19
$\begingroup$

Question:

(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?

$\endgroup$
1
$\begingroup$

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 $$X\left(\rho\left(h^{-1}\right)\right) = -\rho_{*e}\left(\text{Ad}(h)h^*\omega_H\right)\rho\left(h^{-1}\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.