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I’m fairly new to differential geometry. It seems that $\Omega^k(M)$ is the common notation used for differential $k$-form fields on a manifold, $M$. Such that if $\omega \in \Omega^k(M)$ then $\omega \colon x \mapsto \omega_x \in \bigwedge^k T^*_x M$.

Question: But what about some alternating $(k,0)$-tensor field, $\lambda: x \mapsto \lambda_x \in \bigwedge^k T_x M $? I would express this as $\lambda \in (???)$. What is the tangent-bundle analogue of $\Omega^k(M)$?

Example: Let $(M,\omega)$ be a real symplectic manifold with canonical coordinates $(q^i,p_i)$ for $i=1,\dots , n$. The canonical symplectic form is then a differential $2$-form field, $\omega = \text{d}q^i \wedge \text{d}p_i\in \Omega^2(M)$. We can often find the “inverse” of this $2$-form which is then given by $\omega^{-1} = -\partial_{q^i} \wedge \partial_{p_i} $ such that $\omega^{-1}_{x} \in \bigwedge^2 T_x M$. To what space does the bivector field $\omega^{-1}$ belong and what is the notation for it? Somehow, I have never come across it.

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    $\begingroup$ In Poisson geometry this is called the space of multivector fields (or $k$-vector fields) and is sometimes denoted as $\mathfrak X^\bullet(M)$. $\endgroup$
    – T.P.
    Commented Dec 30, 2022 at 18:10
  • $\begingroup$ @T.P. You should convert this to an answer. $\endgroup$ Commented Dec 30, 2022 at 20:14

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I am posting my comment above as an answer here and adding some more details.


In Poisson geometry this is called the space of multivector fields (or $k$-vector fields) and is sometimes denoted as $\mathfrak X^\bullet(M):=\Gamma(\wedge^\bullet TM)$. In particular, a Poisson structure is a 2-vector field $\pi\in \mathfrak X^2(M)$ obeying some structural equation.

In fact, your example with the inverse of the symplectic form is something foundational in Poisson geometry, as it is the basic example of a Poisson structure on a manifold; every symplectic manifold is Poisson but the converse is not true; the Poisson structure is given by $\pi=\omega^{-1}$.

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  • $\begingroup$ ok I like this. But it seems odd I have not come across it before when reading about symplectic and poisson manifolds (granted, I only have entry-level exposure to the subject). $\endgroup$
    – J Peterson
    Commented Dec 31, 2022 at 16:23
  • $\begingroup$ Maybe you saw the Poisson manifold in terms of the Poisson bracket, which is usually the first thing one mentions in Poisson manifolds. The next step would be to prove an equivalence Poisson brackets $\Longleftrightarrow$ Poisson tensors and jump to the language of multivector fields, which is the base for introducing Poisson cohomology. So I guess that if you stay in symplectic-Poisson geometry you’ll eventually see all these things. $\endgroup$
    – T.P.
    Commented Dec 31, 2022 at 16:32
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A differential $k$-form on a smooth manifold $M$ is a section of $\bigwedge^kT^*M$, that is $\Omega^k(M) = \Gamma(M, \bigwedge^kT^*M)$. What you're describing is a section of $\bigwedge^kTM$, which is called a polyvector field. The space of polyvector fields is therefore $\Gamma(M, \bigwedge^kTM)$, but I am not aware of any other notation for this space.

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  • $\begingroup$ ah, now that you mention it, I have seen the $\Gamma$ notation before but have never fully understood what it means. From context, I gather that $\lambda\in \Gamma(M, \bigwedge^k TM)$ means $\lambda: M \to \bigwedge^k TM$? I think I've seen this before but written just as $ \Gamma(\bigwedge^k TM)$ rather than $ \Gamma(M, \bigwedge^kTM)$. Either way, this seems too clunky/long for regular use. $\endgroup$
    – J Peterson
    Commented Dec 30, 2022 at 18:16
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    $\begingroup$ Not quite. It denotes the space of sections, not just maps. If $\pi : \bigwedge^kTM \to M$ denotes the vector bundle projection, then $\lambda : M \to \bigwedge^kTM$ is a section if $\pi\circ\lambda = \operatorname{id}_M$, i.e. $\lambda(m) \in \bigwedge^kT_mM$ for every $m \in M$. $\endgroup$ Commented Dec 30, 2022 at 18:27
  • $\begingroup$ ok I think I see. Does the notation $\Gamma(\cdot)$ for smooth sections always imply a section with respect to the natural projection (rather than some other map)? $\endgroup$
    – J Peterson
    Commented Dec 31, 2022 at 16:36
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    $\begingroup$ Yes. If $\pi : E \to M$ is a smooth vector bundle, then $\Gamma(M, E)$, or just $\Gamma(E)$, denotes the collection of smooth sections of $\pi$. $\endgroup$ Commented Dec 31, 2022 at 20:06

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