Find an example of K, A, B, and I to show $beta_ineq beta_i(A)+beta_i(B)-beta_i(Acap B)$
$begingroup$
I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number
$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?
topological-data-analysis
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
$endgroup$
add a comment |
$begingroup$
I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number
$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?
topological-data-analysis
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
$endgroup$
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18
add a comment |
$begingroup$
I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number
$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?
topological-data-analysis
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
$endgroup$
I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number
$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?
topological-data-analysis
topological-data-analysis
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
edited Nov 30 '18 at 17:15
LexyFidds
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
asked Nov 30 '18 at 16:54
LexyFidds LexyFidds
226
226
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18
add a comment |
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.
Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020317%2ffind-an-example-of-k-a-b-and-i-to-show-beta-i-neq-beta-ia-beta-ib-be%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.
Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
$endgroup$
add a comment |
$begingroup$
Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.
Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
$endgroup$
add a comment |
$begingroup$
Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.
Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
$endgroup$
Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.
Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
answered Dec 8 '18 at 14:56
Henry AdamsHenry Adams
1164
1164
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020317%2ffind-an-example-of-k-a-b-and-i-to-show-beta-i-neq-beta-ia-beta-ib-be%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
$endgroup$
– MPW
Nov 30 '18 at 16:56
$begingroup$
I already fixed it, thanks
$endgroup$
– LexyFidds
Nov 30 '18 at 16:57
$begingroup$
Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
$endgroup$
– Pedro Tamaroff♦
Nov 30 '18 at 17:18