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One can ask

When is $T^1 S^n$, the unit tangent bundle of $S^n$, abstractly diffeomorphic to $S^{n-1}\times S^n$?

For even $n$, the answer is never. This is because $T^1 S^{2n}$ has torsion in its cohomology ring, but $S^{2n-1}\times S^{2n}$ doesn't, so $T^1 S^{2n}$ is not even homotopy equivalent to $S^{2n-1}\times S^{2n}$.

For odd $n$, the answer is more delicate.

Since $S^1$, $S^3$, and $S^7$ are parallelizable, $T^1 S^1$, the unit tangent bundle to $S^1$, is diffeomoprhic to $S^0\times S^1$. Likewise, $T^1 S^3$ is diffeomorphic to $S^2\times S^3$. The same argument shows $T^1 S^7$ is diffeomorphic to $ S^6\times S^7$. In all of these cases, a much stronger statement is true: $T^1 S^{k}\cong S^{k-1}\times S^k$ as bundles over $S^k$ (for $k=1,3,7$.)

So, the first case where I don't know the answer is $T^1 S^5$. The first thing to say is that $S^5$ is not parallelizable, so $T^1 S^5$ is not bundle isomorphic to $S^4\times S^5$. I can prove that $T^1 S^5$ and $S^4\times S^5$ have the same cohomology rings and that their respective tangent bundles have the same Stiefel-Whitney, Pontrjagin, and Euler classes. So none of the "usual" invariants distinguish them. (More generally, I think I can prove that $T^1 S^{2n-1}$ and $S^{2n-2}\times S^{2n-1}$ have isomorphic cohomology rings and the same characteristic classes).

On the other hand, a paper by De Sapio and Walschap, "Diffeomorphism of total space and equivalence of bundles" proves that $TS^n$ and $\mathbb{R}^n\times S^n$ are not abstractly diffeomorphic, unless they are also bundle isomorphic. So, we know that $TS^5$ is not diffeomoprhic to $\mathbb{R}^5\times S^5$. So, if $T^1 S^5$ and $S^4\times S^5$ are diffeomorphic, then no diffeomorphism can extend to a diffeomorphism of $TS^5$ and $\mathbb{R}^5\times S^5$.

I'm not sure how hard the answer for $S^5$ will compared to any other $S^{k}$ with $k$ odd $\neq 1,3,7$, so I'll ask the $S^5$ question separately:

Is $T^1 S^5$ abstractly diffeomorphic to $S^4\times S^5$?

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Nice question!${}$ – Grumpy Parsnip Aug 2 '12 at 2:08
up vote 9 down vote accepted

It turns out, the answer is no. More specifically, $T^1 S^n$ is abstractly diffeomorphic to $S^{n-1}\times S^{n}$ iff $T^1S^n$ is bundle isomorphic to $S^{n-1}\times S^n$.

This is a simple corollary to Theorem 3.6 found in

Wu-Yi Hsiang and J. C. Su. Transactions of the American Mathematical Society , Vol. 130, No. 2 (Feb., 1968), pp. 322-336

And here is a link to (a free version of) their paper.

Theorem 3.6 states that the Stiefel manifolds $V_2(\mathbb{K}^n)$ of orthormal $2$- frames in $\mathbb{K}^n$ (where $\mathbb{K}\in\{\mathbb{R}, \mathbb{C}, \mathbb{H}\}$) are diffeomorphic to products of smaller dimensional manifolds of positive dimension only in the "obvious" cases of $V_2(\mathbb{R}^4)$, $V_2(\mathbb{R}^8)$, $V_2(\mathbb{C}^2)$, $V_2(\mathbb{C}^4)$, $V_2(\mathbb{H}^2)$, which are just various descriptions of $T^1 S^3$ and $T^1 S^7$.

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It only took me a year and a day to find the answer ;-) – Jason DeVito Apr 25 '13 at 18:09

I suspect the answer is no. I have an idea of how to proceed, but nothing close to a full solution. Cut out $X = T^{1}S^{5}$ out of $\mathbb{R}^{10}$ using the polynomial equations:

$q(x,v) = x_{1}v_{1}+\dots+x_{5}v_{5} = 0$

$r(x) = x_{1}^{2}+\dots +x_{5}^{2} - 1 = 0$

$s(v) = v_{1}^{2}+\dots+v_{5}^{2} -1 = 0$

And cut $Y = S^{4}\times S^{5}$ out of $\mathbb{R}^{11}$ using the equations:

$t(u) = u_{1}^{2}+\dots+u_{5}^{2} - 1 = 0$

$w(v) = v_{1}^{2}+\dots+v_{6}^{2} - 1 = 0$

Isomorphic varieties should have isomorphic coordinate rings. I'm trying to compute the coordinate ring of both these algebraic varieties and show that they are not the same. On one hand:

$\mathbb{R}[X] = \mathbb{R}[x_{1},\dots,x_{5},v_{1},\dots,v_{5}]/\langle q(x,v),r(x),s(v)\rangle \stackrel{???}{=} \mathbb{R}[x_{1},\dots,x_{4},v_{1},\dots,v_{4}]\oplus\mathbb{R}[x_{1},\dots,x_{4},v_{1},\dots,v_{4}]x_{5}\oplus\mathbb{R}[x_{1},\dots,x_{4},v_{1},\dots,v_{4}]v_{5}$

On the other hand,

$\mathbb{R}[Y] = \mathbb{R}[u_{1},\dots,u_{5},v_{1},\dots,v_{6}]/\langle t(u),w(v)\rangle \stackrel{???}{=} \mathbb{R}[u_{1},\dots,u_{4},v_{1},\dots,v_{5}]\oplus\mathbb{R}[u_{1},\dots,u_{4},v_{1},\dots,v_{5}]u_{5}\oplus\mathbb{R}[u_{1},\dots,u_{4},v_{1},\dots,v_{5}]v_{6}\oplus\mathbb{R}[u_{1},\dots,u_{4},v_{1},\dots,v_{5}]u_{5}v_{6}$

The question marks are meant to indicate a large amount of uncertainty on my part as to whether or not I wrote down the right answer. I need to put more thought into what that ring really looks like. If what I wrote down is right, then I think that $\mathbb{R}[X]$ is not isomorphic to $\mathbb{R}[Y]$. I'm not sure how to show this concretely, but it seems to be true just by looking at the structure of both rings as direct sums.

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This will, at most, show that the two are not isomorphic as algebraic varieties, but there could well be an diffeomorphism between them which is not polynomial. – Mariano Suárez-Alvarez Jul 10 '12 at 5:19
That was something that crossed my mind: I'm working with morphisms that are too nice. Oh well, had to give it a go! – jmracek Jul 10 '12 at 5:28
Notice that knowing the two varieties are nor isomorphic as such is not uninteresting! – Mariano Suárez-Alvarez Jul 10 '12 at 5:31

This is just a possible way to go, I am not sure if it works.

If you wanted to look at whether or not the bundles were isomorphic you could do a $K$-theory computation, but here you care about the underlying sphere bundles. So instead you would want to look at what happens in $J$-theory. Adams and Atiyah both talk about these groups. There are a couple ways to construct them, one is as the fiber of some power operation on $K$-theory, the other is as the Grothendieck group of something you are asking about, spherical fibrations.

I would think you could compute one of these groups and try and show that stably there are not any differences between the two bundles, but unstably would be another matter.


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Thanks for your reply. The first thing thing I should say is that I know basically nothing about $K$-theory. With that out of the way, I'm not sure how bundle theory can help me with an abstract diffeomorphism question - I already know that they are different as bundles. – Jason DeVito May 2 '12 at 2:00
oh, ok. The thought was that maybe the group vanished so you would get the result as bundles and that would imply the result as smooth manifolds. I guess I did not pay enough attention. – Sean Tilson May 2 '12 at 20:06
Not a problem - you used a lot of words I hope to understand some day ;-). – Jason DeVito May 3 '12 at 1:21

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