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How bad can the intersection of two totally geodesic submanifolds be?
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Let $M$ be a complete Riemannian manifold and let $S_1,S_2 subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totally geodesic, then $S_1 cap S_2$ can be homeomorphic to any closed subset of $mathbbR$ and so is far from being a manifold in general.
manifolds riemannian-geometry geodesic submanifold
asked Aug 2 at 9:36
abenthy
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Let $M$ be a complete Riemannian manifold and let $S_1,S_2 subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totally geodesic, then $S_1 cap S_2$ can be homeomorphic to any closed subset of $mathbbR$ and so is far from being a manifold in general.
manifolds riemannian-geometry geodesic submanifold
asked Aug 2 at 9:36
abenthy
37318
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $M$ be a complete Riemannian manifold and let $S_1,S_2 subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totally geodesic, then $S_1 cap S_2$ can be homeomorphic to any closed subset of $mathbbR$ and so is far from being a manifold in general.
manifolds riemannian-geometry geodesic submanifold
asked Aug 2 at 9:36
abenthy
37318
Let $M$ be a complete Riemannian manifold and let $S_1,S_2 subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totally geodesic, then $S_1 cap S_2$ can be homeomorphic to any closed subset of $mathbbR$ and so is far from being a manifold in general.
manifolds riemannian-geometry geodesic submanifold
asked Aug 2 at 9:36
abenthy
37318
edited Aug 2 at 11:09
asked Aug 2 at 9:36
abenthy
37318
asked Aug 2 at 9:36
abenthy
37318
asked Aug 2 at 9:36
abenthy
37318
37318
add a comment |Â
add a comment |Â
1 Answer
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Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.
Let $p$ be a point in $S_1cap S_2$, and let $(x^1,dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = dim S_i$ for $i=1,2$.
By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1cap U$.
Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2subset U$ such that $L_2 = S_2cap U$.
It follows that $S_1cap S_2cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$.
Since every point of $S_1cap S_2$ has a coordinate neighborhood in which $S_1cap S_2$ is a linear slice, it follows that $S_1cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $square$
answered Aug 2 at 22:47
Jack Lee
25.1k44161
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.
Let $p$ be a point in $S_1cap S_2$, and let $(x^1,dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = dim S_i$ for $i=1,2$.
By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1cap U$.
Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2subset U$ such that $L_2 = S_2cap U$.
It follows that $S_1cap S_2cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$.
Since every point of $S_1cap S_2$ has a coordinate neighborhood in which $S_1cap S_2$ is a linear slice, it follows that $S_1cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $square$
answered Aug 2 at 22:47
Jack Lee
25.1k44161
add a comment |Â
up vote
2
down vote
accepted
Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.
Let $p$ be a point in $S_1cap S_2$, and let $(x^1,dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = dim S_i$ for $i=1,2$.
By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1cap U$.
Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2subset U$ such that $L_2 = S_2cap U$.
It follows that $S_1cap S_2cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$.
Since every point of $S_1cap S_2$ has a coordinate neighborhood in which $S_1cap S_2$ is a linear slice, it follows that $S_1cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $square$
answered Aug 2 at 22:47
Jack Lee
25.1k44161
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.
Let $p$ be a point in $S_1cap S_2$, and let $(x^1,dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = dim S_i$ for $i=1,2$.
By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1cap U$.
Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2subset U$ such that $L_2 = S_2cap U$.
It follows that $S_1cap S_2cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$.
Since every point of $S_1cap S_2$ has a coordinate neighborhood in which $S_1cap S_2$ is a linear slice, it follows that $S_1cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $square$
answered Aug 2 at 22:47
Jack Lee
25.1k44161
Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.
Let $p$ be a point in $S_1cap S_2$, and let $(x^1,dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = dim S_i$ for $i=1,2$.
By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1cap U$.
Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2subset U$ such that $L_2 = S_2cap U$.
It follows that $S_1cap S_2cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$.
Since every point of $S_1cap S_2$ has a coordinate neighborhood in which $S_1cap S_2$ is a linear slice, it follows that $S_1cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $square$
answered Aug 2 at 22:47
Jack Lee
25.1k44161
answered Aug 2 at 22:47
Jack Lee
25.1k44161
answered Aug 2 at 22:47
Jack Lee
25.1k44161
answered Aug 2 at 22:47
Jack Lee
25.1k44161
25.1k44161
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