proof of $d\alpha (X,Y) = X(\alpha(Y))$ $$ $Y(\alpha(X))$ $ $ $\alpha([X,Y])$ (local coordinates)
proof of $d\alpha(X,Y)=X(\alpha(Y))$ $$ $Y(\alpha(X))$ $$ $\alpha([X,Y])$ (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 righthand 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 righthand side of the first equation. Hence, we have
 $d\alpha(X,Y)=X(\alpha(Y))Y(\alpha(X))\alpha([X,Y])$ 

Mathematics Subject Classification
5300
no label found