Evaluation of the derivative of differential forms on vector fields #
In this file we prove the following formula and its corollaries.
If ω is a differentiable k-form and V i are k + 1 differentiable vector fields, then
$$ dω(V_0(x), \dots, V_n(x)) = \left(\sum_{i=0}^k (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_k(x)))(V_i(x)) + \sum_{0 \le i < j\le k} (-1)^{i + j} ω(x; [V_i, V_j](x), V_0(x), …, \widehat{V_i(x)}, …, \widehat{V_j(x)}, …, V_k(x)), $$ where $$[V_i, V_j]$$ is the commutator of the vector fields $$V_i$$ and $$V_j$$. As usual, $$\widehat{V_i(x)}$$ means that this item is removed from the sequence.
There is no convenient way to write the second term in Lean for k = 0,
so we only state this theorem for k = n + 1,
see extDerivWithin_apply_vectorField and extDeriv_apply_vectorField.
In this case, we write the second term as a sum over i j : Fin (n + 1), i ≤ j,
where the indexes (i, j) in our sum currespond to (i, j + 1)
(formally, (Fin.castSucc i, Fin.succ j)) in the formula above.
For this reason, we have - before the sum in our formal statement.
If ω is a differentiable (n + 1)-form and V i are n + 2 differentiable vector fields, then
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)) - \sum_{0 \le i \le j\le n} (-1)^{i + j} ω(x; [V_i, V_{j + 1}](x), V_0(x), …, \widehat{V_i(x)}, …, \widehat{V_{j + 1}(x)}, …, V_k(x)), $$
where $$[V_i, V_{j + 1}]$$ is the commutator of the vector fields $$V_i$$ and $$V_{j + 1}$$. As usual, $$\widehat{V_i(x)}$$ means that this item is removed from the sequence.
In informal texts, this formula is usually written as
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)) - \sum_{0 \le i < j\le n + 1} (-1)^{i + j} ω(x; [V_i, V_j](x), V_0(x), …, \widehat{V_i(x)}, …, \widehat{V_j(x)}, …, V_k(x)). $$
In the sum from our formalization,
each index (i, j) corresponds to the index (Fin.castSucc i, Fin.succ j)
in the sum used in informal texts.
For this reason, i + j in our sum has the opposite parity compared to informal texts,
which changes the sign before the sum from + to -.
If ω is a differentiable (n + 1)-form and V i are n + 2 differentiable vector fields, then
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)) - \sum_{0 \le i \le j\le n} (-1)^{i + j} ω(x; [V_i, V_{j + 1}](x), V_0(x), …, \widehat{V_i(x)}, …, \widehat{V_{j + 1}(x)}, …, V_k(x)), $$
where $$[V_i, V_{j + 1}]$$ is the commutator of the vector fields $$V_i$$ and $$V_{j + 1}$$. As usual, $$\widehat{V_i(x)}$$ means that this item is removed from the sequence.
In informal texts, this formula is usually written as
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)) - \sum_{0 \le i < j\le n + 1} (-1)^{i + j} ω(x; [V_i, V_j](x), V_0(x), …, \widehat{V_i(x)}, …, \widehat{V_j(x)}, …, V_k(x)). $$
In the sum from our formalization,
each index (i, j) corresponds to the index (Fin.castSucc i, Fin.succ j)
in the sum used in informal texts.
For this reason, i + j in our sum has the opposite parity compared to informal texts,
which changes the sign before the sum from + to -.
Let ω be a differentiable n-form and V i be n + 1 differentiable vector fields.
If V i pairwise commute at x, i.e., $$[V_i, V_j](x) = 0$$ for all i ≠ j, then
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)). $$
Let ω be a differentiable n-form and V i be n + 1 differentiable vector fields.
If V i pairwise commute at x, i.e., $$[V_i, V_j](x) = 0$$ for all i ≠ j, then
$$ dω(V_0(x), \dots, V_{n + 1}(x)) = \left(\sum_{i=0}^{n + 1} (-1)^i • D_x(ω(x; V_0(x), \dots, \widehat{V_i(x)}, \dots, V_{n + 1}(x)))(V_i(x)). $$