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Cohomology ring of $mathbb CP^2$ with a 3-cell attached

3

1

$begingroup$

Let $X$ be $mathbb CP^2$ with a 3-cell $e^3$ attached via $S^2to mathbb CP^1subset mathbb CP^2$ of degree $p$.
By cellular cohomology, it is easy to see that $H^k(X,mathbb Z_p)$ is $mathbb Z_p$ when $k=1,2,3,4$ and $0$ otherwise.

I wish to show that the cup product

$$H^2(X,mathbb Z_p)times H^2(X,mathbb Z_p) to H^4(X,mathbb Z_p)$$
is nontrivial. I am also confused about this $e^3$ attached to $mathbb CP^2$. I think it probably doesn't affect this cup product but I don't know how to show it rigorously.

Also, I only know $H^*(mathbb CP^2, mathbb Z)cong (a)/(a^3)$ and I am not sure how it is effected when taking the coefficients to be $mathbb Z_p$.

Thank you.

algebraic-topology homology-cohomology

share|cite|improve this question

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think your cohomology is not correct. If you attach a $3$-cell to $mathbb{C}P^2$ as you describe then you're essentially "filling in" the $2$-skeleton and the resulting space will be a $3$-cell with a $4$-cell attached to its boundary via the Hopf map $S^3 to S^2$. There shouldn't be any cohomology in degree $2$.
  $endgroup$
  – William
  Mar 23 at 22:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you mean for the attaching map to have degree $p$?
  $endgroup$
  – Eric Wofsey
  Mar 23 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EricWofsey Yes... I am sorry. Thanks!
  $endgroup$
  – No One
  Mar 24 at 3:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @William I am sorry... Just updated...
  $endgroup$
  – No One
  Mar 24 at 3:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated my answer for this situation
  $endgroup$
  – William
  Mar 24 at 4:41
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Let $X$ be $mathbb CP^2$ with a 3-cell $e^3$ attached via $S^2to mathbb CP^1subset mathbb CP^2$ of degree $p$.
By cellular cohomology, it is easy to see that $H^k(X,mathbb Z_p)$ is $mathbb Z_p$ when $k=1,2,3,4$ and $0$ otherwise.

I wish to show that the cup product

$$H^2(X,mathbb Z_p)times H^2(X,mathbb Z_p) to H^4(X,mathbb Z_p)$$
is nontrivial. I am also confused about this $e^3$ attached to $mathbb CP^2$. I think it probably doesn't affect this cup product but I don't know how to show it rigorously.

Also, I only know $H^*(mathbb CP^2, mathbb Z)cong (a)/(a^3)$ and I am not sure how it is effected when taking the coefficients to be $mathbb Z_p$.

Thank you.

algebraic-topology homology-cohomology

share|cite|improve this question

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think your cohomology is not correct. If you attach a $3$-cell to $mathbb{C}P^2$ as you describe then you're essentially "filling in" the $2$-skeleton and the resulting space will be a $3$-cell with a $4$-cell attached to its boundary via the Hopf map $S^3 to S^2$. There shouldn't be any cohomology in degree $2$.
  $endgroup$
  – William
  Mar 23 at 22:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you mean for the attaching map to have degree $p$?
  $endgroup$
  – Eric Wofsey
  Mar 23 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EricWofsey Yes... I am sorry. Thanks!
  $endgroup$
  – No One
  Mar 24 at 3:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @William I am sorry... Just updated...
  $endgroup$
  – No One
  Mar 24 at 3:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated my answer for this situation
  $endgroup$
  – William
  Mar 24 at 4:41
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ be $mathbb CP^2$ with a 3-cell $e^3$ attached via $S^2to mathbb CP^1subset mathbb CP^2$ of degree $p$.
By cellular cohomology, it is easy to see that $H^k(X,mathbb Z_p)$ is $mathbb Z_p$ when $k=1,2,3,4$ and $0$ otherwise.

I wish to show that the cup product

$$H^2(X,mathbb Z_p)times H^2(X,mathbb Z_p) to H^4(X,mathbb Z_p)$$
is nontrivial. I am also confused about this $e^3$ attached to $mathbb CP^2$. I think it probably doesn't affect this cup product but I don't know how to show it rigorously.

Also, I only know $H^*(mathbb CP^2, mathbb Z)cong (a)/(a^3)$ and I am not sure how it is effected when taking the coefficients to be $mathbb Z_p$.

Thank you.

algebraic-topology homology-cohomology

share|cite|improve this question

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

$endgroup$

share|cite|improve this question

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

share|cite|improve this question

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

edited Mar 24 at 5:02

Andrews

1,3012423

edited Mar 24 at 5:02

Andrews

1,3012423

asked Mar 23 at 22:20

No OneNo One

2,1101519

asked Mar 23 at 22:20

No OneNo One

2,1101519

1 Answer
1

active

oldest

votes

2

$begingroup$

Edit: I updated my answer for the updated question, where the attaching map $f_pcolon S^2 to mathbb{C}P^2$ has degree $p>1$ and cohomology is taken with $mathbb{Z}/p$ coefficients. My original answer is below.

Don't forget that the cup product is natural, namely if we consider the inclusion $icolon mathbb{C}P^2 to X$ then we get a commutative diagram
$require{AMScd}$
begin{CD}
H^2(X)times H^2(X) @>{cup}>> H^4(X)\
@V{i^*times i^*}VV @V{i^*}VV \
H^2(mathbb{C}P^2)times H^2(mathbb{C}P^2) @>{cup}>> H^4(mathbb{C}P^2)\
end{CD}

for any coefficients. In particular, with $mathbb{Z}/p$ coefficients the long exact sequence for the cofibration $S^2stackrel{f_p}{to} mathbb{C}P^2 to X$ shows that the vertical maps are both isomorphisms, so the cup products are the same.

---

For your second question,

$$ H^*(mathbb{C}P^n;R)cong R[u]/(u^{n+1})$$

for any $R$ and $n$, where $u$ has degree $2$.

---

Old answer

If the attaching map $S^2 to mathbb{C}P^2$ is a degree $pm 1$ map, then $X$ is homotopy equivalent to the quotient of $mathbb{C}P^2$ by its $2$-skeleton, which gives $S^4$.

share|cite|improve this answer

edited Mar 24 at 4:54

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry about the imcompleteness of my original statement. $e^3$ is attached to $mathbb CP^2$ via a degree $p$ map $S^2to mathbb CP^1$. So in the cellular chain with $mathbb Z_p$ coefficients, the map from degree 3 to degree 2 is zero.
  $endgroup$
  – No One
  Mar 24 at 3:23
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Edit: I updated my answer for the updated question, where the attaching map $f_pcolon S^2 to mathbb{C}P^2$ has degree $p>1$ and cohomology is taken with $mathbb{Z}/p$ coefficients. My original answer is below.

Don't forget that the cup product is natural, namely if we consider the inclusion $icolon mathbb{C}P^2 to X$ then we get a commutative diagram
$require{AMScd}$
begin{CD}
H^2(X)times H^2(X) @>{cup}>> H^4(X)\
@V{i^*times i^*}VV @V{i^*}VV \
H^2(mathbb{C}P^2)times H^2(mathbb{C}P^2) @>{cup}>> H^4(mathbb{C}P^2)\
end{CD}

for any coefficients. In particular, with $mathbb{Z}/p$ coefficients the long exact sequence for the cofibration $S^2stackrel{f_p}{to} mathbb{C}P^2 to X$ shows that the vertical maps are both isomorphisms, so the cup products are the same.

---

For your second question,

$$ H^*(mathbb{C}P^n;R)cong R[u]/(u^{n+1})$$

for any $R$ and $n$, where $u$ has degree $2$.

---

Old answer

If the attaching map $S^2 to mathbb{C}P^2$ is a degree $pm 1$ map, then $X$ is homotopy equivalent to the quotient of $mathbb{C}P^2$ by its $2$-skeleton, which gives $S^4$.

share|cite|improve this answer

edited Mar 24 at 4:54

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry about the imcompleteness of my original statement. $e^3$ is attached to $mathbb CP^2$ via a degree $p$ map $S^2to mathbb CP^1$. So in the cellular chain with $mathbb Z_p$ coefficients, the map from degree 3 to degree 2 is zero.
  $endgroup$
  – No One
  Mar 24 at 3:23
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Edit: I updated my answer for the updated question, where the attaching map $f_pcolon S^2 to mathbb{C}P^2$ has degree $p>1$ and cohomology is taken with $mathbb{Z}/p$ coefficients. My original answer is below.

Don't forget that the cup product is natural, namely if we consider the inclusion $icolon mathbb{C}P^2 to X$ then we get a commutative diagram
$require{AMScd}$
begin{CD}
H^2(X)times H^2(X) @>{cup}>> H^4(X)\
@V{i^*times i^*}VV @V{i^*}VV \
H^2(mathbb{C}P^2)times H^2(mathbb{C}P^2) @>{cup}>> H^4(mathbb{C}P^2)\
end{CD}

for any coefficients. In particular, with $mathbb{Z}/p$ coefficients the long exact sequence for the cofibration $S^2stackrel{f_p}{to} mathbb{C}P^2 to X$ shows that the vertical maps are both isomorphisms, so the cup products are the same.

---

For your second question,

$$ H^*(mathbb{C}P^n;R)cong R[u]/(u^{n+1})$$

for any $R$ and $n$, where $u$ has degree $2$.

---

Old answer

If the attaching map $S^2 to mathbb{C}P^2$ is a degree $pm 1$ map, then $X$ is homotopy equivalent to the quotient of $mathbb{C}P^2$ by its $2$-skeleton, which gives $S^4$.

share|cite|improve this answer

edited Mar 24 at 4:54

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry about the imcompleteness of my original statement. $e^3$ is attached to $mathbb CP^2$ via a degree $p$ map $S^2to mathbb CP^1$. So in the cellular chain with $mathbb Z_p$ coefficients, the map from degree 3 to degree 2 is zero.
  $endgroup$
  – No One
  Mar 24 at 3:23
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Edit: I updated my answer for the updated question, where the attaching map $f_pcolon S^2 to mathbb{C}P^2$ has degree $p>1$ and cohomology is taken with $mathbb{Z}/p$ coefficients. My original answer is below.

Don't forget that the cup product is natural, namely if we consider the inclusion $icolon mathbb{C}P^2 to X$ then we get a commutative diagram
$require{AMScd}$
begin{CD}
H^2(X)times H^2(X) @>{cup}>> H^4(X)\
@V{i^*times i^*}VV @V{i^*}VV \
H^2(mathbb{C}P^2)times H^2(mathbb{C}P^2) @>{cup}>> H^4(mathbb{C}P^2)\
end{CD}

for any coefficients. In particular, with $mathbb{Z}/p$ coefficients the long exact sequence for the cofibration $S^2stackrel{f_p}{to} mathbb{C}P^2 to X$ shows that the vertical maps are both isomorphisms, so the cup products are the same.

---

For your second question,

$$ H^*(mathbb{C}P^n;R)cong R[u]/(u^{n+1})$$

for any $R$ and $n$, where $u$ has degree $2$.

---

Old answer

If the attaching map $S^2 to mathbb{C}P^2$ is a degree $pm 1$ map, then $X$ is homotopy equivalent to the quotient of $mathbb{C}P^2$ by its $2$-skeleton, which gives $S^4$.

share|cite|improve this answer

edited Mar 24 at 4:54

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

$endgroup$

share|cite|improve this answer

edited Mar 24 at 4:54

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

answered Mar 23 at 23:08

WilliamWilliam

3,3011228

answered Mar 23 at 23:08

WilliamWilliam

3,3011228