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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













1












$begingroup$


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?










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    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?










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      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?










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 21 at 11:26









      Andrews

      1,2812422




      1,2812422










      asked Mar 20 at 19:20









      DavidDavid

      380211




      380211




















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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













          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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
















          • $begingroup$
            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











          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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













          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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
















          • $begingroup$
            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

















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