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

Solution 1:

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 $T^1 S^7$.

Solution 2:

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.

Solution 3:

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.