Why are $C^infty_pneq C^infty_q$ when $pneq q$?
$begingroup$
Let $C^infty_p$ be the set of all germs of $C^infty(mathbb{R}^n)$ at point $p$.
Why do we always have $C^infty_pneq C^infty_q$ when $pneq q$?
I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.
differential-geometry differential-topology
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
$endgroup$
add a comment |
$begingroup$
Let $C^infty_p$ be the set of all germs of $C^infty(mathbb{R}^n)$ at point $p$.
Why do we always have $C^infty_pneq C^infty_q$ when $pneq q$?
I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.
differential-geometry differential-topology
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
$endgroup$
$begingroup$
They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
$endgroup$
– Amitai Yuval
Nov 3 '18 at 13:14
$begingroup$
@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:42
add a comment |
$begingroup$
Let $C^infty_p$ be the set of all germs of $C^infty(mathbb{R}^n)$ at point $p$.
Why do we always have $C^infty_pneq C^infty_q$ when $pneq q$?
I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.
differential-geometry differential-topology
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
$endgroup$
Let $C^infty_p$ be the set of all germs of $C^infty(mathbb{R}^n)$ at point $p$.
Why do we always have $C^infty_pneq C^infty_q$ when $pneq q$?
I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.
differential-geometry differential-topology
differential-geometry differential-topology
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
asked Nov 3 '18 at 13:06
An old man in the sea.An old man in the sea.
1,61811134
1,61811134
$begingroup$
They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
$endgroup$
– Amitai Yuval
Nov 3 '18 at 13:14
$begingroup$
@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:42
add a comment |
$begingroup$
They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
$endgroup$
– Amitai Yuval
Nov 3 '18 at 13:14
$begingroup$
@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:42
$begingroup$
They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
$endgroup$
– Amitai Yuval
Nov 3 '18 at 13:14
$begingroup$
They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
$endgroup$
– Amitai Yuval
Nov 3 '18 at 13:14
$begingroup$
@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:42
$begingroup$
@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.
If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.
Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
$endgroup$
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
|
show 7 more comments
$begingroup$
They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.
For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.
If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.
Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
$endgroup$
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
|
show 7 more comments
$begingroup$
A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.
If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.
Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
$endgroup$
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
|
show 7 more comments
$begingroup$
A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.
If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.
Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
$endgroup$
A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.
If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.
Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
edited Dec 9 '18 at 6:07
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
answered Dec 9 '18 at 4:28
BrahadeeshBrahadeesh
6,24242361
6,24242361
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
|
show 7 more comments
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
You meant $W ni p$ ? Otherwise I don't see what you mean. If you meant $p in W$ then (thanks to $C^infty_c$) we can find a smooth function agreeing with $f$ around $p$ but disagreeing with $g$ at $q$, even if $f,g$ agree on $mathbb{R}^n$. Thus for analytic functions we have a different situation.
$endgroup$
– reuns
Dec 9 '18 at 5:28
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
@reuns I don't follow. What is $W$? And how do compactly supported functions come into the picture?
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:40
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
Without asking that a fixed $p in W$ then it is not an equivalence relation (not transitive) : we can find $f$ agreeing with $g$ somewhere and with $h$ elsewhere but $g$ doesn't agree with $h$ anywhere
$endgroup$
– reuns
Dec 9 '18 at 5:46
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
$begingroup$
@reuns I still don't follow. Neither the OP nor I have used the notation $W$, so I'm not sure what you mean. And you said "(thanks to $C_c^infty$)" in your previous comment, which I also do not follow.
$endgroup$
– Brahadeesh
Dec 9 '18 at 5:53
1
1
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
$begingroup$
I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)
$endgroup$
– An old man in the sea.
12 hours ago
|
show 7 more comments
$begingroup$
They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.
For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
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– Arthur
Nov 3 '18 at 15:18
add a comment |
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They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.
For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
add a comment |
$begingroup$
They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.
For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
$endgroup$
They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.
For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
edited Nov 3 '18 at 13:27
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
answered Nov 3 '18 at 13:16
ArthurArthur
114k7115197
114k7115197
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
add a comment |
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
Arthur, thanks for your answer. However, I'm not seeing how from the existence of a 'thinner' partition, we conclude both are very distinct...
$endgroup$
– An old man in the sea.
Nov 3 '18 at 14:45
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
$begingroup$
@Anopdmaninthesea. What "thinner partition"? I'm just giving an example of how independent the two partitions are of one another.
$endgroup$
– Arthur
Nov 3 '18 at 15:18
add a comment |
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They are isomorphic to one another, of course, but there are many different isomorphisms. Hence, given $fin C^infty_p$ and $gin C^infty_q$, there is no way to determine whether $f=g$ or not.
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– Amitai Yuval
Nov 3 '18 at 13:14
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@AmitaiYuval Thanks for your comment. So, you're saying that whenever there are 2 or more different isomorphisms, since there is no way to determine whether g=f, we just say that both sets are different?
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– An old man in the sea.
Nov 3 '18 at 14:42