Let $G$ be a Lie group, $M$ be a (smooth) manifold and assume that $G$ acts smoothly on $M$ with a fixed point $p \in M$. (I mean, there is a Lie group homomorphism $\rho : G \to C^{\infty}(M,M)$ where $C^{\infty}(M,M)$ is a mapping space endowed with (possibly infinite dimensional) smooth manifold structure and each element $\rho(g) : M \to M$ fixes $p$.)

Since $\rho(g)(p) = p$ for all $g \in G$, we can consider the action $G \to \mathrm{GL}(T_{p}G)$ on the tangent space at $p$. Then the tangent map of this action at the identity gives a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{End}(T_{p}G)$.

However, we can take the tangent map of $\rho$ directly and obtain $\mathfrak{g} \to T_{\mathrm{Id}}C^{\infty}(M,M)$. My question is, is there any canonical map from $T_{\mathrm{Id}}C^{\infty}(M,M)$ to $\mathrm{End}(T_{p}G)$, so that the two constructions above coincides in some sense?


1 Answer 1


The intuition behind your question does not really work out. The problem is that the isotropy representation of $G$ is already obtained by differentiating the map $x\mapsto \rho(g)(x)$ in the point $p$. Only after that, you have to differentiate the resulting map on $G$ in order to obtain the isotropy representation of $\mathfrak g$. In your proposed alternative way to obtain this, you only consider the differentiation of the map on $G$ but not of the one on $M$.

Second, there is the technical problem that $C^\infty(M,M)$ is certainly not a Lie group, you would have to take the diffeomorphism group $Diff(M)$ here. Form compact $M$ this is indeed an infinite dimensional Lie group with Lie algebra the space $\mathfrak X(M)$ of vector fields on $M$. Already here things are not easy technically since you need manifolds modelled on Frechet spaces. In the non-compact case, things become significantly more complicated.

However, you don't need infinite dimensional manifolds to get the infinitesimal version of a group action: The action of $G$ on $M$ gives rise to a Lie algebra (anti-)homomorphism $\mathfrak g\to M$ described by the so-called fundamental vector fields. Given $X\in\mathfrak g$, you define $\zeta_X\in\mathfrak X(M)$ by $\zeta_X(x):=\tfrac{d}{dt}|_{t=0}\rho(\exp(tX))(x)$. It is easy to see (using only finite dimensional arguments) that $\zeta_X$ is indeed a smooth vector field on $M$ and how this map is compatible with Lie brackets. Lie's second fundamental theorem shows that this is indeed a good infinitesimal version of a smooth group action.

However, as you can see from the definition, if $p\in M$ is a fixed point of $G$, then $\zeta_X(p)=0$ for any $X\in\mathfrak g$ (which is precisely the infinitesimal version of $p$ being a fixed point). To get the isotropy representation of $\mathfrak g$ from this, I believe you have to differentiate (in $M$ directions). Choosing a linear connection $\nabla$ on $TM$, you can consider the map $T_pM\to T_pM$ defined by $\xi\mapsto \nabla_{\xi}\zeta_X(p)\in T_pM$. Since $\zeta_X(p)=0$, this mapping is independent of the choice of $\nabla$ and it should give the action of $X$ on $\xi$ under the isotropy representation of $\mathfrak g$.

  • $\begingroup$ Thank you so much for the detailed answer! This was very helpful. $\endgroup$
    – Luka
    Commented Sep 22, 2023 at 4:39

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