Are $T\mathbb{S}^2$ and $\mathbb{S}^2 \times \mathbb{R}^2$ different?

Question

I have seen the claim that $T\mathbb{S}^2$ and $\mathbb{S}^2 \times \mathbb{R}^2$ are not diffeomorphic, but I have only ever seen the proof that they are not isomorphic as vector bundles (which is a cute application of the hairy ball theorem). How would I prove that they are indeed not diffeomorphic (or better, not homeomorphic)?

Answer 1

Let's start by examining the unit tangent bundle $US^2 \subset TS^2$ (i.e. the set of all tangent vectors of unit length) and its analogue $S^1\times S^2 \subset \mathbb{R}^2 \times S^2$. Observe that the action of $SO(3)$ on $S^2$ extends to $US^2$, and this action is transitive and has trivial stabilizer, so it induces a diffeomorphism $SO(3)\to US^2$. It follows that $$ \pi_1(US^2) \;\cong\; \pi_1\bigl(SO(3)\bigr) \;\cong\; \mathbb{Z}_2. $$ Then $US^2$ is not homeomorphic to $S^1\times S^2$, since $\pi_1(S^1\times S^2)\cong \mathbb{Z}$.

Now, to prove that $X = TS^2$ and $Y = \mathbb{R}^2 \times S^2$ are not homeomorphic, let $X\cup\{\infty\}$ and $Y\cup\{\infty\}$ denote their one point compactifications. I claim that $$ H_2(X\cup\{\infty\},X) \;\cong\; H_1(US^2) \;\cong\; \mathbb{Z}_2\tag*{(1)} $$ and $$ H_2(Y\cup\{\infty\},Y) \;\cong\; H_1(S^1\times S^2) \;\cong\; \mathbb{Z}.\tag*{(2)} $$ To see the isomorphism in (1), let $S \subset TS^2$ denote the canonical copy of $S^2$ in $TS^2$, i.e. the set of zero tangent vectors. By excision $$ H_2(X\cup\{\infty\},X) \;\cong\; H_2(X\cup\{\infty\}-S,X-S) $$ But $X\cup\{\infty\}-S$ is clearly contractible, so $$ H_2(X\cup\{\infty\}-S,X-S) \;\cong\; H_1(X-S) $$ by the long exact sequence for the pair $(X\cup\{\infty\}-S,X-S)$. But $X-S$ deformation retracts onto $US^2$, so $H_1(X-S) \cong H_1(US^2)$, as desired. A similar argument establishes the isomorphism in (2).