Please let me know what is the formula for the Wu class $v_6$ in terms of Stiefel-Whitney classes.
Many thanks.
2026-03-25 11:16:49.1774437409
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What is the formula for the Wu class $v_6$ in terms of Stiefel-Whitney classes?
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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.
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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?
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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
5
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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?
0
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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.
Using the procedure showed by Michael Albanese; and implemented it with Mathematica I am obtaining
$$v_{{6}}={w_{{1}}}^{2}{w_{{2}}}^{2}+{w_{{1}}}^{3}w_{{3}}+w_{{1}}w_{{2}} w_{{3}}+{w_{{3}}}^{2}+{w_{{1}}}^{2}w_{{4}}+w_{{2}}w_{{4}} $$
Please let me know if such result is correct.
Many thanks.
The Mathematica Code: