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Stream: theory: science

Topic: Lawvere metric space from gravitational force

Jean-Baptiste Vienney (Nov 25 2023 at 16:32):

Some days ago, it happened that there were two simalteneous things in my mind: attraction and Lawvere metric spaces. I was not so much thinking to gravitational attraction but to attraction between people: if you have two people $x$ and $y$ and if you denote $d(x,y)$ the attraction generated by $y$ on $x$, then it is clearly not a symmetric function but it still somehow a distance between two people. And I'm always thinking to this fact: human or social sciences are too much complicated for mathematics because they involve less obvious dynamics. But there is some dynamic: maybe attraction between people creates somehow very complicated dynamical systems. People who despise mathematics say it all the time: things are not black or white etc... and they are not wrong but what they don't know is that the mathematics behind human life phenomena are probably even more complicated or as complicated but different than classical mathematics used in physics. For instance, we struggle to understand economics and they generate interesting new topics like game theory etc... But, well, it's easier to think to attraction between planets (for instance) first because it is something that we know better.

Jean-Baptiste Vienney (Nov 25 2023 at 16:32):

Definition: Consider the set $X=\mathbb{R}^3$. Define a function $m:\mathbb{R}^3 \rightarrow [0,\infty)$ which will give the mass
$m_{(x,y,z)}$ of a single object at $(x,y,z)$. If there is no object at $(x,y,z)$, then $m_{(x,y,z)}=0$.

Denote $d:X\times X \rightarrow [0,\infty)$ the euclidean metric.

If $a,b \in \mathbb{R}^3$ are such that $a \neq b$, denote the acceleration (and not the force) generated by the object at $b$ on the object at $a$ by (without the $G$):

$\rho(a,b)=\frac{m_{b}}{d(a,b)^2}$

Jean-Baptiste Vienney (Nov 25 2023 at 16:32):

Define $\rho(a,a)=0$ (the object at $a$ doesn't generate any acceleration on itself).

Proposition: $(X,\rho)$ is a Lawvere metric space.

• $\rho(a,a)=0$ is true by definition (and this definition is the natural one)

• The triangular inequality is verified:

If $a,b,c$ are three distinct points in $X$, then:

$ρ(a,c)=\frac{m_c}{d(a,c)^2}\le ρ(a,b)+ρ(a,c)=\frac{m_b}{d(a,b)^2}+\frac{m_c}{d(b,c)^2}$
​
Proof: To see this, simply note that $d(a,c)^2=d(a,b)^2+d(b,c)^2$ (edit: no, only if the objects are the vertices of a right triangle...), and thus $\rho(a,c)=\frac{m_c}{d(a,b)^2+d(b,c)^2} \le \frac{m_c}{d(b,c)^2}$.

Jean-Baptiste Vienney (Nov 25 2023 at 16:32):

Remarks:

• The separation property is not verified because $\rho(a,b)=0$ if $m_b=0$.
• The symmetry is not verified because $\rho(a,b) \neq \rho(b,a)$ if $a,b$ are two distinct points such that $m_a \neq m_b$.
• The only usual property of a distance which is not required in the definition of a Lawvere metric space and which is verified is that $\rho(a,b) < \infty$. But in fact, I think that maybe the construction can be generalized by replacing $X$ by any "metric space" where the metric $d$ is like a usual metric but with values in $[0,\infty]$ and which verifies not only the triangular inequality but also the stronger $d(a,c)^2 \le d(a,b)^2+d(b,c)^2$. Like this, we would obtain a fully general Lawvere metric.

Jean-Baptiste Vienney (Nov 25 2023 at 16:32):

I'm not sure what I want to ask. Have you already thought to that, read something like that or do you find it interesting? I don't know lot of physics but it sounds interesting to me. I'm sure that by following this idea, we could go much further. Probably I would be happy with any kind of conversation related to that.

Todd Trimble (Nov 25 2023 at 17:07):

Jean-Baptiste Vienney said:

Define $\rho(a,a)=0$ (the object at $a$ doesn't generate any acceleration on itself).

Proposition: $(X,\rho)$ is a Lawvere metric space.

• $\rho(a,a)=0$ is true by definition (and this definition is the natural one)

• The triangular inequality is verified:

If $a,b,c$ are three distinct points in $X$, then:

$ρ(a,c)=\frac{m_c}{d(a,c)^2}\le ρ(a,b)+ρ(a,c)=\frac{m_b}{d(a,b)^2}+\frac{m_c}{d(b,c)^2}$
​
Proof: To see this, simply note that $d(a,c)^2=d(a,b)^2+d(b,c)^2$, and thus $\rho(a,c)=\frac{m_c}{d(a,b)^2+d(b,c)^2} \le \frac{m_c}{d(b,c)^2}$.

I'd check the math here. Suppose the density is uniform with constant $1$, and $a, b, c$ are on a line, parametrized so that $a = 0, b = 10, c = 1$. Then $\rho(a, c) = 1$ and $\rho(a, b) = 1/100$ and $\rho(b, c) = 1/81$.

Jean-Baptiste Vienney (Nov 25 2023 at 17:10):

Hmm. Let me think two minutes.

Jean-Baptiste Vienney (Nov 25 2023 at 17:10):

Oh, yes I see :grinning_face_with_smiling_eyes:

Jean-Baptiste Vienney (Nov 25 2023 at 17:11):

I've considered that the three objects are the vertices of a right triangle, which is obviously not always the case.

Jean-Baptiste Vienney (Nov 25 2023 at 17:12):

Yes, so this inequality is false.

Jean-Baptiste Vienney (Nov 25 2023 at 17:14):

It doesn't work!

Eric Forgy (Nov 25 2023 at 17:18):

For what it's worth, I love to see experimentation like this :hearts:

In my half century+ of experience, if 1 out of 100 ideas work out, you're doing very well. So keep trying :muscle:

Jean-Baptiste Vienney (Nov 25 2023 at 17:22):

Yes, I have much other ideas which work! But it took me a lot of work to get them. Nevertheless, I try new naive things all the time!

David Egolf (Nov 25 2023 at 17:29):

This is fun! It makes me think of a related question. Let $d: X \times X \to [0, \infty]$ be the distance function of a Lawvere metric space, so that:

• $d(x,x)=0$ for all $x \in X$
• $d(x,y)+d(y,z) \geq d(x,z)$ for all $x,y,z \in X$

And let $f: [0, \infty] \to [0, \infty]$ be a function.

Finally, let $\rho: X \times X \to [0, \infty]$ be the following function: $\rho(x,y) = f(d(x,y))$ for all $x,y \in X$. So, $\rho = f \circ d$.

Then, what are some functions $f$ so that $\rho$ is a distance function of a Lawvere metric space?

Jean-Baptiste Vienney (Nov 25 2023 at 17:34):

If $f(0)=0$, $f$ is increasing and $f(x+y) \le f(x) + f(y)$, it will work.

Jean-Baptiste Vienney (Nov 25 2023 at 17:38):

I found thid on Google: Metric Preserving Functions. They don't talk about Lawvere metric spaces but they say:

Jean-Baptiste Vienney (Nov 25 2023 at 17:39):

Screenshot-2023-11-25-at-12.39.07-PM.png

Jean-Baptiste Vienney (Nov 25 2023 at 17:40):

The definition can be generalized to Lawvere metric spaces without any problem!

David Egolf (Nov 25 2023 at 17:45):

Let me try it out. Assume that $f(0)=0$, $f$ is increasing (so $f(x') \geq f(x)$ if $x' \geq x$) and $f(x+y) \leq f(x) + f(y)$.
We first check that $\rho(x,x) = 0$ for all $x \in X$:

• $\rho(x,x) = f(d(x,x)) = f(0) = 0$

Now we check that $\rho(x,y) + \rho(y,z) \geq \rho(x,z)$ for all $x,y,z \in X$:

• $\rho(x,y) + \rho(y,z) = f(d(x,y)) + f(d(y,z)) \geq f(d(x,y) + d(y,z))$
• We know that $d(x,y) + d(y,z) \geq d(x,z)$. Since $f$ is increasing that means that $f(d(x,z) + d(y,z)) \geq f(d(x,z)) = \rho(x,z)$.
• So we conclude $\rho(x,y) + \rho(y,z) \geq \rho(x,z)$, as desired.

David Egolf (Nov 25 2023 at 17:49):

Interestingly, the requirements on $f$ seem at least similar to the definition of a "(lax) monoidal monotone" from here, page 55.

Jean-Baptiste Vienney (Nov 25 2023 at 17:54):

Yes, it a functor from the monoidal category $[0,\infty]$ to it self where an arrow $x \rightarrow y$ is an equality $x \le y$, the identity at $x$ is $x \le x$ and the tensor product is given by addition. However, I'm not sure it is a monoidal functor...

Jean-Baptiste Vienney (Nov 25 2023 at 17:56):

Oh no, an oplax monoidal functor is one such that $f(x+y) \le f(x) + f(y)$ and $f(0)\le 0$ if I'm not mistaken

Jean-Baptiste Vienney (Nov 25 2023 at 17:57):

Jean-Baptiste Vienney said:

If $f(0)=0$, $f$ is increasing and $f(x+y) \le f(x) + f(y)$, it will work.

So this definition is exactly the one of an oplax monoidal functor from $[0,\infty]$ to itself if I'm not mistaken.

Jean-Baptiste Vienney (Nov 25 2023 at 17:58):

David Egolf said:

Interestingly, the requirements on $f$ seem at least similar to the definition of a "(lax) monoidal monotone" from here, page 55.

Yes, we arrived at the same conclusion ahah

David Egolf (Nov 25 2023 at 18:04):

This lecture also seems like a relevant reference. The requirements in the definition of an "oplax monoidal monotone" seem like the requirements we placed on $f$.

Jean-Baptiste Vienney (Nov 25 2023 at 18:24):

Yes, exactly. I'm also wondering about something else: could we generate a topology on a set $X$ from a function $\rho:X \times X \rightarrow [0,\infty]$ which verify even weaker conditions than a Lawvere metric?

From such a function, we can define the open balls $U_\rho(x,\epsilon)=\{y \in X,~\rho(x,y)<\epsilon\}$ for every $x \in X$ and $\epsilon>0$.

An important fact is that in the case where $(X,\rho)$ is a Lawvere metric space, then the open balls form a base of a topology. It means that if $\mathcal{B}=\{U_\rho(x,\epsilon),~x \in X,~\epsilon > 0\}$, then the set $\tau$ constituted by all the unions of elements of $\mathcal{B}$ (including the empty union, i.e. the empty set) is a topology on $X$.

We know that a set $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base of a topology when these two conditions are verified:

• $X \subseteq \underset{B \in \mathcal{B}}{\bigcup}B$
• If $B_1,B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists $B_3 \in \mathcal{B}$ such that $B_3 \subseteq B_1 \cap B_2$

Jean-Baptiste Vienney (Nov 25 2023 at 18:25):

The first condition is implied by the requirement $\rho(x,x)=0$ in the definition of a Lawvere metric space.

Jean-Baptiste Vienney (Nov 25 2023 at 18:25):

The second one is implied by the triangular inequality in this same definition.

Jean-Baptiste Vienney (Nov 25 2023 at 18:27):

I think, maybe we can keep $\rho(x,x)=0$ but ask for less than the triangular inequality while keeping that the open balls form a base of a topology.

Jean-Baptiste Vienney (Nov 25 2023 at 18:29):

Doing a picture illustrating why the triangular inequality implies the second condition above, I would want to ask for this condition to be verified instead of the triangular inequality:

Jean-Baptiste Vienney (Nov 25 2023 at 18:29):

$\forall \epsilon > 0, \exists \eta > 0$ such that for all $a,b,x \in X$ if $\rho(x,b)<\eta$ then $|\rho(a,x)-\rho(a,b)|<\epsilon$

Jean-Baptiste Vienney (Nov 25 2023 at 18:33):

Maybe, we need to ask another similar one in addition, or something slightly different: I'm confused by what happens when $\rho$ is not symmetric.

David Egolf (Nov 25 2023 at 18:54):

I haven't carefully read everything that you said above, but I was wondering if the concept of the "initial topology" could be useful here. If we have a function $f: X \to Y$ where $Y$ is a topological space, then we can induce a topology on $X$ using this function. Namely, we take the induced topology on $X$ to be the coarsest topology so that $f$ becomes continuous.

I was thinking we could set up these maps $X \to_{\Delta} X \times X \to_{\rho} [0, \infty]$, where $\Delta: X \to X \times X$ acts by $\Delta(x) = (x,x)$. Then if we have some topology on $[0, \infty]$, we could put the initial topology on $X \times X$ as induced by $\rho$. Then we could put the initial topology on $X$ as induced by $\Delta$. I don't know what topology to put on $[0, \infty]$ though, to start this process!

Jean-Baptiste Vienney (Nov 25 2023 at 19:21):

It looks very interesting. You can simply put the usual "metric" (slightly more general than in the definition of a metric space because the distance between two points can be equal to $\infty$) topology on $[0,\infty]$. It is the topology such that the open sets are the union of open balls, where open balls are with respect to the "metric" $\sigma(x,y)=|x-y|$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:25):

Let see what is exactly the topology induced on $X$ by this process.

Jean-Baptiste Vienney (Nov 25 2023 at 19:27):

The topology on $X \times X$ will be $\tau_{\rho}^{X \times X}$ defined as the set of all the arbitrary unions of finite intersections of sets $\rho^{-1}(U)$ where $U$ is an open set in $[0,\infty]$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:28):

But the topology $\tau^{[0,\infty]}$ on $[0,\infty]$ is itself the set of all arbitrary unions of open balls.

Jean-Baptiste Vienney (Nov 25 2023 at 19:33):

We can simplify a bit the reasoning by using that the initial topology is transitive (as I read on Wikipedia).

Jean-Baptiste Vienney (Nov 25 2023 at 19:34):

So, the topology we induce on $X$ is the initial topology induced by $\rho \circ \Delta$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:35):

So the topology on $X$ will be $\tau_{\rho}^{X }$ defined as the set of all the arbitrary unions of finite intersections of sets $(\rho \circ \Delta)^{-1}(U)$ where $U$ is an open set in $[0,\infty]$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:36):

Finally $\tau_{\rho}^{X }$ is the set of all arbitrary unions of finite intersections of arbitrary unions of $(\rho \circ \Delta)^{-1}$ of open balls in $[0,\infty]$ :upside_down: (because inverse images preserve arbitrary unions)

Jean-Baptiste Vienney (Nov 25 2023 at 19:38):

I would not be surprised if this description can be slightly simplified.

Jean-Baptiste Vienney (Nov 25 2023 at 19:41):

• We can look at what it is $(\rho \circ \Delta)^{-1}$ of an open ball $U(t,\epsilon)$ for some $t \in [0,\infty]$ and $\epsilon>0$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:43):

Hmm no it doesn't work well

Jean-Baptiste Vienney (Nov 25 2023 at 19:44):

because $(\rho \circ \Delta)^{-1}(U(t,\epsilon))=\{x \in X \text{ such that } \rho(x,x) \in U(t,\epsilon)\}=X$ if $\rho(x,x)=0$ for all $x\in X$.

Jean-Baptiste Vienney (Nov 25 2023 at 19:46):

So the topology on $X$ will be the discrete topology if $\rho(x,x)=0$ for all $x \in X$.

Mike Shulman (Nov 25 2023 at 19:47):

Another thing you could try is the topology jointly induced by the family of functions $\rho(x,-) : X \to [0,\infty]$ for all $x\in X$. Or $\rho(-,x)$, or both of them.

Jean-Baptiste Vienney (Nov 25 2023 at 19:49):

Yes, it looks like a good idea.

David Egolf (Nov 25 2023 at 19:54):

Jean-Baptiste Vienney said:

So the topology on $X$ will be the trivial topology if $\rho(x,x)=0$ for all $x \in X$.

Without thinking carefully about it, I now realize that it makes that sense that this should happen: I think this happens because $(\rho \circ \Delta)(x) = \rho(x,x) = 0$ for all $x \in X$. Doing something like what Mike Shulman suggests sounds more promising!

Jean-Baptiste Vienney (Nov 25 2023 at 20:03):

Yes, I agree. This topology will be given by the unions of finite intersections of sets of the form
$\{u \in X \text{ such that } \rho(x,u) \in B(t,\epsilon)\}$
and
$\{u \in X \text{ such that } \rho(u,x) \in B(t,\epsilon)\}$
for all $x \in X$, $t \in [0,\infty]$ and $\epsilon>0$.

Jean-Baptiste Vienney (Nov 25 2023 at 20:06):

Now, it looks that most of the time the singletons will not be open sets and so the topology will not be discrete which is good.

Jean-Baptiste Vienney (Nov 25 2023 at 20:07):

I mean a singleton could be an open set if the function $\rho$ is somehow singular somewhere on $X \times X$ I think.

Jean-Baptiste Vienney (Nov 25 2023 at 20:14):

It sounds interesting. Now, I can't continue working on this right now because it would require to write things down carefully and would take time to continue on this subject... But maybe another day!

JR (Nov 25 2023 at 23:23):

David Egolf said:

This lecture also seems like a relevant reference. The requirements in the definition of an "oplax monoidal monotone" seem like the requirements we placed on $f$.

See the encyclopedia of distances by Deza and Deza