TITLE: Stiefel-Whitney class of complex projective spaces
QUESTION [8 upvotes]: Let $T\mathbb{C}P^m$ be the tangent bundle of complex projective space. What is the total Stiefel-Whitney class $w(T\mathbb{C}P^m)$? 
Let $a_m$ be the maximal integer such that the $a_m$-th dual Stiefel-Whitney class $\bar w_{a_m}(T\mathbb{C}P^m)$  is nonzero ($\bar w=1/w$, $\bar w_j$ is the degree-$j$-component of $\bar w$). Can we write $a_m$ in terms of $m$?
I do not know how to solve. I even do not know how to solve the case $m=2$...

REPLY [3 votes]: Note that $H^*(\mathbb{CP}^m; \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha]/(\alpha^{m+1})$ where $\deg \alpha = 2$, and 
$$w(T\mathbb{CP}^m) = (1 + \alpha)^{m+1} = 1 + \binom{m+1}{1}\alpha + \binom{m+1}{2}\alpha^2 + \dots + \binom{m+1}{m}\alpha^m.$$
where the coefficients are taken modulo $2$. 
In a (truncated) polynomial ring, $(1 - \beta)^{-1} = 1 + \beta + \beta^2 + \dots$ Setting $\beta = 1 - w(T\mathbb{CP}^m)$, this gives a formula for $\overline{w}(T\mathbb{CP}^m)$ from which we can read off $a_m$.
For example, $w(T\mathbb{CP}^4) = 1 + \binom{5}{1}\alpha + \binom{5}{2}\alpha^2 + \binom{5}{3}\alpha^3 + \binom{5}{4}\alpha^4 = 1 + \alpha + \alpha^4$, so 
$$\overline{w}(T\mathbb{CP}^4) = 1 + (\alpha + \alpha^4) + (\alpha + \alpha^4)^2 + (\alpha  + \alpha^4)^3 + (\alpha + \alpha^4)^4 = 1 + \alpha + \alpha^2 + \alpha^3$$
and hence $a_4 = 6$.
Here are the first few values of $a_m$.
\begin{array}{c|c|c|c}
m & w(T\mathbb{CP}^m) & \overline{w}(T\mathbb{CP}^m) & a_m\\ \hline
1 & 1 & 1 & 0\\
2 & 1 + \alpha + \alpha^2 & 1 + \alpha & 2\\
3 & 1 & 1 & 0\\
4 & 1 + \alpha + \alpha^4 & 1 + \alpha + \alpha^2 + \alpha^3 & 6\\
5 & 1 + \alpha^2 + \alpha^4 & 1 + \alpha^2 & 4\\
6 & 1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 & 1 + \alpha & 2\\
7 & 1 & 1 & 0\\
8 & 1 + \alpha + \alpha^8 & 1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 + \alpha^7 & 14\\
9 & 1 + \alpha^2 + \alpha^8 & 1 + \alpha^2 + \alpha^4 + \alpha^6 & 12\\
10 & 1 + \alpha + \alpha^2 + \alpha^3 = \alpha^8 + \alpha^9 + \alpha^{10} & 1 + \alpha + \alpha^4 + \alpha^5 & 10
\end{array}
It appears as if $a_m = 2s_m$ where $s_m$ is the $m^{\text{th}}$ term of $A035327$ on the Online Encyclopedia of Integer Sequences. That is, $s_m$ is the value obtained by writing $m$ in binary, interchanging zeroes and ones, and then reverting back to decimal. For example, $m = 4 = 100_2$ and $011_2 = 3$ so $s_4 = 3$ and $a_4 = 6$. 
To see why this is the case, write $m = 2^{i_1} + \dots + 2^{i_k}$ with $0 \leq i _1 < \dots < i_k = \lfloor\log_2(m)\rfloor$, then $s_m = 2^{j_1} + \dots + 2^{j_l}$ where $\{j_1, \dots, j_l\} = \{0, \dots, \lfloor\log_2(m)\rfloor\}\setminus\{i_1, \dots, i_k\}$. Then $m + s_m = 2^0 + 2^1 + \dots + 2^{\lfloor\log_2(m)\rfloor - 1} + 2^{\lfloor\log_2(m)\rfloor} = 2^{\lfloor\log_2(m)\rfloor + 1} - 1$ so
$$(1 + \alpha)^{m+1}(1+\alpha)^{s_m} = (1 + \alpha)^{m + 1 + s_m} = (1 + \alpha)^{2^{\lfloor\log_2(m)\rfloor + 1}} = 1 + \alpha^{2^{\lfloor\log_2(m)\rfloor + 1}} = 1$$
as $2^{\lfloor\log_2(m)\rfloor + 1} \geq m + 1$. Therefore 
\begin{align*}
\overline{w}(T\mathbb{CP}^m) &= (1 + \alpha)^{s_m}\\ 
&= (1 + \alpha)^{2^{j_1} + \dots + 2^{j_k}}\\ 
&= (1 + \alpha)^{2^{j_1}}\dots(1 + \alpha)^{2^{j_k}}\\ 
&= \left(1 + \alpha^{2^{j_1}}\right)\dots\left(1 + \alpha^{2^{j_k}}\right)\\ 
&= 1 + \dots + \alpha^{2^{j_1}+\dots + 2^{j_k}}\\
&= 1 + \dots + \alpha^{s_m}.
\end{align*}
Note that $s_m < m$, so $\alpha^{s_m} \neq 0$ and hence $a_m = 2s_m$.