Calculate the Wu class from the Stiefel-Whitney class

Question

The total Stiefel-Whitney class $w=1+w_1+w_2+\cdots$ is related to the total Wu class $u=1+u_1+u_2+\cdots$: The total Stiefel-Whitney class $w$ is the Steenrod square of the Wu class $u$: \begin{align} w=Sq(u),\ \ \ Sq=1+Sq^1+Sq^2 +\cdots . \end{align} The Wu classes can be defined through the Steenrod square (is this right? see nLab). $$ Sq^k(x) = \begin{cases} u_k x & \text{ for any } x \text{ with dim more than } k-1, \\ 0 & \text{ for any } x \text{ with dim less than } k. \end{cases}$$ where $u_k x$ is understood as $u_k\cup x$. Thus we have (dose the second equal sign hold?) \begin{align} w_i=\sum_{k=0}^i Sq^k u_{i-k} = \sum_{k=0}^{i-k-1} u_k u_{i-k} . \end{align}

Now we try to invert the relation. We first expand the above \begin{align} w_1&=u_1, \ \ \ w_2=u_2+u_1^2, \ \ \ w_3=u_3+u_1u_2, \end{align} This allows us to obtain \begin{align} u_1=w_1,\ \ \ u_2=w_2+w_1^2,\ \ \ u_3=w_3+w_1w_2+w_1^3,\ \ \ \end{align}

But on nLab (and several other places), it says $u_3=w_1 w_2$. I must have made an error in my calculation above, but I do not know where. Thank you for help.

Answer 1

First of all, your definition of Wu class is incorrect. If $X$ is a closed connected $n$-manifold, there is a unique class $\nu_k \in H^k(X; \mathbb{Z}_2)$ such that for any $x \in H^{n-k}(X; \mathbb{Z}_2)$, $\operatorname{Sq}^k(x) = \nu_k\cup x$. We call $\nu_k$ the $k^{\text{th}}$ Wu class. If $X$ is also smooth, then the Stiefel-Whitney classes of the tangent bundle of $X$ are related to Steenrod squares and Wu classes by the formula

$$w_i = \sum_{k = 0}^i\operatorname{Sq}^k(\nu_{i-k}).$$

Note $\operatorname{Sq}^k(\nu_{i-k})$ is not simply $\nu_k\cup\nu_{i-k}$ unless $i = n$. So we have

\begin{align*} w_1 &= \operatorname{Sq}^0(\nu_1) = \nu_1\\ w_2 &= \operatorname{Sq}^0(\nu_2) + \operatorname{Sq}^1(\nu_1) = \nu_2 + \nu_1\cup\nu_1\\ w_3 &= \operatorname{Sq}^0(\nu_3) + \operatorname{Sq}^1(\nu_2) = \nu_3 + \operatorname{Sq}^1(\nu_2) \end{align*}

It follows that $\nu_1 = w_1$ and $\nu_2 = w_2 + w_1\cup w_1$. However, at this stage we can only deduce $\nu_3 = w_3 + \operatorname{Sq}^1(\nu_2)$. In order to determine $\nu_3$ in terms of Stiefel-Whitney classes, we need to compute $\operatorname{Sq}^1(\nu_2)$. First note that

\begin{align*} \operatorname{Sq}^1(\nu_2) &= \operatorname{Sq}^1(w_2 + w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^1(w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^0(w_1)\cup\operatorname{Sq}^1(w_1) + \operatorname{Sq}^1(w_1)\cup\operatorname{Sq}^0(w_1) && \text{(by Cartan's formula)}\\ &= \operatorname{Sq}^1(w_2) \end{align*}

so $\nu_3 = w_3 + \operatorname{Sq}^1(w_2)$. To compute Steenrod squares of Stiefel-Whitney classes, we use Wu's formula

$$\operatorname{Sq}^i(w_j) = \sum_{t=0}^k\binom{j-i+t-1}{t}w_{i-t}\cup w_{j+t}.$$

In this case, we see that

$$\operatorname{Sq}^1(w_2) = \binom{0}{0}w_1\cup w_2 + \binom{1}{1}w_0\cup w_3 = w_1\cup w_2 + w_3.$$

Therefore, $\nu_3 = w_3 + \operatorname{Sq}^1(w_2) = w_3 + w_1\cup w_2 + w_3 = w_1\cup w_2$. Suppressing the cup symbol, this agrees with the identity given on nLab.

See this note for more details, as well as the computations for $\nu_4$ and $\nu_5$.

Answer 2

The relation can be written more concretely as $$ w_{k}=\sum_{i+j=k}Sq^{i}(u_{j}) $$ where $u=1+u_{1}+u_{2}\cdots +u_{n}$ is the Wu class. Expanding it we have $$ w_1=Sq^{1}u_{0}+Sq^{0}u_{1}=u_1 $$ because $\mu_0=1\in H^{0}(M)$ and $Sq^{i}(a)=0$ for $i>n$, if $a\in H^{i}(M)$.

The second relation is $$ w_{2}=Sq^{2}u_{0}+Sq^{1}u_{1}+Sq^{0}u_{2}=u_{1}\cup u_{1}+u_{2} $$ Therefore using mod 2 coefficients we have $$ u_{2}=w_{2}+w_{1}\cup w_{1} $$ The third relation is $$ w_{3}=Sq^{3}u_{0}+Sq^{2}u_{1}+Sq^{1}u_{2}+Sq_{0}u_{3}=Sq^{1}u_{2}+u_{3} $$ Therefore it suffice to calculate $Sq^{1}w_{2}+Sq^{1}(w_{1}\cup w_{1})$. The second one can be calculated by the Cartan formula: $$ Sq^{1}(w_{1}\cup w_{1})=Sq^{1}w_{1}\cup w_{1}+w_{1}\cup Sq^{1}w_{1}=0 $$ because of the mod 2 coefficients we are taking. We now have $$ Sq^{1}w_{2}=w_{1}w_{2}+ \binom{-1}{1}w_{3}=w_{1}w_{2}-w_{3}=w_{1}w_{2}+w_{3} $$ where we used Wu's formula: $$ Sq^{k}(w_{m})=w_{k}w_{m}+\binom{k-m}{1}w_{k-1}w_{m+1}\cdots +\binom{k-m}{k}w_{0}w_{m+k} $$ here $\binom{a}{b}=x(x-1)\cdots (x-i+1)/i!$. Therefore we have $$ u_{3}=w_1\cup w_{2} $$ as desired. Thanks for Michael Albanese's answer correcting my mistake.

Reference:

Milnor&Stasheff, page 91, page 132.