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How can a vector field eat a smooth function?
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I was watching this Frederic Schuller's lecture on Grassmann algebra and de Rham cohomology. I don't understand. At 43:40 it looks like a vector field is eating a smooth function.
The $X_i$ should be a vector field, that is a section of a smooth tangent bundle. So it should map a point on the manifold to one of the point's tangent vectors. But omega is a $0$-$n$ tensor field, that is, a function that takes n vector fields and outputs a smooth function (manifold $to mathbb R$). So how can you give the $X_i$ a smooth function when it should eat a point of the manifold? What am I missing?
differential-geometry vector-fields
asked Jul 22 at 16:58
p.h.f.
42
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I was watching this Frederic Schuller's lecture on Grassmann algebra and de Rham cohomology. I don't understand. At 43:40 it looks like a vector field is eating a smooth function.
The $X_i$ should be a vector field, that is a section of a smooth tangent bundle. So it should map a point on the manifold to one of the point's tangent vectors. But omega is a $0$-$n$ tensor field, that is, a function that takes n vector fields and outputs a smooth function (manifold $to mathbb R$). So how can you give the $X_i$ a smooth function when it should eat a point of the manifold? What am I missing?
differential-geometry vector-fields
asked Jul 22 at 16:58
p.h.f.
42
$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
1
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26
 |Â
show 4 more comments
up vote
0
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up vote
0
down vote
favorite
I was watching this Frederic Schuller's lecture on Grassmann algebra and de Rham cohomology. I don't understand. At 43:40 it looks like a vector field is eating a smooth function.
The $X_i$ should be a vector field, that is a section of a smooth tangent bundle. So it should map a point on the manifold to one of the point's tangent vectors. But omega is a $0$-$n$ tensor field, that is, a function that takes n vector fields and outputs a smooth function (manifold $to mathbb R$). So how can you give the $X_i$ a smooth function when it should eat a point of the manifold? What am I missing?
differential-geometry vector-fields
asked Jul 22 at 16:58
p.h.f.
42
I was watching this Frederic Schuller's lecture on Grassmann algebra and de Rham cohomology. I don't understand. At 43:40 it looks like a vector field is eating a smooth function.
The $X_i$ should be a vector field, that is a section of a smooth tangent bundle. So it should map a point on the manifold to one of the point's tangent vectors. But omega is a $0$-$n$ tensor field, that is, a function that takes n vector fields and outputs a smooth function (manifold $to mathbb R$). So how can you give the $X_i$ a smooth function when it should eat a point of the manifold? What am I missing?
differential-geometry vector-fields
asked Jul 22 at 16:58
p.h.f.
42
edited Jul 22 at 17:10
asked Jul 22 at 16:58
p.h.f.
42
asked Jul 22 at 16:58
p.h.f.
42
asked Jul 22 at 16:58
p.h.f.
42
42
$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
1
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26
 |Â
show 4 more comments
$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
1
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26
$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
1
1
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26
 |Â
show 4 more comments
1 Answer
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Given $vecv = (v^1,...,v^n) in mathbbR^n$ and $f: mathbbE^n to mathbbR$. Then we define $vecv[f] = sum_j fracpartial fpartial x^j v^j$. Given a vector field $X: U to mathbbR^n$ we have $X_p in mathbbR^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $mathbbR^n_p$ such that,
$$X_p = sum_i a(i,p) e^i_p$$
- We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) in mathbbR^n$. Then we let $X[f]$ be the vector field such that,
$$ X[f] (p) := X_p[f] = sum_j fracpartial fpartial x^j cdot a(j,p)$$
Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Given $vecv = (v^1,...,v^n) in mathbbR^n$ and $f: mathbbE^n to mathbbR$. Then we define $vecv[f] = sum_j fracpartial fpartial x^j v^j$. Given a vector field $X: U to mathbbR^n$ we have $X_p in mathbbR^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $mathbbR^n_p$ such that,
$$X_p = sum_i a(i,p) e^i_p$$
- We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) in mathbbR^n$. Then we let $X[f]$ be the vector field such that,
$$ X[f] (p) := X_p[f] = sum_j fracpartial fpartial x^j cdot a(j,p)$$
Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
add a comment |Â
up vote
0
down vote
Given $vecv = (v^1,...,v^n) in mathbbR^n$ and $f: mathbbE^n to mathbbR$. Then we define $vecv[f] = sum_j fracpartial fpartial x^j v^j$. Given a vector field $X: U to mathbbR^n$ we have $X_p in mathbbR^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $mathbbR^n_p$ such that,
$$X_p = sum_i a(i,p) e^i_p$$
- We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) in mathbbR^n$. Then we let $X[f]$ be the vector field such that,
$$ X[f] (p) := X_p[f] = sum_j fracpartial fpartial x^j cdot a(j,p)$$
Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given $vecv = (v^1,...,v^n) in mathbbR^n$ and $f: mathbbE^n to mathbbR$. Then we define $vecv[f] = sum_j fracpartial fpartial x^j v^j$. Given a vector field $X: U to mathbbR^n$ we have $X_p in mathbbR^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $mathbbR^n_p$ such that,
$$X_p = sum_i a(i,p) e^i_p$$
- We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) in mathbbR^n$. Then we let $X[f]$ be the vector field such that,
$$ X[f] (p) := X_p[f] = sum_j fracpartial fpartial x^j cdot a(j,p)$$
Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
Given $vecv = (v^1,...,v^n) in mathbbR^n$ and $f: mathbbE^n to mathbbR$. Then we define $vecv[f] = sum_j fracpartial fpartial x^j v^j$. Given a vector field $X: U to mathbbR^n$ we have $X_p in mathbbR^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $mathbbR^n_p$ such that,
$$X_p = sum_i a(i,p) e^i_p$$
- We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) in mathbbR^n$. Then we let $X[f]$ be the vector field such that,
$$ X[f] (p) := X_p[f] = sum_j fracpartial fpartial x^j cdot a(j,p)$$
Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
edited Jul 22 at 17:44
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
answered Jul 22 at 17:34
Faraad Armwood
7,4292619
7,4292619
add a comment |Â
add a comment |Â
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$X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture
– Calvin Khor
Jul 22 at 17:22
@CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector?
– p.h.f.
Jul 22 at 18:13
1
en.wikipedia.org/wiki/Lie_derivative
– Qiaochu Yuan
Jul 22 at 18:20
@p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic.
– Calvin Khor
Jul 22 at 18:25
The X(f)(p) in my comment should be X(p)(f).
– p.h.f.
Jul 22 at 18:26