Möbius function - understanding of relations
I am trying to understand Möbius function from the wikipedia article (and also few others that I have come across so far). This function is defined in posets and so the relations in Special elements section of the wikipedia article confuse me. So what do relations $=$, $leq$, $<$ mean in this section? Do these have the traditional meaning or does these relations mean something different?
relations order-theory mobius-function
asked Dec 10 '18 at 20:31
scarface
456
add a comment |
I am trying to understand Möbius function from the wikipedia article (and also few others that I have come across so far). This function is defined in posets and so the relations in Special elements section of the wikipedia article confuse me. So what do relations $=$, $leq$, $<$ mean in this section? Do these have the traditional meaning or does these relations mean something different?
relations order-theory mobius-function
asked Dec 10 '18 at 20:31
scarface
456
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43
add a comment |
I am trying to understand Möbius function from the wikipedia article (and also few others that I have come across so far). This function is defined in posets and so the relations in Special elements section of the wikipedia article confuse me. So what do relations $=$, $leq$, $<$ mean in this section? Do these have the traditional meaning or does these relations mean something different?
relations order-theory mobius-function
asked Dec 10 '18 at 20:31
scarface
456
I am trying to understand Möbius function from the wikipedia article (and also few others that I have come across so far). This function is defined in posets and so the relations in Special elements section of the wikipedia article confuse me. So what do relations $=$, $leq$, $<$ mean in this section? Do these have the traditional meaning or does these relations mean something different?
relations order-theory mobius-function
relations order-theory mobius-function
asked Dec 10 '18 at 20:31
scarface
456
asked Dec 10 '18 at 20:31
scarface
456
asked Dec 10 '18 at 20:31
scarface
456
asked Dec 10 '18 at 20:31
scarface
456
asked Dec 10 '18 at 20:31
scarface
456
456
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43
add a comment |
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43
add a comment |
1 Answer
1
active
oldest
votes
In the original number-theoretic case, $leq$ would be replaced by $mid$. The integers are partially ordered by divisibility and the Mobius function is the inverse of the zeta function, using the language of the general poset case.
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
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%2f3034420%2fm%25c3%25b6bius-function-understanding-of-relations%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
In the original number-theoretic case, $leq$ would be replaced by $mid$. The integers are partially ordered by divisibility and the Mobius function is the inverse of the zeta function, using the language of the general poset case.
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
add a comment |
In the original number-theoretic case, $leq$ would be replaced by $mid$. The integers are partially ordered by divisibility and the Mobius function is the inverse of the zeta function, using the language of the general poset case.
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
add a comment |
In the original number-theoretic case, $leq$ would be replaced by $mid$. The integers are partially ordered by divisibility and the Mobius function is the inverse of the zeta function, using the language of the general poset case.
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
In the original number-theoretic case, $leq$ would be replaced by $mid$. The integers are partially ordered by divisibility and the Mobius function is the inverse of the zeta function, using the language of the general poset case.
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:33
Matt Samuel
37.2k63465
37.2k63465
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
add a comment |
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
Your answer did not help me. Probably I did not completely get it. Can you expand your answer, please?
– scarface
Dec 10 '18 at 20:38
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.
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%2f3034420%2fm%25c3%25b6bius-function-understanding-of-relations%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
At first $= $ is as usual and $< $ is any partial ordering on a set such that an element has finitely many $<$ and "$1$" is $<$ to every other element. Then $mu(d,n)$ is the function such that $g(n) = sum_{d le n} f(d) implies f(n) = sum_{d le n} g(d)mu(d,n) $. When $<$ is the partial order on $mathbb{Z}_{ge 1}$ defined by $d le n$ if $d |n$ then $mu(d,n) = (-1)^m$ if $n/d$ is a product of $m$ distinct primes, $mu(d,n) = 0$ otherwise.
– reuns
Dec 10 '18 at 21:12
I think it is pretty clearly written here, at page 6. So $leq$ is the relation of poset, $<$ means strong inequality as we know it from primary school.
– scarface
Dec 10 '18 at 21:29
No, they would never use $leq$ and $le$ to refer to different orderings. One is strict and the other one isn't, but otherwise they are the same.
– Matt Samuel
Dec 10 '18 at 23:39
I haven't looked, but it's possible there are some spots where it's ambiguous, where you're expected to know that it's the ordinary ordering because it's referring to something like an index, something that isn't supposed to be an element of the poset.
– Matt Samuel
Dec 10 '18 at 23:43