What is the formula for the Wu class $v_6$ in terms of Stiefel-Whitney classes? Please let me know what is the formula for the Wu class $v_6$ in terms of Stiefel-Whitney classes.
Many thanks.
 A: Using the Mathematica Code I am obtaining
$$v_{{7}}={w_{{1}}}^{2}w_{{2}}w_{{3}}+w_{{1}}{w_{{3}}}^{2}+w_{{1}}w_{{2}
}w_{{4}}
$$
Please let me know if such result is correct.
A: As an application of the formula for $v_7$, we will prove the following lemma:
Let $Y^{15}$ be an orientable fifteen-manifold.  Then we have
$w_{15}(Y^{15}) =w_{14}(Y^{15})=w_{13}(Y^{15}) = 0 $.
Proof.  From the properties of the Wu classes we obtain for $Y^{15}$ that
$$\left\{ v_{{8}}=0,v_{{9}}=0,v_{{10}}=0,v_{{11}}=0,v_{{12}}=0,v_{{13}}=0,v_{{14}}=0,v_{{15}}=0 \right\}$$
Now from the Wu’s formula, we derive that
$$w_{{15}}=0$$
$$w_{{14}}={v_{{7}}}^{2}$$
$$w_{{13}}={\it Sq}^{{6}} \left( v_{{7}} \right) $$
From other side we know that
$$v_{{7}}={w_{{1}}}^{2}w_{{2}}w_{{3}}+w_{{1}}{w_{{3}}}^{2}+w_{{1}}w_{{2}
}w_{{4}}
$$
but given that $Y^{15}$ is orientable, it is to say $w_1 =0$; we obtain that $v_7=0$.
For hence we have that
$$w_{{15}}=0$$
$$w_{{14}}={v_{{7}}}^{2} = 0^2 = 0$$
$$w_{{13}}={\it Sq}^{{6}} \left( v_{{7}} \right)= {\it Sq}^{{6}} \left( 0\right)=0 $$
And then our lemma is proved.
Do you agree?
A: A direct method for the computation of the Wu classes is presented in https://arxiv.org/pdf/1109.4461.pdf
The following paragraph shows the method.

The equation (2.7) is implemented via Maple using the package Schubert, according with the following code:

restart:with(schubert); DIM:=15: b:=bundle(15,c):  eq:=todd(b):
  aux:={c1=w1,c2=w2,c3=w[3],c4=w[4],c5=w[5],c6=w[6],c7=w[7],c8=w[8],c9=w[9],c10=w[10],c11=w[11],c12=w[12],c13=w[13],c14=w[14],c15=w[15]};
  for i from 1 to 11 do
  print(v[i]=subs(w1=w1,subs(aux,coeff(todd(b),t,i)*2^i) mod 2)) end
  do;

Executing such code we obtain

A: As an application of the formula for $v_9$, we will prove the following lemma:
Let $Y^{19}$ be an orientable nineteen-manifold.  Then we have
$w_{19}(Y^{19}) =w_{18}(Y^{19})=w_{17}(Y^{19}) = 0 $.
Proof.  From the properties of the Wu classes we obtain for $Y^{19}$ that
$$\left\{ v_{{10}}=0,v_{{11}}=0,v_{{12}}=0,v_{{13}}=0,v_{{14}}=0,v_{{15}}=0,
 v_{{16}}=0,v_{{17}}=0,v_{{18}}=0,v_{{19}}=0\right\}$$
Now from the Wu’s formula, we derive that
$$w_{{19}}=0$$
$$w_{{18}}={v_{{9}}}^{2}$$
$$w_{{17}}={\it Sq}^{{8}} \left( v_{{9}} \right) $$
From other side we know that
$$v_{{9}}=w_{{1}}w_{{8}}+w_{{7}}{w_{{1}}}^{2}+w_{{6}}{w_{{1}}}^{3}+w_{{5
}}{w_{{1}}}^{4}+w_{{1}}{w_{{4}}}^{2}+w_{{4}}{w_{{1}}}^{5}+w_{{3}}{w_{{
1}}}^{6}+w_{{1}}{w_{{2}}}^{4}+{w_{{2}}}^{2}{w_{{1}}}^{5}+w_{{2}}{w_{{1
}}}^{7}+w_{{6}}w_{{1}}w_{{2}}+w_{{5}}w_{{1}}w_{{3}}+w_{{4}}w_{{3}}{w_{
{1}}}^{2}+w_{{3}}w_{{2}}{w_{{1}}}^{4}
$$
but given that $Y^{19}$ is orientable, it is to say $w_1 =0$; we obtain that $v_9=0$.
For hence we have that
$$w_{{19}}=0$$
$$w_{{18}}={v_{{9}}}^{2} = 0^2 = 0$$
$$w_{{17}}={\it Sq}^{{8}} \left( v_{{9}} \right)= {\it Sq}^{{8}} \left( 0\right)=0 $$
And then our lemma is proved.
Do you agree?
A: As  Aleksandar Milivojevic is noting, Theorem III in Massey's "On the Stiefel-Whitney classes of a manifold", is formulated as
Let $Y^{4k + 3}$ be an orientable (4k+3)-manifold.  Then we have
$w_{4k+3}(Y^{4k+3}) =w_{4k+2}(Y^{4k+3})=w_{4k+1}(Y^{4k+2}) = 0 $.
Proof.  From the properties of the Wu classes we obtain for $Y^{4k+3}$ that
$$\left\{ v_{{2k+2}}=0,v_{{2k+3}}=0,v_{{2k+4}}=0,....,v_{{4k+3}}=0\right\}$$
Now from the Wu’s formula, we derive that
$$w_{{4k+3}}=0$$
$$w_{{4k+2}}={v_{{2k + 1}}}^{2}$$
$$w_{{4k+1}}={\it Sq}^{{2k}} \left( v_{{2k+1}} \right) $$
From other side, given that $Y^{4k+3}$ is orientable, we have that
$$v_{{2k+1}}=0$$.
For hence we have that
$$w_{{4k+3}}=0$$
$$w_{{4k+2}}={v_{{2k+1}}}^{2} = 0^2 = 0$$
$$w_{{4k+1}}={\it Sq}^{{2k}} \left( v_{{2k+1}} \right)= {\it Sq}^{{2k}} \left( 0\right)=0 $$
And then, the Massey`s theorem  is proved.
