Reconstructing unitary representation of Lie group from its generators

The idea behind this equation is the following:

• associate the identity operator $1$ to the identity element of the Lie Group.

$$ U(0)=1 $$

• represent an element infinitesimally close to the identity by

$$ U(\delta\theta)=1+i\delta\theta^aT_a $$

where $\delta\theta^a$ is infinitesimal and $T_a$ are the generators in some representation of the Lie Algebra. Now by repeated applications of $U(\delta\theta)$ we can "walk" on the Lie group from the identity to any element that is connected to the identity. If we hit $U(\delta\theta)$ in some $U(\theta)$ this will "move" $U(\theta)$ to

$$ U(\delta\theta)U(\theta)=U(f(\delta\theta,\theta))= U(\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b}) $$

so if we have a representation $U(\theta)$ of the Lie group element associated with the coordinates $\theta^{a}$, hitting $U(\delta\theta)$ will lead to a representation $U(\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b})$ of the Lie group element associated with the coordinates $\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b}$.

Now, we trace a path in the Lie Group that starts ($s=0$) at the identity and reach the Lie Group element associated to the coordinates $\theta^{a}$ at the end ($s=1$). This path will be in coordinates denoted by $\Theta_{\theta}^a(s)$ and should satisfy

$$ \Theta_{\theta}^{a}(0)=0,\qquad \Theta_{\theta}^{a}(1)=\theta^{a} $$

where $0$ is the identity element of the Lie group in coordinates. We are going to attach at each point in the curve $\Theta_{\theta}(s)^{a}$ a representation given by $U_{\theta}(s)\equiv U(\Theta_{\theta}(s))$ by following the idea presented above, such that at the end of the path we obtain $U[\theta]\equiv U_{\theta}(1)$.

The equation (2.B.2) can be obtained from the composition rule of (2.B.1)

$$ (1+i\delta\theta^{a}T_a)U(\Theta_{\theta}(s))=U(\Theta^a_{\theta}(s) + (h^{-1}) ^a\,_b(\Theta_{\theta}(s))\delta\theta^b) $$

by requiring that $\Theta^a_{\theta}(s) + (h^{-1}) ^a\,_b(\Theta_{\theta}(s))\delta\theta^b$ lies in the path, i.e.

$$ \delta \Theta^a_{\theta}(s)\equiv \Theta^{a}_{\theta}(s+\delta s)-\Theta^{a}_{\theta}(s)= (h^{-1})^a\,_b(\Theta_{\theta}(s))\delta\theta^b $$

which is the same as $\delta\theta^{a}=h^{a}\,_{b}(\Theta_{\theta}(s))\delta\Theta^{a}_{\theta}(s)$. The result is

$$ U(\Theta_{\theta}^a(s+\delta s)) - U(\Theta^a_{\theta}(s))= iT_{a}h^a\,_b(\Theta_{\theta}(s)) U(\Theta_{\theta}(s))\left( \Theta^b_{\theta}(s+\delta s)-\Theta^b_{\theta}(s)\right) $$

If we divide by $\delta s$ we get the equation (2.B.2).

I believe I have found a coordinate-free answer to this question in which only in the end coordinates are invoked. I'm posting it here too, in case anyone also likes this approach. Corrections are highly appreciated.

There will be a slight shift in notation here. The unitary Lie Group representation will be denoted $\pi : G\to {\rm U}(\cal H)$. Its Lie algebra derived representation will be denoted $d\pi : \mathfrak{g}\to {\operatorname{End}}(\cal H)$.

Straight answer to the question: the motivation to use (2.B.2) to look for $\pi$ given $d\pi$ is that it is the "in-representation" version of the equation saying that along any curve in $G$ the Lie algebra element $X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s)$ generates a transformation that infinitesimally moves $\gamma(s)$ towards $\gamma(s+\delta s)$.

Given any curve $\gamma :[0,1]\to G$ starting off of the identity its tangent vector can always be seem as the image of something in the Lie algebra by right-translation: $$\gamma'(s)=[R_{\gamma(s)}]_{\ast e}X(s),\quad X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s).\tag{1}$$

On the contrary to specify any such $\gamma$ we can instead give such $X : [0,1]\to \mathfrak{g}$ and solve (1) with initial condition $\gamma(0)=e$.

The idea is to translate (1) into the representation and find a differential equation for $\pi(\gamma(s))$. We must recall how to define the derivative of $\pi(\gamma(s))$, which is a curve on ${\rm U}(\cal H)$ which has no obvious smooth structure turning it into a Lie group. For that we go with the idea of smooth vectors of a representation. We take ${\cal H}^\infty_\pi$ the space of all $\Psi\in \cal H$ for which $\Pi_\Psi(g)=\pi(g)\Psi$ is smooth. We then define the derivative of $\pi(\gamma(s))$ pointwise on ${\cal H}^\infty_\pi$, i.e.

$$\left[\dfrac{d}{ds}\pi(\gamma(s))\right]\Psi\equiv\dfrac{d}{ds}\left[\pi(\gamma(s))\Psi\right]=\dfrac{d}{ds}\Pi_\Psi(\gamma(s))\tag{2}$$

Since ${\cal H}^\infty_\pi$ may be shown to be dense in the Hilbert space, this defines the derivative of $\pi(\gamma(s))$ everywhere.

Now since $\Pi_\Psi : G\to {\cal H}^\infty_\pi$ we can safely do $$\dfrac{d}{ds}\Pi_\Psi(\gamma(s))=[\Pi_\Psi]_{\ast \gamma(s)}(\gamma'(s))=[\Pi_\Psi]_{\ast \gamma(s)}([R_{\gamma(s)}]_{\ast e} X(s))=(\Pi_\Psi\circ R_{\gamma(s)})_{\ast e}X(s)\tag{3}$$

Now notice that everything happens at fixed $s$, so we are left with the problem of evaluating $(\Pi_\Psi\circ R_g)_{\ast e}Z$ for $Z\in \mathfrak{g}$. To do so we take a short curve $\sigma : (-\epsilon,\epsilon)\to G$ with $\sigma(0)=e$ and $\sigma'(0)=Z$. The obvious such curve is $\sigma(\lambda)=\exp \lambda Z$. We then have $$(\Pi_\Psi\circ R_g)_{\ast e}Z =\dfrac{d}{d\lambda}\bigg|_{\lambda =0}\Pi_\Psi(R_g(\exp \lambda Z))=\dfrac{d}{d\lambda}\bigg|_{\lambda =0}\pi(\exp \lambda Z)\pi(g)\Psi=d\pi(Z)\pi(g)\Psi\tag{4}$$

where the last equality is the definition of the derived representation which may be invoked because if $\Psi$ is a smooth vector so is $\pi(g)\Psi$.

Going back to (3) and (2) this means that $$\left[\dfrac{d}{ds}\pi(\gamma(s))\right]\Psi=d\pi(X(s))\pi(\gamma(s))\Psi,\tag{5}$$

equality for all such smooth vectors then imply equality of the operators and we find $$\dfrac{d}{ds}\pi(\gamma(s))=d\pi(X(s))\pi(\gamma(s)).\tag{6}$$

Eq. (6) is just the "in representation" version of equation (1).

If one introduces coordinates $\theta^a$ centered at the identity it turns out that it is not hard to see that $d\pi(X(s))$ becomes $$d\pi(X(s))=it_ah^a_{\phantom{a}b}(\Theta(s))\dfrac{d\Theta^b(s)}{ds}\tag{7}.$$

Combination of (6) and (7) produces Weinberg's equation (2.B.2) for $U(s)=\pi(\gamma(s))$:

$$\dfrac{d}{ds}U(s)=it_aU(s)h^a_{\phantom{a}b}(\Theta(s))\dfrac{d\Theta^b(s)}{ds}\tag{8}$$

Now Weinberg has the $t_a$, hence he picks standard paths to define each $g\in G$ and uses (8) as a starting point to try defining $\pi(g) = U_g(1)$ where $U_g$ is defined from (8) using the standard path defining $g$.