Mixing
What are the Betti numbers of a double pinched torus?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
What are the betti numbers for a double pinched torus? A intuitive explanation for which holes these betti numbers correspond to would also be much appreciated.
algebraic-topology homology-cohomology betti-numbers
asked Jul 20 at 3:28
Jason Y
362
add a comment |Â
up vote
3
down vote
favorite
What are the betti numbers for a double pinched torus? A intuitive explanation for which holes these betti numbers correspond to would also be much appreciated.
algebraic-topology homology-cohomology betti-numbers
asked Jul 20 at 3:28
Jason Y
362
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What are the betti numbers for a double pinched torus? A intuitive explanation for which holes these betti numbers correspond to would also be much appreciated.
algebraic-topology homology-cohomology betti-numbers
asked Jul 20 at 3:28
Jason Y
362
What are the betti numbers for a double pinched torus? A intuitive explanation for which holes these betti numbers correspond to would also be much appreciated.
algebraic-topology homology-cohomology betti-numbers
asked Jul 20 at 3:28
Jason Y
362
asked Jul 20 at 3:28
Jason Y
362
asked Jul 20 at 3:28
Jason Y
362
asked Jul 20 at 3:28
Jason Y
362
362
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.
Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U simeq S^2$, $V simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:
$... to H_n(U cap V) to H_n(U) oplus H_n(V) to H_n(X) xrightarrowdelta H_n-1(U cap V) to ...$
Note: From the mere fact that $X$ is connected, you know that $H_0(X) = mathbbZ$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.
answered Jul 20 at 4:38
Saad Slaoui
1115
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.
Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U simeq S^2$, $V simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:
$... to H_n(U cap V) to H_n(U) oplus H_n(V) to H_n(X) xrightarrowdelta H_n-1(U cap V) to ...$
Note: From the mere fact that $X$ is connected, you know that $H_0(X) = mathbbZ$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.
answered Jul 20 at 4:38
Saad Slaoui
1115
add a comment |Â
up vote
5
down vote
The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.
Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U simeq S^2$, $V simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:
$... to H_n(U cap V) to H_n(U) oplus H_n(V) to H_n(X) xrightarrowdelta H_n-1(U cap V) to ...$
Note: From the mere fact that $X$ is connected, you know that $H_0(X) = mathbbZ$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.
answered Jul 20 at 4:38
Saad Slaoui
1115
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.
Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U simeq S^2$, $V simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:
$... to H_n(U cap V) to H_n(U) oplus H_n(V) to H_n(X) xrightarrowdelta H_n-1(U cap V) to ...$
Note: From the mere fact that $X$ is connected, you know that $H_0(X) = mathbbZ$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.
answered Jul 20 at 4:38
Saad Slaoui
1115
The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.
Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U simeq S^2$, $V simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:
$... to H_n(U cap V) to H_n(U) oplus H_n(V) to H_n(X) xrightarrowdelta H_n-1(U cap V) to ...$
Note: From the mere fact that $X$ is connected, you know that $H_0(X) = mathbbZ$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.
answered Jul 20 at 4:38
Saad Slaoui
1115
answered Jul 20 at 4:38
Saad Slaoui
1115
answered Jul 20 at 4:38
Saad Slaoui
1115
answered Jul 20 at 4:38
Saad Slaoui
1115
1115
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857251%2fwhat-are-the-betti-numbers-of-a-double-pinched-torus%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password