Exterior Derivative and Lie Derivative

Question

let $\alpha\in\Omega^1(U, \mathbb{R})$ be a 1-form on $U$ and $X, Y : U → \mathbb{R}^n$ vector fields. Verify that the exterior derivative can be calculated via $$d\alpha(X, Y ) = X \cdot\alpha(Y ) − Y \cdot \alpha(X) − \alpha([X, Y ])$$ where the smooth function $\alpha(X): U → \mathbb{R}$ is given by $$\alpha(X)(p) := \alpha_p(X(p))$$ for $p\in U$.

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So for this basically we just need to calculate the exterior derivative of $\alpha$ and then calculate the RHS and show they are equivalent. I have two problems so far; one conceptual and one more conventional/notational.

Conceptually I'm not sure how this will work out as the formula for the exterior derivative of a differential 1-form $\alpha$ = $\alpha_{i}dx_{i}$is given by $$d\alpha = \sum \frac{d\alpha_i}{dx_j} dx_i \wedge dx_j$$

while the RHS involves vector fields multiplied by one forms with a vector field as an argument and the lie bracket which has the formula $$[X,Y](p) = dY_{p}(X(p)) - dX_p(Y(p))$$

i.e., no wedge product "$\wedge$" to be seen. I'm not the most familiar with differential forms or their manipulation so can the wedge product perhapd be expressed as something more conventional?

• Also would the dx$_i$ be dx$_1$ = X and dx$_2$ = Y?

Notationally My main issue is with expressions like "$X \cdot\alpha(Y ) − Y \cdot \alpha(X)$" now to me that looks like "The vector field X times alpha of Y - the vector field Y times alpha of X" however it's seemed implied by my prof that this was saying more like "The derivative with respect to X of Alpha of Y minus the derivative with respect to Y of alpha of X".

• Given what you smart people know about the exterior derivative can you help clarify this for me?

Answer 1

Write $\alpha=\sum_i\alpha_idx_i$ and $X=(X_1,...,X_n)$, $\alpha(X)=\sum_i\alpha_iX_i$. This implies that $Y.\alpha(X)=\sum_i d\alpha_i(Y)X_i+\alpha_idX_i(Y)$.

$[X,Y]=dY.X-dY.X=(dY_1.X-dX_1.Y,...,dY_n.X-dX_n.Y)$ implies that $\alpha([X,Y])= \sum_i\alpha_i.(dY_i.X-dX_i.Y)$

We deduce that $X.\alpha(Y)-Y.\alpha(X)-\alpha([X,Y])=\sum_id\alpha_i(X)Y_i+\alpha_idY_i(X)-d\alpha_i(Y)X_i-\alpha_idX_i(Y)-\alpha_i.(dY_i.X-dX_i.Y)$ $=\sum_id\alpha_i(X)Y_i-d\alpha_i(Y)X_i$.

$d\alpha_i(X)Y_i=\sum_{i,j}d\alpha_i^jX_jY_i$. This implies that $X.\alpha(Y)-Y.\alpha(X)-\alpha([X,Y])=\sum_{i,j}d\alpha_i^j(X_jY_i-X_iY_j)=(\sum\alpha_i^jdx_i\wedge dx_j)(X,Y)$.

Answer 2

When you write $X \alpha(Y)$, for example, it means that you are acting with the vector field $X$ on the function $\alpha(Y)$, which does $$\Bbb R^n \ni p \mapsto \alpha_p(Y_p) \in \Bbb R.$$So this is again a function. Think of $\alpha(Y)$ as something, and $X(\alpha(Y))$ acting on that something.

That being said, you can check the formula in coordinates, since we can benefit from the fact that coordinate vector fields commute. I'll be lazy and use Einstein's convention. Write $\alpha = \alpha_i\,{\rm d}x^i$. Since both sided of the proposed equality are bilinear and anti-symmetric, it suffices to check our formula for $X = \partial_j$ and $Y = \partial_k$, with $j<k$.

On one hand: $$\begin{align}{\rm d}\alpha(\partial_j,\partial_k) &= {\rm d}\alpha_i \wedge {\rm d}x^i(\partial_j,\partial_k) \\ &= (\partial_\ell\alpha_i) {\rm d}x^\ell \wedge {\rm d}x^i(\partial_j,\partial_k) \\ &= (\partial_\ell\alpha_i) \begin{vmatrix}\delta^\ell_j & \delta^\ell_k \\ \delta^i_j & \delta^i_k\end{vmatrix} \\ &= (\partial_\ell \alpha_i)\delta^\ell_j\delta^i_k - (\partial_\ell\alpha_i)\delta^\ell_k\delta^i_j \\ &= \partial_j\alpha_k - \partial_k\alpha_j.\end{align}$$On the other hand: $$\partial_j (\alpha(\partial_k)) - \partial_k(\alpha(\partial_j)) - \require{cancel}\cancelto{0}{[\partial_j,\partial_k]} = \partial_j\alpha_k - \partial_k \alpha_j,$$as wanted.