bio | website | math.berkeley.edu/~sramesh |
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location | Berkeley, CA | |
age | 30 | |
visits | member for | 5 years, 6 months |
seen | yesterday | |
stats | profile views | 3,918 |
I was a graduate student in the Logic program at Berkeley, broadly interested in categorical logic and foundations of mathematics, as well as in applications of category theory to the semantics of programming languages. I work for Google now.
Feb
11 |
revised |
Hat Problem/Hamming Codes
Nitpick about 1-based indexing |
Feb
11 |
comment |
Can we recognize when a category is equivalent to the category of models of a first order theory?
And composition is straightforward: the binary composition of F(X, Y) and G(Y, Z) is the predicate H(X, Z) := There exist Y such that F(X, Y) and G(Y, Z). And the identity morphisms are given by equality predicates [or, technically, given the description as I've stated it so far, the identity morphism on A(X) is the predicate F(X_1, X_2) := A(X_1) & A(X_2) & (X_1 = X_2)]. |
Feb
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Can we recognize when a category is equivalent to the category of models of a first order theory?
As for how to construct the classifying logos of an arbitrary theory T, take as objects the definable predicates (of any arity) in the theory, with morphisms being definable, provably functional relations on the extensions of those predicates up to provable equivalence. I.e., a morphism from D(X) to R(Y) is a predicate F(X, Y) such that the theory proves "For all X and Y, F(X, Y) implies (D(X) and R(Y)), and for all X such that D(X), there exist unique Y such that F(X, Y).", where X is a tuple of variables of any length and similarly for Y. |
Feb
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Can we recognize when a category is equivalent to the category of models of a first order theory?
There are precisely two of these, and neither has a natural transformation into the other; thus, the category lacks a terminal object. |
Feb
11 |
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Can we recognize when a category is equivalent to the category of models of a first order theory?
The category I describe won't necessarily have finite limits: for example, consider the four-element Boolean algebra, viewed as a preorder category. This is a Boolean logos, and the functors from it to Set preserving Boolean logos structure are those which send its top element to 1 (by virtue of being a terminal object), its bottom element to 0 (by virtue of its arrow into the top element being the bottom subobject of the top element), one of the middle elements to 1 and the other to 0 (by virtue of their arrows into the top element being complementary subobjects). |
Feb
11 |
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Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
Oh, that makes sense. Thanks. |
Feb
11 |
revised |
Hat Problem/Hamming Codes
added 21 characters in body; added 25 characters in body; added 1 characters in body |
Feb
11 |
answered | Hat Problem/Hamming Codes |
Feb
11 |
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Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
Is it something like a higher-dimensional analogue of that same phenomenon? (Man, these character limits are brutal... this and the last two posts were meant to be one single comment) |
Feb
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Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
(e.g., if A, B, and C happened to all be identical, in addition to composing the left and right morphisms, there would also be simply projection the left morphism, projecting the right morphism, producing the square of the left morphism, etc.)? That is, the theory of categories doesn't concern itself with such cases as where the three objects in binary composition all happen to accidentally line up; it can't see all these functions from Hom(A, A) x Hom(A, A) to Hom(A, A) for objects A, and so it doesn't impose coherence equalities between them. |
Feb
11 |
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Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
Is the idea here analogous to the fact that, in some sense, the theory of ordinary categories (which I want to describe using a multicategory of some sort, but can't quite, but suppose I could if I had the multicategory live in some fancy category other than Set) sees one and only one function from Hom(A, B) x Hom(B, C) to Hom(A, C) for objects A, B, and C [binary composition], even though, for any particular actual objects in an actual category, there may be many non-equivalent functions of this type which can be built out of the structure of a category |
Feb
11 |
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Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
I'm not sure how to read the diagram you linked to. It looks like a diagram of a plain-vanilla Eckmann-Hilton argument, except for the $l$s and $l^{-1}$ pieces, and I'm not sure what those are. That having been said, I <i>think</i> I understand anyway. At least, the last paragraph above seems like what I was thinking the answer was anyway (as for why there's no demand for a coherence isomorphism between the path all the way around and the identity). But let me make sure I understand: (Question coming in a followup comment with more characters left...) |
Feb
11 |
revised |
Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
Retitled |
Feb
11 |
awarded | Student |
Feb
11 |
revised |
Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments
added 88 characters in body |
Feb
11 |
asked | Effects of “weak” vs. “strict” categories in Eckmann-Hilton arguments |
Feb
11 |
answered | What is information-theoretic lower bound? |
Feb
11 |
awarded | Critic |
Feb
11 |
revised |
Can we recognize when a category is equivalent to the category of models of a first order theory?
Rewording |
Feb
11 |
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Compact Hausdorff and C^*-algebra “objects” in a category.
Actually, it occurs to me that there's a much easier way to describe this than the detour through infinitary Lawvere theories I've been using. Given a monad $M$ on $Set$, an algebra for this in the category $C$ should be an algebra (in the standard sense) for the monad $M^{C^{op}}$ whose carrier is in the range of the Yoneda embedding, where $M^{C^{op}}$ is the monad on presheaves on $C$ induced by postcomposition with $M$. This should be equivalent to what we've both been saying, but seems to me now much clearer (though perhaps others will differ). |