Let $G$ be a connected Lie group and let $\gamma : [0,1]\to G$ be a smooth path starting at the identity $\gamma(0)=e$.

Let $R_g :G\to G$ be the right-translation by $g$, i.e., $$R_g(h)=gh.$$

Then $\gamma$ naturally satisfies a differential equation $$\gamma'(s)=[R_{\gamma(s)}]_{\ast e} X(s),\quad X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s)\in \mathfrak{g}\simeq T_eG.\tag{1}$$

Suppose now $U : G\to {\rm U}({\cal H})$ is a unitary representation of $G$ on a Hilbert space $\cal H$. We also know that $U$ induces a representation of the Lie algebra $\mathfrak{g}$ by means of the derived representation construction. We simply set, for $\Psi\in {\cal H}^\infty_U$ the smooth vectors,

$$D(X)\Psi=\dfrac{d}{ds}\bigg|_{s=0}U(\exp sX) \Psi.\tag{2}$$

My question is: can we translate Eq. (1) in the group $G$ into a differential equation for $U(\gamma(s))$ in the group ${\rm U}({\cal H})$?

My take is that the equation will necessarily involve the Lie algebra representation $D$ and will provide a way to "integrate $D$ to $U$ along paths in $G$".

My Attempt

My initial idea has been to use $[R_{\gamma(s)}]_{\ast e}:\mathfrak{g}\to T_{\gamma(s)}G$ to induce on $T_{\gamma(s)}G$ a Lie algebra structure and induce a representation $D_{\gamma(s)}:T_{\gamma(s)}G\to \operatorname{End}({\cal H})$ by means of $$D_{\gamma(s)}(X_{\gamma(s)})=D([R_{\gamma(s)}]_{\ast e}^{-1}X_{\gamma(s)})$$

and then apply $D_{\gamma(s)}$ to Eq. (1). The issue seems to be how to extract $$\dfrac{d}{ds} U(\gamma(s))$$

out of all of this, which we certainly need to appear if we want a differential equation for it.


Since $U:G\to \operatorname{U}(\mathcal{H})$, we have $U_*:TG\to T\operatorname{U}(\mathcal{H})$, and in particular $U_{*g}:T_gG\to T_{U(g)}\operatorname{U}(\mathcal{H})$. The fact that $U$ is a representation implies that $$ U\circ R_g = R_{U(g)}\circ U $$ ($R$ denotes right multiplication in $G$ on the lhs, and in $\operatorname{U}(\mathcal{H})$ on the rhs). Taking the derivative at $e\in G$ on both sides, and applying to $\xi\in \mathfrak{g}\simeq T_eG$ gives $$ U_{*g}([R_g]_{*e}(\xi)) = [R_{U(g)}]_{*I}(U_{*e}(\xi)) $$ or in more suggestive notation $$ U_{*g}(\xi\cdot g) = U_{*e}(\xi)\cdot U(g). $$ You can use this equation to translate the ODE $\gamma'(s) = X(s)\cdot\gamma(s)$ on $G$ into an ODE on $\operatorname{U}(\mathcal{H})$, namely $$ \frac{d}{ds}U(\gamma(s)) = U_{*\gamma(s)}(\gamma'(s)) = U_{*\gamma(s)}(X(s)\cdot \gamma(s))= U_{*e}(X(s))\cdot U(\gamma(s)). $$ $U_{*e}$ is what you call $D$ in your question, while $U_{*g}(\cdot)\cdot U(g)^{-1}$ is what you call $D_g$.

  • $\begingroup$ Thanks very much, that's all very interesting. In the end of the day, you have just considered ${\rm U}({\cal H})$ as a smooth manifold and $U$ as a standard differentiable map right? The only point is that ${\rm U}({\cal H})$ is a space of operators on a Hilbert space, so it is an infinite dimensional group right? So what is the smooth structure that you are considering here? Are you turning ${\rm U}(\cal H)$ into a Banach space using the operator norm? $\endgroup$ – Gold Dec 27 '19 at 22:32
  • $\begingroup$ I just intended this as a formal argument (ignoring analytic details, which I'm not so familiar with). So yes, I'm thinking of $\operatorname{U}(\mathcal{H})$ as an infinite-dimensional group (in the case that $\mathcal{H}$ is infinite-dimensional), but not too seriously :) $\endgroup$ – user17945 Dec 27 '19 at 23:19
  • $\begingroup$ Thanks for clarifying! I find your argument really interesting. The equation you obtained is indeed what I expected in the end, since it is the coordinate-free version of the equation used by Steven Weinberg in "The Quantum Theory of Fields, Vol. 1" to reconstruct $U$ knowing $D$, but he never says where it came from. Your argument is very compelling towards justifying its use. I think making this fully rigorous could be a challenge, but I might be wrong. This paper (arxiv.org/abs/1309.5891) for instance suggests it is currently unknown how to turn $\rm U(\cal H)$ into a Lie group. $\endgroup$ – Gold Dec 27 '19 at 23:25
  • $\begingroup$ I doubt Weinberg is even thinking of it in this much depth - the argument is essentially just starting with $U(\exp(t\xi)g) = U(\exp(t\xi))U(g)$, and then differentiating both sides wrt $t$ at $t=0$ to obtain $U_{*g}(\xi g) = U_{*e}(\xi)U(g)$. Like I said, I don't know too much about the analytic side of things, but I imagine under suitable conditions (e.g. strong continuity), all equations should strictly be interpreted as applying on the dense subset of smooth (or analytic) vectors. See for example en.wikipedia.org/wiki/Unitary_representation $\endgroup$ – user17945 Dec 27 '19 at 23:45
  • $\begingroup$ I also doubt it, still this way of thinking makes things much clearer. Perhaps, had I ignored analytic details and worked as in the finite-dimensional case, I would have grasped this earlier. As for your final comment, I also believe that this may be turned into rigorous statements by working with the smooth vectors of the representation. Thanks very much again, you helped a lot ! $\endgroup$ – Gold Dec 28 '19 at 2:33

The answer by @user17945 inspired me on finding a seem-to-be rigorous approach and I've decided to outline in this answer. Comments and possible corrections are highly appreciated.

The main issue is that while $G$, being a Lie group, has a natural smooth structure, the same does not happen for ${\rm U}(\cal H)$. In fact, this is the subject of this question on which the paper "The Unitary Group In Its Strong Topology" is cited. The paper comments that endowing ${\rm U}(\cal H)$ with a suitable Lie group structure is still an open question. Something also alluded to in this Phys.SE thread.

In that case, in a rigorous scenario, we can't rely on a smooth structure on ${\rm U}(\cal H)$. Because of that we just shift a little bit the perspective and work with the (dense) subset of smooth vectors of the representation $\pi : G\to {\rm U}(\cal H)$ defined to be $${\cal H}^\infty_\pi = \{\Psi \in {\cal H} : \text{$g\mapsto \pi(g)\Psi$ is smooth}\}.$$

For that, let us fix $\Psi \in {\cal H}^\infty_\pi$ and let us define the map $\Pi_\Psi : G\to \cal{H}^\infty_\pi$ to be $$\Pi_\Psi(g)=\pi(g)\Psi.$$

Take now $\gamma : [0,1]\to G$ and consider $\Pi_\Psi(\gamma(s))$. We differentiate it to obtain $$(\Pi_\Psi\circ \gamma)'(s)=[\Pi_\Psi]_{\ast\gamma(s)}\gamma'(s).$$

Next we use $\gamma'(s) = [R_{\gamma(s)}]_{\ast e}X(s)$ to obtain $$(\Pi_\Psi\circ\gamma)'(s)=[\Pi_\Psi]_{\ast \gamma(s)}[R_{\gamma(s)}]_{\ast e}X(s) = [\Pi_\Psi\circ R_{\gamma(s)}]_{\ast e}X(s).$$

Now we should be worried with the understanding of the pushforward $[\Pi_\Psi\circ R_{\gamma(s)}]_{\ast e}:\mathfrak{g}\to \cal{H}^\infty_\pi$. We understand it from the kinematical viewpoint. This means that to make it act on $Z\in \frak g$ we take some differentiable curve $\sigma : (-\epsilon,\epsilon)\to G$ with $\sigma(0)=e$ and $\sigma'(0)=Z$ and define $$[\Pi_{\Psi}\circ R_{\gamma(s)}]_{\ast e}Z = \dfrac{d}{d\lambda}\bigg|_{\lambda =0}[\Pi_{\Psi}\circ R_{\gamma(s)}](\sigma(\lambda)).$$

We can take such $\sigma(\lambda) = \exp \lambda Z$. Then working through the definitions we find $$\Pi_\Psi\circ R_{\gamma(s)}(\exp \lambda Z) = \Pi_\Psi((\exp \lambda Z)\gamma(s)) =\pi(\exp \lambda Z)\pi (\gamma(s)) \Psi.$$

Now since $\Psi\in {\cal H}^\infty_\pi$ so is $\pi(\gamma(s))\Psi$ for any fixed $s\in [0,1]$. Differentiating at $\lambda =0$ we have just the definition of the derived representation and find $$[\Pi_\Psi\circ R_{\gamma(s)}]_{\ast e}Z = d\pi(Z)[\pi(\gamma(s))\Psi].$$

Finally we do this for $Z = X(s)$. Since we do this for each $s\in [0,1]$ fixed everything is fine and we obtain $$(\Pi_\Psi\circ \gamma)'(s)=d\pi(X(s))[\pi(\gamma(s))\Psi].$$

Reorganizing and using the definition of $\Pi_\Psi$ we find $$\dfrac{d}{ds} \pi(\gamma(s))\Psi = d\pi(X(s))\pi(\gamma(s)) \Psi.$$

Since this differential equation is valid in all of $\cal{H}^\infty_\pi$ and this is a dense subspace we may simply write, assuming extension by continuity, $$\dfrac{d}{ds} \pi(\gamma(s))=d\pi(X(s))\pi(\gamma(s))$$


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