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Going from one notation to another in Yang-Mills
What does the symbol $operatornameTr$ in the Yang-Mills action mean?Formal Definition of Yang Mills LagrangianGauge covariant derivative on principal bundle over $mathbb R^d$Lecture notes on holomorphic Yang-Mills theoryGuidance regarding research in Mathematical PhysicsConcrete example of a gauge transformation of a vector potentialMathematics of Yang-Mills theoryYang–Mills theory and mass gapPrerequisites to Yang-Mills TheoryWhat constitutes a gauge theory? Help me understand electromagnetism as the prototype of all gauge theories
$begingroup$
In many books on Yang-Mills theories, written from a physicist's point of view, the curvature tensor is written as:
$$F_mu nu^a = partial_mu A_nu^a-partial_nu A_mu^a+gf^abcA_mu^bA_nu^c$$
with the covariant derivative as:
$$D_mu=Ipartial_mu-igT^aA^a_mu$$
where $I$ is unit matrix and $T$ are generators of the group and
$$ mathcalL = -frac12operatornameTr(F^2)=- frac14F^amu nu F_mu nu^a$$
this has been basically taken from Wikipedia
But in the book I have been using, with a more mathematical perspective (Baez & Muniain's Gauge Fields, Knots and Gravity) the vector 1-form $A$ is an $textEnd(E)$ valued $1$-form described as
$$A = A_mu i^j e^i otimes e_j otimes dx^mu $$
and the covariant derivative and curvature are
$$D_mu = partial_mu + A_mu,quad F_jk = partial_j A_k - partial_k A_j + [A_j, A_k].$$
How can I go from one notation to the other? I can't do it, I think I'm confused by the superscipts $a$ in the physicist's notation.
And what is $F_munu$ for a non-abelian gauge theory? A matrix?
mathematical-physics quantum-field-theory gauge-theory
$endgroup$
add a comment |
$begingroup$
In many books on Yang-Mills theories, written from a physicist's point of view, the curvature tensor is written as:
$$F_mu nu^a = partial_mu A_nu^a-partial_nu A_mu^a+gf^abcA_mu^bA_nu^c$$
with the covariant derivative as:
$$D_mu=Ipartial_mu-igT^aA^a_mu$$
where $I$ is unit matrix and $T$ are generators of the group and
$$ mathcalL = -frac12operatornameTr(F^2)=- frac14F^amu nu F_mu nu^a$$
this has been basically taken from Wikipedia
But in the book I have been using, with a more mathematical perspective (Baez & Muniain's Gauge Fields, Knots and Gravity) the vector 1-form $A$ is an $textEnd(E)$ valued $1$-form described as
$$A = A_mu i^j e^i otimes e_j otimes dx^mu $$
and the covariant derivative and curvature are
$$D_mu = partial_mu + A_mu,quad F_jk = partial_j A_k - partial_k A_j + [A_j, A_k].$$
How can I go from one notation to the other? I can't do it, I think I'm confused by the superscipts $a$ in the physicist's notation.
And what is $F_munu$ for a non-abelian gauge theory? A matrix?
mathematical-physics quantum-field-theory gauge-theory
$endgroup$
add a comment |
$begingroup$
In many books on Yang-Mills theories, written from a physicist's point of view, the curvature tensor is written as:
$$F_mu nu^a = partial_mu A_nu^a-partial_nu A_mu^a+gf^abcA_mu^bA_nu^c$$
with the covariant derivative as:
$$D_mu=Ipartial_mu-igT^aA^a_mu$$
where $I$ is unit matrix and $T$ are generators of the group and
$$ mathcalL = -frac12operatornameTr(F^2)=- frac14F^amu nu F_mu nu^a$$
this has been basically taken from Wikipedia
But in the book I have been using, with a more mathematical perspective (Baez & Muniain's Gauge Fields, Knots and Gravity) the vector 1-form $A$ is an $textEnd(E)$ valued $1$-form described as
$$A = A_mu i^j e^i otimes e_j otimes dx^mu $$
and the covariant derivative and curvature are
$$D_mu = partial_mu + A_mu,quad F_jk = partial_j A_k - partial_k A_j + [A_j, A_k].$$
How can I go from one notation to the other? I can't do it, I think I'm confused by the superscipts $a$ in the physicist's notation.
And what is $F_munu$ for a non-abelian gauge theory? A matrix?
mathematical-physics quantum-field-theory gauge-theory
$endgroup$
In many books on Yang-Mills theories, written from a physicist's point of view, the curvature tensor is written as:
$$F_mu nu^a = partial_mu A_nu^a-partial_nu A_mu^a+gf^abcA_mu^bA_nu^c$$
with the covariant derivative as:
$$D_mu=Ipartial_mu-igT^aA^a_mu$$
where $I$ is unit matrix and $T$ are generators of the group and
$$ mathcalL = -frac12operatornameTr(F^2)=- frac14F^amu nu F_mu nu^a$$
this has been basically taken from Wikipedia
But in the book I have been using, with a more mathematical perspective (Baez & Muniain's Gauge Fields, Knots and Gravity) the vector 1-form $A$ is an $textEnd(E)$ valued $1$-form described as
$$A = A_mu i^j e^i otimes e_j otimes dx^mu $$
and the covariant derivative and curvature are
$$D_mu = partial_mu + A_mu,quad F_jk = partial_j A_k - partial_k A_j + [A_j, A_k].$$
How can I go from one notation to the other? I can't do it, I think I'm confused by the superscipts $a$ in the physicist's notation.
And what is $F_munu$ for a non-abelian gauge theory? A matrix?
mathematical-physics quantum-field-theory gauge-theory
mathematical-physics quantum-field-theory gauge-theory
edited Mar 21 at 11:26
Andrews
1,2812422
1,2812422
asked Mar 20 at 19:20
DavidDavid
380211
380211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The superscript $a$ in these equations is the group index, that is, it runs over the a set of elements of the representation (I believe the adjoint representation) of the group. The commutator of two quantities $X^b$^ and $Y^c$ becomes $f^abcX^bY^c$ where $f^abc$ are the (fully antisymmetric structure constants for the group. For example, if the YM group is $SU(2)$ which is isomorphic to $SO(3)$ then $f^abc$ is the usual Levi-Civita symbol $varepsilon_abc$, which makes the expression into the familiar commutator.
In the B&M book the group index is suppressed; the implication is that $A_mu$ is formed from group members and is not a group invariant. I think in the expression for $F_jk$ in that book makes the further assumption that the normalization of $A$ is such that the "coupling constant" ($g$ in the physics text) is one.
$endgroup$
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
add a comment |
$begingroup$
This is an answer to your comment on Mark's answer, but a bit long for a comment so here it goes.
Presumably, in BM $A$ takes value in a Lie subalgebra of $mathfrakgl(mathbbR,n)$ and they are taking $e^iotimes e_j$ as a basis for it (think of $e^iotimes e_j$ as the matrix with 1 in the position $(i,j)$ and $0$ elsewhere). So $A$ and $F$ will both carry two "group" indices (components with respect to a basis of the Lie algebra really), and the appropriate number of Greek cotangent space indices.
In other books, especially when dealing with specific Lie algebras, you label a basis of the Lie algebra with a single index, e.g. for $mathfraksu(2)$ you could take $t_k = i sigma_k$ with $sigma_k$ the Pauli matrices. You then expand
$A=A^i_mu t_iotimes dx ^mu$ which is the notation used in Wikipedia.
$endgroup$
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
The superscript $a$ in these equations is the group index, that is, it runs over the a set of elements of the representation (I believe the adjoint representation) of the group. The commutator of two quantities $X^b$^ and $Y^c$ becomes $f^abcX^bY^c$ where $f^abc$ are the (fully antisymmetric structure constants for the group. For example, if the YM group is $SU(2)$ which is isomorphic to $SO(3)$ then $f^abc$ is the usual Levi-Civita symbol $varepsilon_abc$, which makes the expression into the familiar commutator.
In the B&M book the group index is suppressed; the implication is that $A_mu$ is formed from group members and is not a group invariant. I think in the expression for $F_jk$ in that book makes the further assumption that the normalization of $A$ is such that the "coupling constant" ($g$ in the physics text) is one.
$endgroup$
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
add a comment |
$begingroup$
The superscript $a$ in these equations is the group index, that is, it runs over the a set of elements of the representation (I believe the adjoint representation) of the group. The commutator of two quantities $X^b$^ and $Y^c$ becomes $f^abcX^bY^c$ where $f^abc$ are the (fully antisymmetric structure constants for the group. For example, if the YM group is $SU(2)$ which is isomorphic to $SO(3)$ then $f^abc$ is the usual Levi-Civita symbol $varepsilon_abc$, which makes the expression into the familiar commutator.
In the B&M book the group index is suppressed; the implication is that $A_mu$ is formed from group members and is not a group invariant. I think in the expression for $F_jk$ in that book makes the further assumption that the normalization of $A$ is such that the "coupling constant" ($g$ in the physics text) is one.
$endgroup$
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
add a comment |
$begingroup$
The superscript $a$ in these equations is the group index, that is, it runs over the a set of elements of the representation (I believe the adjoint representation) of the group. The commutator of two quantities $X^b$^ and $Y^c$ becomes $f^abcX^bY^c$ where $f^abc$ are the (fully antisymmetric structure constants for the group. For example, if the YM group is $SU(2)$ which is isomorphic to $SO(3)$ then $f^abc$ is the usual Levi-Civita symbol $varepsilon_abc$, which makes the expression into the familiar commutator.
In the B&M book the group index is suppressed; the implication is that $A_mu$ is formed from group members and is not a group invariant. I think in the expression for $F_jk$ in that book makes the further assumption that the normalization of $A$ is such that the "coupling constant" ($g$ in the physics text) is one.
$endgroup$
The superscript $a$ in these equations is the group index, that is, it runs over the a set of elements of the representation (I believe the adjoint representation) of the group. The commutator of two quantities $X^b$^ and $Y^c$ becomes $f^abcX^bY^c$ where $f^abc$ are the (fully antisymmetric structure constants for the group. For example, if the YM group is $SU(2)$ which is isomorphic to $SO(3)$ then $f^abc$ is the usual Levi-Civita symbol $varepsilon_abc$, which makes the expression into the familiar commutator.
In the B&M book the group index is suppressed; the implication is that $A_mu$ is formed from group members and is not a group invariant. I think in the expression for $F_jk$ in that book makes the further assumption that the normalization of $A$ is such that the "coupling constant" ($g$ in the physics text) is one.
answered Mar 20 at 19:38
Mark FischlerMark Fischler
33.9k12552
33.9k12552
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
add a comment |
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
$begingroup$
Then, in BM, the superscipt index $j$ in $A = A_mu i^j e^i otimes e_j otimes dx^mu$ is not the group index, if I understood correctly. Another extra superscipt should be added. Could you please explain more detailedly the fact that the vector potential "is formed from group members and is not a group invariant". And then, if $F_jk$ and $F_jk^a$ what objects are? Are both matrices or the first is a collection of matrices? Thx
$endgroup$
– David
Mar 20 at 19:57
1
1
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
$begingroup$
I'll use $SU(2)$ as an example group. Then the "vector potential" $A_mu^a$ is a $4$-vector under local Lorentz transformations, but the four components are each an element of the adjoint representation of $SU(2)$. That is, each is $2times 2$ unitary complex matrix with determinant $1$. When you apply a gauge transformation, the elements change -- that is what I meant by not being group invariants. Similarly, each of the non-zero elements of the antisymmetric $F_munu$ is a $2times 2$$ $SU(2)$ element.
$endgroup$
– Mark Fischler
Mar 21 at 2:40
add a comment |
$begingroup$
This is an answer to your comment on Mark's answer, but a bit long for a comment so here it goes.
Presumably, in BM $A$ takes value in a Lie subalgebra of $mathfrakgl(mathbbR,n)$ and they are taking $e^iotimes e_j$ as a basis for it (think of $e^iotimes e_j$ as the matrix with 1 in the position $(i,j)$ and $0$ elsewhere). So $A$ and $F$ will both carry two "group" indices (components with respect to a basis of the Lie algebra really), and the appropriate number of Greek cotangent space indices.
In other books, especially when dealing with specific Lie algebras, you label a basis of the Lie algebra with a single index, e.g. for $mathfraksu(2)$ you could take $t_k = i sigma_k$ with $sigma_k$ the Pauli matrices. You then expand
$A=A^i_mu t_iotimes dx ^mu$ which is the notation used in Wikipedia.
$endgroup$
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
add a comment |
$begingroup$
This is an answer to your comment on Mark's answer, but a bit long for a comment so here it goes.
Presumably, in BM $A$ takes value in a Lie subalgebra of $mathfrakgl(mathbbR,n)$ and they are taking $e^iotimes e_j$ as a basis for it (think of $e^iotimes e_j$ as the matrix with 1 in the position $(i,j)$ and $0$ elsewhere). So $A$ and $F$ will both carry two "group" indices (components with respect to a basis of the Lie algebra really), and the appropriate number of Greek cotangent space indices.
In other books, especially when dealing with specific Lie algebras, you label a basis of the Lie algebra with a single index, e.g. for $mathfraksu(2)$ you could take $t_k = i sigma_k$ with $sigma_k$ the Pauli matrices. You then expand
$A=A^i_mu t_iotimes dx ^mu$ which is the notation used in Wikipedia.
$endgroup$
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
add a comment |
$begingroup$
This is an answer to your comment on Mark's answer, but a bit long for a comment so here it goes.
Presumably, in BM $A$ takes value in a Lie subalgebra of $mathfrakgl(mathbbR,n)$ and they are taking $e^iotimes e_j$ as a basis for it (think of $e^iotimes e_j$ as the matrix with 1 in the position $(i,j)$ and $0$ elsewhere). So $A$ and $F$ will both carry two "group" indices (components with respect to a basis of the Lie algebra really), and the appropriate number of Greek cotangent space indices.
In other books, especially when dealing with specific Lie algebras, you label a basis of the Lie algebra with a single index, e.g. for $mathfraksu(2)$ you could take $t_k = i sigma_k$ with $sigma_k$ the Pauli matrices. You then expand
$A=A^i_mu t_iotimes dx ^mu$ which is the notation used in Wikipedia.
$endgroup$
This is an answer to your comment on Mark's answer, but a bit long for a comment so here it goes.
Presumably, in BM $A$ takes value in a Lie subalgebra of $mathfrakgl(mathbbR,n)$ and they are taking $e^iotimes e_j$ as a basis for it (think of $e^iotimes e_j$ as the matrix with 1 in the position $(i,j)$ and $0$ elsewhere). So $A$ and $F$ will both carry two "group" indices (components with respect to a basis of the Lie algebra really), and the appropriate number of Greek cotangent space indices.
In other books, especially when dealing with specific Lie algebras, you label a basis of the Lie algebra with a single index, e.g. for $mathfraksu(2)$ you could take $t_k = i sigma_k$ with $sigma_k$ the Pauli matrices. You then expand
$A=A^i_mu t_iotimes dx ^mu$ which is the notation used in Wikipedia.
answered Mar 20 at 21:40
GFRGFR
3,3241023
3,3241023
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
add a comment |
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
$begingroup$
Thank you. Then, can we say that there is no way to go from one notation to the other? I mean, is the maximum we can say about how to transform one notation to the other one that $A_mu^i t_iotimes dx^mu = A^j_mu ie^iotimes e_j$?
$endgroup$
– David
Mar 20 at 22:20
1
1
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
$begingroup$
Well, the relation is, omitting the spacetime indices, $A^a t_a=A ^a (t_a) ^j_ i e^i otimes e _j$, so $A^j_ i= A ^a (t_a) ^j_ i$. Note that $(t_a) ^j_ i$ is a number, not a matrix.
$endgroup$
– GFR
Mar 21 at 8:45
add a comment |
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