Example of two spaces indisinguishable by their homology modules (with $mathbbZ$ coefficients) but with different cohomology rings

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I'm running a student seminar on cohomology (for masters students) and would like to motivate the dualisation of homology by talking about cup products. So I'm looking for an example of two spaces $X$ and $Y$ with the same homology modules but different cohomology rings. Are there any nice-ish examples which would be reasonable to talk about in a seminar?

I do already have the example of $X=mathbbRP^n$ and $Y=vee_ileq nS^i$ with $mathbbZ/2$ coefficients. The problem with this is that these spaces are distinguished by their homology with $mathbbZ$ coefficients. This might still be sufficient motivation for this kind of seminar, but I'd still prefer to have an example of two spaces where you really need the extra ring structure to tell them apart.

Thanks for any help

algebraic-topology homology-cohomology

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asked Jul 18 at 5:08

Tsein32

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up vote
4
down vote

favorite

2

I'm running a student seminar on cohomology (for masters students) and would like to motivate the dualisation of homology by talking about cup products. So I'm looking for an example of two spaces $X$ and $Y$ with the same homology modules but different cohomology rings. Are there any nice-ish examples which would be reasonable to talk about in a seminar?

I do already have the example of $X=mathbbRP^n$ and $Y=vee_ileq nS^i$ with $mathbbZ/2$ coefficients. The problem with this is that these spaces are distinguished by their homology with $mathbbZ$ coefficients. This might still be sufficient motivation for this kind of seminar, but I'd still prefer to have an example of two spaces where you really need the extra ring structure to tell them apart.

Thanks for any help

algebraic-topology homology-cohomology

share|cite|improve this question

asked Jul 18 at 5:08

Tsein32

1628

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

I'm running a student seminar on cohomology (for masters students) and would like to motivate the dualisation of homology by talking about cup products. So I'm looking for an example of two spaces $X$ and $Y$ with the same homology modules but different cohomology rings. Are there any nice-ish examples which would be reasonable to talk about in a seminar?

I do already have the example of $X=mathbbRP^n$ and $Y=vee_ileq nS^i$ with $mathbbZ/2$ coefficients. The problem with this is that these spaces are distinguished by their homology with $mathbbZ$ coefficients. This might still be sufficient motivation for this kind of seminar, but I'd still prefer to have an example of two spaces where you really need the extra ring structure to tell them apart.

Thanks for any help

algebraic-topology homology-cohomology

share|cite|improve this question

asked Jul 18 at 5:08

Tsein32

1628

I'm running a student seminar on cohomology (for masters students) and would like to motivate the dualisation of homology by talking about cup products. So I'm looking for an example of two spaces $X$ and $Y$ with the same homology modules but different cohomology rings. Are there any nice-ish examples which would be reasonable to talk about in a seminar?

I do already have the example of $X=mathbbRP^n$ and $Y=vee_ileq nS^i$ with $mathbbZ/2$ coefficients. The problem with this is that these spaces are distinguished by their homology with $mathbbZ$ coefficients. This might still be sufficient motivation for this kind of seminar, but I'd still prefer to have an example of two spaces where you really need the extra ring structure to tell them apart.

Thanks for any help

algebraic-topology homology-cohomology

share|cite|improve this question

asked Jul 18 at 5:08

Tsein32

1628

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 5:08

Tsein32

1628

asked Jul 18 at 5:08

Tsein32

1628

asked Jul 18 at 5:08

Tsein32

1628

1628

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
7
down vote



accepted

An easy example is the torus $S^1 times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 vee S^1 vee S^2$ (which has no nontrivial cup products).

A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.

For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $mathbbCP^2 # mathbbCP^2$ and $mathbbCP^1 times mathbbCP^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.

share|cite|improve this answer

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
  – Tsein32
  Jul 18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
  – Mike Miller
  Jul 21 at 4:48
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

A similar example to yours is $mathbb CP^n$ and $bigveeS^i:0<ileq 2n$ with $i$ even$$, as the cohomology ring of $mathbb CP^n$, with coefficients in $Bbb Z$, is

$$mathbb Z[alpha]/alpha^n+1,text deg(alpha)=2.$$

share|cite|improve this answer

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

add a comment | 

up vote
3
down vote

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.

Consider $Sigma mathbbCP^n$ and $bigvee_i=1,dots n S^2i+1$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension $Sigma mathbbCP^n$ are trivial.

How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $Sigma mathbbCP^n$ will admit non-trivial operations for suitable $p$ whilst $bigvee_i=1,dots n S^2i+1$ will not. In particular, if $xin H^2(mathbbCP^2mathbbZ_2)$ is a generator then so is $Sq^2x=x^2in H^4(mathbbCP^2;mathbbZ_2)$. Thus if $sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(sigma x)=sigma(Sq^2x)=sigma (x^2)in H^5(SigmamathbbCP^2;mathbbZ_2)$ is non-trivial.

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answered Jul 18 at 8:59

Tyrone

3,24611125

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

An easy example is the torus $S^1 times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 vee S^1 vee S^2$ (which has no nontrivial cup products).

A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.

For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $mathbbCP^2 # mathbbCP^2$ and $mathbbCP^1 times mathbbCP^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.

share|cite|improve this answer

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
  – Tsein32
  Jul 18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
  – Mike Miller
  Jul 21 at 4:48
  
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

An easy example is the torus $S^1 times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 vee S^1 vee S^2$ (which has no nontrivial cup products).

A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.

For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $mathbbCP^2 # mathbbCP^2$ and $mathbbCP^1 times mathbbCP^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.

share|cite|improve this answer

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
  – Tsein32
  Jul 18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
  – Mike Miller
  Jul 21 at 4:48
  
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

An easy example is the torus $S^1 times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 vee S^1 vee S^2$ (which has no nontrivial cup products).

A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.

For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $mathbbCP^2 # mathbbCP^2$ and $mathbbCP^1 times mathbbCP^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.

share|cite|improve this answer

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

An easy example is the torus $S^1 times S^1$, which has the same homology as but different cohomology ring than the wedge $S^1 vee S^1 vee S^2$ (which has no nontrivial cup products).

A more interesting question is whether there are examples which are both closed manifolds. There might be 3-manifold examples but I don't know how to construct them off the top of my head.

For 4-manifolds we can construct examples by finding simply connected closed 4-manifolds with the same middle Betti number $b_2$ but such that the absolute value of the signature is different, which implies that the cohomology rings are not isomorphic. I think we can take $mathbbCP^2 # mathbbCP^2$ and $mathbbCP^1 times mathbbCP^1$; these both satisfy $b_2 = 2$ but the first one has signature $2$ and the second has signature $0$ (although you don't need to know what signature is to compute that the cohomology rings aren't isomorphic). See this blog post for more background.

share|cite|improve this answer

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 5:47

edited Jul 18 at 5:47

edited Jul 18 at 5:47

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

answered Jul 18 at 5:26

Qiaochu Yuan

269k32564900

269k32564900

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
  – Tsein32
  Jul 18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
  – Mike Miller
  Jul 21 at 4:48
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
  – Tsein32
  Jul 18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
  – Mike Miller
  Jul 21 at 4:48
  
  

  
  
  
  

Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
– Tsein32
Jul 18 at 7:17

Thanks, the example of $mathbbCP^2# mathbbCP^2$ and $mathbbCP^1×mathbbCP^1$ in particular is perfect, as these also can't be distinguished by their fundamental group. This was exactly what I was hoping to find.
– Tsein32
Jul 18 at 7:17

1

1

The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
– Mike Miller
Jul 21 at 4:48

The triple cup product is a good place to distinguish 3-manifolds with their ring structure; the simplest example that comes to mind is $T^3$ vs $#^3 (S^2 times S^1)$.
– Mike Miller
Jul 21 at 4:48

add a comment | 

up vote
4
down vote

A similar example to yours is $mathbb CP^n$ and $bigveeS^i:0<ileq 2n$ with $i$ even$$, as the cohomology ring of $mathbb CP^n$, with coefficients in $Bbb Z$, is

$$mathbb Z[alpha]/alpha^n+1,text deg(alpha)=2.$$

share|cite|improve this answer

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

add a comment | 

up vote
4
down vote

A similar example to yours is $mathbb CP^n$ and $bigveeS^i:0<ileq 2n$ with $i$ even$$, as the cohomology ring of $mathbb CP^n$, with coefficients in $Bbb Z$, is

$$mathbb Z[alpha]/alpha^n+1,text deg(alpha)=2.$$

share|cite|improve this answer

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

add a comment | 

up vote
4
down vote

up vote
4
down vote

A similar example to yours is $mathbb CP^n$ and $bigveeS^i:0<ileq 2n$ with $i$ even$$, as the cohomology ring of $mathbb CP^n$, with coefficients in $Bbb Z$, is

$$mathbb Z[alpha]/alpha^n+1,text deg(alpha)=2.$$

share|cite|improve this answer

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

A similar example to yours is $mathbb CP^n$ and $bigveeS^i:0<ileq 2n$ with $i$ even$$, as the cohomology ring of $mathbb CP^n$, with coefficients in $Bbb Z$, is

$$mathbb Z[alpha]/alpha^n+1,text deg(alpha)=2.$$

share|cite|improve this answer

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 3:04

edited Jul 21 at 3:04

edited Jul 21 at 3:04

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

answered Jul 18 at 6:03

Camilo Arosemena-Serrato

5,50911847

5,50911847

add a comment | 

add a comment | 

up vote
3
down vote

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.

Consider $Sigma mathbbCP^n$ and $bigvee_i=1,dots n S^2i+1$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension $Sigma mathbbCP^n$ are trivial.

How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $Sigma mathbbCP^n$ will admit non-trivial operations for suitable $p$ whilst $bigvee_i=1,dots n S^2i+1$ will not. In particular, if $xin H^2(mathbbCP^2mathbbZ_2)$ is a generator then so is $Sq^2x=x^2in H^4(mathbbCP^2;mathbbZ_2)$. Thus if $sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(sigma x)=sigma(Sq^2x)=sigma (x^2)in H^5(SigmamathbbCP^2;mathbbZ_2)$ is non-trivial.

share|cite|improve this answer

answered Jul 18 at 8:59

Tyrone

3,24611125

add a comment | 

up vote
3
down vote

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.

Consider $Sigma mathbbCP^n$ and $bigvee_i=1,dots n S^2i+1$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension $Sigma mathbbCP^n$ are trivial.

How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $Sigma mathbbCP^n$ will admit non-trivial operations for suitable $p$ whilst $bigvee_i=1,dots n S^2i+1$ will not. In particular, if $xin H^2(mathbbCP^2mathbbZ_2)$ is a generator then so is $Sq^2x=x^2in H^4(mathbbCP^2;mathbbZ_2)$. Thus if $sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(sigma x)=sigma(Sq^2x)=sigma (x^2)in H^5(SigmamathbbCP^2;mathbbZ_2)$ is non-trivial.

share|cite|improve this answer

answered Jul 18 at 8:59

Tyrone

3,24611125

add a comment | 

up vote
3
down vote

up vote
3
down vote

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.

Consider $Sigma mathbbCP^n$ and $bigvee_i=1,dots n S^2i+1$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension $Sigma mathbbCP^n$ are trivial.

How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $Sigma mathbbCP^n$ will admit non-trivial operations for suitable $p$ whilst $bigvee_i=1,dots n S^2i+1$ will not. In particular, if $xin H^2(mathbbCP^2mathbbZ_2)$ is a generator then so is $Sq^2x=x^2in H^4(mathbbCP^2;mathbbZ_2)$. Thus if $sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(sigma x)=sigma(Sq^2x)=sigma (x^2)in H^5(SigmamathbbCP^2;mathbbZ_2)$ is non-trivial.

share|cite|improve this answer

answered Jul 18 at 8:59

Tyrone

3,24611125

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.

Consider $Sigma mathbbCP^n$ and $bigvee_i=1,dots n S^2i+1$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension $Sigma mathbbCP^n$ are trivial.

How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $Sigma mathbbCP^n$ will admit non-trivial operations for suitable $p$ whilst $bigvee_i=1,dots n S^2i+1$ will not. In particular, if $xin H^2(mathbbCP^2mathbbZ_2)$ is a generator then so is $Sq^2x=x^2in H^4(mathbbCP^2;mathbbZ_2)$. Thus if $sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(sigma x)=sigma(Sq^2x)=sigma (x^2)in H^5(SigmamathbbCP^2;mathbbZ_2)$ is non-trivial.

share|cite|improve this answer

answered Jul 18 at 8:59

Tyrone

3,24611125

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 8:59

Tyrone

3,24611125

answered Jul 18 at 8:59

Tyrone

3,24611125

answered Jul 18 at 8:59

Tyrone

3,24611125

3,24611125

add a comment | 

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