If $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are co-oriented, then so are $(e_1, ldots, e_n, u)$ and $(f_1,...
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
$endgroup$
add a comment |
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
$endgroup$
add a comment |
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
$endgroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
edited Dec 6 '18 at 5:54
CuriousKid7
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,669717
1,669717
add a comment |
add a comment |
1 Answer
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If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
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1 Answer
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1 Answer
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$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
11.1k923
add a comment |
add a comment |
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