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multi-vector fields

Mar 19, 20254 min read

Definition

A multivector field of degree on a smooth manifold is a smooth section of the exterior power of the tangent bundle.

Intuition and other identifications

First, we can remember the definition of a [[202404061628|differential form]] is a smooth section of the exterior power of the cotangent bundle.

Another way to think of (identify) a differential form of degree is as a multilinear, alternating map on the space of vector fields

Or restating this pointwise we have

We can define multivector fields dually.

Just like we did for differential forms, we can identify these with multilinear, alternating maps

or written pointwise as

Analogously to the construction of , multivector fields define a graded commutative algebra structure.

(Note that though the sum below says it goes on infinitely, above the dimension of the manifold, it is just adding zeros.)

This graded commutative algebra has the wedge product

for multiplication.

The wedge product follows the formula

Local representation

A -vector field can be represented locally as

where are uniquely determined smooth functions.

Relationship to derivations

Every vector field induces a derivation

We can make a similar multi-derivation for multivector fields defined as

where

Proposition:

On any manifold , the map gives a correspondence

You can't use 'macro parameter character #' in math mode\leftrightarrow \begin{Bmatrix} \text{anti-symmetric multi-derivations} \\ \mathcal{L}:\underbrace{C^{\infty}(M)\times \dots \times C^{\infty}(M)}_{k \text{ times}} \to C^{\infty}(M)\end{Bmatrix}$$ ### Proof It is clear that for a multi-vector field $X$, $\mathcal{L}_{X}$ is a multiderivation that must be antisymmetric since $X$ is identified with an antisymmetric map. __Injective__: Let $\mathcal{L}_{X} = \mathcal{L}_{W}$. Thus, $$X(df_{1}, \dots, df_{k}) = W(df_{1}, \dots, df_{k}) \qquad \forall f_{1}, \dots, f_{k} \in C^{\infty}(M).$$ We can consider functions $g_{1}, \dots, g_{n}$ such that $dg_{1},\dots, dg_{n}$ form a basis for $T_{p}^{*}M$. Since $X$ and $W$ must match for these basis functions, they must match for all of $T_{p}^{*}M$. These can be cycled through each position to show that $X = W$. __Surjective:__ Let $D:C^{\infty}(M) \times C^{\infty}(M) \to C^{\infty}(M)$ be an antisymmetric $k$ multiderivation. If we fix the last $k-1$ functions we get a derivation $$ D(-, f_{2}, \dots, f_{k}):C^{\infty}(M) \longrightarrow C^{\infty}(M).$$ Since this is a derivation, this is isomorphic to some vector field with coefficients $D_{i}$ $$D(-, f_{2}, \dots, f_{k})(p) = \sum D_{i}\frac{\partial}{\partial x^{i}}\bigg|_{p}$$ We can do this with the other positions, but since the derivation is antisymmetric, we can take out the "diagonal" terms and combine the others to get $$D(f_{1},\dots, f_{k}) =\sum_{i_{1}< \dots< i_{k}} D^{i_{1},\dots, i_{k}} \frac{\partial}{\partial x^{i_{1}}} \wedge \dots \wedge \frac{\partial}{\partial X^{i_{k}}}$$ for smooth functions $D^{i_{1}, \dots, i_{k}}$. Therefore, since a $k$-vector field $X \in \mathfrak{X}^{k}(M)$ can be represented locally as $$ X\big|_{U} = \sum_{i_{1}< \dots, i_{k}} X^{i_{1}, \dots, i_{k}}(x) \frac{\partial}{\partial x^{i_{1}}} \wedge \dots \wedge \frac{\partial}{\partial X^{i_{k}}},$$ let $X^{i_{1}, \dots, i_{k}}(x) = D^{i_{1},\dots, i_{k}}(x)$. Therefore, $X \mapsto \mathcal{L}_{X}$ is surjective. Hence, the relationship is 1-1. # References [[@crainic2021]] - Chapter 2

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multi-vector fieldssmooth manifoldsectionexterior powertangent bundlegraded algebraderivationPoisson manifoldSchouten bracket

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