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I would like to ask about an equivalence between two definitions for the Cartan Model. Let $G$ be a connected Lie group, let $\mathfrak{g}$ be its Lie algebra and let $M$ be a $G$-manifold. The Cartan model of $M$ is the $\Omega_G(M) := \{ a \in S(\mathfrak{g}^*) \otimes \Omega(M) | a \text{ is invariant} \}$.

  1. First definition of invariant uses an action of derivations of the Lie algebra $\mathfrak{g}$ on $\mathfrak{g}^*$ and $\Omega(M)$. On $\mathfrak{g}^*$ the action is given by $L_{x_i}(u^k) = - c^k_{i j} u^j$ where $(x_i)$ is any base of the Lie algebra, $(u^i)$ is its dual basis and $c^k_{ij}$ are the structure constants of the Lie bracket for $(x_i)$ basis. We extend on $S(\mathfrak{g}^*)$. On $\Omega(M)$ the action is the Lie derivative $L_X(\omega) := \mathcal{L}_{\overline{X}}(\omega)$ of the induced vector field of the action $\mathfrak{g} \to \mathfrak{X}(M)$. Finally, we form the diagonal action on tensor product. In that case the invariant elements are those that $L_X(a) = 0$ for each $X \in \mathfrak{g}$.
  2. The second definition is by using $G$-actions. The action induced by the Coadjoint representation on $\mathfrak{g}^*$. Namely, $g \cdot u := Ad^*_{g}(u) = u \circ Ad^*_{g^{-1}}$ and we extend on symmetric algebra. On $\Omega(M)$ by the induced action of $G$, the pullback of translations. Namely $g \cdot \omega := (l_{g^{-1}})^*(\omega)$. We also form the diagonal action on tensor product. In that case invariant elements are those that $g \cdot x = x$ fro any $g \in G $

I know that by connectedness a form $\omega \in \Omega(M)$ is $G$-invariant if and only if $L_X(a) = 0$ for each $X \in \mathfrak{g}$.

Also, I think I proved that the $\mathfrak{g}$-action of antiderivations on $\mathfrak{g}^*$ is the coadjoint action of Lie algebra $\mathfrak{g}$ which is the induced Lie algebra morphism of the Lie group homomorphism of the coadjoint representation $ G \xrightarrow{Ad^*} GL(\mathfrak{g}^*)$ which commutes with exponential map. But I am not even sure for the latter proposition.

My problem is that even with these assumptions I struggle proving any direction of equivalence, for elements on the tensor product $ S(\mathfrak{g}^*) \otimes \Omega(M)$.

Fist, is there an equivalence between these definitions? If so, how do I prove this for elements on tensor product?

(I asked the same question on MathStack)

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  • $\begingroup$ To pass from 2 to 1, simply differentiate the action of G, obtaining an action of the Lie algebra of G. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 27, 2023 at 0:07
  • $\begingroup$ @DmitriPavlov, $G$ acts on tensor product $ S(g^*) \otimes \Omega(M)$ which is not a manifold. What do you mean by differentiating the action? $\endgroup$
    – Nash-iOS
    Commented Apr 27, 2023 at 7:05
  • $\begingroup$ Given an element x in the Lie algebra of G, act by the element exp(tx) on the vector space Sg*⊗ΩM (its elements are smooth sections of a vector bundle), then differentiate with respect to t. Sections of vector bundles are differentiated pointwise. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 27, 2023 at 14:47
  • $\begingroup$ I think I need more information. What kind of sections are $S(g^*)$? Also, could you provide more information (or references) on how to differentiate sections? $\endgroup$
    – Nash-iOS
    Commented Apr 27, 2023 at 21:12
  • $\begingroup$ For ΩM, this is just the Lie derivative of differential forms. You can find the definition in any book on differential geometry, see, for example, Section 7.5 in Lee's Manifolds and Differential Geometry. Sg* is a (degreewise) finite-dimensional space, so you can differentiate normally. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 28, 2023 at 1:30

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