Mixing
When is an inclusion map smooth?
$begingroup$
- If $B$ is a manifold, and $Asubseteq B$ is a regular submanifold, then the inclusion map $i:Ato B$ is an embedding and thus smooth.
That $i$ is an embedding seems a bit strong. Is there another way to do this?
I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A to B$ is continuous (assuming subspace topology), but when is $i:A to B$ smooth?
Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A to A times B, i_b(a)=a times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.
Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.
Thanks in advance!
differential-geometry submanifold
edited Jan 20 at 9:25
user26857
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
|
show 1 more comment
$begingroup$
- If $B$ is a manifold, and $Asubseteq B$ is a regular submanifold, then the inclusion map $i:Ato B$ is an embedding and thus smooth.
That $i$ is an embedding seems a bit strong. Is there another way to do this?
I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A to B$ is continuous (assuming subspace topology), but when is $i:A to B$ smooth?
Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A to A times B, i_b(a)=a times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.
Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.
Thanks in advance!
differential-geometry submanifold
edited Jan 20 at 9:25
user26857
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
1
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
1
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
1
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
1
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16
|
show 1 more comment
$begingroup$
- If $B$ is a manifold, and $Asubseteq B$ is a regular submanifold, then the inclusion map $i:Ato B$ is an embedding and thus smooth.
That $i$ is an embedding seems a bit strong. Is there another way to do this?
I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A to B$ is continuous (assuming subspace topology), but when is $i:A to B$ smooth?
Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A to A times B, i_b(a)=a times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.
Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.
Thanks in advance!
differential-geometry submanifold
edited Jan 20 at 9:25
user26857
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
- If $B$ is a manifold, and $Asubseteq B$ is a regular submanifold, then the inclusion map $i:Ato B$ is an embedding and thus smooth.
That $i$ is an embedding seems a bit strong. Is there another way to do this?
I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A to B$ is continuous (assuming subspace topology), but when is $i:A to B$ smooth?
Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A to A times B, i_b(a)=a times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.
Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.
Thanks in advance!
differential-geometry submanifold
differential-geometry submanifold
edited Jan 20 at 9:25
user26857
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 20 at 9:25
user26857
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 20 at 9:25
user26857
39.3k124183
edited Jan 20 at 9:25
user26857
39.3k124183
edited Jan 20 at 9:25
user26857
39.3k124183
39.3k124183
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 18 at 11:17
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
558
1
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
1
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
1
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
1
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16
|
show 1 more comment
1
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
1
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
1
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
1
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16
1
1
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
1
1
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
1
1
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
1
1
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16
|
show 1 more comment
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1
$begingroup$
The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd.
$endgroup$
– Yanko
Jan 18 at 12:09
$begingroup$
@Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A to B$ for $A$ and $B$ smooth manifolds and $A subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 19 at 17:08
1
$begingroup$
I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:31
1
$begingroup$
@Yanko: math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though.
$endgroup$
– Giuseppe Negro
Jan 19 at 17:32
1
$begingroup$
"Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately.
$endgroup$
– Giuseppe Negro
Jan 19 at 18:16