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Suppose $\xi$ and $\eta$ are oriented vector bundles over a CW-complex $B$. Is it possible to express the Thom class (with ${\mathbb Z}$ coefficients) of $\xi\otimes \eta$ or even ${\rm Sym}^2(\xi)$ in terms of the Thom classes of $\xi$ and $\eta$?

If I am correct, there is a difficulty to apply the splitting principle, since one would need that all line bundles of the splitting should be orientable (to secure the existence of the Thom class of each line bundle).

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    $\begingroup$ I don't have an answer, but I have some advice: First think of the case where $B$ is a point. This may help you to clarify the question, and what form of answer you are looking for. $\endgroup$
    – Tom Goodwillie
    Commented Jul 6, 2015 at 21:51
  • $\begingroup$ @James: You sound like you're aware of a corresponding formula for $Z/2$ coefficients (the splitting principle would work there), care to explain? In particular for two line bundles, I am not even sure how to relate the thom spaces to the Thom space of the tensor product (which would be necessary for any meaningful formula on cohomology) $\endgroup$
    – Achim Krause
    Commented Jul 6, 2015 at 23:11
  • $\begingroup$ @Achim: Unfortunately I don't know the (exact) answer for ${\rm mod}\, 2$ coefficients but I would expect (but now the more I think about the less I am convinced about it) that the Thom class $u(L\otimes L')$ would be something like $u(L)+u(L')$ in some reasonable way (since this work for the Euler and Stiefel-Whitney class). But I see some problemes here: In case of the Thom class of $\xi\otimes\eta$ there could be an expression like $u(L)\cup u(L)$ which should make the Thom class of the tensor product zero?! $\endgroup$
    – Panagiotis Konstantis
    Commented Jul 7, 2015 at 9:09
  • $\begingroup$ @Tom: Right! And I think in that case I know the answer: Since every vector bundle over a point is trivial, the Thom space of a trivial bundle $\varepsilon^n$ over a point is the $n$-sphere. Hence the Thom space of $\varepsilon^n \otimes \varepsilon^m$ is $S^{nm}$ and the Thom class may be represented as follows: take a generator of $H^n(S^n;\mathbb Z)$ say $u$, which represents the Thom class of $\varepsilon^n$ over a point. Then the Thom class of $\varepsilon^n \otimes \varepsilon^m$ is $\Sigma^{(m-1)n} u$, the $(m-1)n$-th suspension of $u$ (the roles of $n$ and $m$ can be interchanged). $\endgroup$
    – Panagiotis Konstantis
    Commented Jul 7, 2015 at 10:35
  • $\begingroup$ Basically the fact that one can use the splitting principle is a consequence of the fact that the map $BO(1)^n\rightarrow BO(n)$ induces injection in mod $2$ cohomology, thus to compute the induced map in cohomology by $BO(n)\times BO(m)\rightarrow BO(mn)$, one can replace left hand side by $(BO(1)^m)\times (BO(1)^n)$. Of course, the same kind of arguments work over integers if we place $O$ by $U$, which means that if your bundles happen to be complex, then you will have a formula. As to the your original question, as $H^*(BO(n);Z)$ is known, there might be a way to get a formula. $\endgroup$
    – user43326
    Commented Sep 1, 2015 at 8:46

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