Confusion of sign in cohomology and homology pairing
$begingroup$
Let $(X,A),(Y,B)$ be two pairs of topological spaces. Denote $times$ as external cross product on cohomology level and $E:C_star(X,A)otimes C_star(Y,B)to C_star((X,A)times(Ytimes B))$ as Eilenberg-Zilber map and assume appropriate hypothesis(i.e. excision pairs on $A,B$ and $Atimes Y, Xtimes B$ excision pairs) s.t. $E$ induce homology level isomorphism. Set $<-,->:H^p(X;G_1)otimes H_p(X,G_2)to G_1otimes_Z G_2$ as bilinear pairing where $G_i$ are abelian groups. Whenever abelian group is omitted in homology, it is assumed to be $Z$ coefficients.
$textbf{Q:}$ I am quite confused about the following equation's sign $(-1)^{pq}$. "$ain H_p(X,A),bin H_q(Y,B),uin H^p(X,A;G_1),vin H^q(Y,B;G_2)$, then $(-1)^{pq}<utimes v,E(aotimes b)>=<u,a>otimes<v,b>$." Where is $(-1)^{pq}$ coming from?
The following is my reasoning. First $times$ is supercommutative. Hence $utimes v=(-1)^{pq}vtimes u$. Now I can use $H^q(Y,B,G_2)otimes H^p(X,A,G_1)otimes H_p(X,A)otimes H_q(Y,B)to H^q(Y,B,G_2)otimes G_1otimes H^q(Y,B,G_2)$ where I have used obvious $<-,->$ pairing once. Now I can flip $G_1$ by $Aotimes Bcong Botimes A$ natural. And keep contraction. This gives the equation.
However, I do not see difference between $H^q(X,A,G_1)times H^p(Y,B,G_2)times H_q(X,A)times H_p(Y,B)to G_1otimes G_2$ by directly contracting first slot and thirs slot and contracting second slot with the fourth slot. This reasoning does not give rise to $(-1)^{pq}$.
$textbf{Q':}$ What is the loop hole in the above reasoning? Obviously, I did not use any cohomology ring supercommutativity property.
Say $X,Y$ are smooth orientable manifolds. Essentially I should think $<-,->$ as integration of forms on manifolds. Say I am looking at compact cohomology theory. Now $int_{Utimes V}omega_1wedgeomega_2=int_Uomega_1timesint_Vomega_2$ where I assume $omega_i$ are parametrized by standard coordinates and using canonical orientation induced on $Xtimes Y$ with $U,V$ identified as homology cycle to be integrated. This does give rise to $(-1)^{pq}$ somehow.
$textbf{Q'':}$ What is the loop hole in the reasoning here?
general-topology geometry differential-geometry algebraic-topology
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
$endgroup$
|
show 5 more comments
$begingroup$
Let $(X,A),(Y,B)$ be two pairs of topological spaces. Denote $times$ as external cross product on cohomology level and $E:C_star(X,A)otimes C_star(Y,B)to C_star((X,A)times(Ytimes B))$ as Eilenberg-Zilber map and assume appropriate hypothesis(i.e. excision pairs on $A,B$ and $Atimes Y, Xtimes B$ excision pairs) s.t. $E$ induce homology level isomorphism. Set $<-,->:H^p(X;G_1)otimes H_p(X,G_2)to G_1otimes_Z G_2$ as bilinear pairing where $G_i$ are abelian groups. Whenever abelian group is omitted in homology, it is assumed to be $Z$ coefficients.
$textbf{Q:}$ I am quite confused about the following equation's sign $(-1)^{pq}$. "$ain H_p(X,A),bin H_q(Y,B),uin H^p(X,A;G_1),vin H^q(Y,B;G_2)$, then $(-1)^{pq}<utimes v,E(aotimes b)>=<u,a>otimes<v,b>$." Where is $(-1)^{pq}$ coming from?
The following is my reasoning. First $times$ is supercommutative. Hence $utimes v=(-1)^{pq}vtimes u$. Now I can use $H^q(Y,B,G_2)otimes H^p(X,A,G_1)otimes H_p(X,A)otimes H_q(Y,B)to H^q(Y,B,G_2)otimes G_1otimes H^q(Y,B,G_2)$ where I have used obvious $<-,->$ pairing once. Now I can flip $G_1$ by $Aotimes Bcong Botimes A$ natural. And keep contraction. This gives the equation.
However, I do not see difference between $H^q(X,A,G_1)times H^p(Y,B,G_2)times H_q(X,A)times H_p(Y,B)to G_1otimes G_2$ by directly contracting first slot and thirs slot and contracting second slot with the fourth slot. This reasoning does not give rise to $(-1)^{pq}$.
$textbf{Q':}$ What is the loop hole in the above reasoning? Obviously, I did not use any cohomology ring supercommutativity property.
Say $X,Y$ are smooth orientable manifolds. Essentially I should think $<-,->$ as integration of forms on manifolds. Say I am looking at compact cohomology theory. Now $int_{Utimes V}omega_1wedgeomega_2=int_Uomega_1timesint_Vomega_2$ where I assume $omega_i$ are parametrized by standard coordinates and using canonical orientation induced on $Xtimes Y$ with $U,V$ identified as homology cycle to be integrated. This does give rise to $(-1)^{pq}$ somehow.
$textbf{Q'':}$ What is the loop hole in the reasoning here?
general-topology geometry differential-geometry algebraic-topology
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
$endgroup$
$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
1
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45
|
show 5 more comments
$begingroup$
Let $(X,A),(Y,B)$ be two pairs of topological spaces. Denote $times$ as external cross product on cohomology level and $E:C_star(X,A)otimes C_star(Y,B)to C_star((X,A)times(Ytimes B))$ as Eilenberg-Zilber map and assume appropriate hypothesis(i.e. excision pairs on $A,B$ and $Atimes Y, Xtimes B$ excision pairs) s.t. $E$ induce homology level isomorphism. Set $<-,->:H^p(X;G_1)otimes H_p(X,G_2)to G_1otimes_Z G_2$ as bilinear pairing where $G_i$ are abelian groups. Whenever abelian group is omitted in homology, it is assumed to be $Z$ coefficients.
$textbf{Q:}$ I am quite confused about the following equation's sign $(-1)^{pq}$. "$ain H_p(X,A),bin H_q(Y,B),uin H^p(X,A;G_1),vin H^q(Y,B;G_2)$, then $(-1)^{pq}<utimes v,E(aotimes b)>=<u,a>otimes<v,b>$." Where is $(-1)^{pq}$ coming from?
The following is my reasoning. First $times$ is supercommutative. Hence $utimes v=(-1)^{pq}vtimes u$. Now I can use $H^q(Y,B,G_2)otimes H^p(X,A,G_1)otimes H_p(X,A)otimes H_q(Y,B)to H^q(Y,B,G_2)otimes G_1otimes H^q(Y,B,G_2)$ where I have used obvious $<-,->$ pairing once. Now I can flip $G_1$ by $Aotimes Bcong Botimes A$ natural. And keep contraction. This gives the equation.
However, I do not see difference between $H^q(X,A,G_1)times H^p(Y,B,G_2)times H_q(X,A)times H_p(Y,B)to G_1otimes G_2$ by directly contracting first slot and thirs slot and contracting second slot with the fourth slot. This reasoning does not give rise to $(-1)^{pq}$.
$textbf{Q':}$ What is the loop hole in the above reasoning? Obviously, I did not use any cohomology ring supercommutativity property.
Say $X,Y$ are smooth orientable manifolds. Essentially I should think $<-,->$ as integration of forms on manifolds. Say I am looking at compact cohomology theory. Now $int_{Utimes V}omega_1wedgeomega_2=int_Uomega_1timesint_Vomega_2$ where I assume $omega_i$ are parametrized by standard coordinates and using canonical orientation induced on $Xtimes Y$ with $U,V$ identified as homology cycle to be integrated. This does give rise to $(-1)^{pq}$ somehow.
$textbf{Q'':}$ What is the loop hole in the reasoning here?
general-topology geometry differential-geometry algebraic-topology
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
$endgroup$
Let $(X,A),(Y,B)$ be two pairs of topological spaces. Denote $times$ as external cross product on cohomology level and $E:C_star(X,A)otimes C_star(Y,B)to C_star((X,A)times(Ytimes B))$ as Eilenberg-Zilber map and assume appropriate hypothesis(i.e. excision pairs on $A,B$ and $Atimes Y, Xtimes B$ excision pairs) s.t. $E$ induce homology level isomorphism. Set $<-,->:H^p(X;G_1)otimes H_p(X,G_2)to G_1otimes_Z G_2$ as bilinear pairing where $G_i$ are abelian groups. Whenever abelian group is omitted in homology, it is assumed to be $Z$ coefficients.
$textbf{Q:}$ I am quite confused about the following equation's sign $(-1)^{pq}$. "$ain H_p(X,A),bin H_q(Y,B),uin H^p(X,A;G_1),vin H^q(Y,B;G_2)$, then $(-1)^{pq}<utimes v,E(aotimes b)>=<u,a>otimes<v,b>$." Where is $(-1)^{pq}$ coming from?
The following is my reasoning. First $times$ is supercommutative. Hence $utimes v=(-1)^{pq}vtimes u$. Now I can use $H^q(Y,B,G_2)otimes H^p(X,A,G_1)otimes H_p(X,A)otimes H_q(Y,B)to H^q(Y,B,G_2)otimes G_1otimes H^q(Y,B,G_2)$ where I have used obvious $<-,->$ pairing once. Now I can flip $G_1$ by $Aotimes Bcong Botimes A$ natural. And keep contraction. This gives the equation.
However, I do not see difference between $H^q(X,A,G_1)times H^p(Y,B,G_2)times H_q(X,A)times H_p(Y,B)to G_1otimes G_2$ by directly contracting first slot and thirs slot and contracting second slot with the fourth slot. This reasoning does not give rise to $(-1)^{pq}$.
$textbf{Q':}$ What is the loop hole in the above reasoning? Obviously, I did not use any cohomology ring supercommutativity property.
Say $X,Y$ are smooth orientable manifolds. Essentially I should think $<-,->$ as integration of forms on manifolds. Say I am looking at compact cohomology theory. Now $int_{Utimes V}omega_1wedgeomega_2=int_Uomega_1timesint_Vomega_2$ where I assume $omega_i$ are parametrized by standard coordinates and using canonical orientation induced on $Xtimes Y$ with $U,V$ identified as homology cycle to be integrated. This does give rise to $(-1)^{pq}$ somehow.
$textbf{Q'':}$ What is the loop hole in the reasoning here?
general-topology geometry differential-geometry algebraic-topology
general-topology geometry differential-geometry algebraic-topology
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
asked Dec 15 '18 at 20:54
user45765user45765
2,6792722
2,6792722
$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
1
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45
|
show 5 more comments
$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
1
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45
$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
1
1
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45
|
show 5 more comments
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$begingroup$
Minor side-note while I finish reading it: langle and rangle are how you make inner products look not-awful on here.
$endgroup$
– user3482749
Dec 15 '18 at 20:56
$begingroup$
For $Q)$, either $utimes v=(-1)^{pq}vtimes u$, or $votimes a=(-1)^{qp}aotimes v$, so everything is consistent.
$endgroup$
– Tyrone
Dec 16 '18 at 10:19
$begingroup$
@Tyrone Then why I should care about the order of contraction to produce. As far as I could see, there is difference between order of contractions and conventions adopted for evaluation by chain maps.
$endgroup$
– user45765
Dec 16 '18 at 15:23
$begingroup$
The contractions themselves can actually be performed at the same time, so there is no difficulty ordering them. However you need to pair homology with cohomology, so the $(-1)^{pq}$ sign comes from commuting the homological element $a$ and cohomological element $v$. Once this is done you simultaneously pair $(u,a)$ and $(v,b)$ to get an element in $G_1otimes G_2$. Since the coefficents are concentrated in degree $0$ there are no more signs to worry about.
$endgroup$
– Tyrone
Dec 16 '18 at 16:23
1
$begingroup$
By contraction I mean $H^p(X)otimes H_p(X)rightarrow G_1otimes G_2$. I'm sorry if my notation is a bit different from yours, but hopefully you see what I am trying to convey. The point is that you cannot "directly contract the first and third slots". You must first commute either the second and third, or the first and second. This gives you a sign change either way.
$endgroup$
– Tyrone
Dec 16 '18 at 17:45