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For a vector field $X$ on a manifold there are two ways to define a Lie derivative: an algebraic one using Cartan's formula $\mathcal{L}_X \alpha = i_X d \alpha + d i_X \alpha$ and a dynamical one using the flow $\phi^X_t$, $${\cal L}_X \alpha = \frac{d}{dt}\biggl|_{t=0} \left(\bigl(\Phi^t_X\bigr)^* T\right).$$

Given a multivector field $X$ one can define its Lie derivative by means of Cartan's formula, i.e, $\mathcal{L}_X \alpha = i_X d \alpha + d i_X \alpha$. See, for example [1].

My broad question is if the Lie derivative by a multivector means something dynamical. These are some fuzzy questions for which any help or reference would be welcomed.

  • In the simplest case, if $X = X_1\wedge \dotsb \wedge X_n$ and the $X_i$ span an integrable distribution, is $\alpha$ constant in some sense on the leaves of this distribution? What happens if the multivector is not integrable?

  • Is there a generalization of the concept of flow for a multivector that applies to this situation?

  • My current geometric understanding of general multivectors is that they are linear combinations of hyperplanes modulo the Plücker relations (whose geometric interpretation feels somewhat obscure to me). I would like to have a better interpretation. On question [2] on this site it is stated that they are global sections of a line bundle over the Grasmannian, some reference of this fact would be useful.

EDIT:

  • On [2] it is said that "is known that $Λk(V)$ is isomorphic to the space of global sections of a line bundle over the Grassmanian $Gk(V)$ of k-dimensional subspaces in $V$ - namely of the highest exterior power of the tautological bundle." I am trying to make sense of this. Let $V_k(V^*)$ be the bundle over $G_k(V^*)$ whose elements $(W,a)$ are $k$-dimensional subespaces on $V^*$ and $a$ is a volume form on $W$. A $k$-vector field $X$ can be though of as a section of this bundle: to every subspace we assign the restriction of $X$. If I understand the quotation above and it holds over the reals, $k$-vector fields are presicely algebraic sections of that bundle. Hence, a multivector is an algebraic assignment of a volume to each subespace of $V^*$. Is this true? Any reference would be great.

Also, as a follow-up partially unrelated question. Forms should be then algebraic assignments of volume to subspaces. But why would a differential geometer limit themself to algebraic sections? Has someone studied the sections of this bundle. It is posible to define integrals of submanifolds over such sections?

[1] W. M. Tulczyjew, «The Graded Lie Algebra of Multivector Fields and the Generalized Lie Derivative of Forms».

[2] მამუკა ჯიბლაძე, Cannot multivectors be classified more easily than general tensors?, URL (version: 2017-04-13): Cannot multivectors be classified more easily than general tensors?

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    $\begingroup$ This MO question might be of interest: mathoverflow.net/q/155989/11211 - In that regard, it can be shown that a 2-vector field $X=X_1\wedge X_2$ defines an integrable 2-dimensional distribution iff its Schouten-Nijenhuis bracket with itself vanishes everywhere. This also provides a characterization of when a 2-vector field is a Poisson tensor, check e.g. P. W. Michor, Remarks on the Schouten-Nijenhuis bracket, Rend. Circ. Mat. Palermo Suppl. 16 (1987), pp. 207-215, dml.cz/dmlcz/701423 . I don't know of a similar geometric interpretation of $p$-vector fields for $p>2$. $\endgroup$ Commented Dec 9, 2024 at 19:07
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    $\begingroup$ This does not generalize to $p$-vector fields with $p>2$. One can compute that if $X = X_1 \wedge X_2 \wedge X_3$ then $[X,X] = 0$ always. $\endgroup$ Commented Dec 10, 2024 at 17:42
  • $\begingroup$ This follows from $[X\wedge Y,Z] = X\wedge [Y,Z] \pm [X,Z]\wedge Y $ (see [1]) $\endgroup$ Commented Dec 10, 2024 at 17:45
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    $\begingroup$ They also vanish. Let $Z=X \wedge Y$ be a decomposable $p$-vector, $X$ a descomposable (p-1)-vector and $Y$ a vector. A quick calculation using the previous formula twice $[Z,Z] = \pm 2 Z \wedge [X,Y]$. If we further decompose $X = X_0 \wedge X_1$, and apply the same formula again all the terms contain either $Z \wedge X_0$, $Z \wedge X_1$, $Z \wedge Y$, which all vanish. $\endgroup$ Commented Dec 18, 2024 at 11:32
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    $\begingroup$ @LSpice. Yes, thank you $\endgroup$ Commented Dec 21, 2024 at 16:00

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