On the notion of tensor in Riemannian Geometry
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In DoCarmo’s Riemannian Geometry, a tensor of order $r$ on a Riemannian manifold is defined as a multilinear mapping $$T:Xi^r(M)rightarrow C^{infty}(M)$$
where $M$ is a smooth manifold of dimension $n$, $Xi(M)$ denotes the module of smooth vector fields on $C^{infty}(M)$.
DoCarmo wants to show that $T$ is a pointwise object in the following sense: “Fix a point $pin M$ and let $U$ be a neighborhood of $p$ in $M$ on which it is possible to define vector fields $E_1,...,E_nin Xi(M)$, in such a fashion that at each $qin U$, the vectors ${E_i(q)}$,$i=1,...,n$, form a basis of $T_qM$. ...”
Does the author mean that for each neighborhood of $p$ there exists such vector fields, or what? If so, how can one define such vector fields?
riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 '18 at 15:09
User12239User12239
441216
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add a comment |
$begingroup$
In DoCarmo’s Riemannian Geometry, a tensor of order $r$ on a Riemannian manifold is defined as a multilinear mapping $$T:Xi^r(M)rightarrow C^{infty}(M)$$
where $M$ is a smooth manifold of dimension $n$, $Xi(M)$ denotes the module of smooth vector fields on $C^{infty}(M)$.
DoCarmo wants to show that $T$ is a pointwise object in the following sense: “Fix a point $pin M$ and let $U$ be a neighborhood of $p$ in $M$ on which it is possible to define vector fields $E_1,...,E_nin Xi(M)$, in such a fashion that at each $qin U$, the vectors ${E_i(q)}$,$i=1,...,n$, form a basis of $T_qM$. ...”
Does the author mean that for each neighborhood of $p$ there exists such vector fields, or what? If so, how can one define such vector fields?
riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 '18 at 15:09
User12239User12239
441216
$endgroup$
$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
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– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23
add a comment |
$begingroup$
In DoCarmo’s Riemannian Geometry, a tensor of order $r$ on a Riemannian manifold is defined as a multilinear mapping $$T:Xi^r(M)rightarrow C^{infty}(M)$$
where $M$ is a smooth manifold of dimension $n$, $Xi(M)$ denotes the module of smooth vector fields on $C^{infty}(M)$.
DoCarmo wants to show that $T$ is a pointwise object in the following sense: “Fix a point $pin M$ and let $U$ be a neighborhood of $p$ in $M$ on which it is possible to define vector fields $E_1,...,E_nin Xi(M)$, in such a fashion that at each $qin U$, the vectors ${E_i(q)}$,$i=1,...,n$, form a basis of $T_qM$. ...”
Does the author mean that for each neighborhood of $p$ there exists such vector fields, or what? If so, how can one define such vector fields?
riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 '18 at 15:09
User12239User12239
441216
$endgroup$
In DoCarmo’s Riemannian Geometry, a tensor of order $r$ on a Riemannian manifold is defined as a multilinear mapping $$T:Xi^r(M)rightarrow C^{infty}(M)$$
where $M$ is a smooth manifold of dimension $n$, $Xi(M)$ denotes the module of smooth vector fields on $C^{infty}(M)$.
DoCarmo wants to show that $T$ is a pointwise object in the following sense: “Fix a point $pin M$ and let $U$ be a neighborhood of $p$ in $M$ on which it is possible to define vector fields $E_1,...,E_nin Xi(M)$, in such a fashion that at each $qin U$, the vectors ${E_i(q)}$,$i=1,...,n$, form a basis of $T_qM$. ...”
Does the author mean that for each neighborhood of $p$ there exists such vector fields, or what? If so, how can one define such vector fields?
riemannian-geometry smooth-manifolds vector-fields
riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 '18 at 15:09
User12239User12239
441216
asked Dec 5 '18 at 15:09
User12239User12239
441216
edited Dec 5 '18 at 17:54
User12239
asked Dec 5 '18 at 15:09
User12239User12239
441216
asked Dec 5 '18 at 15:09
User12239User12239
441216
asked Dec 5 '18 at 15:09
User12239User12239
441216
441216
$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23
add a comment |
$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23
$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23
add a comment |
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Such fields do not exist on general neighborhoods of a point $p$ (since for example the whole manifold is a neighborhood of $p$). But for sufficiently small neighborhoods they do exist as the example of the coordinate vector fields of a chart containing $p$ shows.
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
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$begingroup$
Such fields do not exist on general neighborhoods of a point $p$ (since for example the whole manifold is a neighborhood of $p$). But for sufficiently small neighborhoods they do exist as the example of the coordinate vector fields of a chart containing $p$ shows.
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
Such fields do not exist on general neighborhoods of a point $p$ (since for example the whole manifold is a neighborhood of $p$). But for sufficiently small neighborhoods they do exist as the example of the coordinate vector fields of a chart containing $p$ shows.
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
Such fields do not exist on general neighborhoods of a point $p$ (since for example the whole manifold is a neighborhood of $p$). But for sufficiently small neighborhoods they do exist as the example of the coordinate vector fields of a chart containing $p$ shows.
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
$endgroup$
Such fields do not exist on general neighborhoods of a point $p$ (since for example the whole manifold is a neighborhood of $p$). But for sufficiently small neighborhoods they do exist as the example of the coordinate vector fields of a chart containing $p$ shows.
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
11.1k923
answered Dec 5 '18 at 15:18
Andreas CapAndreas Cap
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11.1k923
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$begingroup$
This is the natural follow-up to your previous question. In that case, the Riemann curvature tensor field is the map that describes what happens to a vector if you parallel transport around a loop. That map needs have three inputs: the vector, the shooting direction, the receiving direction. You have a map at each point of the manifold.
$endgroup$
– Giuseppe Negro
Dec 5 '18 at 15:13
$begingroup$
Yes my previous question arose from this question which at first made me think we could define a parallel vector field on any open set. In terms of semantics, DoCarmo’s wording in the translated version of his book is wrong.
$endgroup$
– User12239
Dec 5 '18 at 15:23