Tangent and Cotangent Bundles - Key Proof

We prove Cartan's formula relating the Lie derivative, exterior derivative, and interior product. This fundamental identity reveals the deep structure underlying differential forms.

ProofCartan's Formula

Statement: For any vector field $X$ and differential $k$ -form $\omega$ , we have

\mathcal{L}_X\omega = (i_X \circ d + d \circ i_X)\omega = i_X(d\omega) + d(i_X\omega)

Proof: We proceed by induction on the degree of $\omega$ .

Base case ( $k=0$ ): For $f \in C^\infty(M)$ , we have $i_Xf = 0$ (since $f$ is a 0-form), so

(i_X \circ d + d \circ i_X)f = i_X(df) + d(0) = df(X) = X(f) = \mathcal{L}_Xf

Base case ( $k=1$ ): For a 1-form $\omega$ and vector fields $Y$ , we compute:

(\mathcal{L}_X\omega)(Y) = \mathcal{L}_X(\omega(Y)) - \omega(\mathcal{L}_XY) = X(\omega(Y)) - \omega([X,Y])

On the other hand:

i_X(d\omega)(Y) = (d\omega)(X, Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])

d(i_X\omega)(Y) = Y(\omega(X))

Adding: $(i_X d + d i_X)\omega(Y) = X(\omega(Y)) - \omega([X,Y]) = \mathcal{L}_X\omega(Y)$ .

Inductive step: Suppose the formula holds for forms of degree $< k$ . For a $k$ -form $\omega$ :

First, note that both sides satisfy the Leibniz rule with respect to the wedge product. It suffices to check on decomposable forms $\omega = \alpha \wedge \beta$ where $\alpha$ has degree 1.

Using the Leibniz rule and inductive hypothesis:

\mathcal{L}_X(\alpha \wedge \beta) = \mathcal{L}_X\alpha \wedge \beta + \alpha \wedge \mathcal{L}_X\beta

= (i_Xd\alpha + di_X\alpha) \wedge \beta + \alpha \wedge (i_Xd\beta + di_X\beta)

Similarly computing $(i_Xd + di_X)(\alpha \wedge \beta)$ using the product rules for $d$ and $i_X$ yields the same expression. $\square$

■

Remark

Cartan's formula is more than a computational tool - it reveals that the Lie derivative is built from more fundamental operations. In the language of cohomology, it shows $\mathcal{L}_X$ is a chain homotopy between $i_X \circ d$ and $d \circ i_X$ .

ProofLie Bracket as Commutator

We show that for vector fields $X, Y$ , the Lie bracket $[X,Y]$ is indeed the commutator of differential operators:

For any $f \in C^\infty(M)$ :

[X,Y](f) = X(Y(f)) - Y(X(f))

To verify this is a derivation (satisfies Leibniz rule), compute:

[X,Y](fg) = X(Y(fg)) - Y(X(fg))

= X(Y(f) \cdot g + f \cdot Y(g)) - Y(X(f) \cdot g + f \cdot X(g))

Expanding using the Leibniz rule for $X$ and $Y$ individually:

= XY(f) \cdot g + Y(f) \cdot X(g) + X(f) \cdot Y(g) + f \cdot XY(g)

- YX(f) \cdot g - X(f) \cdot Y(g) - Y(f) \cdot X(g) - f \cdot YX(g)

Canceling terms yields:

= [X,Y](f) \cdot g + f \cdot [X,Y](g)

Thus $[X,Y]$ satisfies the Leibniz rule and is a well-defined vector field. $\square$

■

ExampleVerification in Coordinates

In local coordinates, $X = X^i\frac{\partial}{\partial x^i}$ and $Y = Y^j\frac{\partial}{\partial x^j}$ . Then:

X(Y(f)) = X^i \frac{\partial}{\partial x^i}\left(Y^j \frac{\partial f}{\partial x^j}\right) = X^i Y^j \frac{\partial^2 f}{\partial x^i \partial x^j} + X^i \frac{\partial Y^j}{\partial x^i} \frac{\partial f}{\partial x^j}

By symmetry of mixed partials, the second-order terms cancel in $[X,Y](f)$ , leaving only the first-order terms that define the coordinate expression for $[X,Y]$ .