Mixing
Integral class mod 2 reduction, and Stiefel Whitney class
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
If there is an integral 3rd cohomology class called Y, such that whose mod 2 reduction is the Steenrod square of the 2nd cohomology of Stiefel Whitney class, then we have
$$ Y=Sq^1 w_2 mod 2$$
In 5 dim, is this true that such a lift from $Sq^1 w_2$ to the integral class $Y$
$$
int_5d w_2Y=0?
$$
or maybe
$$
int_5d w_2Y=0 mod 2 (?).
$$Has this relation anything to do with the Euler class or decomposition of Euler class?
My trial: It looks that if $ Y=Sq^1 w_2 mod 2$, we have
$$
Sq^1 (w_2^2)= w_2 Y +Y w_2 mod 2 .
$$
I dont know does it help to relate the above relation to
$
int_5d Sq^1 (w_2^2)=0 mod 2 (?).
$ There is an additional factor of 2 for the later equality which makes the later more trivial, but the earlier nontrivial.
algebraic-topology differential-topology homology-cohomology geometric-topology characteristic-classes
asked Jul 31 at 22:33
annie heart
544616
add a comment |Â
up vote
3
down vote
favorite
If there is an integral 3rd cohomology class called Y, such that whose mod 2 reduction is the Steenrod square of the 2nd cohomology of Stiefel Whitney class, then we have
$$ Y=Sq^1 w_2 mod 2$$
In 5 dim, is this true that such a lift from $Sq^1 w_2$ to the integral class $Y$
$$
int_5d w_2Y=0?
$$
or maybe
$$
int_5d w_2Y=0 mod 2 (?).
$$Has this relation anything to do with the Euler class or decomposition of Euler class?
My trial: It looks that if $ Y=Sq^1 w_2 mod 2$, we have
$$
Sq^1 (w_2^2)= w_2 Y +Y w_2 mod 2 .
$$
I dont know does it help to relate the above relation to
$
int_5d Sq^1 (w_2^2)=0 mod 2 (?).
$ There is an additional factor of 2 for the later equality which makes the later more trivial, but the earlier nontrivial.
algebraic-topology differential-topology homology-cohomology geometric-topology characteristic-classes
asked Jul 31 at 22:33
annie heart
544616
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
Thanks, annie hart.
– amWhy
Jul 31 at 22:47
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If there is an integral 3rd cohomology class called Y, such that whose mod 2 reduction is the Steenrod square of the 2nd cohomology of Stiefel Whitney class, then we have
$$ Y=Sq^1 w_2 mod 2$$
In 5 dim, is this true that such a lift from $Sq^1 w_2$ to the integral class $Y$
$$
int_5d w_2Y=0?
$$
or maybe
$$
int_5d w_2Y=0 mod 2 (?).
$$Has this relation anything to do with the Euler class or decomposition of Euler class?
My trial: It looks that if $ Y=Sq^1 w_2 mod 2$, we have
$$
Sq^1 (w_2^2)= w_2 Y +Y w_2 mod 2 .
$$
I dont know does it help to relate the above relation to
$
int_5d Sq^1 (w_2^2)=0 mod 2 (?).
$ There is an additional factor of 2 for the later equality which makes the later more trivial, but the earlier nontrivial.
algebraic-topology differential-topology homology-cohomology geometric-topology characteristic-classes
asked Jul 31 at 22:33
annie heart
544616
If there is an integral 3rd cohomology class called Y, such that whose mod 2 reduction is the Steenrod square of the 2nd cohomology of Stiefel Whitney class, then we have
$$ Y=Sq^1 w_2 mod 2$$
In 5 dim, is this true that such a lift from $Sq^1 w_2$ to the integral class $Y$
$$
int_5d w_2Y=0?
$$
or maybe
$$
int_5d w_2Y=0 mod 2 (?).
$$Has this relation anything to do with the Euler class or decomposition of Euler class?
My trial: It looks that if $ Y=Sq^1 w_2 mod 2$, we have
$$
Sq^1 (w_2^2)= w_2 Y +Y w_2 mod 2 .
$$
I dont know does it help to relate the above relation to
$
int_5d Sq^1 (w_2^2)=0 mod 2 (?).
$ There is an additional factor of 2 for the later equality which makes the later more trivial, but the earlier nontrivial.
algebraic-topology differential-topology homology-cohomology geometric-topology characteristic-classes
asked Jul 31 at 22:33
annie heart
544616
edited Jul 31 at 22:47
asked Jul 31 at 22:33
annie heart
544616
asked Jul 31 at 22:33
annie heart
544616
asked Jul 31 at 22:33
annie heart
544616
544616
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
Thanks, annie hart.
– amWhy
Jul 31 at 22:47
add a comment |Â
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
Thanks, annie hart.
– amWhy
Jul 31 at 22:47
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
Thanks, annie hart.
– amWhy
Jul 31 at 22:47
Thanks, annie hart.
– amWhy
Jul 31 at 22:47
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
There is always a Bockstein connecting homomorphism $delta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)$ induced by the short exact sequence of coefficients $mathbbZxrightarrowtimes 2 mathbbZxrightarrowred_2 mathbbZ_2$.
The composition $red_2circdelta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)rightarrow H^n+1(X;mathbbZ_2)$ coincides with the operation $Sq^1$.
Since $omega_2in H^2(BO(n);mathbbZ_2)congmathbbZ_2$ is a generator satisfying $Sq^1omega_2=omega_3neq 0$, the class $deltaomega_2in H^3(BO(n);mathbbZ)$ is non-trivial. Since $H^3(BO(n);mathbbZ)cong Ext(H_2(BO(n);mathbbZ)),mathbbZ)cong mathbbZ_2$ by the Universal Coefficient Theorem, it must be that $deltaomega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1omega_2=omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(omega_2^2)=2,omega_2cdot Sq^1omega_2=2omega_2cdot omega_3=0$, which is the same as $Ycdotomega_2+omega_2cdot Y=2,Ycdotomega_2equiv 0mod 2$.
Edit: I'm afraid I don't know what your $int_5domega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=omega_3$? Then you are looking for the evaluation
$$langle omega_2omega_3,[M^5]rangleinmathbbZ_2$$
of $omega_2omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;mathbbZ_2)=mathbbZ_2[x_2,x_3]/(x_2^2,x_3^2),qquad Sq^1x_2=x_3,$$
where $x_iin H^i(M;mathbbZ_2)$. We have $omega_2(M)=x_2$ and therefore $omega_3(M)=Sq^1omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;mathbbZ_2)$ we have
$$langle omega_2omega_3,[M^5]rangleneq 0.$$
answered Aug 1 at 7:47
Tyrone
3,15611025
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
There is always a Bockstein connecting homomorphism $delta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)$ induced by the short exact sequence of coefficients $mathbbZxrightarrowtimes 2 mathbbZxrightarrowred_2 mathbbZ_2$.
The composition $red_2circdelta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)rightarrow H^n+1(X;mathbbZ_2)$ coincides with the operation $Sq^1$.
Since $omega_2in H^2(BO(n);mathbbZ_2)congmathbbZ_2$ is a generator satisfying $Sq^1omega_2=omega_3neq 0$, the class $deltaomega_2in H^3(BO(n);mathbbZ)$ is non-trivial. Since $H^3(BO(n);mathbbZ)cong Ext(H_2(BO(n);mathbbZ)),mathbbZ)cong mathbbZ_2$ by the Universal Coefficient Theorem, it must be that $deltaomega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1omega_2=omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(omega_2^2)=2,omega_2cdot Sq^1omega_2=2omega_2cdot omega_3=0$, which is the same as $Ycdotomega_2+omega_2cdot Y=2,Ycdotomega_2equiv 0mod 2$.
Edit: I'm afraid I don't know what your $int_5domega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=omega_3$? Then you are looking for the evaluation
$$langle omega_2omega_3,[M^5]rangleinmathbbZ_2$$
of $omega_2omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;mathbbZ_2)=mathbbZ_2[x_2,x_3]/(x_2^2,x_3^2),qquad Sq^1x_2=x_3,$$
where $x_iin H^i(M;mathbbZ_2)$. We have $omega_2(M)=x_2$ and therefore $omega_3(M)=Sq^1omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;mathbbZ_2)$ we have
$$langle omega_2omega_3,[M^5]rangleneq 0.$$
answered Aug 1 at 7:47
Tyrone
3,15611025
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
add a comment |Â
up vote
3
down vote
There is always a Bockstein connecting homomorphism $delta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)$ induced by the short exact sequence of coefficients $mathbbZxrightarrowtimes 2 mathbbZxrightarrowred_2 mathbbZ_2$.
The composition $red_2circdelta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)rightarrow H^n+1(X;mathbbZ_2)$ coincides with the operation $Sq^1$.
Since $omega_2in H^2(BO(n);mathbbZ_2)congmathbbZ_2$ is a generator satisfying $Sq^1omega_2=omega_3neq 0$, the class $deltaomega_2in H^3(BO(n);mathbbZ)$ is non-trivial. Since $H^3(BO(n);mathbbZ)cong Ext(H_2(BO(n);mathbbZ)),mathbbZ)cong mathbbZ_2$ by the Universal Coefficient Theorem, it must be that $deltaomega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1omega_2=omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(omega_2^2)=2,omega_2cdot Sq^1omega_2=2omega_2cdot omega_3=0$, which is the same as $Ycdotomega_2+omega_2cdot Y=2,Ycdotomega_2equiv 0mod 2$.
Edit: I'm afraid I don't know what your $int_5domega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=omega_3$? Then you are looking for the evaluation
$$langle omega_2omega_3,[M^5]rangleinmathbbZ_2$$
of $omega_2omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;mathbbZ_2)=mathbbZ_2[x_2,x_3]/(x_2^2,x_3^2),qquad Sq^1x_2=x_3,$$
where $x_iin H^i(M;mathbbZ_2)$. We have $omega_2(M)=x_2$ and therefore $omega_3(M)=Sq^1omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;mathbbZ_2)$ we have
$$langle omega_2omega_3,[M^5]rangleneq 0.$$
answered Aug 1 at 7:47
Tyrone
3,15611025
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There is always a Bockstein connecting homomorphism $delta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)$ induced by the short exact sequence of coefficients $mathbbZxrightarrowtimes 2 mathbbZxrightarrowred_2 mathbbZ_2$.
The composition $red_2circdelta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)rightarrow H^n+1(X;mathbbZ_2)$ coincides with the operation $Sq^1$.
Since $omega_2in H^2(BO(n);mathbbZ_2)congmathbbZ_2$ is a generator satisfying $Sq^1omega_2=omega_3neq 0$, the class $deltaomega_2in H^3(BO(n);mathbbZ)$ is non-trivial. Since $H^3(BO(n);mathbbZ)cong Ext(H_2(BO(n);mathbbZ)),mathbbZ)cong mathbbZ_2$ by the Universal Coefficient Theorem, it must be that $deltaomega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1omega_2=omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(omega_2^2)=2,omega_2cdot Sq^1omega_2=2omega_2cdot omega_3=0$, which is the same as $Ycdotomega_2+omega_2cdot Y=2,Ycdotomega_2equiv 0mod 2$.
Edit: I'm afraid I don't know what your $int_5domega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=omega_3$? Then you are looking for the evaluation
$$langle omega_2omega_3,[M^5]rangleinmathbbZ_2$$
of $omega_2omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;mathbbZ_2)=mathbbZ_2[x_2,x_3]/(x_2^2,x_3^2),qquad Sq^1x_2=x_3,$$
where $x_iin H^i(M;mathbbZ_2)$. We have $omega_2(M)=x_2$ and therefore $omega_3(M)=Sq^1omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;mathbbZ_2)$ we have
$$langle omega_2omega_3,[M^5]rangleneq 0.$$
answered Aug 1 at 7:47
Tyrone
3,15611025
There is always a Bockstein connecting homomorphism $delta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)$ induced by the short exact sequence of coefficients $mathbbZxrightarrowtimes 2 mathbbZxrightarrowred_2 mathbbZ_2$.
The composition $red_2circdelta:H^n(X;mathbbZ_2)rightarrow H^n+1(X;mathbbZ)rightarrow H^n+1(X;mathbbZ_2)$ coincides with the operation $Sq^1$.
Since $omega_2in H^2(BO(n);mathbbZ_2)congmathbbZ_2$ is a generator satisfying $Sq^1omega_2=omega_3neq 0$, the class $deltaomega_2in H^3(BO(n);mathbbZ)$ is non-trivial. Since $H^3(BO(n);mathbbZ)cong Ext(H_2(BO(n);mathbbZ)),mathbbZ)cong mathbbZ_2$ by the Universal Coefficient Theorem, it must be that $deltaomega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1omega_2=omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(omega_2^2)=2,omega_2cdot Sq^1omega_2=2omega_2cdot omega_3=0$, which is the same as $Ycdotomega_2+omega_2cdot Y=2,Ycdotomega_2equiv 0mod 2$.
Edit: I'm afraid I don't know what your $int_5domega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=omega_3$? Then you are looking for the evaluation
$$langle omega_2omega_3,[M^5]rangleinmathbbZ_2$$
of $omega_2omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;mathbbZ_2)=mathbbZ_2[x_2,x_3]/(x_2^2,x_3^2),qquad Sq^1x_2=x_3,$$
where $x_iin H^i(M;mathbbZ_2)$. We have $omega_2(M)=x_2$ and therefore $omega_3(M)=Sq^1omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;mathbbZ_2)$ we have
$$langle omega_2omega_3,[M^5]rangleneq 0.$$
answered Aug 1 at 7:47
Tyrone
3,15611025
edited Aug 1 at 17:07
answered Aug 1 at 7:47
Tyrone
3,15611025
answered Aug 1 at 7:47
Tyrone
3,15611025
answered Aug 1 at 7:47
Tyrone
3,15611025
3,15611025
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
add a comment |Â
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
Thanks +1, I mean, by lifting $Sq^1 w_1$ to $Y$ as an integral cohomology class, then evaluating the $int w_2 Y$, in any 5d manifold within the mod 2 cohomology class
– annie heart
Aug 1 at 14:01
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
@annieheart, as requested....
– Tyrone
Aug 1 at 17:07
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868556%2fintegral-class-mod-2-reduction-and-stiefel-whitney-class%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
the formulation of question contains my thought. It is a relation I suggest may be correct in some cases, maybe generally true.
– annie heart
Jul 31 at 22:36
I add some more thoughts using Wu formula
– annie heart
Jul 31 at 22:46
Thanks, annie hart.
– amWhy
Jul 31 at 22:47