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

+ 3 like - 0 dislike
191 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, 2018
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 (1,105 points)
+ 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, 2018 by (1,105 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?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.