# Is there any standard N-sphere that has non-trivial first Pontryagin class?

I am wondering if there is a standard $N$-sphere that has non-trivial first Pontryagin class on its tangent bundle $TS^n$and frame bundle $FS^n$?

I know that only $S^4$ has non-trivial $H^4(S^n, R)$ cohomology class, while locally I write the first Pontryagin class on $TS^n$ as $Tr(R^2)$, where $R$ is curvature of $S^n$, is trivially zero. Does it mean all sphere have trivial first Pontryagin class?

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Since $p_1(S^n) \in H^4(S^n; \Bbb Z)$, the only possible sphere $S^n$ with $p_1(S^n) \neq 0$ is $S^4$. Now $S^4$ can be considered as the quaternionic projective line $\Bbb H P^1$. There is a general formula for the total Pontryagin class of $\Bbb H P^m$: $$p(\Bbb H P^m) = \frac{(1+u)^{2m+2}}{1+4u},$$ where $u$ is a generator of $H^4(\Bbb H P^m; \Bbb Z)$ (see Exercise 20-A of Milnor and Stasheff's Characteristic Classes). It follows that $$p_1(\Bbb H P^m) = (2m-2)u.$$ Therefore $$p_1(S^4) = p_1(\Bbb H P^1) = 0,$$ and all (standard) spheres have trivial first Pontryagin class.

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Thanks for this instructive answer! So then what about the $p_1$ on Frame Bundle of $S^N$? They are all trivial as well? –  Loliphilia Oct 3 '13 at 23:38

All positive characteristic classes vanish for the tangent bundles of standard spheres, as they are stably trivial. That is, $TS^{n} \oplus \mathbb{R}$ is a trivial bundle. One might see it by looking at the standard embedding $S^{n} \hookrightarrow \mathbb{R}^{n+1}$, where we have

$\mathbb{R}^{n+1} \simeq T\mathbb{R}^{n+1} |_{S^{n}} \simeq N_{\mathbb{R}^{n+1} / S^{n+1}} \oplus TS^{n+1}$.

Now, the normal bundle is trivial as one can orient it consistently in either the "zero-pointing" or "infinity-pointing" directions in $\mathbb{R}^{n+1}$. To see that this implies your result, recall that the Pontryagin classes are the Chern classes of complexification and we have

$\mathbb{C}^{n+1} \simeq (TS^{n} \oplus \mathbb{R}) \otimes \mathbb{C} \simeq (TS^{n} \otimes \mathbb{C}) \oplus \mathbb{C}$.

The Whitney formula implies that

$1 = c(\mathbb{C}^{n+1}) = c((TS^{n} \otimes \mathbb{C}) \oplus \mathbb{C}) \simeq c(TS^{n} \otimes \mathbb{C}) * c(\mathbb{C}) = c(TS^{n} \otimes \mathbb{C})$,

so all the Pontryagin classes vanish.

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