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

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




"proof of $d\alpha (X,Y) = X(\alpha(Y))$ $-$ $Y(\alpha(X))$ $-$ $\alpha([X,Y])$ (local coordinates)" is owned by rspuzio.
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Cross-references: indices, equations, side, sum, terms, expressions, local coordinate, coordinate, restriction, manifold, identity

This is version 5 of proof of $d\alpha (X,Y) = X(\alpha(Y))$ $-$ $Y(\alpha(X))$ $-$ $\alpha([X,Y])$ (local coordinates), born on 2005-11-08, modified 2005-11-08.
Object id is 7471, canonical name is ProofOfDalphaXYXalphaYYalphaXAlphaXY.
Accessed 1219 times total.

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AMS MSC53-00 (Differential geometry :: General reference works )

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