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In physics intros to 2d conformal field theory, people often talk about the "group of conformal transformations". Of course, that's not a group but rather a pseudo-group... that's not what my question is about.

I care about a certain group $G$, which I'll define in a moment. My question is: Is it the trivial group?

Fix, once and for all, a small number $\varepsilon$.
For $j\in\{1,2,3\}$, let $D_j\subset \mathbb C$ be the disc of radius $1+j\varepsilon$ around the origin of the complex plane, so that
$D_1\subset D_2 \subset D_3\subset \mathbb C$.

I'll define my group $G$ by generators and relations:

  • Generators:
    The set $X$ of generators of $G$ is the set of injective holomorphic maps $f:D_2\to D_3$ satisfying $D_1\subset f(D_2)$
    For $f\in X$, I'll write $[f]$ for the corresponding generator of $G$.

  • Relations:
    Note that if $f,g\in X$, then $f\circ g$ is defined on $D_1$.
    Whenever $f,g,h\in X$ satisfy $f|_{D_1}=(g\circ h)|_{D_1}$, then we declare $[f] = [g][h]$ in $G$.

Is $G$ trivial?

What I know so far about this problem:
I know a bunch of elements of $X$ that are trivial in $G$. Here's a proof that for any (small enough) scalar $a$, the element $(z\mapsto z+a)\in X$ is trivial in $G$.

Proof. Consider the map $D_3\to\mathbb CP^1$ given by $z\mapsto \exp\big(2\pi i z\big/\frac a2\big)$.
Pull back the Lie algebra of holomorphic vector fields on $\mathbb CP^1$ along that map to produce a subalgebra of the Lie algebra of vector fields on $D_3$ isomorphic to $\mathfrak{sl}(2,\mathbb C)$.
By small-time-exponentiating these vector fields, one gets an embedding $U\hookrightarrow X$ from some neighbourhood $U$ of the origin of $SL(2,\mathbb C)$ into the set $X$ of generators of $G$.
The universal group associated to the local Lie group $U$ is $SL(2,\mathbb C)$.
So one gets a homomorphism $\psi:SL(2,\mathbb C)\to G$.
The one-parameter group $t\mapsto (z\mapsto z+at)\in G$ is the composite of a certain one-parameter group $\gamma:\mathbb R\to SL(2,\mathbb C)$ with the homomorphism $\psi$.
Since $\gamma$ is not injective, neither is the composite $\psi\circ \gamma$.
The map $(z\mapsto z+a)$ is in the kernel.   q.e.d.

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  • $\begingroup$ Why does $G$ not contain a copy of $[1,\infty)$? Are the maps $z \mapsto cz$ for $c \in [1,\frac{1+3\varepsilon}{1+2\varepsilon}]$ not in $X$ for some obvious reason? $\endgroup$
    – Steve Huntsman
    Commented Aug 6, 2021 at 13:01
  • $\begingroup$ @SteveHuntsman. There is indeed an obvious map from the group of multiplicative reals (indeed, the group of multiplicative complex numbers) to my group $G$. I would be very surprised if this map is injective. $\endgroup$
    – André Henriques
    Commented Aug 6, 2021 at 13:11
  • $\begingroup$ Groups of this flavor come up a lot in inverse semigroup theory where they are called universal groups. Often things are undecidable but you are using a special type of partial injection $\endgroup$
    – Benjamin Steinberg
    Commented Aug 6, 2021 at 14:20
  • $\begingroup$ @BenjaminSteinberg Alternatively, I could have used as generators the set of injective holomorphic maps $f:D_2\to D_3$ that satisfy $|f(z)-z|<\varepsilon$ for all $z\in D_2$. I don't think that the specific choice of set of generators matters much to my question. $\endgroup$
    – André Henriques
    Commented Aug 6, 2021 at 21:51
  • $\begingroup$ What I meant is holonorphic is stronger than partial homeomorphism $\endgroup$
    – Benjamin Steinberg
    Commented Aug 6, 2021 at 22:05

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