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Suppose M is a $2n$-complex dimensional complex manifold.

a) Why is Pontryagin class independent of orientation of the bundle?

 E.g. $p_i(\tau M) = p_i(\bar{\tau M})$

I know that for the top dimension, $p_{2n}(\tau M)=e(\tau M) \cup e(\tau M)$, it can be realized as a square and hence is independent of the actual orientation of $\tau M$, but why is it true for i < 2n?

b) A related question: is $c_{2i}(E) = c_{2i}(\bar{E})$ true in general? Furthermore, is there any general relation between $c(E)$ and $c(\bar{E})$, or $c_i(E)$ and $c_i(\bar{E})$? I know that for line bundle V, V \otimes \bar{V} is trivial, by taking conjugate of the transition function. But is it true for general complex bundle E that E \otimes \bar{E} is trivial? I believe not, I think I read somewhere that it is isomorphic to the bundle Hom(E, E)? What is the implication on characteristic class?

c) T or F: diffeomorphic vector bundles over the same manifold have the same characteristic classes. I think the answer in general is no. For example, the Chern class of the canonical line bundle and the conjugate canonical line bundle over $\mathbb{C}P^1$ are different. However, I remember learning a fact that

 The classifying maps for two isomorphic bundles are homotopic to each other.

And by the functoriality of characteristic class, the characteristic class for the two bundles is just the pullback of the char class of the universal bundle, hence they would be the same...

I am confused. Or maybe the canonical line bundle and its conjugate bundle are not isomorphic after all? I know the scalar multiplication structures on the two bundles are different, but somehow they seem diffeomorphic to me. It would be great if you can help me pick out True and False statements from above. Appreciated!

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at least you should always distinguish real bundles (and you have diffeomorphism here) and complex ones (here you have Chern classes) –  Nikita Kalinin Mar 27 '13 at 21:55
Your second question is answered in Milnor and Stasheff's Characteristic Classes, Lemma 14.9. –  Adam Saltz Mar 28 '13 at 5:02
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1 Answer

up vote 2 down vote accepted

a) How has the Pontryagin class been defined for you? The definition doesn't use an orientation in any way. If $E$ is a real vector bundle (not necessarily oriented) over a manifold $M$, then the Pontryagin classes $p_k(E) \in H^{4k}(M; \Bbb Z)$ can be defined in terms of the Chern classes of the complexification of $E$: $$p_k(E) = (-1)^k c_{2k}(E \otimes \Bbb C).$$ This definition makes no use of any orientation on $E$, so it should be clear that $p_k(E)$ doesn't depend on a choice of orientation.

b) If $E$ is a complex vector bundle and $\bar{E}$ is the conjugate bundle, then we have that $$c_k(\bar{E}) = (-1)^k c_k(E).$$ This is Lemma 14.9 in Milnor and Stasheff's Characteristic Classes.

c) By "diffeomorphic" vector bundles, do you mean isomorphic vector bundles? In that case, yes, characteristic classes are invariants of the isomorphism class of a bundle. One way to see this is by the result you quoted: The classifying maps for two isomorphic bundles are homotopic to each other, and a characteristic class on $E$ is just the pullback of the universal characteristic class on the classifying bundle. Homotopic maps induce the same map on cohomology, so we find that isomorphic bundles have the same characteristic classes.

From b) and c) we can see that the canonical line bundle $\gamma$ and its conjugate bundle $\bar{\gamma}$ are not isomorphic, since $c_1(\gamma) \neq 0$ and $c_1(\gamma) = - c_1(\bar{\gamma})$ ($H^2(\Bbb C P^n; \Bbb Z)$ has no $2$-torsion).

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Henry, Thank you very much for your detailed response. –  Clark Cho Apr 2 '13 at 21:41
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