Prove that $frac{partial^{k} f}{partial x^{i_{1}} cdots partial x^{i_{k}}}$ is a tensor.












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 21:20









Lorenzo B.

1,8602520




1,8602520










asked Dec 17 '18 at 14:14









openspaceopenspace

3,4512822




3,4512822








  • 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




    $begingroup$
    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












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