multi-vector fields

Definition

A multivector field of degree $k$ on a smooth manifold $M$ is a smooth section of the $k^{th}$ exterior power of the tangent bundle.

$\vartheta \in {\mathfrak{X}}^k(M):-\Gamma \left({\bigwedge }^{k}TM\right)$

Intuition and other identifications

First, we can remember the definition of a [[202404061628|differential $k$ form]] is a smooth section of the $k^{th}$ exterior power of the cotangent bundle.

$\alpha \in {\Omega }^k(M):-\Gamma \left({\bigwedge }^k{T}^{\ast }M\right)$

Another way to think of (identify) a differential form of degree $k$ is as a multilinear, alternating map on the space of vector fields $\mathfrak{X}(M)$

$\alpha :\underset{k\text{times}}{\underbrace{\mathfrak{X}(M)\times \cdots \times \mathfrak{X}(M)}}⟶C^{\infty }(M).$

Or restating this pointwise we have ${\alpha }_p:\underset{k\text{times}}{\underbrace{T_{p}M\times \cdots \times T_{p}M}}⟶\mathbb{R}$

We can define multivector fields dually.

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

$\vartheta :\underset{k\text{times}}{\underbrace{{\Omega }^1(M)\times \cdots \times {\Omega }^1(M)}}⟶C^{\infty }(M).$

or written pointwise as ${\vartheta }_p:\underset{k\text{times}}{\underbrace{T_p^{\ast }M\times \cdots \times T_p^{\ast }M}}⟶\mathbb{R}$

Analogously to the construction of ${\Omega }^{\bullet }(M)$, multivector fields define a graded commutative algebra structure.

${\mathfrak{X}}^{\bullet }(M)={⨁}_{k=0}^{\infty }{\mathfrak{X}}^k(M)$

(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

$\wedge :{\mathfrak{X}}^k(M)\times {\mathfrak{X}}^l(M)⟶{\mathfrak{X}}^{k+l}(M)$ for multiplication.

The wedge product follows the formula

$(\alpha \wedge \beta )(v_1,\dots ,v_{k+l})={\sum }_{\sigma \in {\text{Sh}}_k}\text{sgn}(\sigma )\alpha (v_{\sigma (1)},\dots ,v_{\sigma (k)})\beta (v_{\sigma (k+1)},\dots ,v_{\sigma (k+l)}).$

Local representation

A $k$-vector field can be represented locally as

$X{|}_U={\sum }_{i_1<\dots ,i_k}{X}^{i_1,\dots ,i_k}(x)\frac{\partial }{\partial x^{i_1}}\wedge \cdots \wedge \frac{\partial }{\partial X^{i_k}},$

where $X^{i_1,\dots ,i_k}(x)$ are uniquely determined smooth functions.

Relationship to derivations

Every vector field $X\in {\mathfrak{X}}^1(M)$ induces a derivation ${\mathcal{L}}_X:C^{\infty }(M)\to C^{\infty }(M).$

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

${\mathcal{L}}_X:C^{\infty }(M)\times \cdots \times C^{\infty }(M)⟶C^{\infty }(M)$

where

${\mathcal{L}}_X(f_1,\dots ,f_k)=X(\text{d}{f}_1,\dots ,\text{d}{f}_2)$

Proposition:

On any manifold $M$, the map $X\mapsto {\mathcal{L}}_X$ gives a $1-1$ correspondence

\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