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How can I obtain a normal unitary field to higher dimensional surfaces?
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More specifically, say I've got a surface in the form $X(u,v) = (x(u,v), y(u,v),z(u,v),t(u,v))$. The tangent fields are obviously $X_u$ and $X_v$, but I don't know how to give meaning to $X_u times X_v$ in order to obtain a normal unitary field of $X$.
differential-geometry riemannian-geometry
asked Aug 2 at 18:09
Matheus Andrade
587214
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up vote
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More specifically, say I've got a surface in the form $X(u,v) = (x(u,v), y(u,v),z(u,v),t(u,v))$. The tangent fields are obviously $X_u$ and $X_v$, but I don't know how to give meaning to $X_u times X_v$ in order to obtain a normal unitary field of $X$.
differential-geometry riemannian-geometry
asked Aug 2 at 18:09
Matheus Andrade
587214
In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
More specifically, say I've got a surface in the form $X(u,v) = (x(u,v), y(u,v),z(u,v),t(u,v))$. The tangent fields are obviously $X_u$ and $X_v$, but I don't know how to give meaning to $X_u times X_v$ in order to obtain a normal unitary field of $X$.
differential-geometry riemannian-geometry
asked Aug 2 at 18:09
Matheus Andrade
587214
More specifically, say I've got a surface in the form $X(u,v) = (x(u,v), y(u,v),z(u,v),t(u,v))$. The tangent fields are obviously $X_u$ and $X_v$, but I don't know how to give meaning to $X_u times X_v$ in order to obtain a normal unitary field of $X$.
differential-geometry riemannian-geometry
asked Aug 2 at 18:09
Matheus Andrade
587214
asked Aug 2 at 18:09
Matheus Andrade
587214
asked Aug 2 at 18:09
Matheus Andrade
587214
asked Aug 2 at 18:09
Matheus Andrade
587214
587214
In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12
add a comment |Â
In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12
In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12
In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12
add a comment |Â
1 Answer
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Indeed, the expression $X_utimes X_v$ is meaningless in $4$-dimensional Euclidean space.
Let us go back to the $3$-dimensional case and understand what is done there. Let $SsubsetmathbbR^3$ be a regular surface, and let $pin S$. Then $T_pSsubsetmathbbR^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=pmfracX_utimes X_v.$$
Now, if the surface $S$ is embedded in $mathbbR^4$, the orthogonal complement $(T_pS)^perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Indeed, the expression $X_utimes X_v$ is meaningless in $4$-dimensional Euclidean space.
Let us go back to the $3$-dimensional case and understand what is done there. Let $SsubsetmathbbR^3$ be a regular surface, and let $pin S$. Then $T_pSsubsetmathbbR^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=pmfracX_utimes X_v.$$
Now, if the surface $S$ is embedded in $mathbbR^4$, the orthogonal complement $(T_pS)^perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
add a comment |Â
up vote
0
down vote
Indeed, the expression $X_utimes X_v$ is meaningless in $4$-dimensional Euclidean space.
Let us go back to the $3$-dimensional case and understand what is done there. Let $SsubsetmathbbR^3$ be a regular surface, and let $pin S$. Then $T_pSsubsetmathbbR^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=pmfracX_utimes X_v.$$
Now, if the surface $S$ is embedded in $mathbbR^4$, the orthogonal complement $(T_pS)^perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Indeed, the expression $X_utimes X_v$ is meaningless in $4$-dimensional Euclidean space.
Let us go back to the $3$-dimensional case and understand what is done there. Let $SsubsetmathbbR^3$ be a regular surface, and let $pin S$. Then $T_pSsubsetmathbbR^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=pmfracX_utimes X_v.$$
Now, if the surface $S$ is embedded in $mathbbR^4$, the orthogonal complement $(T_pS)^perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
Indeed, the expression $X_utimes X_v$ is meaningless in $4$-dimensional Euclidean space.
Let us go back to the $3$-dimensional case and understand what is done there. Let $SsubsetmathbbR^3$ be a regular surface, and let $pin S$. Then $T_pSsubsetmathbbR^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=pmfracX_utimes X_v.$$
Now, if the surface $S$ is embedded in $mathbbR^4$, the orthogonal complement $(T_pS)^perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
answered Aug 3 at 7:53
Amitai Yuval
14.4k11026
14.4k11026
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
add a comment |Â
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
In the case of the embedding of the Klein bottle in $Bbb R^4$, I doubt you'll be able to choose a global unit normal vector field, although I haven't thought about a proof.
– Ted Shifrin
Aug 3 at 23:13
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
@TedShifrin It seems to me that this specific post is not about the global question; OP is asking about something similar to $X_utimes X_v$, which is a local expression.
– Amitai Yuval
Aug 4 at 5:56
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
Not clear, @AmitaiYuval. After all, we can globally "parametrize" the Klein bottle, just as we can globally parametrize a torus.
– Ted Shifrin
Aug 4 at 6:12
add a comment |Â
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In general, for a $k$-dimensional submanifold in $Bbb R^n$, the tangential Gauss map maps to the Grassmannian of $k$-planes in $Bbb R^n$ and the normal Gauss map maps to the Grassmannian of $(n-k)$-planes in $Bbb R^n$. Only when the manifold is orientable can you map to the Grassmannians of oriented planes.
– Ted Shifrin
Aug 3 at 23:12