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

 $d\alpha(X,Y)=(\alpha_{j,i}-\alpha_{i,j})X^{i}Y^{j}=\alpha_{j,i}X^{i}Y^{j}-% \alpha_{i,j}X^{i}Y^{j}$
 $X(\alpha(Y))=X^{i}\partial_{i}(\alpha_{j}Y^{j})=X^{i}\alpha_{j,i}Y^{j}+X^{i}% \alpha_{j}{Y^{j}}_{,i}$
 $Y(\alpha(X))=Y^{j}\partial_{j}(\alpha_{i}X^{i})=Y^{j}\alpha_{i,j}X^{i}+Y^{j}% \alpha_{i}{X^{i}}_{,j}$
 $\alpha([X,Y])=\alpha_{i}(X^{j}{Y^{i}}_{,j}-Y^{j}{X^{i}}_{,j})=\alpha_{i}X^{j}{% Y^{i}}_{,j}-\alpha_{i}Y^{j}{X^{i}}_{,j}$

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) ProofOfDalphaXYXalphaYYalphaXalphaXYlocalCoordinates 2013-03-22 15:34:01 2013-03-22 15:34:01 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Proof msc 53-00