Lie bracket of vector fields in Penrose's abstract index notationWhich tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields?Lie group actions and f-relatednessIs there much theory of superalgebras acting on manifolds by alternating polyvector fields?Lie bracket of Invariant Vector fields Splitting of the double tangent bundle into vertical and horizontal parts, and defining partial derivativesNormalized Hamiltonian holomorphic vector fields on Sasakian manifoldsDoes for every vector field there always exist a volume form for which the vector field is a homothety?Properties of connection Laplacian on vector fieldsFunctoriality of the formality quasi-isomorphism of E-polydifferential operatorsNotation and geometry facts in a paper on the Diederich-Fornæss index
Lie bracket of vector fields in Penrose's abstract index notation
Which tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields?Lie group actions and f-relatednessIs there much theory of superalgebras acting on manifolds by alternating polyvector fields?Lie bracket of Invariant Vector fields Splitting of the double tangent bundle into vertical and horizontal parts, and defining partial derivativesNormalized Hamiltonian holomorphic vector fields on Sasakian manifoldsDoes for every vector field there always exist a volume form for which the vector field is a homothety?Properties of connection Laplacian on vector fieldsFunctoriality of the formality quasi-isomorphism of E-polydifferential operatorsNotation and geometry facts in a paper on the Diederich-Fornæss index
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In the abstract index notation of Penrose, indicies serve as placeholders to indicate the type of a tensor field. For example, $X^i$ denotes a vector field. What is the commonly accepted notation for the Lie bracket of two vector fields $X^i$ and $Y^j$?
Clearly, $[X^i, Y^j]^k$ does not work, because this would denote a $3$-contravariant tensor. Something like $[ cdot, cdot]_ij^ k X^i Y^j$ would work but looks strange.
dg.differential-geometry
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
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In the abstract index notation of Penrose, indicies serve as placeholders to indicate the type of a tensor field. For example, $X^i$ denotes a vector field. What is the commonly accepted notation for the Lie bracket of two vector fields $X^i$ and $Y^j$?
Clearly, $[X^i, Y^j]^k$ does not work, because this would denote a $3$-contravariant tensor. Something like $[ cdot, cdot]_ij^ k X^i Y^j$ would work but looks strange.
dg.differential-geometry
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
In the abstract index notation of Penrose, indicies serve as placeholders to indicate the type of a tensor field. For example, $X^i$ denotes a vector field. What is the commonly accepted notation for the Lie bracket of two vector fields $X^i$ and $Y^j$?
Clearly, $[X^i, Y^j]^k$ does not work, because this would denote a $3$-contravariant tensor. Something like $[ cdot, cdot]_ij^ k X^i Y^j$ would work but looks strange.
dg.differential-geometry
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
$endgroup$
In the abstract index notation of Penrose, indicies serve as placeholders to indicate the type of a tensor field. For example, $X^i$ denotes a vector field. What is the commonly accepted notation for the Lie bracket of two vector fields $X^i$ and $Y^j$?
Clearly, $[X^i, Y^j]^k$ does not work, because this would denote a $3$-contravariant tensor. Something like $[ cdot, cdot]_ij^ k X^i Y^j$ would work but looks strange.
dg.differential-geometry
dg.differential-geometry
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
asked Jul 3 at 17:41
Tobias DiezTobias Diez
2,75213 silver badges34 bronze badges
2,75213 silver badges34 bronze badges
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2 Answers
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Penrose-Rindler write it $X^inabla_iY^j-Y^inabla_iX^j$.
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
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Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
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– Francois Ziegler
Jul 4 at 16:22
add a comment |
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I think it is standard in the mathematical physics literature to write this as $[X,Y]^k$. The entire expression "$[X,Y]$" is a new vector and $k$ is its index.
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
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$begingroup$
Penrose-Rindler write it $X^inabla_iY^j-Y^inabla_iX^j$.
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
$endgroup$
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
add a comment |
$begingroup$
Penrose-Rindler write it $X^inabla_iY^j-Y^inabla_iX^j$.
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
$endgroup$
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
add a comment |
$begingroup$
Penrose-Rindler write it $X^inabla_iY^j-Y^inabla_iX^j$.
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
$endgroup$
Penrose-Rindler write it $X^inabla_iY^j-Y^inabla_iX^j$.
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
answered Jul 3 at 18:45
Francois ZieglerFrancois Ziegler
21.2k3 gold badges78 silver badges124 bronze badges
21.2k3 gold badges78 silver badges124 bronze badges
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
add a comment |
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
$begingroup$
Well, actually, I wanted to write the (usual) definition of the torsion tensor using indicies ;-). But thanks anyway.
$endgroup$
– Tobias Diez
Jul 4 at 13:49
1
1
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
$begingroup$
@TobiasDiez You can! Above is their formula (4.3.1) which assumes $nabla$ torsion-free (e.g. $nabla_i=partial_i$). In general, their (4.3.29) defines another $tildenabla$’s torsion $T$ by $$[X,Y]^j=X^itildenabla_iY^j-Y^itildenabla_iX^j+T_ik^jX^iY^k.$$
$endgroup$
– Francois Ziegler
Jul 4 at 16:22
add a comment |
$begingroup$
I think it is standard in the mathematical physics literature to write this as $[X,Y]^k$. The entire expression "$[X,Y]$" is a new vector and $k$ is its index.
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
$endgroup$
add a comment |
$begingroup$
I think it is standard in the mathematical physics literature to write this as $[X,Y]^k$. The entire expression "$[X,Y]$" is a new vector and $k$ is its index.
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
$endgroup$
add a comment |
$begingroup$
I think it is standard in the mathematical physics literature to write this as $[X,Y]^k$. The entire expression "$[X,Y]$" is a new vector and $k$ is its index.
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
$endgroup$
I think it is standard in the mathematical physics literature to write this as $[X,Y]^k$. The entire expression "$[X,Y]$" is a new vector and $k$ is its index.
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
answered Jul 3 at 19:46
WillRWillR
434 bronze badges
434 bronze badges
add a comment |
add a comment |
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