Questions on Cartan's magic formula $\mathcal{L}_X=i_X \circ d + d\circ i_X$

Question

Algebra $A$ is called graded algebra if it has a direct sum decomposition $A=\bigoplus_{k\in\Bbb Z} A^k$ s.t. product satisfies $(A^k)(A^l)\subseteq(A^{k+l}) \text{ for each } k, l.$

A differential graded algebra is graded algebra with chain complex structure $d \circ d = 0$.

Derivation of degree $k$ on $A$ means a linear map $D:A \to A$ s.t. $$D(A_j)\subset A_{j+k} \text{ and } D(ab)=(Da)b + (-1)^{ik}a(Db), a\in A_i$$

All smooth forms on $n$-manifold $M$ is a differential graded algebra $\Omega^{\bullet}(M)=\bigoplus_{k=0}^{n} \Omega^k(M)$, with wedge product and exterior derivative.

In proving Cartan's magic formula $\mathcal{L}_X=i_X \circ d + d\circ i_X$ holds for $\Omega^{\bullet}(M)$, we can use the following steps:

1. Prove the lemma: two degree $0$ derivations on $\Omega^{\bullet}(M)$ commuting with $d$ are equal iff they agree on $\Omega^0(M)$.

2. Show that $\mathcal{L}_X$ and $i_X \circ d + d \circ i_X$ are derivations on $\Omega^{\bullet}(M)$ commuting with $d$.

3. Show that $\mathcal{L}_X f = Xf = i_Xdf+ d i_Xf$ for all $f \in C^{\infty}(M)=\Omega^0(M)$.

It's easy to check 2&3, and here're my questions:

1. How to prove this lemma?

2. Why commuting with $d$ in this lemma is so important? Is there any counterexample?

3. Does this lemma still hold without restriction on degree?

4. Does this lemma still hold for general differential graded algebra?

Answer 1

My thoughts in proving this lemma ($\Leftarrow$):

For this special case $\Omega^{\bullet}(M)=\bigoplus_{k=0}^{n} \Omega^k(M)$, let's first consider what $\Omega^0(M)$ and $\Omega^1(M)$ is.

In local chart $(U,(x^i))$, $\Omega^0(U)=C^\infty(U)$ is smooth functions on $U$, and $\Omega^1(U)=\text{span}\{dx^i\}$.

Two degree $0$ derivations $D_1,D_2$ commute with $d$, they agree on product.

Since $\Omega^1(U)=\text{span}\{dx^i\}$ and $x^i \in \Omega^0(U)$, if they agree on $\Omega^0(U)$, they agree on $\Omega^1(U)$.

And other $\Omega^k(U)$ can be generated by elements in $\Omega^1(U)$ via product, so $D_1, D_2$ agree on $\Omega^{\bullet}(U)$ thus $\Omega^{\bullet}(M) \qquad\Box $

I don't know if it holds for general cases, and I'm not sure where is degree $0$ used.

Maybe this proof works only because we already know the structure of $\Omega^{\bullet}(M)$.