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

+ 3 like - 0 dislike
124 views

Please consider the proposition

My questions are : 

1.  How to proof for physicists such proposition?

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

asked Feb 26 in Theoretical Physics by juancho (1,030 points) [ revision history ]
edited Feb 26 by juancho

2 Answers

+ 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 by juancho (1,030 points) [ revision history ]
+ 2 like - 0 dislike

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 by juancho (1,030 points) [ revision history ]
edited Feb 28 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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