Mixing
A manifold admits a nowhere vanishing volume form if and only if it is orientable?
$begingroup$
on this wikipedia article, it is said :
A manifold admits a nowhere vanishing volume form if and only if it is
orientable
I don't really understand why. Isn't $dx^1 wedge ... wedge dx^n$ always a nowhere vanishing? Indeed, for me we should have $dx^1_p wedge ... wedge dx^n_p neq 0 forall p$ since ${dx^1_p,...,dx^n_p}$ is a basis of $T_p^*M$, no?
Also another question :
If $X_1,...,X_n$ is a vector field such that $X_{1p},...,X_{np}$ are linearly independent for all $p$, does this imply that $dx^1_p wedge ... wedge dx^n_p(X_{1p},...,X_{np}) neq 0 forall p$?
differential-geometry manifolds differential-forms
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
$endgroup$
|
show 4 more comments
$begingroup$
on this wikipedia article, it is said :
A manifold admits a nowhere vanishing volume form if and only if it is
orientable
I don't really understand why. Isn't $dx^1 wedge ... wedge dx^n$ always a nowhere vanishing? Indeed, for me we should have $dx^1_p wedge ... wedge dx^n_p neq 0 forall p$ since ${dx^1_p,...,dx^n_p}$ is a basis of $T_p^*M$, no?
Also another question :
If $X_1,...,X_n$ is a vector field such that $X_{1p},...,X_{np}$ are linearly independent for all $p$, does this imply that $dx^1_p wedge ... wedge dx^n_p(X_{1p},...,X_{np}) neq 0 forall p$?
differential-geometry manifolds differential-forms
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
$endgroup$
3
$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
2
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
1
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
2
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37
|
show 4 more comments
$begingroup$
on this wikipedia article, it is said :
A manifold admits a nowhere vanishing volume form if and only if it is
orientable
I don't really understand why. Isn't $dx^1 wedge ... wedge dx^n$ always a nowhere vanishing? Indeed, for me we should have $dx^1_p wedge ... wedge dx^n_p neq 0 forall p$ since ${dx^1_p,...,dx^n_p}$ is a basis of $T_p^*M$, no?
Also another question :
If $X_1,...,X_n$ is a vector field such that $X_{1p},...,X_{np}$ are linearly independent for all $p$, does this imply that $dx^1_p wedge ... wedge dx^n_p(X_{1p},...,X_{np}) neq 0 forall p$?
differential-geometry manifolds differential-forms
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
$endgroup$
on this wikipedia article, it is said :
A manifold admits a nowhere vanishing volume form if and only if it is
orientable
I don't really understand why. Isn't $dx^1 wedge ... wedge dx^n$ always a nowhere vanishing? Indeed, for me we should have $dx^1_p wedge ... wedge dx^n_p neq 0 forall p$ since ${dx^1_p,...,dx^n_p}$ is a basis of $T_p^*M$, no?
Also another question :
If $X_1,...,X_n$ is a vector field such that $X_{1p},...,X_{np}$ are linearly independent for all $p$, does this imply that $dx^1_p wedge ... wedge dx^n_p(X_{1p},...,X_{np}) neq 0 forall p$?
differential-geometry manifolds differential-forms
differential-geometry manifolds differential-forms
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
asked Jan 11 at 19:23
roi_saumonroi_saumon
56438
56438
3
$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
2
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
1
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
2
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37
|
show 4 more comments
3
$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
2
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
1
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
2
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37
3
3
$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
2
2
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
1
1
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
2
2
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37
|
show 4 more comments
1 Answer
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oldest
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To answer your second question:
If you have a finite dimensional vector space $E$ of dimension $n$ then the vector space of all alternating multilinear maps $ E^nrightarrow mathbb{R}$ is one dimensional, see for example here: The determinant function is the only one satisfying the conditions.
That is, if you pick a basis of $E$ you can define the determinant with respect to that basis and each other alternating multilinear map will then be a scalar multiple of the determinant. So a nonzero multilinear alternating map vanishes if and only if all inserted vectors are linear dependent.
answered Jan 14 at 19:32
triitrii
965
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add a comment |
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$begingroup$
To answer your second question:
If you have a finite dimensional vector space $E$ of dimension $n$ then the vector space of all alternating multilinear maps $ E^nrightarrow mathbb{R}$ is one dimensional, see for example here: The determinant function is the only one satisfying the conditions.
That is, if you pick a basis of $E$ you can define the determinant with respect to that basis and each other alternating multilinear map will then be a scalar multiple of the determinant. So a nonzero multilinear alternating map vanishes if and only if all inserted vectors are linear dependent.
answered Jan 14 at 19:32
triitrii
965
$endgroup$
add a comment |
$begingroup$
To answer your second question:
If you have a finite dimensional vector space $E$ of dimension $n$ then the vector space of all alternating multilinear maps $ E^nrightarrow mathbb{R}$ is one dimensional, see for example here: The determinant function is the only one satisfying the conditions.
That is, if you pick a basis of $E$ you can define the determinant with respect to that basis and each other alternating multilinear map will then be a scalar multiple of the determinant. So a nonzero multilinear alternating map vanishes if and only if all inserted vectors are linear dependent.
answered Jan 14 at 19:32
triitrii
965
$endgroup$
add a comment |
$begingroup$
To answer your second question:
If you have a finite dimensional vector space $E$ of dimension $n$ then the vector space of all alternating multilinear maps $ E^nrightarrow mathbb{R}$ is one dimensional, see for example here: The determinant function is the only one satisfying the conditions.
That is, if you pick a basis of $E$ you can define the determinant with respect to that basis and each other alternating multilinear map will then be a scalar multiple of the determinant. So a nonzero multilinear alternating map vanishes if and only if all inserted vectors are linear dependent.
answered Jan 14 at 19:32
triitrii
965
$endgroup$
To answer your second question:
If you have a finite dimensional vector space $E$ of dimension $n$ then the vector space of all alternating multilinear maps $ E^nrightarrow mathbb{R}$ is one dimensional, see for example here: The determinant function is the only one satisfying the conditions.
That is, if you pick a basis of $E$ you can define the determinant with respect to that basis and each other alternating multilinear map will then be a scalar multiple of the determinant. So a nonzero multilinear alternating map vanishes if and only if all inserted vectors are linear dependent.
answered Jan 14 at 19:32
triitrii
965
answered Jan 14 at 19:32
triitrii
965
answered Jan 14 at 19:32
triitrii
965
answered Jan 14 at 19:32
triitrii
965
965
add a comment |
add a comment |
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$begingroup$
Note that $dx^1 wedge ldots wedge dx^n$ is defined locally and depends on the choice of coordinates. And you cannot define coordinates globally except in trivial cases.
$endgroup$
– Mindlack
Jan 11 at 19:29
$begingroup$
@Mindlack Sorry, I don't get your point, could you expand a bit?
$endgroup$
– roi_saumon
Jan 11 at 19:34
2
$begingroup$
How would you define $dx^1 wedge dx^2$ on the 2-sphere for instance?
$endgroup$
– Mindlack
Jan 11 at 19:36
1
$begingroup$
Yes, this is the point: your coordinates are local, they are only defined in charts. And the corresponding $dx^1 wedge dx^2$ at each $p$ will depend of which chart you took. So you will not have one well-defined $2$-form at each point, or if you do, there is no trivial way to make sure it is smooth, or even continuous. If you wish, orientability corresponds to being able to « glue together » enough nowhere-vanishing forms defined in different charts.
$endgroup$
– Mindlack
Jan 11 at 19:53
2
$begingroup$
What you say about $mathbb R^3$ is true. However, it is quite rare for an $n$-manifold to possess an atlas having only a single coordinate chart. In fact that holds if and only if the manifold is diffeomorphic to an open subset of $mathbb R^n$. In particular, it is never true for compact manifolds.
$endgroup$
– Lee Mosher
Jan 11 at 23:37