Isomorphism between torsion subgroups of $H_q (M)$ and $H_{n-q-1} (M)$ where $M$ is a compact oriented...
$begingroup$
It it the exercise in Massey's book
Massey, William S., Singular homology theory, Graduate Texts in Mathematics, 70. New York Heidelberg Berlin: Springer- Verlag. XII, 265 p. DM 49.50; $ 29.20 (1980). ZBL0442.55001.
chapter IX, section 5, exercise 5.1, he states that if $M$ is a compact oriented $n$-manifold, then the torsion subgroup of $H_q (M)$ is isomorphic to the torsion subgroup of $H_{n-q-1} (M)$.
I use the fact that a compact oriented $n$-manifold $M$ must have finitely generated homology groups, $mathrm{Hom} (G,mathbb{Q}/mathbb{Z} )$ is the torsion part of $G$ if $G$ is finitely generated and $mathbb{Q} /mathbb{Z} $ is injective to obtain the isomorphism $T(H_q (M))cong H^q (M;mathbb{Q} /mathbb{Z} )$, and use the Poincare duality to give the isomorphism $T(H_q (M))cong H_{n-q} (M;mathbb{Q} /mathbb{Z} )$. However, by the Universal Coefficient Theorem for Tor, I must state that $H_q (M)otimesmathbb{Q} /mathbb{Z} =0$. Will that be true?
algebraic-topology homology-cohomology homological-algebra
asked Jan 3 at 14:04
CTWCTW
988
$endgroup$
add a comment |
$begingroup$
It it the exercise in Massey's book
Massey, William S., Singular homology theory, Graduate Texts in Mathematics, 70. New York Heidelberg Berlin: Springer- Verlag. XII, 265 p. DM 49.50; $ 29.20 (1980). ZBL0442.55001.
chapter IX, section 5, exercise 5.1, he states that if $M$ is a compact oriented $n$-manifold, then the torsion subgroup of $H_q (M)$ is isomorphic to the torsion subgroup of $H_{n-q-1} (M)$.
I use the fact that a compact oriented $n$-manifold $M$ must have finitely generated homology groups, $mathrm{Hom} (G,mathbb{Q}/mathbb{Z} )$ is the torsion part of $G$ if $G$ is finitely generated and $mathbb{Q} /mathbb{Z} $ is injective to obtain the isomorphism $T(H_q (M))cong H^q (M;mathbb{Q} /mathbb{Z} )$, and use the Poincare duality to give the isomorphism $T(H_q (M))cong H_{n-q} (M;mathbb{Q} /mathbb{Z} )$. However, by the Universal Coefficient Theorem for Tor, I must state that $H_q (M)otimesmathbb{Q} /mathbb{Z} =0$. Will that be true?
algebraic-topology homology-cohomology homological-algebra
asked Jan 3 at 14:04
CTWCTW
988
$endgroup$
$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54
add a comment |
$begingroup$
It it the exercise in Massey's book
Massey, William S., Singular homology theory, Graduate Texts in Mathematics, 70. New York Heidelberg Berlin: Springer- Verlag. XII, 265 p. DM 49.50; $ 29.20 (1980). ZBL0442.55001.
chapter IX, section 5, exercise 5.1, he states that if $M$ is a compact oriented $n$-manifold, then the torsion subgroup of $H_q (M)$ is isomorphic to the torsion subgroup of $H_{n-q-1} (M)$.
I use the fact that a compact oriented $n$-manifold $M$ must have finitely generated homology groups, $mathrm{Hom} (G,mathbb{Q}/mathbb{Z} )$ is the torsion part of $G$ if $G$ is finitely generated and $mathbb{Q} /mathbb{Z} $ is injective to obtain the isomorphism $T(H_q (M))cong H^q (M;mathbb{Q} /mathbb{Z} )$, and use the Poincare duality to give the isomorphism $T(H_q (M))cong H_{n-q} (M;mathbb{Q} /mathbb{Z} )$. However, by the Universal Coefficient Theorem for Tor, I must state that $H_q (M)otimesmathbb{Q} /mathbb{Z} =0$. Will that be true?
algebraic-topology homology-cohomology homological-algebra
asked Jan 3 at 14:04
CTWCTW
988
$endgroup$
It it the exercise in Massey's book
Massey, William S., Singular homology theory, Graduate Texts in Mathematics, 70. New York Heidelberg Berlin: Springer- Verlag. XII, 265 p. DM 49.50; $ 29.20 (1980). ZBL0442.55001.
chapter IX, section 5, exercise 5.1, he states that if $M$ is a compact oriented $n$-manifold, then the torsion subgroup of $H_q (M)$ is isomorphic to the torsion subgroup of $H_{n-q-1} (M)$.
I use the fact that a compact oriented $n$-manifold $M$ must have finitely generated homology groups, $mathrm{Hom} (G,mathbb{Q}/mathbb{Z} )$ is the torsion part of $G$ if $G$ is finitely generated and $mathbb{Q} /mathbb{Z} $ is injective to obtain the isomorphism $T(H_q (M))cong H^q (M;mathbb{Q} /mathbb{Z} )$, and use the Poincare duality to give the isomorphism $T(H_q (M))cong H_{n-q} (M;mathbb{Q} /mathbb{Z} )$. However, by the Universal Coefficient Theorem for Tor, I must state that $H_q (M)otimesmathbb{Q} /mathbb{Z} =0$. Will that be true?
algebraic-topology homology-cohomology homological-algebra
algebraic-topology homology-cohomology homological-algebra
asked Jan 3 at 14:04
CTWCTW
988
asked Jan 3 at 14:04
CTWCTW
988
asked Jan 3 at 14:04
CTWCTW
988
asked Jan 3 at 14:04
CTWCTW
988
asked Jan 3 at 14:04
CTWCTW
988
988
$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54
add a comment |
$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54
$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54
add a comment |
1 Answer
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$begingroup$
It is not true that $text{Tors}(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. You are right that $Bbb Q/Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;Bbb Z) = Bbb Z^{b_q} oplus T$, then $$H^{q}(M;Bbb Q/Bbb Z) cong text{Hom}(H_q(M;Bbb Z), Bbb Q/Bbb Z) cong T oplus (Bbb Q/Bbb Z)^{b_q},$$ noncanonically.
So your approach shows that $T oplus (Bbb Q/Bbb Z)^{b_q} cong H^q(M;Bbb Q/Bbb Z) cong H_{n-q}(M;Bbb Q/Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,
$$0 to H_{n-q}(M;Bbb Z) otimes Bbb Q/Bbb Z to H_{n-q}(M;Bbb Q/Bbb Z) to text{Tor}(H_{n-q-1}(M;Bbb Z), Bbb Q/Bbb Z) to 0.$$
The first term is the divisible subgroup of $H_{n-q}(M;Bbb Q/Bbb Z)$, and is isomorphic to $(Bbb Q/Bbb Z)^{b_{n-q}} = (Bbb Q/Bbb Z)^{b_q}$. The last term is $text{Tors}(H_{n-q-1}(M;Bbb Z))$.
Because we have an isomorphism of groups $H_{n-q}(M;Bbb Q/Bbb Z) cong H^q(M;Bbb Q/Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.
Note that $H_q(M;Bbb Z) otimes Bbb Q/Bbb Z$ is indeed zero if and only if the claim that $H^q(M;Bbb Q/Bbb Z) cong text{Tors}(H_q(M;Bbb Z))$ is true.
$endgroup$
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
add a comment |
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$begingroup$
It is not true that $text{Tors}(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. You are right that $Bbb Q/Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;Bbb Z) = Bbb Z^{b_q} oplus T$, then $$H^{q}(M;Bbb Q/Bbb Z) cong text{Hom}(H_q(M;Bbb Z), Bbb Q/Bbb Z) cong T oplus (Bbb Q/Bbb Z)^{b_q},$$ noncanonically.
So your approach shows that $T oplus (Bbb Q/Bbb Z)^{b_q} cong H^q(M;Bbb Q/Bbb Z) cong H_{n-q}(M;Bbb Q/Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,
$$0 to H_{n-q}(M;Bbb Z) otimes Bbb Q/Bbb Z to H_{n-q}(M;Bbb Q/Bbb Z) to text{Tor}(H_{n-q-1}(M;Bbb Z), Bbb Q/Bbb Z) to 0.$$
The first term is the divisible subgroup of $H_{n-q}(M;Bbb Q/Bbb Z)$, and is isomorphic to $(Bbb Q/Bbb Z)^{b_{n-q}} = (Bbb Q/Bbb Z)^{b_q}$. The last term is $text{Tors}(H_{n-q-1}(M;Bbb Z))$.
Because we have an isomorphism of groups $H_{n-q}(M;Bbb Q/Bbb Z) cong H^q(M;Bbb Q/Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.
Note that $H_q(M;Bbb Z) otimes Bbb Q/Bbb Z$ is indeed zero if and only if the claim that $H^q(M;Bbb Q/Bbb Z) cong text{Tors}(H_q(M;Bbb Z))$ is true.
$endgroup$
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
add a comment |
$begingroup$
It is not true that $text{Tors}(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. You are right that $Bbb Q/Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;Bbb Z) = Bbb Z^{b_q} oplus T$, then $$H^{q}(M;Bbb Q/Bbb Z) cong text{Hom}(H_q(M;Bbb Z), Bbb Q/Bbb Z) cong T oplus (Bbb Q/Bbb Z)^{b_q},$$ noncanonically.
So your approach shows that $T oplus (Bbb Q/Bbb Z)^{b_q} cong H^q(M;Bbb Q/Bbb Z) cong H_{n-q}(M;Bbb Q/Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,
$$0 to H_{n-q}(M;Bbb Z) otimes Bbb Q/Bbb Z to H_{n-q}(M;Bbb Q/Bbb Z) to text{Tor}(H_{n-q-1}(M;Bbb Z), Bbb Q/Bbb Z) to 0.$$
The first term is the divisible subgroup of $H_{n-q}(M;Bbb Q/Bbb Z)$, and is isomorphic to $(Bbb Q/Bbb Z)^{b_{n-q}} = (Bbb Q/Bbb Z)^{b_q}$. The last term is $text{Tors}(H_{n-q-1}(M;Bbb Z))$.
Because we have an isomorphism of groups $H_{n-q}(M;Bbb Q/Bbb Z) cong H^q(M;Bbb Q/Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.
Note that $H_q(M;Bbb Z) otimes Bbb Q/Bbb Z$ is indeed zero if and only if the claim that $H^q(M;Bbb Q/Bbb Z) cong text{Tors}(H_q(M;Bbb Z))$ is true.
$endgroup$
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
add a comment |
$begingroup$
It is not true that $text{Tors}(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. You are right that $Bbb Q/Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;Bbb Z) = Bbb Z^{b_q} oplus T$, then $$H^{q}(M;Bbb Q/Bbb Z) cong text{Hom}(H_q(M;Bbb Z), Bbb Q/Bbb Z) cong T oplus (Bbb Q/Bbb Z)^{b_q},$$ noncanonically.
So your approach shows that $T oplus (Bbb Q/Bbb Z)^{b_q} cong H^q(M;Bbb Q/Bbb Z) cong H_{n-q}(M;Bbb Q/Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,
$$0 to H_{n-q}(M;Bbb Z) otimes Bbb Q/Bbb Z to H_{n-q}(M;Bbb Q/Bbb Z) to text{Tor}(H_{n-q-1}(M;Bbb Z), Bbb Q/Bbb Z) to 0.$$
The first term is the divisible subgroup of $H_{n-q}(M;Bbb Q/Bbb Z)$, and is isomorphic to $(Bbb Q/Bbb Z)^{b_{n-q}} = (Bbb Q/Bbb Z)^{b_q}$. The last term is $text{Tors}(H_{n-q-1}(M;Bbb Z))$.
Because we have an isomorphism of groups $H_{n-q}(M;Bbb Q/Bbb Z) cong H^q(M;Bbb Q/Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.
Note that $H_q(M;Bbb Z) otimes Bbb Q/Bbb Z$ is indeed zero if and only if the claim that $H^q(M;Bbb Q/Bbb Z) cong text{Tors}(H_q(M;Bbb Z))$ is true.
$endgroup$
It is not true that $text{Tors}(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. You are right that $Bbb Q/Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;Bbb Z) = Bbb Z^{b_q} oplus T$, then $$H^{q}(M;Bbb Q/Bbb Z) cong text{Hom}(H_q(M;Bbb Z), Bbb Q/Bbb Z) cong T oplus (Bbb Q/Bbb Z)^{b_q},$$ noncanonically.
So your approach shows that $T oplus (Bbb Q/Bbb Z)^{b_q} cong H^q(M;Bbb Q/Bbb Z) cong H_{n-q}(M;Bbb Q/Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,
$$0 to H_{n-q}(M;Bbb Z) otimes Bbb Q/Bbb Z to H_{n-q}(M;Bbb Q/Bbb Z) to text{Tor}(H_{n-q-1}(M;Bbb Z), Bbb Q/Bbb Z) to 0.$$
The first term is the divisible subgroup of $H_{n-q}(M;Bbb Q/Bbb Z)$, and is isomorphic to $(Bbb Q/Bbb Z)^{b_{n-q}} = (Bbb Q/Bbb Z)^{b_q}$. The last term is $text{Tors}(H_{n-q-1}(M;Bbb Z))$.
Because we have an isomorphism of groups $H_{n-q}(M;Bbb Q/Bbb Z) cong H^q(M;Bbb Q/Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.
Note that $H_q(M;Bbb Z) otimes Bbb Q/Bbb Z$ is indeed zero if and only if the claim that $H^q(M;Bbb Q/Bbb Z) cong text{Tors}(H_q(M;Bbb Z))$ is true.
answered Jan 3 at 15:07
user98602
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
add a comment |
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
Thanks! I have realized of my mistake.
$endgroup$
– CTW
Jan 3 at 15:23
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
$begingroup$
@CTW Sure, and thanks for helping me realize my understanding of which coefficients one may take for Poincare duality was incorrect.
$endgroup$
– user98602
Jan 3 at 15:52
add a comment |
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$begingroup$
You cannot apply Poincare duality like this because $Bbb Q/Bbb Z$ is not a ring. You need to dualize first, then apply UCT.
$endgroup$
– user98602
Jan 3 at 14:21
$begingroup$
But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it.
$endgroup$
– CTW
Jan 3 at 14:39
$begingroup$
Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) cong H^q(M;Bbb Q/Bbb Z)$. I will write an answer.
$endgroup$
– user98602
Jan 3 at 14:54