Mixing
Which is the definition of the set of germs $C_p^{infty}(mathbb R^n)$? Does $C^{infty}(U)$ consist of germs...
$begingroup$
Which one is the set known as $C_p^{infty}(mathbb R^n)$?
The set of germs of smooth real-valued functions defined on $mathbb R^n$
The set of germs of smooth real-valued functions defined on a fixed open subset of $mathbb R^n$ that contains $p$
The set of germs of smooth real-valued functions defined on any open subsets of $mathbb R^n$ that contains $p$
My book sounds like it's saying (1) and then later (3). Is the language of the book actually identifying $C_p^{infty}(mathbb R^n)$ as (1) throughout?
It says
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on $mathbb R^n$ at $p$. $tag{7}$
I think this should be
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on open subsets of $mathbb R^n$ that contain $p$.
Later, my book talks about $C^{infty}(U)$ for an open subset $U$ of $mathbb R^n$. Which one is $C^{infty}(U)$?
The set of germs of smooth real-valued functions defined on $U$
The set of germs of smooth real-valued functions defined on a fixed open subset of $U$ (which in turn is an open subset of $mathbb R^n$)
The set of germs of smooth real-valued functions defined on any open subsets of $U$ (which in turn is an open subset of $mathbb R^n$), thus functions from different germs may have disjoint domains.
My book says
The ring of $C^{infty}$ functions on an open set $U$ is commonly denoted by $C^{infty}(U)$
If $C^{infty}(U)$ is (4) instead of (6), then I think $C_p^{infty}(mathbb R^n)$ should be (1) instead of (3) because of the language in $(7)$.
Edit: Actually, does $C^{infty}(U)$ still consist of germs rather than functions?
My book is An Introduction to Manifolds by Loring W. Tu.
differential-geometry smooth-manifolds smooth-functions germs
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
$endgroup$
|
show 5 more comments
$begingroup$
Which one is the set known as $C_p^{infty}(mathbb R^n)$?
The set of germs of smooth real-valued functions defined on $mathbb R^n$
The set of germs of smooth real-valued functions defined on a fixed open subset of $mathbb R^n$ that contains $p$
The set of germs of smooth real-valued functions defined on any open subsets of $mathbb R^n$ that contains $p$
My book sounds like it's saying (1) and then later (3). Is the language of the book actually identifying $C_p^{infty}(mathbb R^n)$ as (1) throughout?
It says
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on $mathbb R^n$ at $p$. $tag{7}$
I think this should be
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on open subsets of $mathbb R^n$ that contain $p$.
Later, my book talks about $C^{infty}(U)$ for an open subset $U$ of $mathbb R^n$. Which one is $C^{infty}(U)$?
The set of germs of smooth real-valued functions defined on $U$
The set of germs of smooth real-valued functions defined on a fixed open subset of $U$ (which in turn is an open subset of $mathbb R^n$)
The set of germs of smooth real-valued functions defined on any open subsets of $U$ (which in turn is an open subset of $mathbb R^n$), thus functions from different germs may have disjoint domains.
My book says
The ring of $C^{infty}$ functions on an open set $U$ is commonly denoted by $C^{infty}(U)$
If $C^{infty}(U)$ is (4) instead of (6), then I think $C_p^{infty}(mathbb R^n)$ should be (1) instead of (3) because of the language in $(7)$.
Edit: Actually, does $C^{infty}(U)$ still consist of germs rather than functions?
My book is An Introduction to Manifolds by Loring W. Tu.
differential-geometry smooth-manifolds smooth-functions germs
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
$endgroup$
1
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
1
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
1
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
1
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51
|
show 5 more comments
$begingroup$
Which one is the set known as $C_p^{infty}(mathbb R^n)$?
The set of germs of smooth real-valued functions defined on $mathbb R^n$
The set of germs of smooth real-valued functions defined on a fixed open subset of $mathbb R^n$ that contains $p$
The set of germs of smooth real-valued functions defined on any open subsets of $mathbb R^n$ that contains $p$
My book sounds like it's saying (1) and then later (3). Is the language of the book actually identifying $C_p^{infty}(mathbb R^n)$ as (1) throughout?
It says
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on $mathbb R^n$ at $p$. $tag{7}$
I think this should be
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on open subsets of $mathbb R^n$ that contain $p$.
Later, my book talks about $C^{infty}(U)$ for an open subset $U$ of $mathbb R^n$. Which one is $C^{infty}(U)$?
The set of germs of smooth real-valued functions defined on $U$
The set of germs of smooth real-valued functions defined on a fixed open subset of $U$ (which in turn is an open subset of $mathbb R^n$)
The set of germs of smooth real-valued functions defined on any open subsets of $U$ (which in turn is an open subset of $mathbb R^n$), thus functions from different germs may have disjoint domains.
My book says
The ring of $C^{infty}$ functions on an open set $U$ is commonly denoted by $C^{infty}(U)$
If $C^{infty}(U)$ is (4) instead of (6), then I think $C_p^{infty}(mathbb R^n)$ should be (1) instead of (3) because of the language in $(7)$.
Edit: Actually, does $C^{infty}(U)$ still consist of germs rather than functions?
My book is An Introduction to Manifolds by Loring W. Tu.
differential-geometry smooth-manifolds smooth-functions germs
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
$endgroup$
Which one is the set known as $C_p^{infty}(mathbb R^n)$?
The set of germs of smooth real-valued functions defined on $mathbb R^n$
The set of germs of smooth real-valued functions defined on a fixed open subset of $mathbb R^n$ that contains $p$
The set of germs of smooth real-valued functions defined on any open subsets of $mathbb R^n$ that contains $p$
My book sounds like it's saying (1) and then later (3). Is the language of the book actually identifying $C_p^{infty}(mathbb R^n)$ as (1) throughout?
It says
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on $mathbb R^n$ at $p$. $tag{7}$
I think this should be
We write $C_p^{infty}(mathbb R^n)$, or simply $C_p^{infty}$ if there is no possibility of confusion, for the set of all germs of $C^{infty}$ functions on open subsets of $mathbb R^n$ that contain $p$.
Later, my book talks about $C^{infty}(U)$ for an open subset $U$ of $mathbb R^n$. Which one is $C^{infty}(U)$?
The set of germs of smooth real-valued functions defined on $U$
The set of germs of smooth real-valued functions defined on a fixed open subset of $U$ (which in turn is an open subset of $mathbb R^n$)
The set of germs of smooth real-valued functions defined on any open subsets of $U$ (which in turn is an open subset of $mathbb R^n$), thus functions from different germs may have disjoint domains.
My book says
The ring of $C^{infty}$ functions on an open set $U$ is commonly denoted by $C^{infty}(U)$
If $C^{infty}(U)$ is (4) instead of (6), then I think $C_p^{infty}(mathbb R^n)$ should be (1) instead of (3) because of the language in $(7)$.
Edit: Actually, does $C^{infty}(U)$ still consist of germs rather than functions?
My book is An Introduction to Manifolds by Loring W. Tu.
differential-geometry smooth-manifolds smooth-functions germs
differential-geometry smooth-manifolds smooth-functions germs
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
edited Feb 19 at 4:52
Selene Auckland
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
asked Jan 26 at 4:50
Selene AucklandSelene Auckland
7911
7911
1
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
1
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
1
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
1
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51
|
show 5 more comments
1
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
1
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
1
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
1
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51
1
1
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
1
1
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
1
1
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
1
1
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First, $C^infty(U)$ is the set of all smooth functions defined on $U$, not the set of germs at $p$. The set of germs at $p in U$ is denoted by $C_p^infty(U)$.
For the rest, you are correct that the author is being a little lax in his terminology, but the reason he isn't bothering to be exact is that it makes no difference.
One gets an equivalent set of germs with all three definitions.
That is, if we have a point $p$, and a fixed open set $U_p$ with $pin U_p subseteq Bbb R^n$, and define the three equivalences as in the definition of germs for the three collections
$sim_1$ on $C^infty(Bbb R^n)$, with $C_1 = C^infty(Bbb R^n)/ sim_1$
$sim_2$ on $C^infty(U_p)$, with $C_2 = C^infty(U_p) / sim_2$
$sim_3$ on $mathscr F = {f mid V in mathscr O(Bbb R^n) wedge p in V wedge fin C^infty(V)}$, with $C_3 =mathscr F / sim_3$
Then there is a natural one-to-one correspondence between the three sets of germs. Any $f in C^infty(Bbb R^n)$ also is a member of $mathscr F$, and the restriction $f|_{U_p}$ is in $C^infty(U_p)$. And if $g$ is another such function, then it is obvious that
$$f sim_1 g iff f|_{U_p} sim_2 g|_{U_p} iff f sim_3 g$$
which induces injections of $C_1$ into $C_2$ and $C_3$. Similarly, $C^infty(U_p) subseteq mathscr F$, which also induces an injection of $C_2$ into $C_3$. However, for any $f in mathscr F$, it is not hard to show that for some $g sim_3 f, g$ is the restriction of some $g' in C^infty(Bbb R^n)$. This induces an injection of $C_3$ into $C_1$, which is the inverse of the injection of $C_1 to C_3$.
Since the elements of $C_1, C_2, C_3$ are all naturally identifiable with each other, we can consider any of the sets to be the set of germs at $p$.
Now by careful reading of the text, it is apparent that Prof. Tu is actually defining the set of germs $C_p^infty(Bbb R^n)$ to be $C_3$, the one you describe. But because he knows it doesn't matter which is used, he got a little careless in his wording when he actually introduces the set.
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
add a comment |
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$begingroup$
First, $C^infty(U)$ is the set of all smooth functions defined on $U$, not the set of germs at $p$. The set of germs at $p in U$ is denoted by $C_p^infty(U)$.
For the rest, you are correct that the author is being a little lax in his terminology, but the reason he isn't bothering to be exact is that it makes no difference.
One gets an equivalent set of germs with all three definitions.
That is, if we have a point $p$, and a fixed open set $U_p$ with $pin U_p subseteq Bbb R^n$, and define the three equivalences as in the definition of germs for the three collections
$sim_1$ on $C^infty(Bbb R^n)$, with $C_1 = C^infty(Bbb R^n)/ sim_1$
$sim_2$ on $C^infty(U_p)$, with $C_2 = C^infty(U_p) / sim_2$
$sim_3$ on $mathscr F = {f mid V in mathscr O(Bbb R^n) wedge p in V wedge fin C^infty(V)}$, with $C_3 =mathscr F / sim_3$
Then there is a natural one-to-one correspondence between the three sets of germs. Any $f in C^infty(Bbb R^n)$ also is a member of $mathscr F$, and the restriction $f|_{U_p}$ is in $C^infty(U_p)$. And if $g$ is another such function, then it is obvious that
$$f sim_1 g iff f|_{U_p} sim_2 g|_{U_p} iff f sim_3 g$$
which induces injections of $C_1$ into $C_2$ and $C_3$. Similarly, $C^infty(U_p) subseteq mathscr F$, which also induces an injection of $C_2$ into $C_3$. However, for any $f in mathscr F$, it is not hard to show that for some $g sim_3 f, g$ is the restriction of some $g' in C^infty(Bbb R^n)$. This induces an injection of $C_3$ into $C_1$, which is the inverse of the injection of $C_1 to C_3$.
Since the elements of $C_1, C_2, C_3$ are all naturally identifiable with each other, we can consider any of the sets to be the set of germs at $p$.
Now by careful reading of the text, it is apparent that Prof. Tu is actually defining the set of germs $C_p^infty(Bbb R^n)$ to be $C_3$, the one you describe. But because he knows it doesn't matter which is used, he got a little careless in his wording when he actually introduces the set.
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
add a comment |
$begingroup$
First, $C^infty(U)$ is the set of all smooth functions defined on $U$, not the set of germs at $p$. The set of germs at $p in U$ is denoted by $C_p^infty(U)$.
For the rest, you are correct that the author is being a little lax in his terminology, but the reason he isn't bothering to be exact is that it makes no difference.
One gets an equivalent set of germs with all three definitions.
That is, if we have a point $p$, and a fixed open set $U_p$ with $pin U_p subseteq Bbb R^n$, and define the three equivalences as in the definition of germs for the three collections
$sim_1$ on $C^infty(Bbb R^n)$, with $C_1 = C^infty(Bbb R^n)/ sim_1$
$sim_2$ on $C^infty(U_p)$, with $C_2 = C^infty(U_p) / sim_2$
$sim_3$ on $mathscr F = {f mid V in mathscr O(Bbb R^n) wedge p in V wedge fin C^infty(V)}$, with $C_3 =mathscr F / sim_3$
Then there is a natural one-to-one correspondence between the three sets of germs. Any $f in C^infty(Bbb R^n)$ also is a member of $mathscr F$, and the restriction $f|_{U_p}$ is in $C^infty(U_p)$. And if $g$ is another such function, then it is obvious that
$$f sim_1 g iff f|_{U_p} sim_2 g|_{U_p} iff f sim_3 g$$
which induces injections of $C_1$ into $C_2$ and $C_3$. Similarly, $C^infty(U_p) subseteq mathscr F$, which also induces an injection of $C_2$ into $C_3$. However, for any $f in mathscr F$, it is not hard to show that for some $g sim_3 f, g$ is the restriction of some $g' in C^infty(Bbb R^n)$. This induces an injection of $C_3$ into $C_1$, which is the inverse of the injection of $C_1 to C_3$.
Since the elements of $C_1, C_2, C_3$ are all naturally identifiable with each other, we can consider any of the sets to be the set of germs at $p$.
Now by careful reading of the text, it is apparent that Prof. Tu is actually defining the set of germs $C_p^infty(Bbb R^n)$ to be $C_3$, the one you describe. But because he knows it doesn't matter which is used, he got a little careless in his wording when he actually introduces the set.
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
add a comment |
$begingroup$
First, $C^infty(U)$ is the set of all smooth functions defined on $U$, not the set of germs at $p$. The set of germs at $p in U$ is denoted by $C_p^infty(U)$.
For the rest, you are correct that the author is being a little lax in his terminology, but the reason he isn't bothering to be exact is that it makes no difference.
One gets an equivalent set of germs with all three definitions.
That is, if we have a point $p$, and a fixed open set $U_p$ with $pin U_p subseteq Bbb R^n$, and define the three equivalences as in the definition of germs for the three collections
$sim_1$ on $C^infty(Bbb R^n)$, with $C_1 = C^infty(Bbb R^n)/ sim_1$
$sim_2$ on $C^infty(U_p)$, with $C_2 = C^infty(U_p) / sim_2$
$sim_3$ on $mathscr F = {f mid V in mathscr O(Bbb R^n) wedge p in V wedge fin C^infty(V)}$, with $C_3 =mathscr F / sim_3$
Then there is a natural one-to-one correspondence between the three sets of germs. Any $f in C^infty(Bbb R^n)$ also is a member of $mathscr F$, and the restriction $f|_{U_p}$ is in $C^infty(U_p)$. And if $g$ is another such function, then it is obvious that
$$f sim_1 g iff f|_{U_p} sim_2 g|_{U_p} iff f sim_3 g$$
which induces injections of $C_1$ into $C_2$ and $C_3$. Similarly, $C^infty(U_p) subseteq mathscr F$, which also induces an injection of $C_2$ into $C_3$. However, for any $f in mathscr F$, it is not hard to show that for some $g sim_3 f, g$ is the restriction of some $g' in C^infty(Bbb R^n)$. This induces an injection of $C_3$ into $C_1$, which is the inverse of the injection of $C_1 to C_3$.
Since the elements of $C_1, C_2, C_3$ are all naturally identifiable with each other, we can consider any of the sets to be the set of germs at $p$.
Now by careful reading of the text, it is apparent that Prof. Tu is actually defining the set of germs $C_p^infty(Bbb R^n)$ to be $C_3$, the one you describe. But because he knows it doesn't matter which is used, he got a little careless in his wording when he actually introduces the set.
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
First, $C^infty(U)$ is the set of all smooth functions defined on $U$, not the set of germs at $p$. The set of germs at $p in U$ is denoted by $C_p^infty(U)$.
For the rest, you are correct that the author is being a little lax in his terminology, but the reason he isn't bothering to be exact is that it makes no difference.
One gets an equivalent set of germs with all three definitions.
That is, if we have a point $p$, and a fixed open set $U_p$ with $pin U_p subseteq Bbb R^n$, and define the three equivalences as in the definition of germs for the three collections
$sim_1$ on $C^infty(Bbb R^n)$, with $C_1 = C^infty(Bbb R^n)/ sim_1$
$sim_2$ on $C^infty(U_p)$, with $C_2 = C^infty(U_p) / sim_2$
$sim_3$ on $mathscr F = {f mid V in mathscr O(Bbb R^n) wedge p in V wedge fin C^infty(V)}$, with $C_3 =mathscr F / sim_3$
Then there is a natural one-to-one correspondence between the three sets of germs. Any $f in C^infty(Bbb R^n)$ also is a member of $mathscr F$, and the restriction $f|_{U_p}$ is in $C^infty(U_p)$. And if $g$ is another such function, then it is obvious that
$$f sim_1 g iff f|_{U_p} sim_2 g|_{U_p} iff f sim_3 g$$
which induces injections of $C_1$ into $C_2$ and $C_3$. Similarly, $C^infty(U_p) subseteq mathscr F$, which also induces an injection of $C_2$ into $C_3$. However, for any $f in mathscr F$, it is not hard to show that for some $g sim_3 f, g$ is the restriction of some $g' in C^infty(Bbb R^n)$. This induces an injection of $C_3$ into $C_1$, which is the inverse of the injection of $C_1 to C_3$.
Since the elements of $C_1, C_2, C_3$ are all naturally identifiable with each other, we can consider any of the sets to be the set of germs at $p$.
Now by careful reading of the text, it is apparent that Prof. Tu is actually defining the set of germs $C_p^infty(Bbb R^n)$ to be $C_3$, the one you describe. But because he knows it doesn't matter which is used, he got a little careless in his wording when he actually introduces the set.
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 26 at 17:59
Paul SinclairPaul Sinclair
20.5k21543
20.5k21543
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
add a comment |
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
$begingroup$
Thanks. You said "it is not hard". How do you do this without bump functions which are not introduced until 11 sections later?
$endgroup$
– Selene Auckland
Feb 19 at 2:42
1
1
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
$begingroup$
I assume by "bump functions" you mean functions of compact support? Or partitions of unity? Why would I do this without using such functions? They exist, no matter when you're told about them. I was just explaining why Prof. Tu was careless with his language. He was well aware of the equivalences I discussed. Authors often choose $C_1$ for their definition of germs instead of $C_3$. He chose $C_3$, but then carelessly used wording suggestive of $C_1$ It was a minor mistake on his part, not something readers were supposed to follow and understand.
$endgroup$
– Paul Sinclair
Feb 19 at 3:06
add a comment |
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1
$begingroup$
All definitions are equivalent.
$endgroup$
– Dante Grevino
Jan 26 at 5:13
$begingroup$
@DanteGrevino Thanks! Can you prove without bump functions?
$endgroup$
– Selene Auckland
Jan 26 at 8:46
1
$begingroup$
In the three definitions there are germs of functions instead of functions. What is your definition of germ? For me it is clear from definition of germ that the three definitions of stalk in $p$ agree. (Two germ of functions are equal if and only if the functions are equal in some open neghborhood around $p$.)
$endgroup$
– Dante Grevino
Jan 26 at 13:47
1
$begingroup$
I mean on one hand 1. 2. 3. are equivalent. And on the other hand $C^infty(U)$ is the set of smooth functions on $U$ and NOT germs.
$endgroup$
– Dante Grevino
Jan 26 at 13:52
1
$begingroup$
@DanteGrevino Thanks!
$endgroup$
– Selene Auckland
Feb 19 at 4:51