Saturday, 8 June 2013

Category of all models isomorphic to a given one

This is a question for aficionados of category theory.

Is there, for any given model $\mathcal{A} = (A, \vec{R})$, a category, say $Mod(\mathcal{A})$, of all models isomorphic to $\mathcal{A}$? So, the objects are these models, and the morphisms are the isomorphisms.

If so, what properties does this category $Mod(\mathcal{A})$ have?

11 comments:

  1. I don't understand your notation. In your email you said A is a 'domain' and R is a 'special relation', but I don't know what you mean by those terms, either.

    Nonetheless, there is clearly a category of all models isomorphic to a given one, for the simple reason that for any given blah-di-blah, whatever it might be, there's always a category of all blah-di-blahs isomorphic to that one, as long as

    1) you have defined the concept of isomorphism of blah-di-blahs,

    2) you can compose isomorphisms in an associative way, and

    3) the identity morphism is an isomorphism.

    (And if these 3 conditions don't hold, you shouldn't be talking about isomorphisms of blah-di-blahs!)

    Whenever you look at category of 'all blah-di-blahs isomorphic to a given one', with isomorphisms as morphism, you're looking a groupoid with one connected component. Such a groupoid is equivalent to a group. And, this group is simply the automorphism group of your given blah-di-blah.

    An automorphism is an isomorphism from something to itself. In plain English, it's a symmetry of that thing.

    So: studying the category you're talking about is just a glorified way of studying the symmetry group of your blah-di-blah. So, you might as well just study that symmetry group and drop the category-theoretic terminology.

    But from a category-theoretic viewpoint, it's much more interesting to study the category of all blah-di-blahs, and not just with isomorphisms between them, but all morphisms. That's when category theory really becomes useful.

    This is far as I can go without knowing what you're talking about. :-)

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  2. John, many thanks!

    Yes, that's what I have in mind indeed. (Since we lack a latex enabled preview here, let's see if the latex works.)

    The notation is, e.g., $(X,\leq)$ for an order, where $X$ is the domain/carrier set, and $\leq \subseteq X^2$. Or $(X, \circ)$ for a group; etc. I don't want to be to fussy about the signature of the model.

    Suppose we have, say, an 2-element total strict order $(A, R)$, where $A = \{0,1\}$ and $R = \{(0,1)\}$. The category here is the category of all models $(B,S)$ isomorphic to $(A,R$. That is, all $(B,S)$ such that, for some bijection $f:A \to B$, we have, for all $a_1, a_2 \in A$, $(f(a_1), f(a_2)) \in S$ iff $(a_1, a_2) in R$.

    But I think this is not quite the automorphism group $Aut(\mathcal{A})$, because automorphisms keep the carrier set fixed. In a sense, it's much bigger, because the carrier set can be changed arbitrarily. It seems like the automorphism group is a kind of canonical subcategory.

    Cheers,

    Jeff

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    1. Okay, now I understand your terminology.

      Yes, the groupoid of all things isomorphic to a given thing is much bigger than the group of automorphisms of that thing, but they are equivalent in the usual technical sense in category theory. And this means that anything truly interesting you can say about can be translated into a statement about the other. So, you might as well just study whatever is simpler, and that's the group.

      (This sort of fact is precisely the sort of thing that aficionados of category theory know, that most other people don't.)

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    2. John,

      Brilliant - thanks!

      But I don't really mean the category (i.e., groupoid) of automorphisms of the model $\mathcal{A}$ to itself. I mean the category of all models isomorphic to $\mathcal{A}$. And one can get an isomorphic copy of $\mathcal{A}$ simply by trivially permuting the carrier set ...

      Consider the 2-element order $(A,R)$ above, with $A = \{0,1\}$ and $R = \{(0,1)\}$. It's rigid, because the transposition swapping $0$ and $1$ flips $R$ to $S = \{(1,0)\}$. So, $R$ is not invariant under this swap.

      Still, I can simply consider instead the new model $(A, S)$, and this is (trivially) isomorphic to the original! So, any permutation $\pi : A \to A$ gives me an isomorphic copy. So it seems like, at least when we fix a specific domain/carrier set $A$, the group here is $Sym(A)$ not the automorphisms...

      Cheers,

      Jeff

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  3. [Oops - typos.]

    John,

    If I get the picture right, when we fix the domain/carrier set $A$ of our given model $\mathcal{A}$, we get a groupoid of all models, with domain $A$, isomorphic to $\mathcal{A}$.

    But surely this corresponds to any permutation $\pi: A \to A$? In other words, the relevant group is $Sym(A)$, rather than $Aut(\mathcal{A})$?

    Cheers,

    Jeff

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    1. This comment has been removed by the author.

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    2. No. The main problem here is that you're saying "this corresponds to" without saying precisely what "this" is or precisely how it's supposed to correspond to a permutation of the set A.

      Given a groupoid where all objects are isomorphic, it's equivalent, in the sense of category theory, to the groupoid containing just one of those objects, and all its automorphisms.

      So, for example, the groupoid of all 9-element ordered sets is equivalent to the groupoid with any one 9-element ordered set as its only object, and all order-preserving isomorphisms of this ordered set as morphisms. There is just one such isomorphism, the identity. So, the groupoid of all 9-element ordered sets is equivalent to the groupoid with just one object and just one morphism. The latter groupoid is another way of talking about the trivial group.

      The 'triviality' of the trivial group here is a way of noticing that the structure of a 9-element ordered set is completely 'rigid': it doesn't admit any automorphisms.

      This is an atypical example: most structures have automorphisms, aka symmetries, and that's when this subject becomes interesting.

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  4. John,

    So, what I've got in mind is clearer with the example, say, for the 2-element order example above, $\mathcal{A} = (A,R)$.

    - there are two isomorphisms (keeping $A$ fixed);
    - but only one automorphism, as $\mathcal{A}$ is rigid.

    Cheers,

    Jeff

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  5. John,
    Here's the more exact definition for "corresponds", from a few weeks ago,

    http://m-phi.blogspot.co.uk/2013/05/quine-transform-of-model.html

    (This is analogous to applying a diff $\phi : M \to M$ to a spacetime $(M, g, \dots)$, "moving" all the fields along under $\phi$.)

    Cheers,

    Jeff

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  6. It seems that what is necessary is exclusive isomorphism to give an exclusive set, which in turn requires exclusive variables. Otherwise you end up with an infinite set, e.g. automorphisms of automorphisms, with numerous permutations of boundedness.

    One of the most concise variations of this kind of thing is discussed in David K. Lewis' <>, but I think what he reached was basically a philosophical Klein Bottle. To move beyond that there are several options:

    1. Find the exclusive categories of (A, R) however they are defined, in a format similar to (AA, AB, BB, BA), but only if the variables are exclusive. In my view this means that AA is the opposite of BB and AB is the opposite of BA. Other cases are more difficult or else trivial, in a modal sense referring to specific contexts rather than the most general ones.

    2. Define an infinite set in a different way than exclusivity.

    3. Define a local-boundary set, such as a modal set, e.g. a local permutation of 'available' as opposed to 'definitive' properties. This may be something like 'measurements', or linear descriptions super-imposed on one another.

    Maybe some of this changes some of the thinking.

    My strictly qualia approach can be found in the Dimensional Philosopher's Toolkit (2013). If it seems basic, it is also much more useful than average. It has a real method for 'operating' the data, which I believe goes beyond mere permutation.

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  7. Thanks for your comment, Nathan.

    Jeff

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