3
$\begingroup$

Let $S^m$ be the $m$-sphere and $\tau (S^m)$ the sphere bundle consisting of unit tangent vectors in the tangent bundle $TS^m$. Then we have a fibration $$ S^{m-1}\longrightarrow \tau(S^m)\longrightarrow S^m. $$ The Serre spectral sequence has $E_2$-page $$ E_2^{p,q}=H^p(S^{m};\mathbb{Z})\otimes H^q(S^{m-1};\mathbb{Z})=\mathbb{Z}[x_{m}]/(x_m^2)\otimes \mathbb{Z}[x_{m-1}]/(x_{m-1}^2) $$ where the dimensions of the generators are $|x_{m-1}|=m-1$, $|x_m|=m$, and converges to $$ H^*(\tau(S^m);\mathbb{Z}). $$ My unknown part is the differential $$d_{m}: \mathbb{Z}x_{m-1}\longrightarrow \mathbb{Z}x_m. $$

Question: Suppose $m$ is even. Is it possible that $$ H^{m-1}(\tau(S^m);\mathbb{Z}) $$ and $$ H^{m}(\tau(S^m);\mathbb{Z}) $$ are both torsion (do not have $\mathbb{Z}$-part) and both of their torsions are prime to $3$? Are there any references or results?

Note: When $m=3$, $S^3$ is a Lie group hence $TS^3$ is a trivial bundle. Hence $\tau (S^3)$ is a trivial bundle and $$ H^*(\tau(S^3);\mathbb{Z})=H^*(S^2;\mathbb{Z})\otimes H^*(S^3;\mathbb{Z}). $$

$\endgroup$
4
$\begingroup$

If $m$ is even, $d_m$ is multiplication by $2$:

For every unit sphere bundle of an $m$-dimendional vector bundle over a space $X$, $d_m$ sends $[S^{m-1}] \otimes 1$ to the euler class $e(x) \in H^m(X)$. In your case, we know that $$\langle e, [S^m]\rangle = \chi(S^m) = 1+(-1)^m$$ which is $2$ if $m$ is even. You can read off all the cohomology of $\tau(S^m)$ from that.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.