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For any smooth manifold $M$, the tangent bundle $TM$ as a manifold is always orientable. In other words, the first Stiefel-Whitney class $w_1$ of the manifold $TM$ always vanishes.

Question: Does the manifold $TM$ also have vanishing higher Stiefel-Whitney classes $w_{i>1}$? If not, how can we compute them provided we know the Stiefel-Whitney classes of $M$?

p.s. I'm concerned about $w_2$ in particular.

Leo
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  • Potentially relevant class of examples: complex projective spaces. See https://math.stackexchange.com/questions/1241359/examples-with-zero-first-stiefel-whitney-class-and-nonzero-second-stiefel-whitne?rq=1 (see the answer), https://math.stackexchange.com/questions/1206867/stiefel-whitney-class-of-complex-projective-spaces (general calculation) – leslie townes Dec 22 '21 at 16:04
  • @leslietownes Thanks for the examples! – Leo Dec 22 '21 at 16:16

1 Answers1

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The characteristic classes of a tangent bundle $TM$, treated as a manifold itself, are defined using the double tangent bundle $TTM$. Given the projection $\pi: TM \to M$, the double tangent bundle splits as $\pi^* TM \oplus \pi^* TM$, and we have $$w_1(TTM) = 2 w_1(\pi^* TM) \equiv 0,$$ so $TM$ is orientable as you've asserted.

Similarly, $$w_2(TTM) = 2w_2(\pi^* TM) + w_1(\pi^* TM)^2 \equiv \pi^* w_1(TM)^2,$$ so whether it vanishes depends on $w_1(TM)^2$ in the cohomology ring of $M$.

You can work out what the other Stiefel-Whitney classes $w_i(TTM)$ are, using the formula $w(TTM) = \pi^* w(TM)^2$, where $w = 1 + w_1 + w_2 + \cdots$ is the total Stiefel-Whitney class.

JHF
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