proof of $d\alpha (X,Y)=X(\alpha (Y))$ $-$ $Y(\alpha (X))$ $-$ $\alpha ([X,Y])$ (local coordinates)

Since this result is local (in other words, the identity holds on the whole manifold if and only if its restriction to every coordinate patch of the manifold holds), it suffices to demonstrate it in a local coordinate system. To do this, we shall compute coordinate expressions for the various terms and verify that the sum of terms on the right-hand side equals the left-hand side.

$$d\alpha (X,Y)=({\alpha }_{j,i}-{\alpha }_{i,j})X^i{Y}^j={\alpha }_{j,i}{X}^i{Y}^j-{\alpha }_{i,j}{X}^i{Y}^j$$

$$X(\alpha (Y))=X^i{\partial }_i({\alpha }_j{Y}^j)=X^i{\alpha }_{j,i}{Y}^j+X^i{\alpha }_{j}Y^j{}_{,i}$$

$$Y(\alpha (X))=Y^j{\partial }_j({\alpha }_i{X}^i)=Y^j{\alpha }_{i,j}{X}^i+Y^j{\alpha }_{i}X^i{}_{,j}$$

$$\alpha ([X,Y])={\alpha }_i(X^{j}Y^i{}_{,j}-Y^{j}X^i{}_{,j})={\alpha }_i{X}^{j}Y^i{}_{,j}-{\alpha }_i{Y}^{j}X^i{}_{,j}$$

Upon combining the right-hand sides of the last three equations and cancelling common terms, we obtain

$$X^i{\alpha }_{j,i}{Y}^j+X^i{\alpha }_{j}Y^j{}_{,i}-Y^j{\alpha }_{i,j}{X}^i-{\alpha }_i{X}^{j}Y^i{}_{,j}$$

Upon renaming dummy indices (switching $i$ with $j$), the second and fourth terms cancel. What remains is exactly the right-hand side of the first equation. Hence, we have

$$d\alpha (X,Y)=X(\alpha (Y))-Y(\alpha (X))-\alpha ([X,Y])$$

Title	proof of $d\alpha (X,Y)=X(\alpha (Y))$ $-$ $Y(\alpha (X))$ $-$ $\alpha ([X,Y])$ (local coordinates)
Canonical name	ProofOfDalphaXYXalphaYYalphaXalphaXYlocalCoordinates
Date of creation	2013-03-22 15:34:01
Last modified on	2013-03-22 15:34:01
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	8
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 53-00