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How to determine the homology groups of $(v,w)inmathbbC^2mid vw neq 0$
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I want to determine the homology groups of $(v,w)inmathbbC^2mid vw neq 0$. I googled this problem and found that there's a general theory called arrangements of hyperplanes. But this theory is quite involved. So I'm wondering if there's some simpler way to do this.
I was considering Mayer-Vietoris sequence, but I was sort of confused. Since $(v,w)inmathbbC^2$ is the complement of two complex planes, I need to find an open cover with good intersections. But I don't know how to do this. Can anyone enlighten me or give me some hints? Thanks!
algebraic-topology homology-cohomology complex-geometry
edited Jul 20 at 17:28
Michael Hardy
204k23186462
asked Jul 20 at 16:46
Johnny
13811
add a comment |Â
up vote
1
down vote
favorite
I want to determine the homology groups of $(v,w)inmathbbC^2mid vw neq 0$. I googled this problem and found that there's a general theory called arrangements of hyperplanes. But this theory is quite involved. So I'm wondering if there's some simpler way to do this.
I was considering Mayer-Vietoris sequence, but I was sort of confused. Since $(v,w)inmathbbC^2$ is the complement of two complex planes, I need to find an open cover with good intersections. But I don't know how to do this. Can anyone enlighten me or give me some hints? Thanks!
algebraic-topology homology-cohomology complex-geometry
edited Jul 20 at 17:28
Michael Hardy
204k23186462
asked Jul 20 at 16:46
Johnny
13811
1
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
1
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to determine the homology groups of $(v,w)inmathbbC^2mid vw neq 0$. I googled this problem and found that there's a general theory called arrangements of hyperplanes. But this theory is quite involved. So I'm wondering if there's some simpler way to do this.
I was considering Mayer-Vietoris sequence, but I was sort of confused. Since $(v,w)inmathbbC^2$ is the complement of two complex planes, I need to find an open cover with good intersections. But I don't know how to do this. Can anyone enlighten me or give me some hints? Thanks!
algebraic-topology homology-cohomology complex-geometry
edited Jul 20 at 17:28
Michael Hardy
204k23186462
asked Jul 20 at 16:46
Johnny
13811
I want to determine the homology groups of $(v,w)inmathbbC^2mid vw neq 0$. I googled this problem and found that there's a general theory called arrangements of hyperplanes. But this theory is quite involved. So I'm wondering if there's some simpler way to do this.
I was considering Mayer-Vietoris sequence, but I was sort of confused. Since $(v,w)inmathbbC^2$ is the complement of two complex planes, I need to find an open cover with good intersections. But I don't know how to do this. Can anyone enlighten me or give me some hints? Thanks!
algebraic-topology homology-cohomology complex-geometry
edited Jul 20 at 17:28
Michael Hardy
204k23186462
asked Jul 20 at 16:46
Johnny
13811
edited Jul 20 at 17:28
Michael Hardy
204k23186462
edited Jul 20 at 17:28
Michael Hardy
204k23186462
edited Jul 20 at 17:28
Michael Hardy
204k23186462
204k23186462
asked Jul 20 at 16:46
Johnny
13811
asked Jul 20 at 16:46
Johnny
13811
asked Jul 20 at 16:46
Johnny
13811
13811
1
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
1
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55
add a comment |Â
1
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
1
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55
1
1
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
1
1
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
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Define the map $Gamma:X to X$ by $Gamma(v,w)=(fracvv, fracw)$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (fracvv, fracw)+(1-t)(v,w)$$
By direct calculation you can show that for $t in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 times S^1$. Now compute this by Kunneth if you know, else you can use a Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= begincases
mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
Observe that $U cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_alpha $there is an isomorphism of $H_n(X)$ with the direct sum $oplus_alpha X_alpha $.
We get: $$H_n(U cap V)= begincases
mathbbZ oplus mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U cap V$ to $U ,V$ $k,l$ and be the inclusions from $U,V$ to the torus):
$$dots longrightarrow H_n(U)oplus H_n(V) longrightarrow H_n(S^1 times S^1) longrightarrow H_n-1(U cap V) longrightarrow dots $$
So for $n > 2 $ We immediately have $H_n(S^1 times S^1)=0$. For $n=2$ We have the sequence:
$$dots 0 longrightarrow H_2(S^1 times S^1) longrightarrow mathbbZ oplus mathbbZ longrightarrow mathbbZ oplus mathbbZ longrightarrow dots $$
Observe that $H(U) cong H(v) cong mathbbZ $ $a$ and $b$ are different basis element in homology class $U cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 times S^1)$ is $0$ so the Kernal from $H_2(S^1 times S^1)$ is $0$. This implies that $H_2(S^1 times S^1) cong mathbbZ<a-b> cong mathbbZ $.
As for $n=1$ the Mayer-Vietoris sequence will be:
$$dots longrightarrow mathbbZoplusmathbbZ stackrel(i_*, j_*)longrightarrow mathbbZoplusmathbbZ stackrelk_* - l_*longrightarrow H_1(S^1 times S^1) stackrelpartiallongrightarrow mathbbZoplusmathbbZ stackrel(i_* , j_*)longrightarrowmathbbZoplusmathbbZ longrightarrow dots $$
So we get a short exact sequence:
$$0longrightarrow mathrmker partial longrightarrow H_1(S^1 times S^1) longrightarrow
mathrmIm partial
longrightarrow 0$$
Now $mathrmIm partial = mathrmKer (i_* , j_*)$ which we have seen that is $mathbbZ$. To compute $mathrmKer partial$ We use:
$$mathrmker partial = mathrmIm (k_*-l_*) =(mathbbZoplus mathbbZ) / mathrmKer(k_*-l_* ) = (mathbbZoplus mathbbZ)/mathrmIm(i_*, j_*) = mathbbZ$$
Using the fact that the long sequence is exact for the identity's in the formula.
So we Get:
$$H_1(S^1 times S^1)/ mathbbZ cong mathbbZ$$
This implies that $H_1(S^1 times S^1) cong mathbbZ oplus mathbbZ $
For $n=0$ we know that the torus is connected so $H_0(S^1 times S^1) cong mathbbZ$
answered Jul 24 at 9:07
Elad
54929
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Define the map $Gamma:X to X$ by $Gamma(v,w)=(fracvv, fracw)$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (fracvv, fracw)+(1-t)(v,w)$$
By direct calculation you can show that for $t in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 times S^1$. Now compute this by Kunneth if you know, else you can use a Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= begincases
mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
Observe that $U cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_alpha $there is an isomorphism of $H_n(X)$ with the direct sum $oplus_alpha X_alpha $.
We get: $$H_n(U cap V)= begincases
mathbbZ oplus mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U cap V$ to $U ,V$ $k,l$ and be the inclusions from $U,V$ to the torus):
$$dots longrightarrow H_n(U)oplus H_n(V) longrightarrow H_n(S^1 times S^1) longrightarrow H_n-1(U cap V) longrightarrow dots $$
So for $n > 2 $ We immediately have $H_n(S^1 times S^1)=0$. For $n=2$ We have the sequence:
$$dots 0 longrightarrow H_2(S^1 times S^1) longrightarrow mathbbZ oplus mathbbZ longrightarrow mathbbZ oplus mathbbZ longrightarrow dots $$
Observe that $H(U) cong H(v) cong mathbbZ $ $a$ and $b$ are different basis element in homology class $U cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 times S^1)$ is $0$ so the Kernal from $H_2(S^1 times S^1)$ is $0$. This implies that $H_2(S^1 times S^1) cong mathbbZ<a-b> cong mathbbZ $.
As for $n=1$ the Mayer-Vietoris sequence will be:
$$dots longrightarrow mathbbZoplusmathbbZ stackrel(i_*, j_*)longrightarrow mathbbZoplusmathbbZ stackrelk_* - l_*longrightarrow H_1(S^1 times S^1) stackrelpartiallongrightarrow mathbbZoplusmathbbZ stackrel(i_* , j_*)longrightarrowmathbbZoplusmathbbZ longrightarrow dots $$
So we get a short exact sequence:
$$0longrightarrow mathrmker partial longrightarrow H_1(S^1 times S^1) longrightarrow
mathrmIm partial
longrightarrow 0$$
Now $mathrmIm partial = mathrmKer (i_* , j_*)$ which we have seen that is $mathbbZ$. To compute $mathrmKer partial$ We use:
$$mathrmker partial = mathrmIm (k_*-l_*) =(mathbbZoplus mathbbZ) / mathrmKer(k_*-l_* ) = (mathbbZoplus mathbbZ)/mathrmIm(i_*, j_*) = mathbbZ$$
Using the fact that the long sequence is exact for the identity's in the formula.
So we Get:
$$H_1(S^1 times S^1)/ mathbbZ cong mathbbZ$$
This implies that $H_1(S^1 times S^1) cong mathbbZ oplus mathbbZ $
For $n=0$ we know that the torus is connected so $H_0(S^1 times S^1) cong mathbbZ$
answered Jul 24 at 9:07
Elad
54929
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
add a comment |Â
up vote
2
down vote
Define the map $Gamma:X to X$ by $Gamma(v,w)=(fracvv, fracw)$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (fracvv, fracw)+(1-t)(v,w)$$
By direct calculation you can show that for $t in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 times S^1$. Now compute this by Kunneth if you know, else you can use a Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= begincases
mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
Observe that $U cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_alpha $there is an isomorphism of $H_n(X)$ with the direct sum $oplus_alpha X_alpha $.
We get: $$H_n(U cap V)= begincases
mathbbZ oplus mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U cap V$ to $U ,V$ $k,l$ and be the inclusions from $U,V$ to the torus):
$$dots longrightarrow H_n(U)oplus H_n(V) longrightarrow H_n(S^1 times S^1) longrightarrow H_n-1(U cap V) longrightarrow dots $$
So for $n > 2 $ We immediately have $H_n(S^1 times S^1)=0$. For $n=2$ We have the sequence:
$$dots 0 longrightarrow H_2(S^1 times S^1) longrightarrow mathbbZ oplus mathbbZ longrightarrow mathbbZ oplus mathbbZ longrightarrow dots $$
Observe that $H(U) cong H(v) cong mathbbZ $ $a$ and $b$ are different basis element in homology class $U cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 times S^1)$ is $0$ so the Kernal from $H_2(S^1 times S^1)$ is $0$. This implies that $H_2(S^1 times S^1) cong mathbbZ<a-b> cong mathbbZ $.
As for $n=1$ the Mayer-Vietoris sequence will be:
$$dots longrightarrow mathbbZoplusmathbbZ stackrel(i_*, j_*)longrightarrow mathbbZoplusmathbbZ stackrelk_* - l_*longrightarrow H_1(S^1 times S^1) stackrelpartiallongrightarrow mathbbZoplusmathbbZ stackrel(i_* , j_*)longrightarrowmathbbZoplusmathbbZ longrightarrow dots $$
So we get a short exact sequence:
$$0longrightarrow mathrmker partial longrightarrow H_1(S^1 times S^1) longrightarrow
mathrmIm partial
longrightarrow 0$$
Now $mathrmIm partial = mathrmKer (i_* , j_*)$ which we have seen that is $mathbbZ$. To compute $mathrmKer partial$ We use:
$$mathrmker partial = mathrmIm (k_*-l_*) =(mathbbZoplus mathbbZ) / mathrmKer(k_*-l_* ) = (mathbbZoplus mathbbZ)/mathrmIm(i_*, j_*) = mathbbZ$$
Using the fact that the long sequence is exact for the identity's in the formula.
So we Get:
$$H_1(S^1 times S^1)/ mathbbZ cong mathbbZ$$
This implies that $H_1(S^1 times S^1) cong mathbbZ oplus mathbbZ $
For $n=0$ we know that the torus is connected so $H_0(S^1 times S^1) cong mathbbZ$
answered Jul 24 at 9:07
Elad
54929
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Define the map $Gamma:X to X$ by $Gamma(v,w)=(fracvv, fracw)$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (fracvv, fracw)+(1-t)(v,w)$$
By direct calculation you can show that for $t in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 times S^1$. Now compute this by Kunneth if you know, else you can use a Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= begincases
mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
Observe that $U cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_alpha $there is an isomorphism of $H_n(X)$ with the direct sum $oplus_alpha X_alpha $.
We get: $$H_n(U cap V)= begincases
mathbbZ oplus mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U cap V$ to $U ,V$ $k,l$ and be the inclusions from $U,V$ to the torus):
$$dots longrightarrow H_n(U)oplus H_n(V) longrightarrow H_n(S^1 times S^1) longrightarrow H_n-1(U cap V) longrightarrow dots $$
So for $n > 2 $ We immediately have $H_n(S^1 times S^1)=0$. For $n=2$ We have the sequence:
$$dots 0 longrightarrow H_2(S^1 times S^1) longrightarrow mathbbZ oplus mathbbZ longrightarrow mathbbZ oplus mathbbZ longrightarrow dots $$
Observe that $H(U) cong H(v) cong mathbbZ $ $a$ and $b$ are different basis element in homology class $U cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 times S^1)$ is $0$ so the Kernal from $H_2(S^1 times S^1)$ is $0$. This implies that $H_2(S^1 times S^1) cong mathbbZ<a-b> cong mathbbZ $.
As for $n=1$ the Mayer-Vietoris sequence will be:
$$dots longrightarrow mathbbZoplusmathbbZ stackrel(i_*, j_*)longrightarrow mathbbZoplusmathbbZ stackrelk_* - l_*longrightarrow H_1(S^1 times S^1) stackrelpartiallongrightarrow mathbbZoplusmathbbZ stackrel(i_* , j_*)longrightarrowmathbbZoplusmathbbZ longrightarrow dots $$
So we get a short exact sequence:
$$0longrightarrow mathrmker partial longrightarrow H_1(S^1 times S^1) longrightarrow
mathrmIm partial
longrightarrow 0$$
Now $mathrmIm partial = mathrmKer (i_* , j_*)$ which we have seen that is $mathbbZ$. To compute $mathrmKer partial$ We use:
$$mathrmker partial = mathrmIm (k_*-l_*) =(mathbbZoplus mathbbZ) / mathrmKer(k_*-l_* ) = (mathbbZoplus mathbbZ)/mathrmIm(i_*, j_*) = mathbbZ$$
Using the fact that the long sequence is exact for the identity's in the formula.
So we Get:
$$H_1(S^1 times S^1)/ mathbbZ cong mathbbZ$$
This implies that $H_1(S^1 times S^1) cong mathbbZ oplus mathbbZ $
For $n=0$ we know that the torus is connected so $H_0(S^1 times S^1) cong mathbbZ$
answered Jul 24 at 9:07
Elad
54929
Define the map $Gamma:X to X$ by $Gamma(v,w)=(fracvv, fracw)$. This is a continues map as a composition of continues maps. This is a deformation retraction by the homotopy:
$$t (fracvv, fracw)+(1-t)(v,w)$$
By direct calculation you can show that for $t in [0,1]$ the norm of each coordinate is different then $0$. This shows that the homotopy is well defined. Let the retract subspace be $A$ then observe that it’s homeomorphic to $S^1 times S^1$. Now compute this by Kunneth if you know, else you can use a Mayer-Vietoris but it’s harder.
To make this answer more complete here is how you can compute the torus($S^1 times S^1$) using Mayer-Vietoris. Fi
I will Cover the Torus by 2 Open sets of the Torus $A=U$ and $B=V$ as seen in the figure }. Observe that $U$ and $V$ are cylinders and by the retraction trick their homology groups are the same as $S^1$. Hence will get:
$$H_n(V)=H_n(U)= begincases
mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
Observe that $U cap V$ is a disjoint union of $S^1$ so by the following theorem :
Corresponding to the decomposition of a space $X$ into its path components
$X_alpha $there is an isomorphism of $H_n(X)$ with the direct sum $oplus_alpha X_alpha $.
We get: $$H_n(U cap V)= begincases
mathbbZ oplus mathbbZ & textif n=0,1 \
0 & textotherwise
endcases$$
So by Mayer-Vietoris the following sequence is exact ( $i,j$ be the inclusions from $U cap V$ to $U ,V$ $k,l$ and be the inclusions from $U,V$ to the torus):
$$dots longrightarrow H_n(U)oplus H_n(V) longrightarrow H_n(S^1 times S^1) longrightarrow H_n-1(U cap V) longrightarrow dots $$
So for $n > 2 $ We immediately have $H_n(S^1 times S^1)=0$. For $n=2$ We have the sequence:
$$dots 0 longrightarrow H_2(S^1 times S^1) longrightarrow mathbbZ oplus mathbbZ longrightarrow mathbbZ oplus mathbbZ longrightarrow dots $$
Observe that $H(U) cong H(v) cong mathbbZ $ $a$ and $b$ are different basis element in homology class $U cap V$ they will be both sent to the same homology class element in the inclusion function. So the kernel of $(i_*, j_*)$ in the exact sequence is $<a-b>$. This is an exact sequence and the image to $H_2(S^1 times S^1)$ is $0$ so the Kernal from $H_2(S^1 times S^1)$ is $0$. This implies that $H_2(S^1 times S^1) cong mathbbZ<a-b> cong mathbbZ $.
As for $n=1$ the Mayer-Vietoris sequence will be:
$$dots longrightarrow mathbbZoplusmathbbZ stackrel(i_*, j_*)longrightarrow mathbbZoplusmathbbZ stackrelk_* - l_*longrightarrow H_1(S^1 times S^1) stackrelpartiallongrightarrow mathbbZoplusmathbbZ stackrel(i_* , j_*)longrightarrowmathbbZoplusmathbbZ longrightarrow dots $$
So we get a short exact sequence:
$$0longrightarrow mathrmker partial longrightarrow H_1(S^1 times S^1) longrightarrow
mathrmIm partial
longrightarrow 0$$
Now $mathrmIm partial = mathrmKer (i_* , j_*)$ which we have seen that is $mathbbZ$. To compute $mathrmKer partial$ We use:
$$mathrmker partial = mathrmIm (k_*-l_*) =(mathbbZoplus mathbbZ) / mathrmKer(k_*-l_* ) = (mathbbZoplus mathbbZ)/mathrmIm(i_*, j_*) = mathbbZ$$
Using the fact that the long sequence is exact for the identity's in the formula.
So we Get:
$$H_1(S^1 times S^1)/ mathbbZ cong mathbbZ$$
This implies that $H_1(S^1 times S^1) cong mathbbZ oplus mathbbZ $
For $n=0$ we know that the torus is connected so $H_0(S^1 times S^1) cong mathbbZ$
answered Jul 24 at 9:07
Elad
54929
edited Jul 24 at 11:19
answered Jul 24 at 9:07
Elad
54929
answered Jul 24 at 9:07
Elad
54929
answered Jul 24 at 9:07
Elad
54929
54929
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
add a comment |Â
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
$S^1times S^1$ is a CW-complex, so its homology may be computed with cellular homology
– Max
Jul 24 at 9:44
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
Thanks! I understand what you and Miller have suggested. By considering stereographic projection, we may map $S^3-=1-=1$ to $mathbbR^3$. This case is easy to calculate. But when the complex planes get more, say $v=0 $, $w=0 $ and $v=w $, the calculation becomes harder and it's hard to see through. Could you please give me some further suggestions?
– Johnny
Jul 24 at 9:49
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
I will gladly give further suggestions but for what? Did you understand why you only need to calculate the Homology of $S^1 times S^1$ to compute the Homology of your space? Is your problem calculating $H_n(S^1 times S^1)$?
– Elad
Jul 24 at 10:57
add a comment |Â
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1
Here's a hint to get you started: $Bbb C^2 - z=0$ deformation retracts to $|z|=varepsilon$, which is $S^1timesBbb C$.
– Ted Shifrin
Jul 20 at 17:18
@TedShifrin So $mathbbC^2- v=0- w=0$ deformation retracts to $ S^1 vee S^1 times mathbbC$? I'm not sure since $mathbbC^2$ is sort of difficult for me to imagine.
– Johnny
Jul 20 at 17:27
1
No, I don't think that's right. You have those two intersecting cylinders and you have to remove the stuff "near the origin." So you ought to be able to deformation retract to something more tractable.
– Ted Shifrin
Jul 20 at 17:32
It deformation retracts onto its intersection with $S^3$, the set of points where $v^2+w^2=1$. Can you figure out what taking the intersection deletes from $S^3$? And can you translate that to a subspace of $Bbb R^3$?
– Mike Miller
Jul 21 at 22:55