René Thom in 1952 proved the formula $$ Sq^i(U_2)=\Phi_2(w_i), $$ which in modern parlance says that the Steenrod squares of the mod $2$ Thom class of an orthogonal bundle are the images under the mod $2$ Thom isomorphism of the Stiefel-Whitney classes.

Apparently there is a mod $p$ analogue of this formula for oriented bundles, in which the $Sq^i$ are replaced by Steenrod powers $P^i_p$ and the $w_i$ are replaced by polynomials in the Pontryagin classes. Thom refers to this in his famous paper 1954 Commentarii paper; on page 49 he uses the instances $$ P^1_3(U_3)=\Phi_3(p_1),\qquad P^2_3(U_3)=\Phi_3(p_1^2+2p_2),\qquad P^1_5(U_5)=\Phi_5(p_1^2-2p_2), $$ and in a footnote at the bottom of page 49 claims that the calculations appear in the article

Borel, Armand; Serre, Jean-Pierre, Groupes de Lie et puissances réduites de Steenrood, Am. J. Math. 75, 409-448 (1953). ZBL0050.39603.

as well as some mimeographed notes bu Wu Wen Tsun with the title "Sur les puissances de Steenrod". Unfortunately I can't access Wu's notes, and the Borel-Serre paper only(!) gives the calculations of $P^i_p(p_k)$ from which I'm having trouble deducing the Thom class result. Hence my question:

Question: Does anyone know the general formula for $\Phi_p^{-1}P^i_p(U_p)$ in terms of the mod $p$ reductions of Pontryagin classes, where $U_p$ is the mod $p$ reduction of the Thom class for oriented bundles and $\Phi_p$ is the mod $p$ Thom isomorphism?

  • 3
    $\begingroup$ If memory serves, this is described in Milnor and Stasheff's book Characteristic Classes... yes it looks like it is Theorem 19.7, attributed to Wu. $\endgroup$ Commented May 21 at 19:45
  • $\begingroup$ @ChrisSchommer-Pries You're right! I read the first 18 chapters 20 years ago...maybe in another 20 years I'll read chapter 20. $\endgroup$
    – Mark Grant
    Commented May 22 at 8:52

1 Answer 1


I don't know a reference, but you can proceed as follows. By the splitting principle, it suffices to give the formula for vector bundles which are sums of complex line bundles, and we may as well then suppose it is an external sum of line bundles. So let $L_i \to B$ be a collection of complex line bundles, and $$E = L_1 \boxplus \cdots \boxplus L_n \to B \times \cdots \times B$$ be the external direct sum. The Thom space is then $$Th(E) = Th(L_1) \wedge \cdots \wedge Th(L_n)$$ and the Thom class of $E$ is $u_E = u_{L_1} \otimes \cdots \otimes u_{L_n}$.

Let $P = Id + P^1 + P^2 + \cdots$ be the total Steenrod power. Then by the axioms of Steenrod powers we have $$P(u_{L_i}) = u_{L_i} + u_{L_i}^{p} = \Phi_{L_i}(1+ e(L_i)^{p-1}).$$ Thus $$\Phi_E^{-1}P(u_E) = \bigotimes_{i=1}^n(1+e(L_i)^{p-1}) = \bigotimes_{i=1}^n(1+p_1(L_i)^{(p-1)/2}).$$ Pulling back along the diagonal to get the internal direct sum $E' = \Delta^*E = L_1 \oplus \cdots \oplus L_n$ we get $$\Phi_{E'}^{-1}P(u_{E'}) = \prod_{i=1}^n(1+p_1(L_i)^{(p-1)/2}).$$ This is a symmetric function in the $p_1(L_i)$'s, so can be expressed in terms of $p_1(E'), ..., p_n(E')$. This expression is then valid for all oriented bundles, by the splitting principle.

Concretely, we have $$\prod_{i=1}^n(1+p_1(L_i)) = \sum_j \sigma_j(p_1(L_1), \ldots, p_1(L_n))$$ with $\sigma_j(x_1, \ldots, x_n)$ the elementary symmetric polynomials, so the algebraic problem is to express the symmetric polynomial $\sigma_j(x_1^{(p-1)/2}, \ldots, x_n^{(p-1)/2})$ in terms of elementary symmetric polynomials. This is a standard operation on symmetric polynomials, called the $(p-1)/2$nd Frobenius. It is for example implemented in SAGE. I don't know a closed formula, probably there isn't one.

  • $\begingroup$ Thanks Oscar, this was super helpful. $\endgroup$
    – Mark Grant
    Commented May 23 at 9:22
  • $\begingroup$ You strike me as someone familiar with Thom's paper...am I right in thinking that there is a mistake on p.49 in the "Calcul mod $3$", where he has "$F^*(St^8_3 \iota)=U\cdot ((P_4)^2+2P_8)$"? I'm interpreting this as $P^2_3(U_3)=\Phi_3((p_1)^2 + 2p_2)$ above, but now it seems to me that $P^2_3(U_3)=\Phi_3(p_2)$. This wouldn't affect any of his later arguments. $\endgroup$
    – Mark Grant
    Commented May 23 at 9:27
  • 1
    $\begingroup$ I would guess it is a typo. The later calculation $F^*(St^8_5 \iota) = U \cdot ((P_4)^2 - 2 P_8)$ for p=5 fits with what I wrote. $\endgroup$ Commented May 25 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.