Stiefel-Whitney classes and Steenrod squares

August 14, 2026

Stiefel-Whitney classes and Steenrod squares

I recently thought a bit about Steenrod squares and Stiefel-Whitney classes, and found that I was forced to organise the jumbled mix of tidbits I knew about them. I've written this up since I found it formed a very interesting picture. All of this is basically a compilation of answers I found useful on MO, and any credit goes to the posters of said answers.

First recall the axiomatic definition of Stiefel-Whitney classes (I leave discussion of the extended properties to Milnor-Stasheff or these notes).

For all finite rank real vector bundles $E$ on a paracompact base space $X$, define the Stiefel-Whitney classes to be the unique classes $w(E)\in H^{*}(X;\mathbb {Z} /2\mathbb {Z} )$ such that the following axioms are fulfilled:

Normalization: The Whitney class of the tautological line bundle over the real projective space $\mathbf {P} ^{1}(\mathbb {R} )$ is nontrivial, i.e. $w(\gamma _{1}^{1})=1+a\in H^{*}(\mathbf {P} ^{1}(\mathbb {R} );\mathbb {Z} /2\mathbb {Z} )=(\mathbb {Z} /2\mathbb {Z} )[a]/(a^{2})$.
Rank: $w_{0}(E)=1\in H^{0}(X)$, and for $i$ above the rank of $E$, $w_{i}=0\in H^{i}(X)$, that is, $w(E)\in H^{\leqslant \mathrm {rank} (E)}(X)$.
Whitney product formula: $w(E\oplus F)=w(E)\smile w(F)$, that is, the Whitney class of a direct sum is the cup product of the summands' classes.
Naturality: $w(f^{*}E)=f^{*}w(E)$ for any real vector bundle $E\to X$ and map $f\colon X'\to X$, where $f^{*}E$ denotes the pullback vector bundle.

The reason to give an axiomatic definition is that there are many equivalent constructions but it is hard to understand the relation of different constructions to each other (and sometimes annoying to prove their equivalence), and ultimately the construction is not the point (although different viewpoints have different advantages when actually thinking about them).

With that said, let me give a concrete, geometric definition (which is also the one I learned as a master's student).

Let $V\to X$ be a real vector bundle. Let $S^\infty=\bigcup S^n$, the infinite dimension sphere. Taking product with $S^\infty$ gives a vector bundle $V\times S^\infty\to X\times S^\infty$. I produce a vector bundle $V'\to X\times \mathbb{RP}^\infty$ by dividing out by an action of the cyclic group of order $2$ on both base and total space:

on the base $X\times S^\infty$, the involution is $(x,y)\mapsto (x,-y)$
on the total space $V\times S^\infty$, the involution is $(v,y)\mapsto (-v,-y)$.

The Euler class $e(V')$ of $V'$ is an element of degree $n$ in $H^*(X\times \mathbb{RP}^\infty; \mathbb{Z}/2) = H^*(X;\mathbb{Z}/2)[t]$. The following formula holds:
$$ e(V') = t^n + w_1(V)t^{n-1}+\cdots + w_n(V).$$
So if you have an Euler class, then you can use this as the *definition* of the Stiefel-Whitney classes. The mod-2 Euler class is fairly easy to define from Milnor-Stasheff's point of view: $e(V')$ is the pullback along the $0$ section of the orientation class in the cohomology of the Thom space of $V'$.

It's easy to check the axioms in this case. It's certainly natural, since $V\mapsto V'$ and $e$ are functorial. Whitney sum follows from $(V\oplus W)'\approx V'\oplus W'$ and the Whitney sum formula for the Euler class. If $R\to *$ is the trivial bundle, then $R'\to \mathbb{RP}^\infty$ is the canonical line, so $e(R')=t$ and so $w_0(R)=1$ and $w_1(R)=0$. You can use this to show that $w_0(V)\in H^0X$ is equal to $1$ for *any* bundle over $X$, by pulling back $V$ over any point of $X$ (where it becomes trivial). If $L\to \mathbb{RP}^\infty$ is the canonical line, then $L'\to \mathbb{RP}^\infty\times \mathbb{RP}^\infty$ is $L_1\otimes L_2$, the tensor product of the canonical line bundles over each factor. So $e(L')=s+t= 1\cdot t^1 + s\cdot t^0$, giving $w_0(L)=1$ and $w_1(L)=s$ (where $s\in H^1\mathbb{RP}^\infty$ is the generator).

Stiefel-Whitney classes have a much richer structure, but to describe them we need to introduce Steenrod powers.

Steenrod powers

Decades ago, people proved that in addition to cup products, there exist a series of interesting maps between the cohomology groups of spaces called Steenrod power operations. By using completely abstract nonsense, Epstein generalised these to very general module categories.

Let $\mathcal{M}$ be the category of modules over a ring $R$ of characteristic $p$. Suppose $R$ has a commutative associative diagonal $\Delta\colon R\to R\otimes R$ and $\mathcal{M}$ has a tensor product. Given $R$-modules $A,B\in \mathcal{M}$ we define may consider their tensor product as abelian groups $A\otimes_{\mathbb{Z}} B$ as an $R$-module via the diagonal $\Delta$.

Let $\mathbf{Ab}$ denote the category of abelian groups equipped with its usual tensor product $-\otimes_{\mathbb{Z}}-$. There are adjoint functors
\[\mathrm{frgt}\colon \mathcal{M} \to \mathbf{Ab} \quad\textrm{and}\quad \mathrm{indt}\colon \mathbf{Ab} \to \mathcal{M},\]
where $\mathrm{frgt}$ is the forgetful functor, and for $A\in\mathbf{Ab}$ we have $\mathrm{indt}(A)=R \otimes_{\mathbb{Z}} A$.

Theorem (Epstein): If $A$ is a commutative associative algebra in $\mathcal{M}$ and $C$ is a commutative associative coalgebra in $\mathcal{M}$, then $\mathrm{Ext}_R^\ast(C,A)$ admits reduced power operations satisfying the following properties when $p=2$:

$\mathrm{Sq}^i\colon \mathrm{Ext}^n(C,A)\to \mathrm{Ext}^{n+i}(C,A)$ is a homomorphism;
$\mathrm{Sq}^n\colon\mathrm{Ext}^n(C,A)\to \mathrm{Ext}^{2n}(C,a)$ is given by $x\mapsto x^2$;
if $n>i$, then $\mathrm{Sq}^i\colon\mathrm{Ext}^n(C,A)\to \mathrm{Ext}^{n+i}(C,A)$ is the zero map;
(Cartan formulae) $\mathrm{Sq}^n(xy)=\sum_{i+j=n}\mathrm{Sq}^i(x)\mathrm{Sq}^j(y)$;
(Adem's relations) suppose $\mathrm{Sq}^i=0$ for $i<0$ in $\mathrm{Ext}_R^\ast(C,A)$, if $a<2b$ then \[\mathrm{Sq}^a\mathrm{Sq}^b=\sum_{j}\binom{b-1-j}{a-2j}\mathrm{Sq}^{a+b-j}\mathrm{Sq}^j\]
if $i<0$, then $\mathrm{Sq}^i=0$.

when $p>2$ there exist power operations $P^i$ and $Q^i$ satisfying

\begin{align*} &P^i\colon \mathrm{Ext}^n(C,A)\to \mathrm{Ext}^{n+2i(p-1)}(C,A) \text{ and} \\ &Q^i\colon \mathrm{Ext}^n(C,A)\to \mathrm{Ext}^{n+2i(p-1)+1}(C,A) \end{align*}are homomorphisms
$P^n \colon\mathrm{Ext}^{2n}(C,A)\to \mathrm{Ext}^{2np}(C,a)$ is given by $x\mapsto x^p$
if $n>2i$, then $P^i$ is the zero map, and if $n \geq 2i$ then $Q^i$ is the zero map;
(Cartan formulae) \begin{align*}&P^n(xy)=\sum_{i+j=n}P^i(x)P^j(y); \\&Q^n(xy)=\sum_{i+j=n}Q^i(x)P^j(y) + (-1)^{\rm{dim}(x)}\sum_{i+j=n}P^i(x)Q^j(y) \end{align*}
if $i<0$, then $P^i,Q^i=0$.
If $ a < pb $ then

\[P^a P^b = \sum_t (-1)^{a+t} \binom{(b-1)(b-t)-1}{a-pt} P^{a+b-t} P^t. \]
If $ a \leq pb $ then \[ Q^a P^b = \sum_t (-1)^{a+t} \binom{(p-1)(b-t)-1}{a-pt} Q^{a+b-t} P^t \]
If $ a \leq pb $ then \[ Q^a Q^b = \sum_t (-1)^{a+t+1} \binom{(p-1)(b-t)-1}{a-pt-1} Q^{a+b-t} Q^t. \]
If $ a < pb $ then \[ Q^a P^b = \sum_t (-1)^{a+t} \binom{(p-1)(b-t)-1}{a-pt} Q^{a+b-t} P^t.\]

As before, there are good reasons to give abstract axiomatic definitions, but the reader might very reasonably have symbol overload, so let's see an alternative and rather explicit description, due to Hatcher, in the case $p=2$ (It is quite explicit and geometric, which appeals to some people like me, but purists might dislike this for the same reason they dislike his textbooks). Hatcher's construction uses the fact that elements of $H^n(X; G)$ are the same as homotopy classes of maps from X to the Eilenberg-Maclane space $K(G, n)$, a connected CW-complex whose nth homotopy group is G and others are trivial. Then you can understand any cohomology operation $H^n(-; G)\to H^m(-;H)$ as a map on two spaces, $K(G,n)\to K(H,m)$. This makes it easy to see why cohomology operations on the same group have to increase dimension, and it gives you an actual space to do Steenrod squares in.

The thing that Hatcher does that is different from, say, Steenrod and Epstein, is that rather than construct anything on cohomology rings, he constructs an operation on Eilenberg-Maclane spaces. The $n$th cohomology group of any space $X$ with coefficients in $G$ is naturally isomorphic to the group of homotopy classes of maps from $X$ to an Eilenberg-Maclane space $K(G, n)$: this identification is nice and functorial because if you map $X\rightarrow Y$ you can map $X\rightarrow Y\rightarrow K(G, n)$ to get a pullback map on cohomology groups. Now if you want to make a cohomology operation, which can be defined naturally on cohomology groups for any $X$, all you need is an operation $K(G, n)\rightarrow K(H, m)$ and by composing with that you get a natural map $H^n(X; G)\rightarrow H^m(X;H)$.

So Hatcher's idea is to take a kind of (smash) product of $K(\mathbb Z/2, n)$ (which he calls $K_n$) with itself, and map that to $K_{2n}$ in a way that will be like the cup-product square in dimension $n$, since $\mathrm{Sq}^n(\alpha)=\alpha^2$ when $\alpha$ is of dimension $n$. More explicitly, since $H^n(X\wedge X)$ is isomorphic to $H^n(X)\otimes H^n(X)$ by the Kunneth formula, the point is to find an element of $H^{2n}(X\wedge X)$ which is the cup-product square. (His notation is NOT standard.)

Intuitively, the idea is to do this for $K_n$ so that it works on all $X$, and to put the copies of $K_n$ back together by quotienting by a $\mathbb Z/2$ action. But the $\mathbb Z/2$-action on $K_n\wedge K_n$ given by switching is not free, so to make it free he takes $S^\infty\times K_n\wedge K_n$ and uses the antipodal map on $S^\infty$ to map $(x, y, z)$ to $(-x, z, y)$ and give a free $\mathbb Z/2$ action. (The coordinates here are really in $S^\infty\times K_n\times K_n$). Then the whole rest of the construction is about dealing with the extra things that $\mathbb RP^\infty$ gives you, and from there he can calculate what happens to homology elements of $\wedge X$ when it is mapped to $K_{2n}$. That tells you the action of $\mathrm{Sq}^{n-i}$ on $H^i(X)$.

Ok, fine. What does this mean? (e.g. homology represents 'holes' of a certain dimension)

The external cup square $a \otimes a \in H^{2n}(X \times X)$ of $a \in H^n(X)$ induces a map $f:X \times X \to K(\mathbb{Z}/2\mathbb{Z}, 2n)$. It can be shown that this map factors through a map $g:(X \times X) \times_{\mathbb{Z}/2\mathbb{Z}} E\mathbb{Z}/2\mathbb{Z} \to K_{2n}$ (retaining the above notation), where $\mathbb{Z}/2\mathbb{Z}$ acts on the product by permuting the factors and $E\mathbb{Z}/2\mathbb{Z}$ can be taken to just be $S^\infty$. If you unravel what this means, it says that our original map $f$ was homotopic to the map obtained by first switching the coordinates and then applying $f$. It also says that this homotopy, when applied twice to get a homotopy from $f$ to itself, is homotopic to the identity homotopy, and we similarly have a whole series higher "coherence" homotopies. Now $X \times B\mathbb{Z}/2\mathbb{Z}$ maps to $(X \times X) \times_{\mathbb{Z}/2\mathbb{Z}} E\mathbb{Z}/2\mathbb{Z}$ as the diagonal, so we get a map $X \times B\mathbb{Z}/2\mathbb{Z} \to K(2n)$. But $B\mathbb{Z}/2\mathbb{Z}$'s cohomology is just $\mathbb{Z}/2\mathbb{Z}[t]$, so this gives a cohomology class $Sq(a) \in H^*(X)[t]$ of degree $2n$. If we write $Sq(a)=\sum s(i) t^i$, it can be shown that $s(i)=\mathrm{Sq}^{n-i}a$.

Now note that if our map $f$ actually was invariant under switching the factors (which you might think it ought to be, given that it appears to be defined symmetrically in the two factors), we could take $g$ to be just the projection onto $X \times X$ followed by $f$. This would mean that $Sq(a)$ comes from just projecting away the $B\mathbb{Z}/2\mathbb{Z}$ and then using $a^2$, i.e. $\mathrm{Sq}^n(a)=a^2$ and $\mathrm{Sq}^i(a)=0$ for all other $i$. Thus the nonvanishing of the lower Steenrod squares somehow measures how the cup product, while homotopy-commutative (in terms of the induced maps to Eilenberg-MacLane spaces), cannot be straightened to be actually commutative. Indeed, in the universal example $X=K(\mathbb{Z}/2\mathbb{Z},n)$, the map $f$ is exactly the universal map representing the cup product of two cohomology classes of degree $n$.

We'll return to this below in the discussion on cohomology operations. Note that this answers a very common question: Why is the formula for the cup product so goofy and asymmetric? The answer is that it has to be, because the Steenrod operations obstruct the possibility of doing better.

There are a huge number of uses of Steenrod squares. For instance, they commute with suspension, but suspension kills cup products (even though $\mathrm{Sq}^n$ is the cup product square on $H^n(X)$, after suspension the cup product square would be $\mathrm{Sq}^{n+1}$ but $\mathrm{Sq}^n$ is still defined). It's also useful that the sum $\mathrm{Sq}^*$ of the various maps forms a ring operation of the cohomology ring.

Wu classes

Let me now elaborate on the relationship between Stiefel-Whitney classes and Steenrod squares.

Definition: Let $M$ be a Poincare complex and let
\[\langle.,.\rangle : H^i(M) \times H_i(M) \rightarrow \mathbb{F}_2 \]
be the pairing of cohomology and homology. $ M $ satisfies Poincaré duality for $ \mathbb{F}_2$ (co)homology, hence we have
\[\mathrm{Hom}(H^{n-k}(M), \mathbb{F}_2) \cong H_{n-k}(M) \cong H^k(M),\]
where a cohomology class $ y \in H^k(M) $ corresponds to the homomorphism $ x \mapsto \langle y \cup x, [M] \rangle $. In particular, one element of $ \mathrm{Hom}(H^{n-k}(M), \mathbb{F}_2)$ is uniquely determined by the expression $ x \mapsto \langle \mathrm{Sq}^k(x), [M] \rangle $.
Under the above isomorphism, the unique class corresponds to a degree $ k $ cohomology class $ v_k $. We call $v_k$ the $k$th Wu class of $M$ and $1+\sum_{i=1}^n v_k$ the total Wu class.

Note that we have the following:
\begin{align*}
& \langle v_k \cup x, [M] \rangle = \langle \text{Sq}^k(x), [M] \rangle, \text{for all } x \in H^{n-k}(M), \text{and} \\
& \langle v \cup x, [M] \rangle = \langle \text{Sq}(x), [M] \rangle, \text{for all } x \in H^*(M)
\end{align*}

For a closed connected smooth manifold, the Wu classes are related to the Stiefel-Whitney classes by the following theorem of Wu.

Theorem 2 (Wu formula): Let $M$ be a closed connected smooth $n$-manifold. Then $w_k=\sum_{i=0}^k\mathrm{Sq}^{k-i}(v_i)$.

The definition of the Wu classes only requires the existence of the Poincare duality isomorphisms. Thus, we may turn the Wu formula on its head and instead take this as the definition of the Stiefel--Whitney classes for a general Poincare complex.

Definition: For a compact connected Poincare complex $M$ of dimension $n$ we define the Stiefel--Whitney classes $w_k(M)$ of $M$ by the formula $w_k=\sum_{i=0}^k\mathrm{Sq}^{k-i}(v_i)$.

Geometric interpretation

A short digression on a possible geometric interpretation of Wu classes. First, if $X$ is a manifold of dimension $d$ then one can produce classes in $H^n(X)$ by proper maps $f: V \to X$ where $V$ is a manifold of dimension $d-n$ through many possible formalisms - eg. intersection theory (the value on a transverse $i$-cycle is the count of intersection points), or using the fundamental class in locally finite homology and duality, or Thom classes, or as the pushforward $ f_*(1) $ where $1$ is the unit class in $H^0(V)$. Taking this last approach, suppose $f$ is an immersion and thus has a normal bundle $\nu$. If $x = f_*(1) \in H^n(X)$ then $\mathrm{Sq}^i(x) = f_*(w_i(\nu))$. This is essentially the Wu formula.

That is, if cohomology classes are represented by submanifolds, and for example cup product reflects intersection data, then Steenrod squares remember normal bundle data.

Orientation classes

We have seen above that Steenrod operations come from an "extended square" construction on cohomology classes.
If $X$ is a space, let $DX=(X\times X \times S^\infty)/(Z/2)$, where I divide by the involution $(x_1,x_2,y)\to (x_2,x_1,-y)$. The "extended square" is a function
$$P: H^n(X) \to H^{2n}(DX).$$
Cohomology is with mod-2 coefficients. If you restict along the "diagonal" embedding $d: X\times RP^\infty \to DX$, you get Steenrod squares:
$$d^*(P(a)) = t^{n}\mathrm{Sq}^0(a) + t^{n-1}\mathrm{Sq}^1(a) + \cdots + \mathrm{Sq}^n(a).$$
There's a relative version of this: if $V\to X$ is a vector bundle, so is $DV\to DX$; write $T(V)$ for the Thom space of $V$, and write $f: T(V) \to T(DV)$ for the map induced by diagonal inclusion. If $u\in H^nT(V)$ is the orientation class, then
$$f^*(P(u))= t^{n}\mathrm{Sq}^0(u)+t^{n-1}\mathrm{Sq}^1(u)+\cdots +\mathrm{Sq}^n(u).$$
Recall that by the above discussion on Wu classes, we have $\mathrm{Sq}^i(u)=u\cup w_i(V)$.

The neat fact is that $P(u)\in H^{2n}T(DV)$ has to be the orientation class $u'$ of $DV\to DX$! So as long as I can describe the orientation class, I don't need to know about Steenrod operations! Thus, $f^*(u')\in H^*TV[t]$ is the polynomial whose coefficients are the SW classes. To get the formula I gave originally, observe that $f^*(u')=u\, e(V')$; this is because the pullback of the bundle $TV\to TX$ along $d: X\to DX$ is the same as the bundle $V+V' \to X$.

Why is $P(u)$ the orientation class of $DV$? The orientation class of a bundle in ordinary cohomology mod-2 is the unique element which restricts to the fundamental class of the sphere when you restrict to each fibre, so you just have to check that $P(u)$ has this property. And this is pretty easy (the operation $P$ is natural, and it's easy to understand how $P$ works when you have a discrete space, or a bundle over a discrete space.)

Cohomology operations

Steenrod operations are an example of what's known as a *power operation*. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.

A cohomology class on $X$ amounts to a map $a: X\to R:= \prod_{n\geq0} K(\mathbb{F}_2,n)$. So the cup product of $a$ and $b$ is given by
$$X\times X \to R\times R \xrightarrow{\mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes *addition* of cohomology classes.)

You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/\Sigma_n$, where $\Sigma_n$ is the symmetric group, i.e.,
$$X^n \xrightarrow{a^n} R^n \rightarrow R$$
should factor through the quotient $X^n/\Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $\Sigma_n$, and a product map:
$$\mu_n' : E(n)\times R^n\to R$$
which is $\Sigma_n$ invariant, so it factors through $(E(n)\times R^n)/\Sigma_n$. Thus, given $a: X\to R$, you get
$$P'(a): (E(n)\times X^n)/\Sigma_n \to (E(n)\times R^n)/\Sigma_n \to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/\Sigma_n \times X\to R.$$

More generally, the Steenrod square is an example of a cohomology operation, i.e. a natural transformation from the cohomology functor to itself. There are a few different types of cohomology operations, but the most general is an **unstable** cohomology operation. This is simply a natural transformation from $E^k(-)$ to $E^l(-)$ for some fixed $k$ and $l$. Here, one regards the graded cohomology functors as a family of set-valued functors so the functions induced by these unstable operations do not necessarily respect any of the structure of $E^k(X)$.

Some do, however. In particular, there are **additive** cohomology operations. These are unstable operations which are homomorphisms of abelian groups.

In particular, for any multiplicative cohomology theory (in particular, ordinary cohomology or ordinary cohomology with $\mathbb{Z}/2\mathbb{Z}$ coefficients) there are the \textit{power} operations: $x \to x^k$. These are additive if the coefficient ring has the right characteristic (the so-called freshman's dream). In particular, squaring is additive in $\mathbb{Z}/2\mathbb{Z}$ cohomology.

Given an unstable cohomology operation $r: E^k(-) \to E^l(-)$ there is a way to manufacture a new operation $\Omega r: \tilde{E}^{k-1} \to \tilde{E}^{l-1}(-)$ using the suspension isomorphism (where the tilde denotes that these are reduced groups):

$E^{k-1}(X) \cong E^k(\Sigma X) \to E^l(\Sigma X) \cong E^{l-1}(X)$

This is quite straightforward and is a cheap way of producing more operations. When applied to the power operations it produces almost nothing since the ring structure on the cohomology of a suspension is trivial: apart from the inclusion of the coefficient ring all products are zero.

What is an interesting question is whether or not this looping can be reversed. Namely, if $r$ is an unstable operation, when is there another operation $s$ such that $\Omega s = r$? And how many such are there? Most interesting is the question of when there is an infinite chain of operations, $(r_k)$ such that $\Omega r_k = r_{k-1}$. When this happens, we say that $r$ comes from a **stable** operation (there is a slight ambiguity here as to when the sequence $(r_k)$ \textit{is} a stable operation or merely comes from a stable operation).

One necessary condition is that $r$ be additive. This is not, in general, sufficient. For example, the Adams operations in $K$-theory are additive but all but two are not stable.

However, for ordinary cohomology with coefficients in a field, additive is sufficient for an operation to come from a stable operation. Moreover, there is a unique sequence for each additive operation. This means that the squaring operation in $\mathbb{Z}/2\mathbb{Z}$ cohomology has a sequence of "higher" operations which loop down to squaring. These are the Steenrod squares.

The sequence stops with the actual squaring (rather, becomes zero after that point) because, as remarked above, the power operations loop to zero.

One important feature of these operations is that they give necessary conditions for a spectrum to be a suspension spectrum of a space. If a spectrum is such a suspension spectrum then its $\mathbb{Z}/2\mathbb{Z}$-cohomology must be a ring. That's not enough, however, it must also have the property that, in the right dimensions, the Steenrod operations act by squaring. (Of course, this is necessary but not sufficient.)

Here are my sources:
https://mathoverflow.net/questions/461/understanding-steenrod-squares/2783#2783
https://mathoverflow.net/questions/6377/why-does-one-think-to-steenrod-squares-and-powers/6384#6384
https://amathew.wordpress.com/tag/steenrod-algebra/page/2/
https://amathew.wordpress.com/2011/10/18/thoms-construction-of-the-stiefel-whitney-classes/

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