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

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?

• 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). – user10354138 Apr 21 '19 at 15:25
• @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 $\Omega^{\bullet}$ – 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 $$\Omega^{\bullet}(M)$$.