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Please let me know what is the formula for the Wu class $v_6$ in terms of Stiefel-Whitney classes.

Many thanks.

Juan Ospina
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    The first three Wu classes are computed here. To see the calculation of the fourth and fifth Wu classes, follow the link in my answer. If you understand these, you can calculate $\nu_n$ for any $n$ although it becomes computationally more intense as $n$ increases. – Michael Albanese Sep 11 '18 at 02:34
  • @MichaelAlbanese, I am implementing your procedure using Mathematica. I am obtaining an expression for $v_6$. Please, could you checked it? Many thanks. – Juan Ospina Sep 12 '18 at 13:01
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    How did you use Mathematica to compute this? – Michael Albanese Sep 12 '18 at 13:50
  • @MichaelAlbanese, many thanks for your indications and checks. The Mathematica Code was added in my answer. Please let me know what do you think. All the best. – Juan Ospina Sep 12 '18 at 13:56
  • Does this really need a new answer every time you observe something new? That seems more appropriate for personal notes or a blog. –  Nov 08 '18 at 03:00

6 Answers6

5

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:

rule1 = {Sq[i_, a_ + b_] :> Sq[i, a] + Sq[i, b]};

rule2 = {Sq[0, x_] :> x};
rule3 = {Sq[i_, v[i_]] :> v[i]^2};
rule4 = {Sq[i_, 0] :> 0};



W[k_] := Simplify[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(13\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(13\)]If[i > j, 0, 1]*
     If[i + j == k, 1, 0]*Sq[i, v[j]]\)\)]


W[1] /. rule2 /. rule3


rule5 := {v[1] :> w[1]}


W[2] /. rule2 /. rule3 /. rule5

rule6 := {v[2] :> w[2] + w[1]^2}



rule7 := {Sq[1, x_^2] :> 0};


rule7A := {Sq[1, x_^4] :> 0};




Sq[i_, w[j_]] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(t = 0\), \(i\)]\(w[i - t]*w[j + t]*
   Mod[Binomial[j - i + t - 1, t], 2]\)\)

Sq[k_, x_*y_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(13\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(13\)]If[i + j == k, 1, 
     0]*Sq[i, x]*Sq[j, y]\)\)


rule7B := {Sq[k_, x_^(n_)] :> \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(13\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(13\)]If[i + j == k, 1, 
        0]*Sq[i, x^\((n - 1)\)]*Sq[j, x]\)\)};

w[0] := 1;

W[3] /. rule2 /. rule6 /. rule1 /. rule7

Solve[w[3] == v[3] + w[1] w[2] + w[3], v[3]] /. -1 -> 1


rule8 := {v[3] :> w[1]*w[2]};


rule9 := {(x_ + y_)^2 :> x^2 + y^2}





rule10 := {(2*x_) :> 0}

rule10A := {(3*x_) :> x}
rule10B := {(7*x_) :> x}
rule10C := {(12*x_) :> 0}

Simplify[W[4] /. rule2 /. rule3 /. rule8 /. rule6 /. rule9] /. rule10

Solve[w[4] == v[4] + w[1]^4 + w[2]^2 + w[1] w[3], v[4]] /. -1 -> 1


rule11 := {v[4] :> w[1]^4 + w[2]^2 + w[1] w[3] + w[4]}



Expand[W[5] /. rule11 /. rule8 /. rule2 /. rule1 /. rule1 /. rule1 /. 
     rule10 /. rule10A] /. rule7 /. rule7A

Solve[w[5] == 
   v[5] + w[1]^3 w[2] + w[1] w[2]^2 + w[1]^2 w[3] + w[1] w[4] + w[5], 
  v[5]] /. -1 -> 1

rule12 := {v[5] :> w[1]^3 w[2] + w[1] w[2]^2 + w[1]^2 w[3] + w[1] w[4]}

Expand[Expand[
             W[6] /. rule12 /. rule11 /. rule8 /. rule2 /. rule1 /. 
                  rule1 /. rule1 /. rule2 /. rule7 /. rule10] /. 
            rule10A /. rule10 /. rule10B /. rule7B /. rule7B /. 
       rule7 /. rule2 /. rule7B] /. rule10 /. rule10A /. rule10C

Solve[w[6] == 
   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] + w[6]] /. -1 -> 1
Juan Ospina
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    I just did it by hand and I got the same answer. – Michael Albanese Sep 12 '18 at 13:49
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    Now that you have $\nu_6$, it seems like you can add this as a new rule and then solve for $\nu_7$. Is that correct? – Michael Albanese Sep 12 '18 at 14:42
  • @MichaelAlbanese, you are right. In the Mathematica notebook, ·$v_1, v_2,v_3,v_4,v_5, v_6$ were computed successively and then them were added successively as new rules. In order to compute $v_7$ is necessary to know $v_6$. I will try to compute $v_7$ and I will post the result. – Juan Ospina Sep 12 '18 at 15:02
  • @MichaelAlbanese, the expression for $v_7$ was posted. Please check it. Many thanks. – Juan Ospina Sep 12 '18 at 17:22
  • @MichaelAlbanese, in the preprint https://arxiv.org/pdf/1109.4461.pdf, in the equation (2.8), appears the expression for $v_6$; but the equation (2.8) has a typo. In the equation (2.8) appears $w_3w_1 ^2$ but it must be $w_3w_1^3$ . Do you agree? – Juan Ospina Sep 14 '18 at 14:36
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    Yes. I think it is just a typo as $w_3w_1^2$ does not even have degree six. – Michael Albanese Sep 14 '18 at 18:16
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    @MichaelAlbanese, please look the equation (2.7) in https://arxiv.org/pdf/1109.4461.pdf, it gives a very nice method to compute all the Wu classes. – Juan Ospina Sep 14 '18 at 19:03
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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.

Juan Ospina
  • 2,257
1

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?

Juan Ospina
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  • I don't think this is correct. Recall that $w_{13} = \operatorname{Sq}^6(\nu_7) + \operatorname{Sq}^5(\nu_8) + \operatorname{Sq}^4(\nu_9) + \operatorname{Sq}^3(\nu_{10}) + \operatorname{Sq}^2(\nu_{11}) + \operatorname{Sq}^1(\nu_{12}) + \operatorname{Sq}^0(\nu_{13})$. – Michael Albanese Sep 13 '18 at 13:59
  • @MichaelAlbanese, using the property
    $v_i = 0$ for $2i > 15$ we have that $w_{13}= Sq^6(v_7)$. Do you agree?     – Juan Ospina Sep 13 '18 at 14:16
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    Of course. I missed that, then yes, this looks right. – Michael Albanese Sep 13 '18 at 14:33
  • @MichaelAlbanese, a similar lemma is valid in dimensions 3, 7, 11, 15. In the case of dim = 19, it is necessary to compute $v_9$, but I think that $v_9 =0$ when $w_1 = 0$. I am conjecturing that the lemma is valid in general for dim = 4 i + 3 for i =0,1,2,3.... Do you agree? – Juan Ospina Sep 13 '18 at 14:38
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    Yes. Your conjecture is true. – Michael Albanese Sep 13 '18 at 15:15
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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.

enter image description here 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

enter image description here

Juan Ospina
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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?

Juan Ospina
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    Looks fine, and this result corresponds to Theorem III in Massey's "On the Stiefel-Whitney classes of a manifold", https://www.jstor.org/stable/2372878?seq=1#metadata_info_tab_contents. (The result states that $w_n = w_{n-1} = w_{n-2} = 0$ for an oriented $n$-manifold if $n = 4k+3$.) It's also useful to note that every odd-indexed Wu class of an orientable manifold is 0. – Aleksandar Milivojević Nov 06 '18 at 17:44
  • @AleksandarMilivojevic many thanks for the link, the paper by Massey looks very interesting. Your observations are very worthwhile. – Juan Ospina Nov 06 '18 at 22:27
  • @AleksandarMilivojevic, I am observing that Massey presents the mentioned theorem as "Theorem III" on page 94. Massey says that the proof of theorem III is presented at section 3 but I can not find it. Please can you say me where is the proof? Many thanks. – Juan Ospina Nov 06 '18 at 22:49
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    See the last paragraph of Section 3. He says he is proving Theorem II, but it is really Theorem III being proved. – Aleksandar Milivojević Nov 06 '18 at 23:46
  • @AleksandarMilivojevic, yes, you are right. There is a typo in the last paragraph of Section 3. Many thanks. – Juan Ospina Nov 07 '18 at 00:51
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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.

Juan Ospina
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