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.
Best Answer
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 $\nu_5$.