Is $S^2times S^2$ homeomorphic to $mathbb{CP}^2#mathbb{CP}^2$?
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Is $S^2times S^2$ homeomorphic to $mathbb{CP}^2#mathbb{CP}^2$?
My idea: by the product formula for the Euler characteristic, we have $chi(S^2times S^2)=chi(S^2)^2=4$. By the sum formula for Euler characteristic we have $chi(mathbb{CP}^2#mathbb{CP}^2)=2chi(mathbb{CP}^2)-chi(S^4)=4$. So both manifolds have the same Euler characteristic.
By the sum formula for the signature, $sigma(mathbb{CP}^2#mathbb{CP}^2)=2$. Is the signature of $S^2times S^2$ also 2? I'm not sure how to compute it.
If so, then I think we can conclude that the two manifolds must be homeomorphic by Friedman's classification of simply connected manifolds.
Is this reasoning correct?
general-topology differential-geometry algebraic-topology differential-topology
asked yesterday
srp
1627
add a comment |
up vote
2
down vote
favorite
Is $S^2times S^2$ homeomorphic to $mathbb{CP}^2#mathbb{CP}^2$?
My idea: by the product formula for the Euler characteristic, we have $chi(S^2times S^2)=chi(S^2)^2=4$. By the sum formula for Euler characteristic we have $chi(mathbb{CP}^2#mathbb{CP}^2)=2chi(mathbb{CP}^2)-chi(S^4)=4$. So both manifolds have the same Euler characteristic.
By the sum formula for the signature, $sigma(mathbb{CP}^2#mathbb{CP}^2)=2$. Is the signature of $S^2times S^2$ also 2? I'm not sure how to compute it.
If so, then I think we can conclude that the two manifolds must be homeomorphic by Friedman's classification of simply connected manifolds.
Is this reasoning correct?
general-topology differential-geometry algebraic-topology differential-topology
asked yesterday
srp
1627
1
For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
1
The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
1
To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is $S^2times S^2$ homeomorphic to $mathbb{CP}^2#mathbb{CP}^2$?
My idea: by the product formula for the Euler characteristic, we have $chi(S^2times S^2)=chi(S^2)^2=4$. By the sum formula for Euler characteristic we have $chi(mathbb{CP}^2#mathbb{CP}^2)=2chi(mathbb{CP}^2)-chi(S^4)=4$. So both manifolds have the same Euler characteristic.
By the sum formula for the signature, $sigma(mathbb{CP}^2#mathbb{CP}^2)=2$. Is the signature of $S^2times S^2$ also 2? I'm not sure how to compute it.
If so, then I think we can conclude that the two manifolds must be homeomorphic by Friedman's classification of simply connected manifolds.
Is this reasoning correct?
general-topology differential-geometry algebraic-topology differential-topology
asked yesterday
srp
1627
Is $S^2times S^2$ homeomorphic to $mathbb{CP}^2#mathbb{CP}^2$?
My idea: by the product formula for the Euler characteristic, we have $chi(S^2times S^2)=chi(S^2)^2=4$. By the sum formula for Euler characteristic we have $chi(mathbb{CP}^2#mathbb{CP}^2)=2chi(mathbb{CP}^2)-chi(S^4)=4$. So both manifolds have the same Euler characteristic.
By the sum formula for the signature, $sigma(mathbb{CP}^2#mathbb{CP}^2)=2$. Is the signature of $S^2times S^2$ also 2? I'm not sure how to compute it.
If so, then I think we can conclude that the two manifolds must be homeomorphic by Friedman's classification of simply connected manifolds.
Is this reasoning correct?
general-topology differential-geometry algebraic-topology differential-topology
general-topology differential-geometry algebraic-topology differential-topology
asked yesterday
srp
1627
asked yesterday
srp
1627
asked yesterday
srp
1627
asked yesterday
srp
1627
asked yesterday
srp
1627
1627
1
For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
1
The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
1
To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday
add a comment |
1
For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
1
The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
1
To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday
1
1
For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
1
1
The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
1
1
To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday
To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
10
down vote
accepted
The intersection form on $mathbb{CP}^2# mathbb{CP^2}$ is $pmatrix{1 & 0 \ 0 & 1}$, while that of $S^2 times S^2$ is $begin{pmatrix}0 & 1 \ 1 & 0end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.
answered yesterday
anomaly
16.9k42662
add a comment |
up vote
5
down vote
It is not even true that $S^2times S^2$ is homotopy equivalent to $mathbb{C}P^2#mathbb{C}P^2$.
Pinching to one factor gives a map $mathbb{C}P^2#mathbb{C}P^2rightarrow mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)rightarrow H^4(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2times S^2)$ vanish identically.
In fact it is not difficult to show that
$$Sigma(mathbb{C}P^2#mathbb{C}P^2)simeq Sigmamathbb{C}P^2vee S^3$$
and
$$Sigma(S^2times S^2)simeq S^3vee S^3vee S^5$$
which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.
answered yesterday
Tyrone
4,04011125
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
The intersection form on $mathbb{CP}^2# mathbb{CP^2}$ is $pmatrix{1 & 0 \ 0 & 1}$, while that of $S^2 times S^2$ is $begin{pmatrix}0 & 1 \ 1 & 0end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.
answered yesterday
anomaly
16.9k42662
add a comment |
up vote
10
down vote
accepted
The intersection form on $mathbb{CP}^2# mathbb{CP^2}$ is $pmatrix{1 & 0 \ 0 & 1}$, while that of $S^2 times S^2$ is $begin{pmatrix}0 & 1 \ 1 & 0end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.
answered yesterday
anomaly
16.9k42662
add a comment |
up vote
10
down vote
accepted
up vote
10
down vote
accepted
The intersection form on $mathbb{CP}^2# mathbb{CP^2}$ is $pmatrix{1 & 0 \ 0 & 1}$, while that of $S^2 times S^2$ is $begin{pmatrix}0 & 1 \ 1 & 0end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.
answered yesterday
anomaly
16.9k42662
The intersection form on $mathbb{CP}^2# mathbb{CP^2}$ is $pmatrix{1 & 0 \ 0 & 1}$, while that of $S^2 times S^2$ is $begin{pmatrix}0 & 1 \ 1 & 0end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.
answered yesterday
anomaly
16.9k42662
answered yesterday
anomaly
16.9k42662
answered yesterday
anomaly
16.9k42662
answered yesterday
anomaly
16.9k42662
16.9k42662
add a comment |
add a comment |
up vote
5
down vote
It is not even true that $S^2times S^2$ is homotopy equivalent to $mathbb{C}P^2#mathbb{C}P^2$.
Pinching to one factor gives a map $mathbb{C}P^2#mathbb{C}P^2rightarrow mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)rightarrow H^4(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2times S^2)$ vanish identically.
In fact it is not difficult to show that
$$Sigma(mathbb{C}P^2#mathbb{C}P^2)simeq Sigmamathbb{C}P^2vee S^3$$
and
$$Sigma(S^2times S^2)simeq S^3vee S^3vee S^5$$
which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.
answered yesterday
Tyrone
4,04011125
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
|
show 2 more comments
up vote
5
down vote
It is not even true that $S^2times S^2$ is homotopy equivalent to $mathbb{C}P^2#mathbb{C}P^2$.
Pinching to one factor gives a map $mathbb{C}P^2#mathbb{C}P^2rightarrow mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)rightarrow H^4(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2times S^2)$ vanish identically.
In fact it is not difficult to show that
$$Sigma(mathbb{C}P^2#mathbb{C}P^2)simeq Sigmamathbb{C}P^2vee S^3$$
and
$$Sigma(S^2times S^2)simeq S^3vee S^3vee S^5$$
which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.
answered yesterday
Tyrone
4,04011125
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
|
show 2 more comments
up vote
5
down vote
up vote
5
down vote
It is not even true that $S^2times S^2$ is homotopy equivalent to $mathbb{C}P^2#mathbb{C}P^2$.
Pinching to one factor gives a map $mathbb{C}P^2#mathbb{C}P^2rightarrow mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)rightarrow H^4(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2times S^2)$ vanish identically.
In fact it is not difficult to show that
$$Sigma(mathbb{C}P^2#mathbb{C}P^2)simeq Sigmamathbb{C}P^2vee S^3$$
and
$$Sigma(S^2times S^2)simeq S^3vee S^3vee S^5$$
which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.
answered yesterday
Tyrone
4,04011125
It is not even true that $S^2times S^2$ is homotopy equivalent to $mathbb{C}P^2#mathbb{C}P^2$.
Pinching to one factor gives a map $mathbb{C}P^2#mathbb{C}P^2rightarrow mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)rightarrow H^4(mathbb{C}P^2#mathbb{C}P^2;mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2times S^2)$ vanish identically.
In fact it is not difficult to show that
$$Sigma(mathbb{C}P^2#mathbb{C}P^2)simeq Sigmamathbb{C}P^2vee S^3$$
and
$$Sigma(S^2times S^2)simeq S^3vee S^3vee S^5$$
which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.
answered yesterday
Tyrone
4,04011125
answered yesterday
Tyrone
4,04011125
answered yesterday
Tyrone
4,04011125
answered yesterday
Tyrone
4,04011125
4,04011125
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
|
show 2 more comments
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
1
1
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
The intersection form and thus signature already showed that the two are not homotopy equivalent, but it's interesting to upgrade this to stable homotopy.
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
(Rather the absolute value of signature.)
– Mike Miller
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
Can you elaborate on why $Sigma(CP^2 # CP^2) simeq Sigma CP^2 vee S^3$? The left hand side has two distinct elements whose $Sq^2$ is nonzero, while the right hand side only has one.
– Aleksandar Milivojevic
yesterday
2
2
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
@AleksandarMilivojevic The connected sum is formed by attaching a $4$-cell to the wedge $S^2vee S^2$ by a Hopf map into each summand of the wedge (ignoring orientations here). Suspend and consider the composite of the (suspended attaching map) with the self-equivalence $S^3vee S^3rightarrow S^3vee S^3$ given by $(a,b)mapsto (a,b-a)$, written in terms of $pi_3$. Note that you can only do this after suspension, since the Hopf map, with non-trivial Hopf invariant, is not a co-H-map.
– Tyrone
yesterday
1
1
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
You can choose generators $x,y$ for $H^2(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}^2$ such that $x^2=y^2=Z$ generates $H^4(mathbb{C}P^2#mathbb{C}P^2)congmathbb{Z}$, and $xcdot y=0$. Then the suspension splitting is induced by the choice of basis $sigma(x+y)$, $sigma x$ for $H^3(Sigma(mathbb{C}P^2#mathbb{C}P^2))$. Evaluate $Sq^2$ on (the mod 2 reductions of) these classes and you see that only one evaluates non-trivially.
– Tyrone
yesterday
|
show 2 more comments
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For computing the signature: It's just the signature of the underlying intersection form $H^2otimes H^2 to H^4$ (with coefficients in $mathbb{Z}$). For $S^2times S^2$, that form comes from the Kuenneth formula.
– anomaly
yesterday
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The classification is significantly more complicated than just Euler characteristic and signature.
– Mike Miller
yesterday
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To elaborate, Freedman's theorem classifies smoothable, simply-connected $4$-manifolds up to homeomorphism by their intersection form. Rank (which is functionally equivalent to $chi$, since the manifolds here are simply-connected) and signature are enough to classify forms over $mathbb{Q}$, but the situation over $mathbb{Z}$ is extremely complicated in the definite case (but see Donaldson's theorem).
– anomaly
yesterday