The second fundamental form on a submaniford
Given $M$ a Riemannian manifold with metric the levi-civita connection $nabla$, $N$ embedded submanifold of $M$, and vector fields $X$, $Y$ on $N$, define $f(X,Y)$ to be the orthogonal projection of $nabla_{bar Y}bar X$ onto the orthogonal complement of $TN$, where $bar X$ and $bar Y$ are extensions of $N$.
I wonder whether this is well-defined. First, I have no idea whether it is always possible to extend vector fields from an embedded submanifold. Second, I do not know why the definition is independent of the extensions of $Y$.
FOR THE FIRST QUESTION: I know that it is always possible to extend from a properly embedded submanifold. What about arbitrary embedded submanifold then?
FOR THE SECOND QUESTION: I know that $nabla$ only cares about local data of $Y$, but here extending from $N$ does not guarantee us local data(there isn't any neighborhood of $M$ contained in $N$, if $N$ has dimension strictly smaller).
differential-geometry
asked Nov 28 '18 at 3:08
Keith
410316
add a comment |
Given $M$ a Riemannian manifold with metric the levi-civita connection $nabla$, $N$ embedded submanifold of $M$, and vector fields $X$, $Y$ on $N$, define $f(X,Y)$ to be the orthogonal projection of $nabla_{bar Y}bar X$ onto the orthogonal complement of $TN$, where $bar X$ and $bar Y$ are extensions of $N$.
I wonder whether this is well-defined. First, I have no idea whether it is always possible to extend vector fields from an embedded submanifold. Second, I do not know why the definition is independent of the extensions of $Y$.
FOR THE FIRST QUESTION: I know that it is always possible to extend from a properly embedded submanifold. What about arbitrary embedded submanifold then?
FOR THE SECOND QUESTION: I know that $nabla$ only cares about local data of $Y$, but here extending from $N$ does not guarantee us local data(there isn't any neighborhood of $M$ contained in $N$, if $N$ has dimension strictly smaller).
differential-geometry
asked Nov 28 '18 at 3:08
Keith
410316
add a comment |
Given $M$ a Riemannian manifold with metric the levi-civita connection $nabla$, $N$ embedded submanifold of $M$, and vector fields $X$, $Y$ on $N$, define $f(X,Y)$ to be the orthogonal projection of $nabla_{bar Y}bar X$ onto the orthogonal complement of $TN$, where $bar X$ and $bar Y$ are extensions of $N$.
I wonder whether this is well-defined. First, I have no idea whether it is always possible to extend vector fields from an embedded submanifold. Second, I do not know why the definition is independent of the extensions of $Y$.
FOR THE FIRST QUESTION: I know that it is always possible to extend from a properly embedded submanifold. What about arbitrary embedded submanifold then?
FOR THE SECOND QUESTION: I know that $nabla$ only cares about local data of $Y$, but here extending from $N$ does not guarantee us local data(there isn't any neighborhood of $M$ contained in $N$, if $N$ has dimension strictly smaller).
differential-geometry
asked Nov 28 '18 at 3:08
Keith
410316
Given $M$ a Riemannian manifold with metric the levi-civita connection $nabla$, $N$ embedded submanifold of $M$, and vector fields $X$, $Y$ on $N$, define $f(X,Y)$ to be the orthogonal projection of $nabla_{bar Y}bar X$ onto the orthogonal complement of $TN$, where $bar X$ and $bar Y$ are extensions of $N$.
I wonder whether this is well-defined. First, I have no idea whether it is always possible to extend vector fields from an embedded submanifold. Second, I do not know why the definition is independent of the extensions of $Y$.
FOR THE FIRST QUESTION: I know that it is always possible to extend from a properly embedded submanifold. What about arbitrary embedded submanifold then?
FOR THE SECOND QUESTION: I know that $nabla$ only cares about local data of $Y$, but here extending from $N$ does not guarantee us local data(there isn't any neighborhood of $M$ contained in $N$, if $N$ has dimension strictly smaller).
differential-geometry
differential-geometry
asked Nov 28 '18 at 3:08
Keith
410316
asked Nov 28 '18 at 3:08
Keith
410316
asked Nov 28 '18 at 3:08
Keith
410316
asked Nov 28 '18 at 3:08
Keith
410316
asked Nov 28 '18 at 3:08
Keith
410316
410316
add a comment |
add a comment |
1 Answer
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I am not quite sure what you mean with the issue of embedded vs. properly embedded. For an embedded submanifold $Nsubset M$, you should always have local charts $(U,u)$ such that $u(Ucap N)$ is the intersection of $u(U)$ with a linear subspace of $mathbb R^n$. Under these assumptions, you can certainly extend a local vector field on $N$ defined on $Ucap N$ to a local vector field on $M$ defined on $U$, which certainly is sufficient for what you need.
Concerning the question of being well-defined, the situation is very simple in one variable. The fact that $(bar X,bar Y)mapsto nabla_{bar Y}bar X$ is linear over smooth functions in the second variable easily implies that for any point $x$, the values $nabla_{bar Y}bar X(x)$ depends only on $bar Y(x)=Y(x)$. In the other variable and additional argument is needed. You have assumed from the beginning that $Y$ and $X$ are tangent to $N$. In particular, this implies that for any local extension $bar Y$ of $Y$, any flow line of $bar Y$ that starts in a point of $N$ remains in $N$. Thus you can invoke another well known fact about linear connections (that is used for example to define the covariant derivatives along curves): The value $nabla_{bar Y}bar X(x)$ depends only on the values of $bar X$ along the flow line of $bar Y$ through $x$. Since for $xin N$ that flow line remains in $N$, the restriction of $bar X$ to the flow line is independent of the extension.
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
add a comment |
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I am not quite sure what you mean with the issue of embedded vs. properly embedded. For an embedded submanifold $Nsubset M$, you should always have local charts $(U,u)$ such that $u(Ucap N)$ is the intersection of $u(U)$ with a linear subspace of $mathbb R^n$. Under these assumptions, you can certainly extend a local vector field on $N$ defined on $Ucap N$ to a local vector field on $M$ defined on $U$, which certainly is sufficient for what you need.
Concerning the question of being well-defined, the situation is very simple in one variable. The fact that $(bar X,bar Y)mapsto nabla_{bar Y}bar X$ is linear over smooth functions in the second variable easily implies that for any point $x$, the values $nabla_{bar Y}bar X(x)$ depends only on $bar Y(x)=Y(x)$. In the other variable and additional argument is needed. You have assumed from the beginning that $Y$ and $X$ are tangent to $N$. In particular, this implies that for any local extension $bar Y$ of $Y$, any flow line of $bar Y$ that starts in a point of $N$ remains in $N$. Thus you can invoke another well known fact about linear connections (that is used for example to define the covariant derivatives along curves): The value $nabla_{bar Y}bar X(x)$ depends only on the values of $bar X$ along the flow line of $bar Y$ through $x$. Since for $xin N$ that flow line remains in $N$, the restriction of $bar X$ to the flow line is independent of the extension.
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
add a comment |
I am not quite sure what you mean with the issue of embedded vs. properly embedded. For an embedded submanifold $Nsubset M$, you should always have local charts $(U,u)$ such that $u(Ucap N)$ is the intersection of $u(U)$ with a linear subspace of $mathbb R^n$. Under these assumptions, you can certainly extend a local vector field on $N$ defined on $Ucap N$ to a local vector field on $M$ defined on $U$, which certainly is sufficient for what you need.
Concerning the question of being well-defined, the situation is very simple in one variable. The fact that $(bar X,bar Y)mapsto nabla_{bar Y}bar X$ is linear over smooth functions in the second variable easily implies that for any point $x$, the values $nabla_{bar Y}bar X(x)$ depends only on $bar Y(x)=Y(x)$. In the other variable and additional argument is needed. You have assumed from the beginning that $Y$ and $X$ are tangent to $N$. In particular, this implies that for any local extension $bar Y$ of $Y$, any flow line of $bar Y$ that starts in a point of $N$ remains in $N$. Thus you can invoke another well known fact about linear connections (that is used for example to define the covariant derivatives along curves): The value $nabla_{bar Y}bar X(x)$ depends only on the values of $bar X$ along the flow line of $bar Y$ through $x$. Since for $xin N$ that flow line remains in $N$, the restriction of $bar X$ to the flow line is independent of the extension.
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
add a comment |
I am not quite sure what you mean with the issue of embedded vs. properly embedded. For an embedded submanifold $Nsubset M$, you should always have local charts $(U,u)$ such that $u(Ucap N)$ is the intersection of $u(U)$ with a linear subspace of $mathbb R^n$. Under these assumptions, you can certainly extend a local vector field on $N$ defined on $Ucap N$ to a local vector field on $M$ defined on $U$, which certainly is sufficient for what you need.
Concerning the question of being well-defined, the situation is very simple in one variable. The fact that $(bar X,bar Y)mapsto nabla_{bar Y}bar X$ is linear over smooth functions in the second variable easily implies that for any point $x$, the values $nabla_{bar Y}bar X(x)$ depends only on $bar Y(x)=Y(x)$. In the other variable and additional argument is needed. You have assumed from the beginning that $Y$ and $X$ are tangent to $N$. In particular, this implies that for any local extension $bar Y$ of $Y$, any flow line of $bar Y$ that starts in a point of $N$ remains in $N$. Thus you can invoke another well known fact about linear connections (that is used for example to define the covariant derivatives along curves): The value $nabla_{bar Y}bar X(x)$ depends only on the values of $bar X$ along the flow line of $bar Y$ through $x$. Since for $xin N$ that flow line remains in $N$, the restriction of $bar X$ to the flow line is independent of the extension.
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
I am not quite sure what you mean with the issue of embedded vs. properly embedded. For an embedded submanifold $Nsubset M$, you should always have local charts $(U,u)$ such that $u(Ucap N)$ is the intersection of $u(U)$ with a linear subspace of $mathbb R^n$. Under these assumptions, you can certainly extend a local vector field on $N$ defined on $Ucap N$ to a local vector field on $M$ defined on $U$, which certainly is sufficient for what you need.
Concerning the question of being well-defined, the situation is very simple in one variable. The fact that $(bar X,bar Y)mapsto nabla_{bar Y}bar X$ is linear over smooth functions in the second variable easily implies that for any point $x$, the values $nabla_{bar Y}bar X(x)$ depends only on $bar Y(x)=Y(x)$. In the other variable and additional argument is needed. You have assumed from the beginning that $Y$ and $X$ are tangent to $N$. In particular, this implies that for any local extension $bar Y$ of $Y$, any flow line of $bar Y$ that starts in a point of $N$ remains in $N$. Thus you can invoke another well known fact about linear connections (that is used for example to define the covariant derivatives along curves): The value $nabla_{bar Y}bar X(x)$ depends only on the values of $bar X$ along the flow line of $bar Y$ through $x$. Since for $xin N$ that flow line remains in $N$, the restriction of $bar X$ to the flow line is independent of the extension.
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
answered Dec 5 '18 at 15:40
Andreas Cap
11k923
11k923
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
add a comment |
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
Thank you for your answer, just one more question. I know that the connection only cares about the local data of a vector field, so it "suffices" to extend a vector field locally. However, I still need to be able to extend this local thing to a global vector field because formally the connection takes value as global vector fields. Is it always possible to extend a vector field on an open subset?
– Keith
Dec 6 '18 at 3:42
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
This can be done using a bump function. Given a nieghborhood $U$ of $x$ on which a local vector field is defined, choose an open neighborhood $V$ of $x$ such that $bar Vsubset U$. Then there is a smooth function $f:Mto [0,1]$ with support contained in $U$ that is identically $1$ on $bar V$. Now for any local vector field $xi$ defined on $U$, take $fxi$. This coincides with $xi$ on $V$ (and hence locally around $x$) and it can be smoothly extended by zero to all of $M$.
– Andreas Cap
Dec 6 '18 at 7:51
add a comment |
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