Mixing
Germs: Why is it sensible to define a function on a collection of equivalence classes by its action on each...
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I am following Loring W. Tu in his second edition of 'An introduction to manifolds'. Here is a pdf-copy of the book. On page 87 he defines $C^infty_p(M)$ as the set of germs of $C^infty$-functions at $pin M$. He then defines a derivation as a map $D : C^infty_p(M) to mathbb{R}$. In contrast, half way down the page he defines the partial derivative in coordinates as a map $partial/partial x^i|_p : C^infty(M) to mathbb{R}$, and claims that it is easy to check that this is a derivation.
I do not understand this because $partial/partial x^i|_p$ cannot be a derivation when it's domain is $C^infty(M)$ and not $C^infty_p(M)$. He also casually seems to talk of functions as if they are equivalence classes. To me it seems too simple that the author has mistaken the set of equivalence classes for its elements. Is there a principle regarding equivalence classes or germs that are being applied implicitly that I am missing? Could you explain please?
smooth-manifolds equivalence-relations tangent-spaces germs
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
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add a comment |
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I am following Loring W. Tu in his second edition of 'An introduction to manifolds'. Here is a pdf-copy of the book. On page 87 he defines $C^infty_p(M)$ as the set of germs of $C^infty$-functions at $pin M$. He then defines a derivation as a map $D : C^infty_p(M) to mathbb{R}$. In contrast, half way down the page he defines the partial derivative in coordinates as a map $partial/partial x^i|_p : C^infty(M) to mathbb{R}$, and claims that it is easy to check that this is a derivation.
I do not understand this because $partial/partial x^i|_p$ cannot be a derivation when it's domain is $C^infty(M)$ and not $C^infty_p(M)$. He also casually seems to talk of functions as if they are equivalence classes. To me it seems too simple that the author has mistaken the set of equivalence classes for its elements. Is there a principle regarding equivalence classes or germs that are being applied implicitly that I am missing? Could you explain please?
smooth-manifolds equivalence-relations tangent-spaces germs
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
$endgroup$
1
$begingroup$
Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39
add a comment |
$begingroup$
I am following Loring W. Tu in his second edition of 'An introduction to manifolds'. Here is a pdf-copy of the book. On page 87 he defines $C^infty_p(M)$ as the set of germs of $C^infty$-functions at $pin M$. He then defines a derivation as a map $D : C^infty_p(M) to mathbb{R}$. In contrast, half way down the page he defines the partial derivative in coordinates as a map $partial/partial x^i|_p : C^infty(M) to mathbb{R}$, and claims that it is easy to check that this is a derivation.
I do not understand this because $partial/partial x^i|_p$ cannot be a derivation when it's domain is $C^infty(M)$ and not $C^infty_p(M)$. He also casually seems to talk of functions as if they are equivalence classes. To me it seems too simple that the author has mistaken the set of equivalence classes for its elements. Is there a principle regarding equivalence classes or germs that are being applied implicitly that I am missing? Could you explain please?
smooth-manifolds equivalence-relations tangent-spaces germs
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
$endgroup$
I am following Loring W. Tu in his second edition of 'An introduction to manifolds'. Here is a pdf-copy of the book. On page 87 he defines $C^infty_p(M)$ as the set of germs of $C^infty$-functions at $pin M$. He then defines a derivation as a map $D : C^infty_p(M) to mathbb{R}$. In contrast, half way down the page he defines the partial derivative in coordinates as a map $partial/partial x^i|_p : C^infty(M) to mathbb{R}$, and claims that it is easy to check that this is a derivation.
I do not understand this because $partial/partial x^i|_p$ cannot be a derivation when it's domain is $C^infty(M)$ and not $C^infty_p(M)$. He also casually seems to talk of functions as if they are equivalence classes. To me it seems too simple that the author has mistaken the set of equivalence classes for its elements. Is there a principle regarding equivalence classes or germs that are being applied implicitly that I am missing? Could you explain please?
smooth-manifolds equivalence-relations tangent-spaces germs
smooth-manifolds equivalence-relations tangent-spaces germs
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
edited Feb 2 at 21:24
Mikkel Rev
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
asked Feb 2 at 19:53
Mikkel RevMikkel Rev
705414
705414
1
$begingroup$
Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39
add a comment |
1
$begingroup$
Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39
1
1
$begingroup$
Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39
$begingroup$
Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39
add a comment |
2 Answers
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The key idea is that germs of functions are those on which the partials agree, no matter which function from the germ you choose.
Let me give a very concrete example: we can say that for functions from the real to the reals, there's an equivalence relation: $f$ and $g$ are equivalent if and only if $f(0) = g(0)$.
Now I define the "zero-square" $Z(f)$ of a function $f$ to be $f(0)^2$. Then I claim that a very slight generalization of "zero-square" makes sense as a function on equivalence classes. Why? Because if $fsim g$, then $Z(f) = f(0)^2 = g(0)^2= Z(g)$, so $Z$, evaluated on ANY member of an equivalence class, gets you the same number. We say that this slight generalization (which usually is given some typographically similar name, like $bar{Z}$, but is sometimes just denoted with the same letter (alas) is "well defined on equivalence classes".
That's what Tu is doing here. The partial-derivative-at-$p$ map acts on all $C^infty$ functions defined in a neighborhood of $p$; for any two elements of the same germ at $p$, you get the same result. Hence the partial-derivative-at-$p$ map ends up being "well defined on germs".
You're right that this is, to some degree, an abomination -- it's yet another of those mathematical puns, like writing $dz/dx = dz/du cdot du/dx$, where the $z$ on the left and the $z$ on the right are two completely different functions. But it's also really commonplace...
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
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add a comment |
$begingroup$
In general if $mathcal A$ is a commutative algebra and $M$ is an $mathcal A$-module, we will say that a linear map $D : mathcal A to M$ is a derivation if $D(fg)=f Dg + g Df $ for $f, g in mathcal A$. Operation $left. frac{partial}{partial x^i} right|_p$ satisfies this property, no matter if we consider $C^{infty}(M)$ or the space of germs as its domain.
Secondly, given an element of $C^{infty}(M)$ we can always consider its germ at $p$. Even though partial derivative at $p$ was initially defined on functions on $M$, its value actually depends only on the germ at $p$. Moreover given any germ at $p$ you can always extend it to a smooth function on whole $M$ using bump functions, so in the end when evaluating derivatives of any finite order at a point it doesn't make much difference if you work with functions on $M$, functions on some fixed neighbourhood of $p$ or germs at $p$.
answered Feb 2 at 21:39
BlazejBlazej
1,632620
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add a comment |
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The key idea is that germs of functions are those on which the partials agree, no matter which function from the germ you choose.
Let me give a very concrete example: we can say that for functions from the real to the reals, there's an equivalence relation: $f$ and $g$ are equivalent if and only if $f(0) = g(0)$.
Now I define the "zero-square" $Z(f)$ of a function $f$ to be $f(0)^2$. Then I claim that a very slight generalization of "zero-square" makes sense as a function on equivalence classes. Why? Because if $fsim g$, then $Z(f) = f(0)^2 = g(0)^2= Z(g)$, so $Z$, evaluated on ANY member of an equivalence class, gets you the same number. We say that this slight generalization (which usually is given some typographically similar name, like $bar{Z}$, but is sometimes just denoted with the same letter (alas) is "well defined on equivalence classes".
That's what Tu is doing here. The partial-derivative-at-$p$ map acts on all $C^infty$ functions defined in a neighborhood of $p$; for any two elements of the same germ at $p$, you get the same result. Hence the partial-derivative-at-$p$ map ends up being "well defined on germs".
You're right that this is, to some degree, an abomination -- it's yet another of those mathematical puns, like writing $dz/dx = dz/du cdot du/dx$, where the $z$ on the left and the $z$ on the right are two completely different functions. But it's also really commonplace...
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
$endgroup$
add a comment |
$begingroup$
The key idea is that germs of functions are those on which the partials agree, no matter which function from the germ you choose.
Let me give a very concrete example: we can say that for functions from the real to the reals, there's an equivalence relation: $f$ and $g$ are equivalent if and only if $f(0) = g(0)$.
Now I define the "zero-square" $Z(f)$ of a function $f$ to be $f(0)^2$. Then I claim that a very slight generalization of "zero-square" makes sense as a function on equivalence classes. Why? Because if $fsim g$, then $Z(f) = f(0)^2 = g(0)^2= Z(g)$, so $Z$, evaluated on ANY member of an equivalence class, gets you the same number. We say that this slight generalization (which usually is given some typographically similar name, like $bar{Z}$, but is sometimes just denoted with the same letter (alas) is "well defined on equivalence classes".
That's what Tu is doing here. The partial-derivative-at-$p$ map acts on all $C^infty$ functions defined in a neighborhood of $p$; for any two elements of the same germ at $p$, you get the same result. Hence the partial-derivative-at-$p$ map ends up being "well defined on germs".
You're right that this is, to some degree, an abomination -- it's yet another of those mathematical puns, like writing $dz/dx = dz/du cdot du/dx$, where the $z$ on the left and the $z$ on the right are two completely different functions. But it's also really commonplace...
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
$endgroup$
add a comment |
$begingroup$
The key idea is that germs of functions are those on which the partials agree, no matter which function from the germ you choose.
Let me give a very concrete example: we can say that for functions from the real to the reals, there's an equivalence relation: $f$ and $g$ are equivalent if and only if $f(0) = g(0)$.
Now I define the "zero-square" $Z(f)$ of a function $f$ to be $f(0)^2$. Then I claim that a very slight generalization of "zero-square" makes sense as a function on equivalence classes. Why? Because if $fsim g$, then $Z(f) = f(0)^2 = g(0)^2= Z(g)$, so $Z$, evaluated on ANY member of an equivalence class, gets you the same number. We say that this slight generalization (which usually is given some typographically similar name, like $bar{Z}$, but is sometimes just denoted with the same letter (alas) is "well defined on equivalence classes".
That's what Tu is doing here. The partial-derivative-at-$p$ map acts on all $C^infty$ functions defined in a neighborhood of $p$; for any two elements of the same germ at $p$, you get the same result. Hence the partial-derivative-at-$p$ map ends up being "well defined on germs".
You're right that this is, to some degree, an abomination -- it's yet another of those mathematical puns, like writing $dz/dx = dz/du cdot du/dx$, where the $z$ on the left and the $z$ on the right are two completely different functions. But it's also really commonplace...
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
$endgroup$
The key idea is that germs of functions are those on which the partials agree, no matter which function from the germ you choose.
Let me give a very concrete example: we can say that for functions from the real to the reals, there's an equivalence relation: $f$ and $g$ are equivalent if and only if $f(0) = g(0)$.
Now I define the "zero-square" $Z(f)$ of a function $f$ to be $f(0)^2$. Then I claim that a very slight generalization of "zero-square" makes sense as a function on equivalence classes. Why? Because if $fsim g$, then $Z(f) = f(0)^2 = g(0)^2= Z(g)$, so $Z$, evaluated on ANY member of an equivalence class, gets you the same number. We say that this slight generalization (which usually is given some typographically similar name, like $bar{Z}$, but is sometimes just denoted with the same letter (alas) is "well defined on equivalence classes".
That's what Tu is doing here. The partial-derivative-at-$p$ map acts on all $C^infty$ functions defined in a neighborhood of $p$; for any two elements of the same germ at $p$, you get the same result. Hence the partial-derivative-at-$p$ map ends up being "well defined on germs".
You're right that this is, to some degree, an abomination -- it's yet another of those mathematical puns, like writing $dz/dx = dz/du cdot du/dx$, where the $z$ on the left and the $z$ on the right are two completely different functions. But it's also really commonplace...
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
edited Feb 2 at 22:41
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
answered Feb 2 at 21:40
John HughesJohn Hughes
65.5k24293
65.5k24293
add a comment |
add a comment |
$begingroup$
In general if $mathcal A$ is a commutative algebra and $M$ is an $mathcal A$-module, we will say that a linear map $D : mathcal A to M$ is a derivation if $D(fg)=f Dg + g Df $ for $f, g in mathcal A$. Operation $left. frac{partial}{partial x^i} right|_p$ satisfies this property, no matter if we consider $C^{infty}(M)$ or the space of germs as its domain.
Secondly, given an element of $C^{infty}(M)$ we can always consider its germ at $p$. Even though partial derivative at $p$ was initially defined on functions on $M$, its value actually depends only on the germ at $p$. Moreover given any germ at $p$ you can always extend it to a smooth function on whole $M$ using bump functions, so in the end when evaluating derivatives of any finite order at a point it doesn't make much difference if you work with functions on $M$, functions on some fixed neighbourhood of $p$ or germs at $p$.
answered Feb 2 at 21:39
BlazejBlazej
1,632620
$endgroup$
add a comment |
$begingroup$
In general if $mathcal A$ is a commutative algebra and $M$ is an $mathcal A$-module, we will say that a linear map $D : mathcal A to M$ is a derivation if $D(fg)=f Dg + g Df $ for $f, g in mathcal A$. Operation $left. frac{partial}{partial x^i} right|_p$ satisfies this property, no matter if we consider $C^{infty}(M)$ or the space of germs as its domain.
Secondly, given an element of $C^{infty}(M)$ we can always consider its germ at $p$. Even though partial derivative at $p$ was initially defined on functions on $M$, its value actually depends only on the germ at $p$. Moreover given any germ at $p$ you can always extend it to a smooth function on whole $M$ using bump functions, so in the end when evaluating derivatives of any finite order at a point it doesn't make much difference if you work with functions on $M$, functions on some fixed neighbourhood of $p$ or germs at $p$.
answered Feb 2 at 21:39
BlazejBlazej
1,632620
$endgroup$
add a comment |
$begingroup$
In general if $mathcal A$ is a commutative algebra and $M$ is an $mathcal A$-module, we will say that a linear map $D : mathcal A to M$ is a derivation if $D(fg)=f Dg + g Df $ for $f, g in mathcal A$. Operation $left. frac{partial}{partial x^i} right|_p$ satisfies this property, no matter if we consider $C^{infty}(M)$ or the space of germs as its domain.
Secondly, given an element of $C^{infty}(M)$ we can always consider its germ at $p$. Even though partial derivative at $p$ was initially defined on functions on $M$, its value actually depends only on the germ at $p$. Moreover given any germ at $p$ you can always extend it to a smooth function on whole $M$ using bump functions, so in the end when evaluating derivatives of any finite order at a point it doesn't make much difference if you work with functions on $M$, functions on some fixed neighbourhood of $p$ or germs at $p$.
answered Feb 2 at 21:39
BlazejBlazej
1,632620
$endgroup$
In general if $mathcal A$ is a commutative algebra and $M$ is an $mathcal A$-module, we will say that a linear map $D : mathcal A to M$ is a derivation if $D(fg)=f Dg + g Df $ for $f, g in mathcal A$. Operation $left. frac{partial}{partial x^i} right|_p$ satisfies this property, no matter if we consider $C^{infty}(M)$ or the space of germs as its domain.
Secondly, given an element of $C^{infty}(M)$ we can always consider its germ at $p$. Even though partial derivative at $p$ was initially defined on functions on $M$, its value actually depends only on the germ at $p$. Moreover given any germ at $p$ you can always extend it to a smooth function on whole $M$ using bump functions, so in the end when evaluating derivatives of any finite order at a point it doesn't make much difference if you work with functions on $M$, functions on some fixed neighbourhood of $p$ or germs at $p$.
answered Feb 2 at 21:39
BlazejBlazej
1,632620
answered Feb 2 at 21:39
BlazejBlazej
1,632620
answered Feb 2 at 21:39
BlazejBlazej
1,632620
answered Feb 2 at 21:39
BlazejBlazej
1,632620
1,632620
add a comment |
add a comment |
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Are you sure he defines the partial derivative as acting on $C^infty(M)$? He says $f$ is s smooth function in a nhood of $p$. On a different note, it’s standard in maths to identify an equivalence class with one of its elements/representatives (why would this be so?) and then use this element throughout to denote the equivalence class.
$endgroup$
– g.s
Feb 2 at 21:39