# A proof for physicists of the proposition 5.1 in "Foundations of Differential Geometry Vol 1" by Kobayashi and Nomizu

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328 views My questions are :

1.  How to proof for physicists such proposition?

2.  What could be a physical application of that proposition?

edited Feb 26, 2018

+ 2 like - 0 dislike  Then, we must to prove that The proof is as follows:

$$R_{a}^{*} D \phi (X,Y)= D \phi (R_{a *} X , R_{a *} Y)=( (d \phi ) h) (R_{a *} X , R_{a *} Y)$$

$$R_{a}^{*} D \phi (X,Y)=d \phi( h(R_{a *} X),h(R_{a *} Y) )=d \phi( (h \circ R_{a *} )(X) , (h \circ R_{a *} )(Y) ) )$$

Using that we obtain

$$R_{a}^{*} D \phi (X,Y)=d \phi( ( R_{a *} \circ h )(X) , ( R_{a *} \circ h )(Y) ) ) =d \phi( R_{a *} ( h (X) ), R_{a *} (h (Y) )) )$$

it is to say

$$R_{a}^{*} D \phi (X,Y) = R_{a}^{*} d\phi (h(X),h(Y)) = (R_{a}^{*} \circ d)(\phi (h(X),h(Y)) )$$

Now using We obtain

$$R_{a}^{*} D \phi (X,Y) = ( d \circ R_{a}^{*} )(\phi (h(X),h(Y)) ) = d ( R_{a}^{*} \phi (h(X),h(Y)))$$

Finally, using we obtain

$$R_{a}^{*} D \phi (X,Y) = d ( \rho \left( a ^{-1} \right) \phi (h(X),h(Y))) =\rho \left( a ^{-1} \right) d \phi (h(X),h(Y))$$

$$R_{a}^{*} D \phi (X,Y) = \rho \left( a ^{-1} \right) D \phi (X,Y)$$

it is to say

$$R_{a}^{*} D \phi = \rho \left( a ^{-1} \right) D \phi$$

answered Feb 26, 2018 by (1,130 points)
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Consider an elementary particle with an internal degree of freedom described by a Lie group $G$.  Such particle is associated with a field which is represented by the pseudotensorial  r-form $\phi$ of type $( \rho,V )$.  The corresponding lagrangian has the form

$$L = \overline{(D \phi)}(D \phi)$$

and such lagrangian is invariant according with the following computation.

$$L ' = \overline{(D \phi)}'(D \phi)'$$

where the gauge transformation is given by

$$(D \phi)'= R_{a}^{*} D \phi = \rho \left( a ^{-1} \right) D \phi$$

Then we have that

$$L ' = \overline{(D \phi)} ' (D \phi) ' =\overline{(R_{a}^{*} D \phi )}(R_{a}^{*} D \phi ) =\overline{( \rho \left( a ^{-1} \right) D \phi )}( \rho \left( a ^{-1} \right) D \phi )$$

$$L ' =\overline{D \phi}( \rho \left( a ^{-1} \right))^{-1} ( \rho \left( a ^{-1} \right) D \phi )= \overline{D \phi} \rho \left( (a ^{-1})^{-1} \right) \rho \left( a ^{-1} \right) D \phi$$

$$L ' =\overline{D \phi} \rho(a)\rho \left( a ^{-1} \right) D \phi = \overline{D \phi} \rho(aa ^{-1}) D \phi = \overline{D \phi} \rho(e) D \phi$$

$$L ' =\overline{D \phi}.Id.D \phi= \overline{D \phi}.D \phi = L$$

answered Feb 28, 2018 by (1,130 points)
edited Feb 28, 2018 by juancho

How does one say pseudo-tensor $r$-form in more modern language? It's an r-form valued in the orientation line?

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