Acyclic Chain Complex
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
add a comment |
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
edited Dec 4 '18 at 0:58
LexyFidds
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
226
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
0
active
oldest
votes
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%2f3024957%2facyclic-chain-complex%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3024957%2facyclic-chain-complex%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
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
– BWW
Dec 4 '18 at 1:03
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
– LexyFidds
Dec 4 '18 at 1:09
Aaaahhh you’re right, I think I’ve been indexing incorrectly
– LexyFidds
Dec 4 '18 at 1:12