Let $Spin(3)$ be embedded in $Spin(5)$ by the spin embedding then we have a fibration:

$$Spin(3) \rightarrow Spin(5) \rightarrow Spin(5)/Spin(3)$$

Where $Spin(5)/Spin(3)$ is $V_{5,2}$, the Stiefel manifold of orthonormal 2-frames in $\mathbb{R}^5$. Now we have $Spin(3)\cong S^3$ and $V_{5,3}$ is the sphere bundle in the tangent bundle of $S^4$ so the integral cohomology $H^*(V_{5,3})$ is $\mathbb{Z}$ at degree 0,7 and $\mathbb{Z}_2$ at degree 4.

Let $E$ be the Serre spectral sequence of the fibration, then $E^2_{4,3}=\mathbb{Z}_2$ but there can be no non-zero differentials to or from here. So $E^\infty_{4,3}\cong\mathbb{Z}_2$ and I get that $H^7(Spin(5))$ has some torsion. On the other hand $Spin(5)\cong Sp(2)$ and $H^*(Sp(2))\cong\Lambda(e_3,e_7)$ as it is a 3-sphere bundle over $S^7$ so $H^7(Spin(5))\cong \mathbb{Z}$

This is clearly a contradiction and I have messed up somewhere I would appriciate any help in finding my error.

  • $\begingroup$ For the spin embedding, why is $Spin(5)/Spin(3)$ the Stiefel manifold $V_{5,2}$? If you have the usual embedding, I believe you. Also, look out for extension problems - a $\mathbb{Z}/2\mathbb{Z}$ on the $E^\infty$ page doesn't mean a $\mathbb{Z}/2\mathbb{Z}$ in the actual cohomology, if I recall correctly. $\endgroup$ – Jason DeVito May 27 '12 at 0:36
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    $\begingroup$ (A message from the indexing police: If you want the spectral sequence for cohomology, the indices should look like $E_\infty^{p,q}$ :) ) $\endgroup$ – Mariano Suárez-Álvarez May 27 '12 at 0:44
  • $\begingroup$ Indeed, in the same (anti)diagonal as the $E_2^{4,3}=\mathbb Z/2\mathbb Z$ there is $E_2^{7,0}=\mathbb Z$, so convergence only gives you a short exact sequence $0\to\mathbb Z\to H^7(\operatorname{Spin(5)})\to\mathbb Z/2\mathbb Z\to0$. $\endgroup$ – Mariano Suárez-Álvarez May 27 '12 at 0:49
  • $\begingroup$ Isn't $E_2^{4,3}=\mathbb{Z}_2$ as opposed to $E_2^{4,2}$? $\endgroup$ – cipher May 27 '12 at 0:53
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    $\begingroup$ Notice that the map $e:H^7(V_{5,3})=\mathbb Z\to H^7(\operatorname{Spin(5)})$ in that short exact sequence is just the map induced by the projection of the bundle. One might be able to compute it in some other way. (For example, the map $Spin(5)\to V_{5,3}$ factors through $Spin(5)\to SO(5)$, and the latter is a double covering, so one should be able to show that $e(1)$ is twice some other class: this resolves the ambiguity in the extension) $\endgroup$ – Mariano Suárez-Álvarez May 27 '12 at 1:06

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