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Stream: learning: questions

Topic: pairwise relationships of three things

David Egolf (Nov 24 2022 at 03:19):

I was thinking about the situation where we have three things with asymmetric pairwise relationships. Drawing arrows between each pair to indicate the direction of the relationship, I think we get two basic shapes: (1) a shape where the tip of each arrow leads to the tail of the next and (2) a shape like the previous except where one of the arrows are reversed, so it looks like the diagram for a morphism in a slice category.

The shape (2) corresponds to a morphism in a slice category.
Does the more symmetric shape (1) correspond to a morphism or object in some famous category? Intuitively, it seems unlikely to be nice for relating two things, because of how symmetric it is. But maybe it's really handy for talking about certain kinds of relationships between three things?

Ralph Sarkis (Nov 24 2022 at 03:50):

A morphism in a coslice category? I don't understand what you mean, can you try drawing your shapes with quiver ?

David Egolf (Nov 24 2022 at 04:06):

Sure, @Ralph Sarkis. Here's the slice category morphism shape:
shape (2)

and here's the other shape I had in mind:
shape (1)

I was just learning about how shape (2) is handy near the start of "Algebra: Chapter 0" (in the context of introducing morphisms corresponding to diagrams), and it made me wonder how shape (1) gets used in category theory. Does shape (1) describe a morphism or object in some famous category, I wonder?

Ralph Sarkis (Nov 24 2022 at 16:33):

I am still not sure I totally get what you mean. In particular, when you say shape (2) "corresponds" to a morphism in a slice category, I don't see a precise meaning. It is true that when defining morphisms in a slice category, we can use a diagram of this shape, but this shape can appear in many places without a specific link to slice categories. For instance, we can define the composition of two morphisms using that diagram, the hypotenuse is the composite of the other two sides. Due to how central composition is to category theory, shape (1) will inevitably appear less often.

You have probably learned that CTists love to say that diagrams commute. For shape (1), it can be ambiguous. We usually say that if two paths have the same source and target, then their composite should coincide. In shape (1), it means that if you go around the triangle once, twice, or n times, the composite will be the same. But it could also mean that this is the same thing as going around the triangle 0 times, i.e. taking the empty path whose composite is the identity. Naming the arrows $f$, $g$, and $h$, shape (1) commuting implies $(hgf)^n = hgf$ and $(fhg)^n = fhg$ and $(gfh)^n = gfh$, and depending on your point of view, $n$ can range over all of $\N$ or only the positive integers ($n\neq 0$).

If you allow $n=0$, then all the morphisms in the triangle are isomorphisms. Puzzle: find morphisms that make the triangle commute in the weaker sense (not allowing $n\neq 0$), such that none are isomorphisms.

David Egolf (Nov 24 2022 at 18:35):

You're not seeing a precise meaning because I don't have one in mind. :upside_down: I'm just interested in thinking about ways in which three things can be related, in terms of pairwise relationships - it seems pretty fundamental. I was reading a little in a philosophical commentary on the Tao Te Ching, and the emphasis there (to my reading) on "things" achieving their "thing-ness" by means of interactions got me in a philosophical mood, I guess.

I had not thought about what it would mean for shape (1) to commute! To talk about that, I had assumed I would need to define some endomorphisms on the objects to provide a second path other than going around the triangle. But I notice that the identity morphisms are already defined, corresponding to the $n=0$ case you describe. It also seems interesting how commuting in this case requires three conditions (for each $n$), to respect the symmetry of the diagram!

David Egolf (Nov 24 2022 at 18:42):

Thinking about the puzzle, my first intuition is to have each morphism be a "projection" of some kind. That way, when we loop back around to apply that morphism again, it won't do anything the second time around. That should ensure that successive loopings don't change anything, which is what we need.
Let's say all three objects in the triangle are the same object, $A$. And let me also assume that the three morphisms $h$, $g$, $f$ commute with one another. Finally, let me assume that each morphism is like a projection, so that $h^2 = h, g^2 = g, f^2 = f$. Then $(hgf)^2 = (hgf)(hgf) = h^2 g^2 f^2 = hgf$. I think our conditions for the triangle commuting will all be satisfied now, for $n > 0$. As an example, let's work in the category of finite dimensional vector spaces, let $A$ be a three dimensional vector space made from real numbered triples $(a,b,c)$, let $h$ set the first coordinate to zero, let $g$ set the second coordinate to zero, and let $f$ set the third coordinate to zero. None of these are isomorphisms, and they satisfy the conditions I needed above.

David Egolf (Nov 24 2022 at 18:49):

Thinking a bit more about the shapes. In shape two, if we have commutativity, the hypotenuse can be "computed" from the other sides of the triangle, as it can be produced by composition. There is no guarantee that the vertical right leg can be formed from the hypotenuse by composition, however, so this computation often loses information.

In shape 1, it seems like the three morphisms are on more even grounds, with none of them being "computable" from one another even if we have commutativity (we can't necessarily form one morphism in the triangle from another by means of composition).
I'm not sure in what applied contexts we would have three objects "on even footing" like in shape 1, but it seems like there should be some.

David Egolf (Nov 24 2022 at 18:57):

I think shape 1 would show up, for example, when modelling a process that has three steps that are endlessly repeated. Maybe something like a very simple robot that has three parts to its algorithm that loop endlessly, where each part of the algorithm determines a change to its state.

David Egolf (Nov 24 2022 at 19:02):

Anyways, thanks for sharing your thoughts, I appreciate it!