# Transform differential equation on Lie group to differential equation on representation of the Lie group

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 $$D_g$$.

• 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? – Gold Dec 27 '19 at 22:32
• 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 :) – user17945 Dec 27 '19 at 23:19
• 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. – Gold Dec 27 '19 at 23:25
• 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 – user17945 Dec 27 '19 at 23:45
• 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 ! – 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, $$\dfrac{d}{ds} \pi(\gamma(s))=d\pi(X(s))\pi(\gamma(s))$$