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Stream: learning: questions

Topic: Free metric group on a metric space

fosco (Aug 04 2022 at 09:06):

A "metric group" is a group $G$, whose underlying set is a metric space, and whose metric is bi-invariant, i.e. each translation is an isometry, and inversion $x\mapsto x^{-1}$ is also an isometry.

One can define a functor $U$ from the category of metric groups to the category of metric spaces that forgets the group structure. I am convinced $U$ is a right adjoint (nevertheless it is a very abstract argument that convinced me so: see theorem 3.8 here, so I might be missing something, or I might be considering the wrong choice of morphisms to apply this theorem).

How to describe the left adjoint to $U$?

In simple terms: given a metric space $(X,d)$ I would like to put a metric on the free group on $X$ with the property that whenever $G$ is a metric group, nonexpansive maps $f : X \to G$ correspond bijectively to nonexpansive group homomorphisms $\bar f : FX \to G$.

Morgan Rogers (he/him) (Aug 04 2022 at 10:30):

You might think that there is a canonical choice for the restriction of the metric to the generators, but suppose that the original metric space is unbounded (take the affine real line with its usual metric, say). Suppose $d(x,e) = c$. Then in order for the metric on the generators to coincide with the usual metric on $\R$, we must have $d(x,y) - c \leq d(e,y)$ by the triangle inequality, which means that elements which are more than $2c$ away from $x$ must be increasingly distant from the identity element. This asymmetry seems strange to me. Maybe you're considering a more flexible class of metric spaces (eg one where infinite distance is allowed?)

Morgan Rogers (he/him) (Aug 04 2022 at 10:32):

If you do allow infinite distances, then I expect the identity element will be infinitely far from the generators and the metric structure on the remaining elements will be determined by the free multiplication and inversion structure on the group.

Ralph Sarkis (Aug 04 2022 at 10:51):

Using the equations $x^{-1}\cdot x \cdot y = y = y \cdot x\cdot x^{-1}$ and $(x^{-1})^{-1} = x$, one has that all translations and inversions must be isomorphisms in $\mathbf{Met}$. Hence, a metric group is just a group in $\mathbf{Met}$.

Then, Mardare et al. give an almost explicit construction of the free object using quantitative equational logic. In short, you construct the free group as usual andthe distance between any two words is given by the smallest distance that can be derived in their logic with the assumption $x =_{\varepsilon} y$ for any $x,y \in X$ at distance $\varepsilon$. Morgan's last comment is in the right direction but you should note that this "determined by" can be quite complicated. For instance, it took us a few months to realize that one axiom we added to a theory made the metric structure collapse (i.e. all elements are at distance 0 from one another, hence the free object is a singleton space).

In the group example, I don't think $1$ is at infinite distance from everything else. In deed, since $1 = x\cdot x^{-1}$ and translation is an isometry, the distance beween $1$ and $x\cdot y$ is equal to the distance between $x^{-1}$ and $y$. More generally, $d(uxv,uyv) = d(x,y)$. Let me think more about this to find the distance (I have a hunch it will be something simple).

fosco (Aug 04 2022 at 11:09):

And yet posing $d(x,1)=\infty$ for every $x$ seems the only reasonable choice most of the times: $d(x,y)\le d(x,e) + d(e,y)$, so it must be infinite when (say) the space has infinite diameter. When it does not have infinite diameter, the natural choice would be to metrize $X \cup \{e\}$, i.e. the disjoint union of $X$ and a point that I want to use as empty word/identity element

fosco (Aug 04 2022 at 11:10):

Graev produced an example of a metric on a free group that seems to go in the right direction, but it requires you to be able to etend the metric on $X$ to a metric on $X\cup \{e\}\cup X^{-1}$, see here image.png

fosco (Aug 04 2022 at 11:11):

What bothers me is that the extension of the metric on $X$ to a metric on $\bar X$ is far from being a canonical choice: it's any metric $D$ extending the one on $X$ and such that $D(x^{-1},y^{-1})=d(x,y)$ and $D(x^{-1},y)=D(x, y^{-1})$

fosco (Aug 04 2022 at 11:13):

in principle there might be none or many such D's and making a "trivial" choice, e.g. $D(x^{-1},y)=\infty=D(x, y^{-1})$ trivializes everything, because $D(x,y)=D(x, (y^{-1})^{-1})=\infty$...

Ralph Sarkis (Aug 04 2022 at 11:15):

fosco said:

in principle there might be none or many such D's and making a "trivial" choice, e.g. $D(x^{-1},y)=\infty=D(x, y^{-1})$ trivializes everything, because $D(x,y)=D(x, (y^{-1})^{-1})=\infty$...

I am not sure what you mean, $y^{-1}$ is not in the orginal set $X$, so you don't know what $D(x,(y^{-1})^{-1})$ is right? (well, it is d(x,y) but the other rules on D don't say anything else)

Ralph Sarkis (Aug 04 2022 at 11:19):

What bothers me more is that $\rho$ is computed using sums, so $d(xy,ab) = d(x,a) + d(y,b)$ which is in contradiction with nonexpansiveness of the multiplication which states $d(xy,ab) = \max\{d(x,a), d(y,b)\}$.

Ugh... I just realized I assumed multiplication is nonexpansive from the start because that is what I am used to. Do you not require that multiplication is nonexpansive? Then, my first comment is wrong.

fosco (Aug 04 2022 at 11:20):

mmh, maybe I am confused (I certainly am, I have thought about this problem for 2 days straight and I am starting to have physical pain about it!), what I mean is that in a metric group the distances $d(x,y)$ are fixed by the distance to/from the identity e: $d(x,y)=d(xy^{-1},e)$ (and by the way also $=d(y^{-1}x,e)$), so $d(x,y^{-1})=d(xy,e)=d(yx,e)$

fosco (Aug 04 2022 at 11:22):

but now this gives you very little choice for defining $d(x,y^{-1})$, it's $d(yx,e)$, and this is infinite if you make a "boring" choice of the coproduct metric on $X\cup \{e\}$

fosco (Aug 04 2022 at 11:23):

Ralph Sarkis said:

What bothers me more is that $\rho$ is computed using sums, so $d(xy,ab) = d(x,a) + d(y,b)$ which is in contradiction with nonexpansiveness of the multiplication which states $d(xy,ab) = \max\{d(x,a), d(y,b)\}$.

Ugh... I just realized I assumed multiplication is nonexpansive from the start because that is what I am used to. Do you not require that multiplication is nonexpansive? Then, my first comment is wrong.

Yes, this is another subtlety: I am considering $G\times G$ not with the product metric (=with max) but with the taxicab metric (=with sum)

Morgan Rogers (he/him) (Aug 04 2022 at 11:27):

So in summary we have $d(u,v) = \infty$ unless $\ell(u) = \ell(v)$, in which case the distance is the sum of the distances between the parallel positions. Or at least that works for the free monoid construction, you probably need to check that inverses don't mess that up too much.

fosco (Aug 04 2022 at 11:30):

Yes, the free monoid is just $+X+X\otimes X+\dots$ and on this coproduct of powers there is a canonical choice of metric: two lists are distant $\sum d(x_i,y_i)$ if they have the same length, and infinity otherwise

Ralph Sarkis (Aug 04 2022 at 11:33):

fosco said:

Yes, this is another subtlety: I am considering $G\times G$ not with the product metric (=with max) but with the taxicab metric (=with sum)

Ok, fortunately this does not affect us too much.

You still have that translations and inversion are isometries for free: translation is $G \xrightarrow{(x,id)} G\otimes G \xrightarrow{\cdot} G$ and its inverse is the same with $x^{-1}$, and inversion is its own inverse.

Now for the explicit construction, we need my more recent paper which lets you require that multiplication is nonexpansive with respect to the taxicab metric. So we do have an adjoint, and we still need to characterize the distance between words (with something simpler than the logic in our paper).

Ralph Sarkis (Aug 04 2022 at 11:34):

fosco said:

Yes, the free monoid is just $+X+X\otimes X+\dots$ and on this coproduct of powers there is a canonical choice of metric: two lists are distant $\sum d(x_i,y_i)$ if they have the same length, and infinity otherwise

Then why doesn't the Graev metric work with the trivial choice of $D$: the coproduct metric on $X+X+1$?

fosco (Aug 04 2022 at 11:36):

Ralph Sarkis said:

Then why doesn't the Graev metric work with the trivial choice of $D$: the coproduct metric on $X+X+1$?

in that case, $d(x,y^{-1})=d(x,e)=d(x^{-1},e)=\infty$ for every x,y. But then $d(a,b)=d(ab^{-1},e)$ must be infinite; or am I using the axioms wrong?

Reid Barton (Aug 04 2022 at 11:37):

What are $x$, $y$, $a$, $b$ ranging over?

fosco (Aug 04 2022 at 11:38):

all over X

Reid Barton (Aug 04 2022 at 11:39):

Then $ab^{-1}$ isn't in $X$, so the conclusion doesn't follow

fosco (Aug 04 2022 at 11:42):

There's some clash of notation that does not allow me to see what I'm getting wrong

fosco (Aug 04 2022 at 11:42):

probably I keep calling invariably "d" different things

Morgan Rogers (he/him) (Aug 04 2022 at 11:43):

$X$ isn't a group on it's own, just a space, so $ab^{-1}$ isn't an element of $X$.

Ralph Sarkis (Aug 04 2022 at 11:44):

What you have is $d(a,b) =D(a,b)= \rho(a,b) = \rho(ab^{-1},e)$ but neither $d$ nor $D$ are defined on $ab^{-1}$ ($\rho$ is the Graev metric).

fosco (Aug 04 2022 at 11:45):

so $d(a,b)=D(ab^{-1},e)$ (capital = in the free group), ad then this latter distance is infinite if $ab^{-1}$ does not reduce to 1, and the infimum of blah blah blah otherwise

Well, D is defined on $ab^{-1}$, because it's defined on words, through rho!

Tobias Fritz (Aug 04 2022 at 12:13):

I don't know about the (non-abelian) group case, but there's an analogous free construction in the linear setting: The forgetful functor from Banach spaces to pointed metric spaces has a left adjoint, known as the formation of the Arens-Eells space or Lipschitz-free space. Every pointed metric space embeds isometrically into its Arens-Eells space. See Nik Weaver's book "Lipschitz Algebras" for details, and in particular Theorem 3.6.

fosco (Aug 04 2022 at 12:57):

That's interesting, thanks. Do you believe the linear case would give some insight on how to reason more in general? I think a normed group is way more strict...

Tobias Fritz (Aug 04 2022 at 13:18):

What do you mean by a normed group? If you want to require the condition that $d(e,a^n) = n d(e,a)$, then yes, it's quite strict: DHJ Polymath proved a couple of years ago that every metric group satisfying this type of homogeneity of the metric is abelian!

Morgan Rogers (he/him) (Aug 04 2022 at 13:22):

and torsion free, I suppose.

Tobias Fritz (Aug 04 2022 at 13:22):

Right, the groups that allow for such a metric are exactly the torsion-free abelian ones. (Where proving abelianness is the hard part.)

Tobias Fritz (Aug 04 2022 at 13:26):

So as for how much the construction of the Arens-Eells space is relevant in a non-abelian setting, I'm not sure. Is the Graev metric the one which corresponds to a left adjoint from pointed metric spaces to metric groups, necessarily turning basepoints into neutral elements? In that case, I guess that this Graev metric is already strictly closer to what you want than the Arens-Eells space, right?