Calculating the Lie algebra representation of the regular representation on subspace of functions on $\mathbb R$.

Question

Let $G = \mathbb R$ and let $\pi$ be the regular representation of $G$ on $L^2(\mathbb R)$, that is, $\pi(g)(f)(x) = f(x-g)$ for $g \in G$. Let $V = \{f \in \mathcal C_c^\infty | supp f \subseteq [0,1]\}$. I am a bit confused at how to calculate the corresponding representation of the Lie algebra $\mathfrak g$ of $G$. That should be for $X \in \mathfrak g$: $(Xf)(y) = \frac{d}{dt}|_{t=0} (\pi(exp(tX))f))(y) = [f'(y-e^{tX}) \cdot (-e^{tX} X)] _{t=0} = -f'(y-1) X$. But if this were correct, it wouldnt make sense, since now the support has moved and the above $V$ is supposed to be invariant under $\mathfrak g$. Where do I go wrong?

Answer 1

If $G = \mathbb R$, then $\mathfrak g = \mathbb R$ and $\exp\colon \mathfrak g = \mathbb R \rightarrow \mathbb R = G$ is actually the identity map. Indeed, since $L_g(v) = g + v$ for $v,g\in G$, we have $dL_g(X) = X$ for $g\in G$, $X\in \mathfrak g$ and hence $\tilde X(g) \equiv \tilde X(0)$ for all $g\in G$ and all left-invariant vector fields $\tilde X\in \mathfrak g$. Now, if $\tilde X\in \mathfrak g$ (corresponding to $X\in T_0G$) and $\gamma\colon \mathbb R\rightarrow G$ is the integral curve for $\tilde X$ through $0$, i. e. $\gamma(0) = 0$ and $\dot\gamma(t) = \tilde X(\gamma(t))$ for $t\in \mathbb R$, then $\dot\gamma(t) = \tilde X(\gamma(t)) = \tilde X(0) = X$ and thus $\exp(X) = \gamma(1) = X$.

Now, for $f\in V$ and $X\in T_0G$ your above calculation yields $$(Xf)(y) = f'(y)X\qquad \text{$y\in \mathbb R$}$$ and hence $Xf\in V$ as desired.