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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 $$ H^*(\tau(S^3);\mathbb{Z})=H^*(S^2;\mathbb{Z})\otimes H^*(S^3;\mathbb{Z}). $$

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1 Answer 1

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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.

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