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

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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.

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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 (1,105 points)
edited Mar 26 by juancho
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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

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