Let $M^n$ be a parallelizable manifold with the nowhere dependent vector fields $X_1,\ldots, X_n$ forming a basis for the tangent space at each point of $M$. The Lie brackets of these fields are closed under their span so there are functions $c^k_{ij}$ such that $$ [X_i,X_j] = \sum_{k=1}^n c^k_{ij} X_k $$

Let $\theta^i$ be the dual 1-form to $X_i$. I want to express the 2-form $d\theta^i$ in terms of the $c^k_{ij}$ and the basis of 2-forms $\theta^j \wedge \theta^k$. The trick is supposedly to take any smooth function $f:M \to \mathbb{R}$ and the fact that $$df = \sum_{i=1}^n X_i(f) \theta^i $$ to compute $0=d^2 f$ to arrive at $$ 0=\sum_{j=1}^n \left( \sum_{k=1}^n X_k X_j(f) \theta^k \wedge \theta^j + X_j(f) d\theta^j \right)$$ From there you can introduce the $c^k_{ij}$ by noting $$ \sum_{j=1}^n \sum_{k=1}^n X_k X_j(f) \theta^k \wedge \theta^j = \sum_{1 \leq k < j \leq n} (X_k X_j(f) - X_j X_k(f)) \theta^k \wedge \theta^j = \sum_{1 \leq k < j \leq n} [X_k,X_j](f) \theta^k \wedge \theta^j \\ = \sum_{1 \leq k < j \leq n} \left( \sum_{p=1}^n c^p_{kj} X_p(f) \right) \theta^k \wedge \theta^j$$ to get $$ 0 = \sum_{1 \leq k < j \leq n} \left( \sum_{p=1}^n c^p_{kj} X_p(f) \right) \theta^k \wedge \theta^j + \sum_{j=1}^n X_j(f) d \theta^j $$

Naively, I don't think you can recover information about $d\theta^j$ from this alone since the sum over all $j$ is required for the expression on the right to equal zero, and because of the dependence on $f$. I would think the only way to decouple this information would be to find a function $f_i$ such that $X_j(f_i) = \delta_{ij}$. I'm trying to look at functions like $g(X_i,X_i)$, where $g$ is the metric, but I haven't been able to make this work. So I'm looking for HINTS as to how to find such a function and/or to see if there is an error in my computation. Thanks.


2 Answers 2


Hint: Use the following invariant formula for the exterior derivative of a $1$-form $\theta$: $$ d\theta(X,Y) = X(\theta(Y)) - Y(\theta(X)) - \theta([X,Y]). $$


Hint: It is possible to find $f_i$ locally and show the identity. You may assume that on a contractible open subset $U$ of $M$, $d\theta_i=0$, the Poincare Lemma implies that on $U$, $\theta_i=df_i$, on $U$, $X_j(f_i)=\delta_{ij}$.

  • $\begingroup$ I see. But if locally $d\theta^i =0$ then does that not make the computation locally meaningless ($X_i(f_i) d\theta^i = 1 \cdot 0$)? Do you mean to treat $d\theta^i$ as symbolically nontrivial(as it by definition must be) and continuing the computation with $f_i$? Or to patch together the $f_i$ via a partition of unity? $\endgroup$ May 13, 2016 at 23:55
  • $\begingroup$ $X_i(f_i)=df_i.X_i=\theta_i(X_i)=1$. I believe it is all what you need. $\endgroup$ May 14, 2016 at 0:05
  • $\begingroup$ Oh yes now I see. $\endgroup$ May 14, 2016 at 0:09
  • $\begingroup$ Since the $X_i$'s are given and the $\theta^i$'s are their dual $1$-forms, the $\theta^i$'s are completely determined, and you cannot assume that $d\theta^i=0$. $\endgroup$
    – Jack Lee
    May 14, 2016 at 15:21
  • $\begingroup$ Really ? This basic example shows it is possible. Suppose that $X=R^n$ and the $X_i$ are the constant vectors defined by the base $(e_1,...,e_n)$, then $\theta_i$ is the dual basis. $\endgroup$ May 14, 2016 at 18:16

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