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?

  • $\begingroup$ Take any Riemannian metric and the associated Levi-Civita $\nabla_X$. All such $\nabla_X$ obviously agree on $\Omega^0(M)$ because it is just the usual derivation $X$, but they won't necessarily agree on forms. So commuting with $d$ is there to stop this silly example. Since $\Omega^\bullet$ is generated by $df$ for $f\in\Omega^0$, some form of Leibniz rule allow you to conclude (1). $\endgroup$ – user10354138 Apr 21 '19 at 15:25
  • $\begingroup$ @user10354138 Thank you. And do you have any ideas on question 3&4? As I said in the answer below, I wonder this method works only because we already know the structure og $\Omega^{\bullet}$ $\endgroup$ – Andrews Apr 21 '19 at 15:59

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 $\Omega^{\bullet}(M)$.

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