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

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 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 by (1,105 points)
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 (1,105 points)
edited Mar 26 by juancho

 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$\hbar$ysic$\varnothing$OverflowThen 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.