(Exercise 3.4.12 - Sharpe) Let $H$ be a Lie group, $V$ a vector space, and $\rho: H \to Gl(V)$ a representation. Let $U$ be a manifold, $X$ a vector field on $U$, and $h: U \to H$ and $f: U \to V$ smooth functions. Show that $$X(\rho (h^{-1})f) = \rho(h^{-1})X(f) - \rho_{*e}(Ad(h)(h^*\omega_H(H) )) f$$ where $\omega_H$ is a Maurer-Cartan form.

Attempt: Here are some of the results I have used so far

$(i)$ If $\iota : G \to G$ is the inverse function of $G$ and $\omega_H$ is a Maurer-Cartan form then $$\iota^* \omega_H(v) = - Ad(g)\omega_H(v) \,\,\, \text{for}\,\,\, v \in T_g(G)$$

$(ii)$ A smooth map $f:N \to M$ induces a pullback map $f^*: A^p(N,V) \to A^p(M,V)$ defined by $$(f^*\omega)_x (X_1, \ldots, X_p) = \omega_{f(x)} (f_*X_1, \ldots,f_*X_p)$$ where $X_1, \ldots, X_p \in T_x (M)$.

By the Leibniz rule we have

$$X(\rho (h^{-1})f) = \rho (h^{-1})X(f) + X(\rho(h^{-1}))f$$

Now the idea was to show that $X(\rho(h^{-1})) = - \rho_{*e}(Ad(h)(h^*\omega_H(H) ))$. Thus, using the chain rule, $(i)$ and $(ii)$ we get

$$\begin{align}X(\rho(h^{-1}))|_h &\underset{\text{def}}= \rho(h^{-1})_*(X_h)\\&\underset{\text{C.H.}}= (\rho_{*e}h^{-1}_*)(X)\\&=[\rho_{*e}(\iota h)_*](X)\\&\underset{\text{C.H.}}=\rho_{*e}[\iota_* h_*(X)]\\&=\rho_{*e}[L_{h*}\circ L_{h^{-1}*}\iota_* h_*(X)]\\&=\rho_{*e}[L_{h*}\omega_H(\iota_* h_*(X))]\\&\underset{\text(ii)}=\rho_{*e}[L_{h*}\iota^*(\omega_H(h_*X))]\\&\underset{\text{(ii)}}= \rho_{*e}[L_{h*} \iota^*(h^*\omega_H(X))]\\&\underset{\text{(i)}}= -\rho_{*e}[L_{h*}Ad(h)h^*\omega_H(X) ]\end{align}$$

this is what I got so far.

$1)$ Is it correct?

$2)$ What can be improved?

$3)$ Where to go from here?

4) Should the vector field be considered left-invariant?


1 Answer 1


I will proceed in a slightly different way than you did. The gist of the computations is of course the same.

Let $u\in U$, let $u(t):\mathbb{R}\to U$ be a smooth path such that $u(0) = u$ and $\dot{u}(0) = X(u)$. Then \begin{align} X\left(\rho\left(h^{-1}\right)\right)(u) = & \left.\frac{d}{dt}\right|_{t=0}\rho\left(h^{-1}(u(t))\right)\\ = & \left.\frac{d}{dt}\right|_{t=0}\rho\left(\iota(h(u)^{-1}h(u(t)))\right)\rho(h(u))^{-1}\\ = & (\rho_*)_e\left(\left.\frac{d}{dt}\right|_{t=0}\iota\left(L_{h(u)^{-1}}h(u(t))\right)\right)\rho(h(u))^{-1}\\ = & (\rho_*)_e\left(\iota_*\left(\left(L_{h(u)^{-1}}\right)_*\left.\frac{d}{dt}\right|_{t=0}h(u(t))\right)\right)\rho(h(u))^{-1}\\ = & (\rho_*)_e\left(\iota^*\left(\omega_{h(u)}\left(\left.\frac{d}{dt}\right|_{t=0}h(u(t))\right)\right)\right)\rho(h(u))^{-1}\\ = & (\rho_*)_e\left(\iota^*\left(\omega_{h(u)}\left(h_*X(u)\right)\right)\right)\rho(h(u))^{-1}\\= & -(\rho_*)_e\left(\text{Ad}_{h(u)}\omega_{h(u)}\left(h_*X(u)\right)\right)\rho(h(u))^{-1}\\ \end{align} In the fifth line I used the fact that $\iota_*=\iota^*$ and the definition of the Maurer-Cartan form, and in the last line I used the identity you mentioned. in your notation, this reads $$X\left(\rho\left(h^{-1}\right)\right) = -\rho_{*e}\left(\text{Ad}(h)h^*\omega_H\right)\rho\left(h^{-1}\right).$$


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