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Stream: learning: questions

Topic: Pcoh and biproducts


view this post on Zulip Alejandro Díaz-Caro (Mar 21 2024 at 21:24):

A quick question: Does the category of Probabilistic Coherent Spaces (Pcoh) have biproducts? I was checking Danos and Ehrhard's paper and I am not completelly sure. There is certainly a Cartesian product, noted as the additive conjunction &\&, and it is said that iXi=(&iXi)\oplus_{i} X_{i} = (\&_{i} X_i^{\bot})^{\bot}. But I am not sure if that implies that the coproduct is the same as the Cartesian product.

view this post on Zulip Morgan Rogers (he/him) (Mar 22 2024 at 09:29):

It seems like they're different: the \oplus notation should not be interpreted as a biproduct unless they explicitly say so.

view this post on Zulip Damiano Mazza (Mar 22 2024 at 09:52):

Yes, they are different. For example, if 1\mathbf 1 is the space ({},[0,1])(\{\ast\},[0,1]) (the monoidal unit), then both 1&1\mathbf 1\mathrel{\&}\mathbf 1 and 111\oplus\mathbf 1 have web {0,1}\{0,1\}, but P(1&1)=[0,1]×[0,1]\mathsf P(\mathbf 1\mathrel{\&}\mathbf 1)=[0,1]\times[0,1] (the unit square), whereas P(11)\mathsf P(\mathbf 1\oplus\mathbf 1) is the set of all pairs (α,β)[0,1]×[0,1](\alpha,\beta)\in[0,1]\times[0,1] such that α+β1\alpha+\beta\leq 1 (the "unit triangle").

view this post on Zulip Damiano Mazza (Mar 22 2024 at 09:55):

Notice that the type 11\mathbf 1\oplus\mathbf 1 is the type of Booleans, and in fact the space 11\mathbf 1\oplus\mathbf 1 reflects this intuition in the probabilistic case: an element of P(11)\mathsf P(\mathbf 1\oplus\mathbf 1) is a sub-probability distribution on {0,1}\{0,1\}, i.e., a probabilistic Boolean. On the other hand, since 1\mathbf 1 is the space of sub-probability distributions on the singleton (intuitively representing the probability that a program of unit type terminates), 1&1\mathbf 1\mathrel{\&}\mathbb 1 is just the space of pairs of such things.

view this post on Zulip Alejandro Díaz-Caro (Mar 22 2024 at 10:23):

Thank you both!
Damiano: your explanation is exactly what I was looking for. Now I think I understand the intuition much better.