induced connection on open sets
$begingroup$
Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.
For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.
differential-geometry
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.
For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.
differential-geometry
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.
For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.
differential-geometry
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
$endgroup$
Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.
For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.
differential-geometry
differential-geometry
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
asked Dec 8 '12 at 17:19
DubiousDubious
3,36472675
3,36472675
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
$endgroup$
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
add a comment |
$begingroup$
Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).
For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
$$nabla^U_XY=sumnabla_{phi_j X}Y$$
and
$$nabla^U_YX=sumnabla_Yphi_j X$$
The definition is locally meaningful, because the covering is locally finite, and so are the sums.
Hope it helps.
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
$endgroup$
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
1
$begingroup$
@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:16
1
$begingroup$
No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
$endgroup$
– wisefool
Dec 8 '12 at 18:18
1
$begingroup$
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
$endgroup$
– wisefool
Dec 8 '12 at 18:21
|
show 2 more comments
$begingroup$
For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:
If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.
If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.
Furthermore:
Any $X_pin T_pM$ can be extended to a section of $TM$.
For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).
Now you can define the induced connection $nabla'$ as follows:
$$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$
where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.
answered Jan 15 at 16:58
triitrii
80817
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
$endgroup$
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
add a comment |
$begingroup$
The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
$endgroup$
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
add a comment |
$begingroup$
The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
$endgroup$
The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
answered Dec 8 '12 at 17:32
levaplevap
47.9k33274
47.9k33274
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
add a comment |
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
$begingroup$
The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
$endgroup$
– Dubious
Dec 8 '12 at 17:58
add a comment |
$begingroup$
Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).
For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
$$nabla^U_XY=sumnabla_{phi_j X}Y$$
and
$$nabla^U_YX=sumnabla_Yphi_j X$$
The definition is locally meaningful, because the covering is locally finite, and so are the sums.
Hope it helps.
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
$endgroup$
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
1
$begingroup$
@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:16
1
$begingroup$
No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
$endgroup$
– wisefool
Dec 8 '12 at 18:18
1
$begingroup$
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
$endgroup$
– wisefool
Dec 8 '12 at 18:21
|
show 2 more comments
$begingroup$
Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).
For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
$$nabla^U_XY=sumnabla_{phi_j X}Y$$
and
$$nabla^U_YX=sumnabla_Yphi_j X$$
The definition is locally meaningful, because the covering is locally finite, and so are the sums.
Hope it helps.
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
$endgroup$
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
1
$begingroup$
@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:16
1
$begingroup$
No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
$endgroup$
– wisefool
Dec 8 '12 at 18:18
1
$begingroup$
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
$endgroup$
– wisefool
Dec 8 '12 at 18:21
|
show 2 more comments
$begingroup$
Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).
For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
$$nabla^U_XY=sumnabla_{phi_j X}Y$$
and
$$nabla^U_YX=sumnabla_Yphi_j X$$
The definition is locally meaningful, because the covering is locally finite, and so are the sums.
Hope it helps.
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
$endgroup$
Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).
For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
$$nabla^U_XY=sumnabla_{phi_j X}Y$$
and
$$nabla^U_YX=sumnabla_Yphi_j X$$
The definition is locally meaningful, because the covering is locally finite, and so are the sums.
Hope it helps.
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
answered Dec 8 '12 at 17:33
wisefoolwisefool
3,179711
3,179711
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
1
$begingroup$
@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:16
1
$begingroup$
No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
$endgroup$
– wisefool
Dec 8 '12 at 18:18
1
$begingroup$
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
$endgroup$
– wisefool
Dec 8 '12 at 18:21
|
show 2 more comments
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
1
$begingroup$
@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:16
1
$begingroup$
No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
$endgroup$
– wisefool
Dec 8 '12 at 18:18
1
$begingroup$
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
$endgroup$
– wisefool
Dec 8 '12 at 18:21
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
$begingroup$
why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
$endgroup$
– Dubious
Dec 8 '12 at 17:42
1
1
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
$endgroup$
– Jason DeVito
Dec 8 '12 at 18:01
$begingroup$
The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
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– Jason DeVito
Dec 8 '12 at 18:01
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@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
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– Jason DeVito
Dec 8 '12 at 18:16
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@Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
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– Jason DeVito
Dec 8 '12 at 18:16
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No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
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– wisefool
Dec 8 '12 at 18:18
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No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
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– wisefool
Dec 8 '12 at 18:18
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Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
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– wisefool
Dec 8 '12 at 18:21
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Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
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– wisefool
Dec 8 '12 at 18:21
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show 2 more comments
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For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:
If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.
If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.
Furthermore:
Any $X_pin T_pM$ can be extended to a section of $TM$.
For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).
Now you can define the induced connection $nabla'$ as follows:
$$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$
where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.
answered Jan 15 at 16:58
triitrii
80817
$endgroup$
add a comment |
$begingroup$
For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:
If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.
If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.
Furthermore:
Any $X_pin T_pM$ can be extended to a section of $TM$.
For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).
Now you can define the induced connection $nabla'$ as follows:
$$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$
where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.
answered Jan 15 at 16:58
triitrii
80817
$endgroup$
add a comment |
$begingroup$
For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:
If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.
If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.
Furthermore:
Any $X_pin T_pM$ can be extended to a section of $TM$.
For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).
Now you can define the induced connection $nabla'$ as follows:
$$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$
where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.
answered Jan 15 at 16:58
triitrii
80817
$endgroup$
For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:
If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.
If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.
Furthermore:
Any $X_pin T_pM$ can be extended to a section of $TM$.
For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).
Now you can define the induced connection $nabla'$ as follows:
$$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$
where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.
answered Jan 15 at 16:58
triitrii
80817
answered Jan 15 at 16:58
triitrii
80817
answered Jan 15 at 16:58
triitrii
80817
answered Jan 15 at 16:58
triitrii
80817
80817
add a comment |
add a comment |
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