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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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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 at 17:22
    
@JasonStarr The total Chern class, but I guess the only nontrivial one here would be $c_3$. –  jlv Jul 22 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 at 21:34

1 Answer 1

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 at 17:26
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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 mathoverflow.net/questions/153289 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 at 20:52

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