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

+ 2 like - 0 dislike
162 views

Please consider the following theorem

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

asked Mar 24 in Mathematics by juancho (1,105 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

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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 by juancho (1,105 points) [ revision history ]
edited Mar 26 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 by juancho (1,105 points) [ revision history ]
edited Mar 26 by juancho

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