# Wu classes for physicists: a question about the preprint " Ninebrane Structures "

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I am reading the very interesting pre-print  " Ninebrane Structures"  (https://arxiv.org/pdf/1405.7686.pdf). In such pre-print on page 3 and 4 appears the following lemma and its proof: My questions  are:

1.  Provide a more direct proof (algorithmical or automatic) of the lemma 2.1.

2.  How  is the corresponding lemma for a  manifold with 13 dimensions ?

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An automated proof of the lemma 2.1 can be performed using Maple as follows:

aux:=seq(sq(v[i])=v[i],i=0..11);

aux1:=seq(sq[i](v[i])=v[i]^2,i=1..11);

aux2:=seq(sq[i](0)=0,i=0..11);

aux3:=seq(v[i]=0,i=6..11);

W:=(k)->subs(aux2,subs(aux3,aux1,aux,sum([seq(seq(sq[i](v[j])*if(i+ j= k,1,0)*if(i>j,0,1),j=0..11),i=0..11)][n],n=1..12*12)));

Then we obtain that:

$$w_{{11}}=0$$

$$w_{{10}}={v_{{5}}}^{2}$$

$$w_{{9}}={\it Sq}^{{4}} \left( v_{{5}} \right)$$

Now using that and given that $w_{{1}}=0$, we deduce that  $v_5 =0$; for hence we finally obtain that

$$w_{{11}}=0$$

$$w_{{10}}={v_{{5}}}^{2} = 0$$

$$w_{{9}}={\it Sq}^{{4}} \left( v_{{5}} \right) = 0$$

Do you agree?

answered Sep 10, 2018 by (1,130 points)
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Respect to the second question, there is not analogous lemma for dim =13; but there is a lemma for dim = 15, as follows:

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?

answered Sep 13, 2018 by (1,130 points)

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