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I am trying to calculate the Chern class of the tangent bundle of the sphere $S^6$. I am told that this is an interesting case, since $S^6$ is not a complex manifold, but it has an almost complex structure on it, induced by the octonions (and I would like to learn more abut complex structures and the octonions). The Chern class of $S^2$ can be found by using its complex structure, but that can't be extended to $S^6$.

Apparently, one way of solving this is viewing $S^6$ in $\mathbb R^7$, then $\mathbb R^7$ as the "imaginary" part of the octonions, and the octonions as $\mathbb R^8$. With such a decomposition we could write $T\mathbb R^8|_{S^6}=TS^6\oplus L$ for $L$ a line bundle (the "real" part of the octonions), but I'm not sure how that works out. So my question is how I would find the Chern class through this complex structure approach.

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closed as off-topic by Qiaochu Yuan, Chris Gerig, abx, Stefan Kohl, S. Carnahan Jul 24 '14 at 13:02

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Which Chern class? Obviously the first and second Chern classes of every almost complex structure equal 0, since $H^2(S^6)$ and $H^4(S^6)$ both vanish. – Jason Starr Jul 22 '14 at 17:22
@JasonStarr The total Chern class, but I guess the only nontrivial one here would be $c_3$. – jlv Jul 22 '14 at 17:25
A curious detail is that the decomposition $T\mathbb R^8|_{S^6} = TS^6 \oplus L$ is not that of complex vector bundles. This is pretty obvious because the standard complex structure on $\mathbb R^8$ doesn't map the normal to $S^6$ to the last coordinate vector (that is the real one), so $L$ is not preserved under multiplication by $i$. A slightly odd way to see this is to note that if this were a complex direct sum, the Chern class of $TS^6$ would be trivial. It also follows from this that it's probably impossible to compute $c(TS^6)$ from this decomposition without additional input. – Frol Zapolsky Jul 22 '14 at 21:34

Since $H^2(S^6;\mathbb Z) = H^4(S^6;\mathbb Z) = 0$, we get $c_1(TS^6) = c_2(TS^6) = 0$. As for $c_3(TS^6)$, it equals the Euler class $e(TS^6)$, which is the Euler characteristic of $S^6$ (which equals $2$) times the generator of $H^6(S^6;\mathbb Z) = \mathbb Z$.

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Is there a way to do this without using the Euler class? – jlv Jul 22 '14 at 17:26
Does the following count: "$c_{top}(E)$ is the number of zeroes of a section of the vector bundle $E$ (counted with appropriate multiplicity). When $E$ is the tangent bundle, and your base space is a smooth compact manifold $X$, then the number of zeroes is the Euler characteristic $\chi(X)$. This is the Poincare-Hopf theorem and is deducible from the Lefschetz fixed point theorem or Morse theory." To me, this is just describing some ways to think about Euler class without using that word, but maybe you disagree. – David Speyer Jul 22 '14 at 20:52

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