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Are properties of maps between manifolds seen in $mathbbC^n$ or in $mathbbR^n$ equivalent?
Are definitions for smooth map between manifolds are equivalent?Second (and higher) derivatives of maps between manifoldsequivalent characterizations of differentiability of maps between smooth manifolds (a la Warner)Smooth functions between manifolds and subsets of manifoldssmooth manifolds, equivalent statementsChange of variables formula for diffeomorphisms between manifoldsAre diffeomorphic smooth manifolds truly equivalent?smooth submersions and maps with local sectionsDefinition and clarification of submanifold, immersion and submersionCriterion for smoothness of maps between submanifolds of euclidean spaces
$begingroup$
Let me elaborate:
By properties, I mean qualities such as "is a submersion", "is an immersion", "is an embedding", etc.
Let's say I have a function $f:S^1 rightarrow S^3$ between the spheres viewed as smooth manifolds in $mathbbC$ and $mathbbC^2$ respectively.
In class we saw that to prove for example that $f$ is a submersion, we should first define maps (or "charts") $varphi:mathbbR^2rightarrow S_1$, $psi:S^3rightarrow mathbbR^4$ and check that the differential $D_x(psi circ f circ varphi)$ is onto.
In other words, to show $f$ has one of the properties, we first have to view it as a function from $mathbbR^2$ to $mathbbR^4$.
I would like to know if it is equivalent to view $f$ as a function from $mathbbC$ to $mathbbC^2$ instead when it makes things much more convenient and simply differentiate with respect to the complex variable as if it were real since it would simplify a lot the calculations (here $varphi$ and $psi$ would simply be identity! or rather, inclusion from the spheres to the complex plane)
In particular, would this work for all complex functions? Or only those that are "analytic" (e.g. which do not include terms in $bar z$)?
To give a concrete example, let's consider:
$f(z)=(cz^n,c^m/ne^ipi/nz^m)$ where $c$ is real positive chosen such that $f$ is from $S_1$ to $S_3$ (i.e. if $|z|=1$ then $|f(z)|=1)$
Say I want to know whether it is a immersion. The "safe" way would be to convert $z=x+iy$ to $(x,y)$ in $mathbbR^2$ and get a function sending $(x,y)$ to $(a,b,c,d)$ from $mathbbR^2$ to $mathbbR^4$ where $a,b,c,d$ are functions of $x$ and $y$. But computing this could be tedious depending on the function involved, so can I simply compute the complex differential with respect to $z$ and determine whether it is injective? (from $mathbbC$ to $mathbbC^2$)
For example here I would get $D_zf(h)=(cnz^n-1,c^m/ne^ipi/nmz^m-1).h$ ($hin mathbbC)$
and $D_zf(h)=0 Leftrightarrow h=0$ assuming $c>0$ and $n>0$ (and $zneq0$ but here $zin S_1$ so it is automatically true)
so $KerD_zf=0$ and $f$ is an immersion. Is this correct? Would this still work if there were non-analytic terms (e.g. $bar z$ instead of $z$)?
differential-geometry manifolds smooth-manifolds
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
$endgroup$
add a comment |
$begingroup$
Let me elaborate:
By properties, I mean qualities such as "is a submersion", "is an immersion", "is an embedding", etc.
Let's say I have a function $f:S^1 rightarrow S^3$ between the spheres viewed as smooth manifolds in $mathbbC$ and $mathbbC^2$ respectively.
In class we saw that to prove for example that $f$ is a submersion, we should first define maps (or "charts") $varphi:mathbbR^2rightarrow S_1$, $psi:S^3rightarrow mathbbR^4$ and check that the differential $D_x(psi circ f circ varphi)$ is onto.
In other words, to show $f$ has one of the properties, we first have to view it as a function from $mathbbR^2$ to $mathbbR^4$.
I would like to know if it is equivalent to view $f$ as a function from $mathbbC$ to $mathbbC^2$ instead when it makes things much more convenient and simply differentiate with respect to the complex variable as if it were real since it would simplify a lot the calculations (here $varphi$ and $psi$ would simply be identity! or rather, inclusion from the spheres to the complex plane)
In particular, would this work for all complex functions? Or only those that are "analytic" (e.g. which do not include terms in $bar z$)?
To give a concrete example, let's consider:
$f(z)=(cz^n,c^m/ne^ipi/nz^m)$ where $c$ is real positive chosen such that $f$ is from $S_1$ to $S_3$ (i.e. if $|z|=1$ then $|f(z)|=1)$
Say I want to know whether it is a immersion. The "safe" way would be to convert $z=x+iy$ to $(x,y)$ in $mathbbR^2$ and get a function sending $(x,y)$ to $(a,b,c,d)$ from $mathbbR^2$ to $mathbbR^4$ where $a,b,c,d$ are functions of $x$ and $y$. But computing this could be tedious depending on the function involved, so can I simply compute the complex differential with respect to $z$ and determine whether it is injective? (from $mathbbC$ to $mathbbC^2$)
For example here I would get $D_zf(h)=(cnz^n-1,c^m/ne^ipi/nmz^m-1).h$ ($hin mathbbC)$
and $D_zf(h)=0 Leftrightarrow h=0$ assuming $c>0$ and $n>0$ (and $zneq0$ but here $zin S_1$ so it is automatically true)
so $KerD_zf=0$ and $f$ is an immersion. Is this correct? Would this still work if there were non-analytic terms (e.g. $bar z$ instead of $z$)?
differential-geometry manifolds smooth-manifolds
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
$endgroup$
2
$begingroup$
The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
$endgroup$
– tomasz
Mar 17 at 13:38
$begingroup$
@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
$endgroup$
– Evariste
Mar 17 at 13:58
2
$begingroup$
As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
$endgroup$
– tomasz
Mar 17 at 14:16
$begingroup$
@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
$endgroup$
– Evariste
Mar 17 at 19:57
$begingroup$
Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
$endgroup$
– tomasz
Mar 18 at 4:14
add a comment |
$begingroup$
Let me elaborate:
By properties, I mean qualities such as "is a submersion", "is an immersion", "is an embedding", etc.
Let's say I have a function $f:S^1 rightarrow S^3$ between the spheres viewed as smooth manifolds in $mathbbC$ and $mathbbC^2$ respectively.
In class we saw that to prove for example that $f$ is a submersion, we should first define maps (or "charts") $varphi:mathbbR^2rightarrow S_1$, $psi:S^3rightarrow mathbbR^4$ and check that the differential $D_x(psi circ f circ varphi)$ is onto.
In other words, to show $f$ has one of the properties, we first have to view it as a function from $mathbbR^2$ to $mathbbR^4$.
I would like to know if it is equivalent to view $f$ as a function from $mathbbC$ to $mathbbC^2$ instead when it makes things much more convenient and simply differentiate with respect to the complex variable as if it were real since it would simplify a lot the calculations (here $varphi$ and $psi$ would simply be identity! or rather, inclusion from the spheres to the complex plane)
In particular, would this work for all complex functions? Or only those that are "analytic" (e.g. which do not include terms in $bar z$)?
To give a concrete example, let's consider:
$f(z)=(cz^n,c^m/ne^ipi/nz^m)$ where $c$ is real positive chosen such that $f$ is from $S_1$ to $S_3$ (i.e. if $|z|=1$ then $|f(z)|=1)$
Say I want to know whether it is a immersion. The "safe" way would be to convert $z=x+iy$ to $(x,y)$ in $mathbbR^2$ and get a function sending $(x,y)$ to $(a,b,c,d)$ from $mathbbR^2$ to $mathbbR^4$ where $a,b,c,d$ are functions of $x$ and $y$. But computing this could be tedious depending on the function involved, so can I simply compute the complex differential with respect to $z$ and determine whether it is injective? (from $mathbbC$ to $mathbbC^2$)
For example here I would get $D_zf(h)=(cnz^n-1,c^m/ne^ipi/nmz^m-1).h$ ($hin mathbbC)$
and $D_zf(h)=0 Leftrightarrow h=0$ assuming $c>0$ and $n>0$ (and $zneq0$ but here $zin S_1$ so it is automatically true)
so $KerD_zf=0$ and $f$ is an immersion. Is this correct? Would this still work if there were non-analytic terms (e.g. $bar z$ instead of $z$)?
differential-geometry manifolds smooth-manifolds
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
$endgroup$
Let me elaborate:
By properties, I mean qualities such as "is a submersion", "is an immersion", "is an embedding", etc.
Let's say I have a function $f:S^1 rightarrow S^3$ between the spheres viewed as smooth manifolds in $mathbbC$ and $mathbbC^2$ respectively.
In class we saw that to prove for example that $f$ is a submersion, we should first define maps (or "charts") $varphi:mathbbR^2rightarrow S_1$, $psi:S^3rightarrow mathbbR^4$ and check that the differential $D_x(psi circ f circ varphi)$ is onto.
In other words, to show $f$ has one of the properties, we first have to view it as a function from $mathbbR^2$ to $mathbbR^4$.
I would like to know if it is equivalent to view $f$ as a function from $mathbbC$ to $mathbbC^2$ instead when it makes things much more convenient and simply differentiate with respect to the complex variable as if it were real since it would simplify a lot the calculations (here $varphi$ and $psi$ would simply be identity! or rather, inclusion from the spheres to the complex plane)
In particular, would this work for all complex functions? Or only those that are "analytic" (e.g. which do not include terms in $bar z$)?
To give a concrete example, let's consider:
$f(z)=(cz^n,c^m/ne^ipi/nz^m)$ where $c$ is real positive chosen such that $f$ is from $S_1$ to $S_3$ (i.e. if $|z|=1$ then $|f(z)|=1)$
Say I want to know whether it is a immersion. The "safe" way would be to convert $z=x+iy$ to $(x,y)$ in $mathbbR^2$ and get a function sending $(x,y)$ to $(a,b,c,d)$ from $mathbbR^2$ to $mathbbR^4$ where $a,b,c,d$ are functions of $x$ and $y$. But computing this could be tedious depending on the function involved, so can I simply compute the complex differential with respect to $z$ and determine whether it is injective? (from $mathbbC$ to $mathbbC^2$)
For example here I would get $D_zf(h)=(cnz^n-1,c^m/ne^ipi/nmz^m-1).h$ ($hin mathbbC)$
and $D_zf(h)=0 Leftrightarrow h=0$ assuming $c>0$ and $n>0$ (and $zneq0$ but here $zin S_1$ so it is automatically true)
so $KerD_zf=0$ and $f$ is an immersion. Is this correct? Would this still work if there were non-analytic terms (e.g. $bar z$ instead of $z$)?
differential-geometry manifolds smooth-manifolds
differential-geometry manifolds smooth-manifolds
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
edited Mar 17 at 13:59
Evariste
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
asked Mar 17 at 12:25
EvaristeEvariste
1,1842618
1,1842618
2
$begingroup$
The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
$endgroup$
– tomasz
Mar 17 at 13:38
$begingroup$
@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
$endgroup$
– Evariste
Mar 17 at 13:58
2
$begingroup$
As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
$endgroup$
– tomasz
Mar 17 at 14:16
$begingroup$
@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
$endgroup$
– Evariste
Mar 17 at 19:57
$begingroup$
Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
$endgroup$
– tomasz
Mar 18 at 4:14
add a comment |
2
$begingroup$
The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
$endgroup$
– tomasz
Mar 17 at 13:38
$begingroup$
@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
$endgroup$
– Evariste
Mar 17 at 13:58
2
$begingroup$
As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
$endgroup$
– tomasz
Mar 17 at 14:16
$begingroup$
@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
$endgroup$
– Evariste
Mar 17 at 19:57
$begingroup$
Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
$endgroup$
– tomasz
Mar 18 at 4:14
2
2
$begingroup$
The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
$endgroup$
– tomasz
Mar 17 at 13:38
$begingroup$
The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
$endgroup$
– tomasz
Mar 17 at 13:38
$begingroup$
@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
$endgroup$
– Evariste
Mar 17 at 13:58
$begingroup$
@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
$endgroup$
– Evariste
Mar 17 at 13:58
2
2
$begingroup$
As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
$endgroup$
– tomasz
Mar 17 at 14:16
$begingroup$
As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
$endgroup$
– tomasz
Mar 17 at 14:16
$begingroup$
@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
$endgroup$
– Evariste
Mar 17 at 19:57
$begingroup$
@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
$endgroup$
– Evariste
Mar 17 at 19:57
$begingroup$
Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
$endgroup$
– tomasz
Mar 18 at 4:14
$begingroup$
Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
$endgroup$
– tomasz
Mar 18 at 4:14
add a comment |
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The way you write it does not really make sense: $S^1$ and $S^3$ are not complex manifolds. $S^1$ is one-dimensional, and $S^3$ is 3-dimensional. What you can do is show that your map extend to an immersion of the punctured complex line to the punctured complex plane (as real manifolds), and then argue that the restriction of an immersion to a submanifold is an immersion. Now, punctured complex line and punctured complex plane are complex manifolds, so you can show that the extension is a complex immersion, and hence a real immersion.
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– tomasz
Mar 17 at 13:38
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@tomasz Okay well the class I follow is very elementary I have not seen the properties of complex manifolds, only real manifolds (well I guess, we just called them "manifolds" and the charts were to $mathbbR^n$ instead of $mathbbC^n$). Come to think of it, it makes sense that complex manifolds must have even dimension since $mathbbC$ has dimension $2$. I'll remove that tag then since it did not mean what I thought it meant! I'm not supposed to use this I think... I was wondering whether my method works because I saw some exercises where they just differentiate the complex variable...
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– Evariste
Mar 17 at 13:58
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As I have said, you can do that, but only for complex manifolds. A complex smooth map between complex manifolds is real smooth (but not the opposite). If a complex smooth map is a complex immersion, it is a real immersion and vice versa. You can do this, but it needs to be justified. This is not very hard when you have a grasp of the basics, but it may well be that it is a bit above your current level.
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– tomasz
Mar 17 at 14:16
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@tomasz Is there a similar trick for submersions? It seems like the restriction of a submersion to a submanifold isn't necessarily a submersion
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– Evariste
Mar 17 at 19:57
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Well, for submersions it is a bit more subtle. For immersions it's straightforward, since being firstly, being an immersion is preserved when you enlarge the codomain, and secondly, because the inclusion of a submanifold is an immersion. What you can do is to use the real differential (which, as a function, is the same as the complex differential) and look at the image of the tangent space of the submanifold (which will be a linear subspace of the tangent space of the ambient manifold) and see if it is mapped onto the tangent space of the submanifold of the ambient codomain.
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– tomasz
Mar 18 at 4:14