Prove that $frac{partial^{k} f}{partial x^{i_{1}} cdots partial x^{i_{k}}}$ is a tensor.
$begingroup$
Consider a function : $f(x^{1} dots x^{n}) ($$x^{i}- i$-th coordinate
$) in C^{infty}$ and $P$ such point : $forall 0<l <k$ $dfrac{partial ^{l} f}{partial x^{i_{1}} dots partial x^{i_{l}}}$ is zero at point $P$.
Now consider $A_{i_1 dots i_k} =dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$ we want to prove that it's a tensor. To prove this we need to check tensor-law:
$A_{i'_1 dots i'_k} =dfrac{partial x^{i_{1}}}{partial x^{i'_{1}}} dots dfrac{partial x^{i_{k}}}{partial x^{i'_{k}}} dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$.
I thought about considering $f = sum_{n_1 dots j_n} c(j_1 dots j_n) x_{i_1}^{j_1} dots x_{i_n}^{j_n}$ then condition about derivatives gives us some properties about coefficients.
Maybe there is a better way to prove it?
differential-geometry tensors
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
$endgroup$
|
show 1 more comment
$begingroup$
Consider a function : $f(x^{1} dots x^{n}) ($$x^{i}- i$-th coordinate
$) in C^{infty}$ and $P$ such point : $forall 0<l <k$ $dfrac{partial ^{l} f}{partial x^{i_{1}} dots partial x^{i_{l}}}$ is zero at point $P$.
Now consider $A_{i_1 dots i_k} =dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$ we want to prove that it's a tensor. To prove this we need to check tensor-law:
$A_{i'_1 dots i'_k} =dfrac{partial x^{i_{1}}}{partial x^{i'_{1}}} dots dfrac{partial x^{i_{k}}}{partial x^{i'_{k}}} dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$.
I thought about considering $f = sum_{n_1 dots j_n} c(j_1 dots j_n) x_{i_1}^{j_1} dots x_{i_n}^{j_n}$ then condition about derivatives gives us some properties about coefficients.
Maybe there is a better way to prove it?
differential-geometry tensors
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
$endgroup$
1
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41
|
show 1 more comment
$begingroup$
Consider a function : $f(x^{1} dots x^{n}) ($$x^{i}- i$-th coordinate
$) in C^{infty}$ and $P$ such point : $forall 0<l <k$ $dfrac{partial ^{l} f}{partial x^{i_{1}} dots partial x^{i_{l}}}$ is zero at point $P$.
Now consider $A_{i_1 dots i_k} =dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$ we want to prove that it's a tensor. To prove this we need to check tensor-law:
$A_{i'_1 dots i'_k} =dfrac{partial x^{i_{1}}}{partial x^{i'_{1}}} dots dfrac{partial x^{i_{k}}}{partial x^{i'_{k}}} dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$.
I thought about considering $f = sum_{n_1 dots j_n} c(j_1 dots j_n) x_{i_1}^{j_1} dots x_{i_n}^{j_n}$ then condition about derivatives gives us some properties about coefficients.
Maybe there is a better way to prove it?
differential-geometry tensors
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
$endgroup$
Consider a function : $f(x^{1} dots x^{n}) ($$x^{i}- i$-th coordinate
$) in C^{infty}$ and $P$ such point : $forall 0<l <k$ $dfrac{partial ^{l} f}{partial x^{i_{1}} dots partial x^{i_{l}}}$ is zero at point $P$.
Now consider $A_{i_1 dots i_k} =dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$ we want to prove that it's a tensor. To prove this we need to check tensor-law:
$A_{i'_1 dots i'_k} =dfrac{partial x^{i_{1}}}{partial x^{i'_{1}}} dots dfrac{partial x^{i_{k}}}{partial x^{i'_{k}}} dfrac{partial ^{k} f}{partial x^{i_{1}} dots partial x^{i_{k}}}$.
I thought about considering $f = sum_{n_1 dots j_n} c(j_1 dots j_n) x_{i_1}^{j_1} dots x_{i_n}^{j_n}$ then condition about derivatives gives us some properties about coefficients.
Maybe there is a better way to prove it?
differential-geometry tensors
differential-geometry tensors
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
edited Dec 17 '18 at 21:20
Lorenzo B.
1,8602520
1,8602520
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
asked Dec 17 '18 at 14:14
openspaceopenspace
3,4242822
3,4242822
1
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41
|
show 1 more comment
1
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41
1
1
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41
|
show 1 more 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%2f3043992%2fprove-that-frac-partialk-f-partial-xi-1-cdots-partial-xi-k%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.
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%2f3043992%2fprove-that-frac-partialk-f-partial-xi-1-cdots-partial-xi-k%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
1
$begingroup$
Have you tried doing the case $k=2$ to see what's going on?
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:03
$begingroup$
@TedShifrin for case $k=2$ we have that all partial derivatives at point $P$ are zero. That's means that function behaves like : $dfrac{partial f}{partial x^{i}}$~$g(bar{x})(bar{x} -P)$
$endgroup$
– openspace
Dec 17 '18 at 19:13
$begingroup$
I would think more about how the second derivatives transform when you make a change of coordinates. I don't quite understand your statement.
$endgroup$
– Ted Shifrin
Dec 17 '18 at 19:33
$begingroup$
@TedShifrin you mean: $A_{i'_1, i'_2} = dfrac{partial x^{i_1}}{partial x^{i'_1}} dfrac{partial x^{i_2}}{partial x^{i'_2}} dfrac{partial^{2} f}{partial x^{i_1}partial x^{i_2}}$. That's part you are talking about ?
$endgroup$
– openspace
Dec 17 '18 at 20:30
$begingroup$
@TedShifrin the general question is to prove that $A_{i_1 dots i_k} = dfrac{partial^{k} f}{partial x^{i_1} dots partial x^{i_{k}}}$ is a tensor under conditions I've mentioned.
$endgroup$
– openspace
Dec 17 '18 at 20:41