$O(n)$ is an extension of $\mathbb{Z}_2$ by $SO(n)$, $$1\to SO(n) \to O(n)\to \mathbb{Z}_2 \to 1.$$

Below we denote the Stiefel-Whitney class of real vector bundle $V_G$ of the group $G$ as: $$ w_j(V_{G}) : =w_j({G}). $$

My question is that how do the "generalized" Stiefel-Whitney class of $O(n)$ and $SO(n)$ relate to each other? What are related conversion formulas for $$ w_j(O(n)) = w_j(SO(n)) + ...? $$

What I have known are that:

  1. $$ w_3({O(3)})=w_1({O(3)})^3+w_1({O(3)})w_2({SO(3)})+w_3({SO(3)}) $$ $$ {=w_1(\mathbb{Z}_2)^3+w_1(\mathbb{Z}_2)w_2(SO(3))+w_3(SO(3)) } $$ $$ =w_1(\mathbb{Z}_2)^3+w_1(\mathbb{Z}_2)w_2(SO(3))+w_3(SO(3)) $$ $$ {=w_1(\mathbb{Z}_2)^3+w_1(\mathbb{Z}_2)w_2(SO(3))+w_1(TM)w_2(O(3))} $$ $$ {=w_1(\mathbb{Z}_2)^3+w_1(\mathbb{Z}_2)w_2(SO(3))+w_1(TM)w_2(SO(3))} $$

2. $$ w_2({O(3)})=w_1({O(3)})^2+w_1({O(3)})w_2({SO(3)}) $$ $$ =w_1(\mathbb{Z}_2)^2+w_1(\mathbb{Z}_2)w_2({SO(3)}) $$

  1. When $n=1 \mod 4$, $$w_2(O(n)) = w_2(SO(n)) \mod 2, $$ When $n=3 \mod 4$,

$$ w_2(O(n)) = w_2(SO(n)) + w_1 \cup w_1 \mod 2, $$

The $w_1 \cup w_1$ is an obstruction to lifting $w_1$ to $\mathbb Z_4$ cohomology class.

Again, do we have

$$ w_j(O(n)) = w_j(SO(n)) + ...? $$

also, do we have

$$ w_2(O(2)) = w_2(SO(2))? $$ $$ w_j(O(2)) = w_j(SO(2))? $$

Edit for clarification: p.s. See eq. 2.5 of this journal free-access online article -- I am using the same definition as theirs of “generalized” Stiefel-Whitney class of real vector bundles: $w_j(O(n))$ and $w_j(SO(n))$.

  • 1
    $\begingroup$ I'm sorry I cannot understand what the "SW classes of $O(n)$ or $SO(n)$" are. The Stiefel-Whitney classes are what they are (the indecomposable in the mod 2 homology of $BO(n)$). Are you asking for a description of their restriction to $BSO(n)$? $\endgroup$ Jan 5, 2019 at 22:31
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    $\begingroup$ @annieheart I'm sorry to be obtuse, but I cannot find an answer to my question in your link. Can you be more specific? $\endgroup$ Jan 5, 2019 at 22:45
  • 3
    $\begingroup$ See eq. 2.5 of this paper, scipost.org/SciPostPhys.4.4.021/pdf -- I am using the same definition as theirs of “generalized” Stiefel-Whitney class of real vector bundles: O(n) and SO(n) $\endgroup$
    – wonderich
    Jan 5, 2019 at 23:37
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    $\begingroup$ The question is, as I understand it: if $n$ is odd, one can use the splitting $O(n) = SO(n)\times\mathbb Z/2$ to, given a principal $O(n)$-bundle $P\to M$, construct a principal $SO(n)$-bundle $Q\to M$. Then, what are the Stiefel-Whitney classes of $Q$ in terms of those of $P$? $\endgroup$ Jan 6, 2019 at 0:14
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    $\begingroup$ I cannot actually think of a worse choice of notation. $\endgroup$
    – mme
    Jan 6, 2019 at 0:22

1 Answer 1


First I will write up what your question is asking in terms of Arun Debray's comment. I strongly suggest that when discussing questions like this, you use precise notation as in the following; I found your question impossible to understand until that comment.

First, the Stiefel-Whitney classes are a set of generators of the cohomology ring of $BO(n)$: we have the presentation $H^*(BO(n);\Bbb F_2) = \Bbb F_2[w_1, \cdots, w_n]$. Via the inclusion map $i: SO(n) \to O(n)$, we have a map $BSO(n) \to BO(n)$, and the induced map $i^*: H^*(BO(n);\Bbb F_2) \to H^*(BSO(n);\Bbb F_2)$ is well-known to have kernel $\langle w_1\rangle$; in particular, we may use this homomorphism to write $H^*(BSO(n);\Bbb F_2) = \Bbb F_2[w_2, \cdots, w_n]$, where these classes are the images of the correspinding $w_i$ under restriction. In particular, one computes the Stiefel-Whitney classes of an oriented bundle by forgetting the orientation.

You have a group homomorphism $p: O(n) \to SO(n)$ given by $p(A) = \det A \cdot A$; this induces a map of classifying spaces $BO(n) \to BO(n)$, with image inside $BSO(n)$ when $N$ is odd.

Your question is: in terms of the generators specified above, what is the induced map $f^*: H^*(BO(n);\Bbb F_2) \to H^*(BO(n);\Bbb F_2)$?

The way to do this is to investigate what this does at the level of vector bundles; it takes an unoriented vector bundle $E$ to the corresponding canonically oriented bundle $E \otimes \det(E)$; that is, $$(f^*w_i)(E) = w_i(E \otimes \det(E)).$$

I claim that if $E$ has rank $k$, then $$w_i\left(E \otimes \det(E)\right) = \sum_{j=0}^i \binom{k-j}{k-i} w_1(E)^{i-j} w_j(E),$$ and in fact a more general formula is true for any even-rank bundle $V$ replacing $E$ and any real line bundle replacing $\det(E)$.

This follows from the splitting principle: if $V$ splits as a direct sum $V = \eta_1 \oplus \cdots \oplus \eta_{k}$, then $$V \otimes \lambda = \bigoplus (\eta_i \otimes \lambda),$$ and taking Stiefel-Whitney classes before and after taking the tensor product, we have $$w(V) = \prod (1 + w_1(\eta_i)) = \sum_{i=0}^{k}\sigma_i(w_1(\eta_1), \cdots, w_1(\eta_{k}))$$ and $$w(V \otimes \lambda) = \prod (1 + w_1(\eta_i) + w_1(\lambda)) = \sum_{i=0}^k \sigma_i(w_1(\eta_i) + w_1(\lambda), \cdots, w_1(\eta_k) + w_1(\lambda))),$$ where $\sigma_i$ is the $i$th symmetric polynomial in $k$ variables. Expanding the latter out, we obtain $$\sum_{i=0}^k \sum_{j=0}^i \binom{k-j}{k-i} w_1(\lambda)^{i-j} \sigma_j(w_1(\eta_1), \cdots, w_1(\eta_k)) = \sum_{i=0}^k \sum_{j=0}^i \binom{k-j}{k-i} w_1(\lambda)^{i-j}w_j(V);$$ the binomial appears because when counting terms of this form, we first fix the $j$-element of $w_1(\eta)$'s that arise, and then choose from the remaining $(k-j)$-element set a collection of $(i-j)$ copies of $w_1(\lambda)$.

The splitting principle says that any formula of characteristic classes which is true for direct sums of line bundles is true for all bundles, so the general result follows. This is your desired formula.

Plugging in, we find that your desired formulas in (3) are correct: we have $f^*w_2 = w_2 + (k-1) w_1^2 + \binom{k-1}{k-3} w_1^2 = w_2 + \frac{(k-1)(k+2)}{2} w_1^2$, and this is equal to $w_2$ for $k \equiv 1, 2 \pmod 4$ and equal to $w_2 + w_1^2$ for $k \equiv 0, 3 \pmod 4$, as desired.

Note for your last displayed formula that $E \otimes \det(E)$ is not oriented when $E$ has even rank. However, it is true that in rank 2, $E \otimes \det(E) \cong E$, which gives $f^*w_j = w_j$ in this case.

  • $\begingroup$ The original version of this answer had some rather drastically incorrect claims about $E \otimes \det(E)$. They have been fixed. $\endgroup$
    – mme
    Jan 6, 2019 at 22:36
  • $\begingroup$ This is wrong again: the formula for the more complicated symmetric polynomial should have a binomial coefficient $\binom{k-j}{k-i}$ in the formula, I think. But there is a more serious issue. When $k$ is even, the formula should just reduce to $w_j(E \otimes \det(E)) = w_j(E)$, as there are no relations between Stiefel-Whitney classes. But it doesn't seem to. I won't edit until I understand what's going on. In particular the formula does depend on various parity facts about the rank. $\endgroup$
    – mme
    Jan 6, 2019 at 23:22
  • $\begingroup$ for the peanut gallery, my mistake was that the formula $E \otimes \det(E) \cong E$ only holds for all $E$ of a given rank when $E$ is of rank 2, and in this case the formulas do indeed collapse! Sorry for the many edits! $\endgroup$
    – mme
    Jan 6, 2019 at 23:50
  • $\begingroup$ This is very nice -- I will accept it as an answer -- but may ask more details later. Thanks! +1 $\endgroup$
    – wonderich
    Jan 7, 2019 at 17:44
  • $\begingroup$ (please feel free to edit to make sure they are correct.) $\endgroup$
    – wonderich
    Jan 7, 2019 at 17:44

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