Mixing
Poincaré duality and quantum (co)homology for $S^2 times S^2$
Consider the symplectic manifold $(M,omega):=(S^2 times S^2, omega_{FS}oplus omega_{FS})$. The homology $H_*(M;mathbb{C})$ has as a basis the 4 non-trivial classes: $[pt],[M],A$ and $B$, where
$$
A=[pt times S^2] quad text{and}quad B=[S^2 times pt].
$$
Denoting by $mathcal{F}$ the field of Laurent polynomials $mathcal{F}:=mathbb{C}[[s]$ in a formal variable $s$, the quantum homology $QH(M):=H(M;mathbb{C})otimes_{mathbb{C}}mathcal{F}$ is equipped with the quantum product $ast$. An element of $mathcal{F}$ is written $sum_{jin mathbb{Z}}z_js^j$ and $z_j=0$ for all $j$ large enough. I know that
$$
Aast B=[pt] quad text{and} quad Aast A=Bast B=[M]s^2.
$$
I would like to compute Poincaré duality in this picture, but for some reason I keep getting it wrong. I would be very grateful if someone could explain to me how it works. What I would like to know is that are the dual classes to each of the classes $[pt],[M],A$ and $B$?
symplectic-geometry poincare-duality
asked Nov 19 '18 at 23:44
MBIS
1555
add a comment |
Consider the symplectic manifold $(M,omega):=(S^2 times S^2, omega_{FS}oplus omega_{FS})$. The homology $H_*(M;mathbb{C})$ has as a basis the 4 non-trivial classes: $[pt],[M],A$ and $B$, where
$$
A=[pt times S^2] quad text{and}quad B=[S^2 times pt].
$$
Denoting by $mathcal{F}$ the field of Laurent polynomials $mathcal{F}:=mathbb{C}[[s]$ in a formal variable $s$, the quantum homology $QH(M):=H(M;mathbb{C})otimes_{mathbb{C}}mathcal{F}$ is equipped with the quantum product $ast$. An element of $mathcal{F}$ is written $sum_{jin mathbb{Z}}z_js^j$ and $z_j=0$ for all $j$ large enough. I know that
$$
Aast B=[pt] quad text{and} quad Aast A=Bast B=[M]s^2.
$$
I would like to compute Poincaré duality in this picture, but for some reason I keep getting it wrong. I would be very grateful if someone could explain to me how it works. What I would like to know is that are the dual classes to each of the classes $[pt],[M],A$ and $B$?
symplectic-geometry poincare-duality
asked Nov 19 '18 at 23:44
MBIS
1555
add a comment |
Consider the symplectic manifold $(M,omega):=(S^2 times S^2, omega_{FS}oplus omega_{FS})$. The homology $H_*(M;mathbb{C})$ has as a basis the 4 non-trivial classes: $[pt],[M],A$ and $B$, where
$$
A=[pt times S^2] quad text{and}quad B=[S^2 times pt].
$$
Denoting by $mathcal{F}$ the field of Laurent polynomials $mathcal{F}:=mathbb{C}[[s]$ in a formal variable $s$, the quantum homology $QH(M):=H(M;mathbb{C})otimes_{mathbb{C}}mathcal{F}$ is equipped with the quantum product $ast$. An element of $mathcal{F}$ is written $sum_{jin mathbb{Z}}z_js^j$ and $z_j=0$ for all $j$ large enough. I know that
$$
Aast B=[pt] quad text{and} quad Aast A=Bast B=[M]s^2.
$$
I would like to compute Poincaré duality in this picture, but for some reason I keep getting it wrong. I would be very grateful if someone could explain to me how it works. What I would like to know is that are the dual classes to each of the classes $[pt],[M],A$ and $B$?
symplectic-geometry poincare-duality
asked Nov 19 '18 at 23:44
MBIS
1555
Consider the symplectic manifold $(M,omega):=(S^2 times S^2, omega_{FS}oplus omega_{FS})$. The homology $H_*(M;mathbb{C})$ has as a basis the 4 non-trivial classes: $[pt],[M],A$ and $B$, where
$$
A=[pt times S^2] quad text{and}quad B=[S^2 times pt].
$$
Denoting by $mathcal{F}$ the field of Laurent polynomials $mathcal{F}:=mathbb{C}[[s]$ in a formal variable $s$, the quantum homology $QH(M):=H(M;mathbb{C})otimes_{mathbb{C}}mathcal{F}$ is equipped with the quantum product $ast$. An element of $mathcal{F}$ is written $sum_{jin mathbb{Z}}z_js^j$ and $z_j=0$ for all $j$ large enough. I know that
$$
Aast B=[pt] quad text{and} quad Aast A=Bast B=[M]s^2.
$$
I would like to compute Poincaré duality in this picture, but for some reason I keep getting it wrong. I would be very grateful if someone could explain to me how it works. What I would like to know is that are the dual classes to each of the classes $[pt],[M],A$ and $B$?
symplectic-geometry poincare-duality
symplectic-geometry poincare-duality
asked Nov 19 '18 at 23:44
MBIS
1555
asked Nov 19 '18 at 23:44
MBIS
1555
edited Nov 26 '18 at 9:23
asked Nov 19 '18 at 23:44
MBIS
1555
asked Nov 19 '18 at 23:44
MBIS
1555
asked Nov 19 '18 at 23:44
MBIS
1555
1555
add a comment |
add a comment |
1 Answer
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To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of
$$[pt] in QH_0(M), quad [M] in QH_4(M), quad A in QH_2(M), quad B in QH_2(M)$$
are just
$$PD[pt] in QH^4(M), quad PD[M] = 1 in QH^0(M), quad PD(A) in QH^2(M), quad PD(B) in QH^2(M)$$
where $PD$ denotes the Poincare dual for the ordinary homology class.
By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $mathcal{F}$. According to your expression, you seem to take $mathcal{F} = mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C otimes s^k in QH_{c -2k}(M)$ is $PD(C) otimes s^{-k} in QH^{(4-c) + 2k}(M)$.
I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.
answered Nov 20 '18 at 2:28
Hwang
320110
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
add a comment |
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1 Answer
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To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of
$$[pt] in QH_0(M), quad [M] in QH_4(M), quad A in QH_2(M), quad B in QH_2(M)$$
are just
$$PD[pt] in QH^4(M), quad PD[M] = 1 in QH^0(M), quad PD(A) in QH^2(M), quad PD(B) in QH^2(M)$$
where $PD$ denotes the Poincare dual for the ordinary homology class.
By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $mathcal{F}$. According to your expression, you seem to take $mathcal{F} = mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C otimes s^k in QH_{c -2k}(M)$ is $PD(C) otimes s^{-k} in QH^{(4-c) + 2k}(M)$.
I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.
answered Nov 20 '18 at 2:28
Hwang
320110
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
add a comment |
To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of
$$[pt] in QH_0(M), quad [M] in QH_4(M), quad A in QH_2(M), quad B in QH_2(M)$$
are just
$$PD[pt] in QH^4(M), quad PD[M] = 1 in QH^0(M), quad PD(A) in QH^2(M), quad PD(B) in QH^2(M)$$
where $PD$ denotes the Poincare dual for the ordinary homology class.
By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $mathcal{F}$. According to your expression, you seem to take $mathcal{F} = mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C otimes s^k in QH_{c -2k}(M)$ is $PD(C) otimes s^{-k} in QH^{(4-c) + 2k}(M)$.
I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.
answered Nov 20 '18 at 2:28
Hwang
320110
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
add a comment |
To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of
$$[pt] in QH_0(M), quad [M] in QH_4(M), quad A in QH_2(M), quad B in QH_2(M)$$
are just
$$PD[pt] in QH^4(M), quad PD[M] = 1 in QH^0(M), quad PD(A) in QH^2(M), quad PD(B) in QH^2(M)$$
where $PD$ denotes the Poincare dual for the ordinary homology class.
By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $mathcal{F}$. According to your expression, you seem to take $mathcal{F} = mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C otimes s^k in QH_{c -2k}(M)$ is $PD(C) otimes s^{-k} in QH^{(4-c) + 2k}(M)$.
I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.
answered Nov 20 '18 at 2:28
Hwang
320110
To my understanding, the quantum homology and the quantum cohomology are basically the same thing. So the Poincare duals of
$$[pt] in QH_0(M), quad [M] in QH_4(M), quad A in QH_2(M), quad B in QH_2(M)$$
are just
$$PD[pt] in QH^4(M), quad PD[M] = 1 in QH^0(M), quad PD(A) in QH^2(M), quad PD(B) in QH^2(M)$$
where $PD$ denotes the Poincare dual for the ordinary homology class.
By the way, the structure of the quantum homology highly depends on choice of the coefficient ring $mathcal{F}$. According to your expression, you seem to take $mathcal{F} = mathbb{C}[s]$ with $s = e^{-A/2} = e^{-B/2}$. Since $M$ is monotone, we don't need to worry about infinite sums. The variable $s$ has degree $-2$, so if $C$ is a homology class of degree $c$, the Poincare dual of $C otimes s^k in QH_{c -2k}(M)$ is $PD(C) otimes s^{-k} in QH^{(4-c) + 2k}(M)$.
I recommend the book $J$-holomorphic Curves and Symplectic Topology by McDuff and Salamon. In the 2nd edition, Example 11.1.13 explains exactly the same example and the Poincare duality is explained in Remark 11.1.20.
answered Nov 20 '18 at 2:28
Hwang
320110
edited Nov 20 '18 at 2:44
answered Nov 20 '18 at 2:28
Hwang
320110
answered Nov 20 '18 at 2:28
Hwang
320110
answered Nov 20 '18 at 2:28
Hwang
320110
320110
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
add a comment |
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
Great, thanks a lot. This was exactly the reference I needed!
– MBIS
Nov 25 '18 at 14:21
add a comment |
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