Mixing
Relation of complete homogeneous symmetric polynomials and the elementary symmetric polynomials
$begingroup$
I was reading about the symmetric polynomials and saw the following relation:
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$
The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.
(Question) I was wondering if this relation still holds for any length $k$, i.e.
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$
P.S. I tried cases and it holds for any case.
abstract-algebra commutative-algebra symmetric-polynomials
asked Feb 3 at 10:11
MaxMax
1,0041319
$endgroup$
add a comment |
$begingroup$
I was reading about the symmetric polynomials and saw the following relation:
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$
The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.
(Question) I was wondering if this relation still holds for any length $k$, i.e.
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$
P.S. I tried cases and it holds for any case.
abstract-algebra commutative-algebra symmetric-polynomials
asked Feb 3 at 10:11
MaxMax
1,0041319
$endgroup$
add a comment |
$begingroup$
I was reading about the symmetric polynomials and saw the following relation:
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$
The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.
(Question) I was wondering if this relation still holds for any length $k$, i.e.
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$
P.S. I tried cases and it holds for any case.
abstract-algebra commutative-algebra symmetric-polynomials
asked Feb 3 at 10:11
MaxMax
1,0041319
$endgroup$
I was reading about the symmetric polynomials and saw the following relation:
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$
The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.
(Question) I was wondering if this relation still holds for any length $k$, i.e.
$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$
P.S. I tried cases and it holds for any case.
abstract-algebra commutative-algebra symmetric-polynomials
abstract-algebra commutative-algebra symmetric-polynomials
asked Feb 3 at 10:11
MaxMax
1,0041319
asked Feb 3 at 10:11
MaxMax
1,0041319
edited Mar 17 at 11:06
Max
asked Feb 3 at 10:11
MaxMax
1,0041319
asked Feb 3 at 10:11
MaxMax
1,0041319
asked Feb 3 at 10:11
MaxMax
1,0041319
1,0041319
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have
$$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
$$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
$$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
$endgroup$
add a comment |
Your Answer
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%2f3098386%2frelation-of-complete-homogeneous-symmetric-polynomials-and-the-elementary-symmet%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
$begingroup$
We have
$$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
$$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
$$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
$endgroup$
add a comment |
$begingroup$
We have
$$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
$$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
$$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
$endgroup$
add a comment |
$begingroup$
We have
$$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
$$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
$$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
$endgroup$
We have
$$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
$$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
$$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
edited Feb 3 at 19:11
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
answered Feb 3 at 18:48
Matt SamuelMatt Samuel
39.3k63870
39.3k63870
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%2f3098386%2frelation-of-complete-homogeneous-symmetric-polynomials-and-the-elementary-symmet%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
