Geometry of a space of specific unitary operators on $L^2(\mathbb{R})$ I'm a grad student working in harmonic analysis and, upon asking the question, "What operators have such and such a property?" ended up with a space that I think might have some interesting geometry. Unfortunately, I have to first know about geometry to determine whether it's worth it for me to pursue that idea, so I am hoping for some confirmation (or a specific statement to the contrary) for a couple of relevant questions before I invest the time it will take to relearn the relevant material.
I'm working through do Carmo's Riemannian Geometry, but both it and Google have not been helpful in finding examples of spaces of operators as manifolds. Search results always return 'Oh, you want functions on smooth manifolds!' Useful for learning about geometry, but less apparently applicable to what I'm looking at. Once upon a time, I had a decent grounding in geometry but I haven't used it since before Covid, so please pardon my rather eclectic surviving knowledge base.
For a similar space to what I'm working with, let's take the space generated by modulation, translation, and a phase factor, ie $x(a,b,c)=e^{2\pi ia}M_bT_c$, in the sense that, for $g\in L^2(\mathbb{R})$, $(e^{2\pi ia}M_bT_cg)(u)=e^{2\pi ia}e^{2\pi ibu}g(u-c)$. It seems immediate that it's a manifold, but not necessarily one with any interesting properties.

*

*Is there any room for something of interest to arise or does having the chart from all of $\mathbb{R}^3$ already say everything about the space?


*In particular, I'm trying to relate the derivatives resulting from the various real coordinates (a,b,c) to the infinitesimal generators of the various operators, but I'm taking derivatives of the operators themselves, ie the points of the manifold. I'm not really sure what the dual / space of linear functionals on this space of operators would look like, if I absolutely have to work with that to talk about the tangent space. Phrased a bit differently, rather than functions $f:Manifold\to\mathbb{R}$, I'm trying to take functions $\tilde{g}:Manifold\to L^2(\mathbb{R})$ by $\tilde{g}(x(a,b,c))(u):=x(a,b,c)g(u)$. Does this idea even make sense for differential geometry, since the result is not in $\mathbb{R}$? See below for an explicit example of what I mean.
The generators of $e^{2\pi ia},e^{2\pi ibu},T_c$ are $2\pi i,2\pi iu,-\frac{d}{du}$, in the sense that $exp(-c\frac{d}{du})g(u)=g(u-c)$ for appropriate $g$. Let's say that test functions are translated and modulated Gaussians, since the linear span of those is dense in $L^2$ and they are very, very nice.
In the order specified by $x(a,b,c)=e^{2\pi ia}M_bT_c$, we can take a test function $g$ and then compute $\frac{d}{da},\frac{d}{db},\frac{d}{dc}$.
$\frac{d}{da}\tilde{g}(e^{2\pi ia}M_bT_c)(u)=\frac{d}{da}e^{2\pi ia}M_bT_cg(u)=[2\pi i][e^{2\pi ia}M_bT_cg(u)]=[2\pi i][x(a,b,c)g(u)]=[2\pi i][\tilde{g}(x(a,b,c))(u)]$
$\frac{d}{db}e^{2\pi ia}M_bT_cg(u)=[2\pi iu][e^{2\pi ia}M_bT_cg(u)]=[2\pi iu][x(a,b,c)g(u)]$
$\frac{d}{dc}e^{2\pi ia}M_bT_cg(u)=[2\pi i-\frac{d}{du}][e^{2\pi ia}M_bT_cg(u)]=[2\pi i-\frac{d}{du}][x(a,b,c)g(u)]$
So we get a nice (presumed) basis for the tangent space from the generators of the functions comprising the space, though it's a change of basis away from $\{\frac{d}{da},\frac{d}{db},\frac{d}{dc}\}$.
Is this sensible to do?


*Based on #2: The full space I'm working with also has a very nice basis for the tangent space in terms of infinitesimal generators, so I can take the usual bracket (eg $[u,\frac{d}{du}]=-1$) with these vectors, but I can't compute the exponential map or easily take derivatives of the actual operators, despite having a closed form for what they look like (ie $\frac{d}{dx_i}$ is very hard to compute for the naive chart I have in the same vein as the example above). Would there be any differences here when computing the Levi Civita connection or is it still just 'pick a metric, compute the Christoffel symbols, and hope it lines up with what I want' as it was in my real differential geometry courses?

Assuming that this isn't meaningless drivel, if someone knows of and could recommend a relevant text, that would also be helpful, as I haven't been able to find anything that deals with spaces of operators.
 A: The thing you are probably looking for is the Heisenberg group. This is group consists of matrices of the form
$$
\begin{pmatrix}1 & a & c \\ 0 & 1 & b \\ 0 & 0 &1 \end{pmatrix}
$$
with $x,y,z \in \mathbb{R}$. This gives rise to the same representation on $L^2(\mathbb{R})$ as you described it in your post. Take any matrix $H$ of the form above and define its representation by
$$
\pi(H)f(t)=e^{ic}e^{itb}f(t+a).
$$
You can check that this indeed a unitary representation.
But our Heisenberg group is also a Lie group (group + manifold), meaning we can pick generators. These generators are then given by matrices of the form
$$
X=\begin{pmatrix}0 & a & 0 \\ 0 & 0 & 0 \\ 0 & 0 &0 \end{pmatrix} \; Y=\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & b \\ 0 & 0 &0 \end{pmatrix} \;
Z=\begin{pmatrix}0 & 0 & c \\ 0 & 0 & 0 \\ 0 & 0 &0 \end{pmatrix},
$$
which satisfy certain commutation relations and these should be somewhat in line with the ones your derived. This is what we call the Lie algebra $\mathfrak{h}$ of the Heisenberg group $H$. The matrix exponential is a one-to-one map between these two spaces. So instead of studying connection and curvature, I would recommend studying Lie groups and their corresponding Lie-algebras and their representations. For example, how the infinitesimal action of $\mathfrak{h}$ will look like. You will immediately learn about the Stone-von-Neumann theorem then, if you havent done so yet.
I am not an expert, but there are usually some heavy obstructions for representations of Lie groups on infinite-dimensional spaces and you might rely on either reducing your space to a finite-dimensional one or taking a dense subspace of $L^2(\mathbb{R})$.
Geometrically, you can do all the standard constructions, however, I do not see how this would yield any interesting information, but maybe someone else can chip in here. It is, of course, a Riemannian manifold, you can even realize it as a Riemannian sub-manifold. If you want to take a look, check out this paper: https://mwwong.info.yorku.ca/files/2014/09/EwertowskiGGA.pdf?x45988
