Question

We would like to compare the following two real 4 dimensional manifolds:

1)$M$=The tangent bundle of $S^{2}$

2)$N$= The total space of the canonical line bundle over $\mathbb{C}P^{1}\simeq S^{2}$

Are these two manifolds homeomorphic? Are they diffeomorphic?

Some remarks: They have the same homotopy type. So there is no an obvious obstruction for $M$ and $N$ to be homeomorphic. On the other hand $M$ is parallelizable. So it is natural to ask: Is $N$ parallelizable, too?

Is the later equivalent to ask " Is $TS^{2}\oplus \ell$ a trivial real 4 dimensional bundle. Where $\ell$ is the realification of the canonical line bundle?"?

Answer 1

You can prove that they're not homeomorphic by employing the same strategy as in the answer here.

If they were homeomorphic, so would be their one-point compactifications, and hence they would have the same cohomology. $TS^2$ has no embedded spheres with self-intersection number 1; the canonical line bundle does (the zero section!)

Another way of phrasing this: the intersection form $H_2(M) \otimes H_2(M) \to \Bbb Z$ is well-defined for any 4-manifold, compact or not, and is preserved by homeomorphisms. (If you want to be careful about this, you want to think about it in terms of non-compact Poincare duality. We can avoid that in the above situation b/c it's a special case, where the cohomology of the 1-pt compactification is that of $H^*(BV,SV)$.) So one just checks that for $TS^2$, $\iota(\sigma_0,\sigma_0) = 2$ and for $\kappa$, $\iota(\sigma_0, \sigma_0) = 1$, despite the zero section $\sigma_0$ being a generator of $H_2(M)$ in both cases.

Answer 2

The second stiefel whitney class of your bundle $TS^2\oplus l$ is non-zero. For the chern classes you can see

$$c(TS^2\oplus l)=(1+2a)(1-a)=1+a$$

where $a$ is the generator of $H^2(S^2;\mathbb{Z})$ (here I see $l$ with the complex structure still). Taking mod 2 reduction shows that the second stiefel whitney class is non-zero.

Answer 3

Here's another (probably unnecessarily elaborate) proof that they are not diffeomorphic using Eliashberg's amazing theorem about which open $2$-handlebodies admit Stein structures and the adjunction inequality for Stein surfaces. I am sorry if this is not really accessible for the asker.

$M$ is the 4-manifold given by gluing an open $2$-handle along the unknot with framing coefficient $2$, and $N$ is the $4$-manifold given by gluing an open $2$-handle to the unknot with framing coefficient $1$. Therefore $\bar M$ ($M$ with the opposite orientation) is the $4$-manifold given by gluing a $2$-handle with framing $-2$ to the mirror of the unknot (which is still the unknot), and $\bar N$ is given by gluing a $2$-handle with framing $-1$ to the (mirror of the) unknot.

Now the unknot $K$ has a Legendrian embedding with $tb(K)=-1$ (with the Legendrian projection the simple one with only two cusps), hence by Eliashberg's construction of Stein manifolds $\bar M$ admits a Stein structure.

Now we need to show $N$ and $\bar N$ cannot admit a Stein structure. If we take a disk bounded by the attaching unknot in $\Bbb R^3$ union the core of the $2$-handle we get an embedded $2$-sphere $\Sigma$ with $[\Sigma]^2= \pm1$. By the adjunction inequality for Stein surfaces, we have $$[\Sigma]^2 + |\langle c_1,[\Sigma]\rangle |\leq2g(\Sigma)-2$$

In our case this is just $\pm1+|\langle c_1,[\Sigma]\rangle | \leq -2$ which is a contradiction. So neither $N$ nor $\bar N$ admit Stein structures.