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

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 $S^4\times S^5$?

-
Nice question!${}$ – Grumpy Parsnip Aug 2 '12 at 2:08

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 $T^1 S^7$.

-
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.

-
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.

Goodluck.

-
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