Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  A proof for physicists of Theorem 5.2 in "Foundations of Differential Geometry Vol 1" by Kobayashi and Nomizu

+ 2 like - 0 dislike
1156 views

Please consider the following theorem

My question is:  How to prove this theorem for physicists?

asked Mar 24, 2019 in Mathematics by juancho (1,130 points) [ revision history ]

hi Juancho! What editon do you cite ( and what page ) ? I always wonder what is the exact definition of 'sectional curvature' and its geometrical meaning.

Hi igael.  I am studying the edition of 1963. The theorem 5.2 is on page 77.  I am trying to get a proof for physicists as an alternative respect to the proofs by Kobayashi-Nomizu and Nakahara.

Kobayashi-Nomizu and Nakahara take the formula for the curvature as a given expression and then they proceed to verify it. I am trying to derive the formula itself from the scratch. Please let me know if you consider that I am obtaining my goal.

2 Answers

+ 2 like - 0 dislike

A possible proof for physicists of the theorem 5.2  is as follows:

We will use the following basic facts:

1.  $ X = X _H + X_V$

2.  $ \omega \left( X_{{H}} \right) =0$

3.  $\omega \left( X_{{V}} \right) =X_{V}^* = constant$

4.  $ Z  \omega \left( X_{{V}} \right) =0 $

5.  $  [X_H , Y_V] = Z_H$

5.a.   $ \omega ([X_H , Y_V]) = \omega(Z_H) = 0$

6.  $ \omega ([X_V , Y_V]) = [\omega(X_V) ,  \omega(Y_V) ] $

Proof :  Given that   $[X_V , Y_V]^* = [X_V  ^*,Y_V  ^*]$ then according with the fact 3 : $ \omega ([X_V , Y_V])  =  [X_V , Y_V]^* $ and then : $ \omega ([X_V , Y_V])  =  [X_V ^* , Y_V^* ] $; finally using again the fact 3 we obtain : $ \omega ([X_V , Y_V]) =  [\omega(X_V) ,  \omega(Y_V) ] $.

7.  $2 d\omega(X,Y)=  X \omega(Y) - Y \omega(X) -  \omega ([X,Y])$

8.   $\Omega \left( X,Y \right)  = d\omega(X_H,Y_H)  $

From the fact 8 we write:

$$\Omega \left( X,Y \right)  = d\omega(X_H,Y_H) $$

Using  fact 1 we have: that

$$\Omega \left( X,Y \right)  = d\omega( X - X_V,Y - Y_V) $$

which by bi-linearity is rewritten as

$$\Omega \left( X,Y \right)  = d\omega(X , Y) -  d\omega(X , Y_V) -  d\omega(X_V , Y) +  d\omega(X_V , Y_V) $$

Applying the fact 7  respectively to the three last terms of the right hand side of the last equation we have that

$$\Omega \left( X,Y \right)  = d\omega(X , Y) -  {\frac {1}{2}} X \omega(Y_V) + {\frac {1}{2}} Y_V \omega(X) + {\frac {1}{2}}\omega ([X,Y_V])   - $$

$${\frac {1}{2}} X_V \omega(Y) + {\frac {1}{2}} Y \omega(X_V) + {\frac {1}{2}}\omega ([X_V,Y]) + $$

$${\frac {1}{2}} X_V \omega(Y_V) - {\frac {1}{2}} Y_V \omega(X_V) -  {\frac {1}{2}}\omega ([X_V,Y_V]) $$

Applying the fact 4 we have that :   $X \omega(Y_V)  =  Y \omega(X_V) =  X_V \omega(Y_V) =Y_V \omega(X_V) = 0 $.  Using such results the main equation is reduced to

$$\Omega \left( X,Y \right)  = d\omega(X , Y)  + {\frac {1}{2}} Y_V \omega(X) + {\frac {1}{2}}\omega ([X,Y_V])   - $$

$${\frac {1}{2}} X_V \omega(Y) + {\frac {1}{2}}\omega ([X_V,Y]) - {\frac {1}{2}}\omega ([X_V,Y_V]) $$

Now, using the fact 1 in the second, third, fourth and fifth terms of the right hand side of the last equation we obtain that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)  + {\frac {1}{2}} Y_V \omega(X_H  + X_V) + {\frac {1}{2}}\omega ([X_H + X_V,Y_V])   - $$

$${\frac {1}{2}} X_V \omega(Y_H + Y_V) + {\frac {1}{2}}\omega ([X_V,Y_H+Y_V]) - {\frac {1}{2}}\omega ([X_V,Y_V]) $$

By linearity we have that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)  + {\frac {1}{2}} Y_V \omega(X_H)  + {\frac {1}{2}} Y_V \omega(X_V) + {\frac {1}{2}}\omega ([X_H ,Y_V]+[ X_V,Y_V])   - $$

$${\frac {1}{2}} X_V \omega(Y_H) - {\frac {1}{2}} Y_V \omega(Y_V) + {\frac {1}{2}}\omega ([X_V,Y_H]+[X_V,Y_V]) - {\frac {1}{2}}\omega ([X_V,Y_V]) $$

Using the facts 2 and 4 the last equation is reduced to

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}\omega ([X_H ,Y_V]+[ X_V,Y_V])  + $$

$${\frac {1}{2}}\omega ([X_V,Y_H]+[X_V,Y_V]) - {\frac {1}{2}}\omega ([X_V,Y_V]) $$

Using again linearity we obtain that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}\omega ([X_H ,Y_V])+ {\frac {1}{2}}\omega([ X_V,Y_V])  + $$

$${\frac {1}{2}}\omega ([X_V,Y_H])+ {\frac {1}{2}}\omega([X_V,Y_V]) - {\frac {1}{2}}\omega ([X_V,Y_V]) $$

Simplifying the last equation we have that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}\omega ([X_H ,Y_V])+ {\frac {1}{2}}\omega([ X_V,Y_V])  + {\frac {1}{2}}\omega ([X_V,Y_H])$$

Using the fact 5 we obtain that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}\omega (Z_H)+ {\frac {1}{2}}\omega([ X_V,Y_V])  + {\frac {1}{2}}\omega (W_H)$$

Using again the fact 2, the last equation is reduced to

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}\omega([ X_V,Y_V]) $$

Now using the fact 6 the last equation is transformed to

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}[\omega(X_V) ,  \omega(Y_V) ] $$

Using the fact 1 in the second term of the right hand side of the last equation we have that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}[\omega(X - X_H) ,  \omega(Y - Y_H) ] $$

By linearity we obtain that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}[\omega(X) - \omega(X_H) ,  \omega(Y) -\omega (Y_H) ] $$

Finally using the fact 2 we derive that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}[\omega(X) ,  \omega(Y) ] $$

and the Theorem 5.2 is proved.

The structure equation (often called "the structure equation of Elie Cartan") is sometimes written, for the sake of simplicity,  as follows:

$$\Omega   = d\omega  + {\frac {1}{2}}[\omega,  \omega] $$

The corresponding expression for the Yang-Mills field is

$$F \left( X,Y \right)  = dA(X , Y)   + {\frac {1}{2}}[A(X) ,  A(Y) ] $$

or, for the sake of simplicity,  as follows:

$$F   = dA + {\frac {1}{2}}[A, A] $$

answered Mar 24, 2019 by juancho (1,130 points) [ revision history ]
edited Mar 26, 2019 by juancho
+ 1 like - 0 dislike

As an application of the theorem 5.2 we will prove the following

According with the theorem  5.2 we have that

$$\Omega \left( X,Y \right)  = d\omega(X , Y)   + {\frac {1}{2}}[\omega(X) ,  \omega(Y) ] $$

Then

$$\Omega \left( X_H,Y_H \right)  = d\omega(X_H , Y_H)   + {\frac {1}{2}}[\omega(X_H) ,  \omega(Y_H) ] $$

Using the fact 2 we obtain

$$\Omega \left( X_H,Y_H \right)  = d\omega(X_H , Y_H)   + {\frac {1}{2}}[0 , 0] $$

which is reduced to

$$\Omega \left( X_H,Y_H \right)  = d\omega(X_H , Y_H)  $$

Now, using the fact 7 l the last equation is transformed to

$$\Omega \left( X_H,Y_H \right)  = {\frac {1}{2}} X_H \omega(Y_H) - {\frac {1}{2}} Y_H \omega(X_H) - {\frac {1}{2}}\omega ([X_H,Y_H]) $$

Using again the fact 2, the last equation is reduced to

$$\Omega \left( X_H,Y_H \right)  = - {\frac {1}{2}}\omega ([X_H,Y_H]) $$

and then the corollary 5.3 is proved.

answered Mar 26, 2019 by juancho (1,130 points) [ revision history ]
edited Mar 26, 2019 by juancho

Your answer

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):
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$\hbar$ysicsOverflo$\varnothing$
Then 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).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...