Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: are measures functors?

David Egolf (Jan 28 2022 at 18:57):

I was learning a little about measures. The definition of a measure kind of looks a bit like a functor (?), in the sense that the measure of a disjoint union of two sets is the sum of the measure of the two sets. Is a measure "just" a functor between some appropriate categories?
For easy reference here is the definition of a measure I'm working with:
measure

(image from the book "Measure Theory and Probability Theory")

Matteo Capucci (he/him) (Jan 29 2022 at 09:02):

Yes, they are maps of quantales, i.e. cocomplete ordered monoids

Matteo Capucci (he/him) (Jan 29 2022 at 09:04):

The 'functoriality' is in the order-preserving property, and then you ask them to also preserve the quantale operation (set union on the algebra, + on the positive reals)

Matteo Capucci (he/him) (Jan 29 2022 at 09:04):

The two are closely related here

Matteo Capucci (he/him) (Jan 29 2022 at 09:06):

In fact from these you can derive that a measure is also cocontinuous, i.e. it sends the union of a 'growing' sequence of sets to the supremum (= join operation on the positive reals) of their union

Matteo Capucci (he/him) (Jan 29 2022 at 09:08):

Mmh actually... Maybe this is not enough to get a map of quantales bc that preserves arbitrary unions, not just directed ones
So maybe 'map of ordered monoid' with a bunch of extra niceness is the best you can do

Nathaniel Virgo (Jan 29 2022 at 09:08):

What do Markov kernels correspond to in this picture?

Matteo Capucci (he/him) (Jan 29 2022 at 09:40):

Profunctors, I guess..?

John Baez (Jan 29 2022 at 18:50):

Matteo Capucci (he/him) said:

In fact from these you can derive that a measure is also cocontinuous, i.e. it sends the union of a 'growing' sequence of sets to the supremum (= join operation on the positive reals) of their union.

Mmh actually... Maybe this is not enough to get a map of quantales bc that preserves arbitrary unions, not just directed ones

John Baez (Jan 29 2022 at 18:51):

It preserves countable directed colimits.

John Baez (Jan 29 2022 at 18:54):

The theory of measurable cardinals introduces a fun variation on the standard concept of measure. You consider a set $X$ and a map $\mu$ from the power set $P(X)$ to $\{0,1\}$, and you require that this is monotone and sends disjoint unions of collections $X_i \in P(X)$ of size $\kappa$ to sums:

$\displaystyle{ \mu(\bigcup_{i = 1}^\kappa X_i) = \sum_{i = 1}^\kappa \mu(X_i) }$

John Baez (Jan 29 2022 at 18:56):

Here addition in $\{0,1\}$ is defined by $0+0 = 0, 0+1 = 1+0 = 1+1 = 1$, so it's just "or" for booleans.

John Baez (Jan 29 2022 at 18:56):

And here $\kappa$ is some cardinal, that you get to choose.

David Egolf (Jan 30 2022 at 01:53):

Matteo Capucci (he/him) said:

Yes, they are maps of quantales, i.e. cocomplete ordered monoids

I've never heard of quantales before! I have heard of monoids, but what does it mean for a monoid to be "ordered"?

David Egolf (Jan 30 2022 at 01:55):

Wikipedia says: "An ordered semigroup is a semigroup (S,•) together with a partial order ≤ that is compatible with the semigroup operation. An ordered monoid and an ordered group are, respectively, a monoid or a group that are endowed with a partial order that makes them ordered semigroups."
I'm wondering how this corresponds to a category - what are the objects and morphisms?

Zhen Lin Low (Jan 30 2022 at 01:55):

It means the monoid has an order and the multiplication is order-preserving (in each variable and jointly). Quantales are additionally cocomplete and the multiplication preserves sups (in each variable separately).

Zhen Lin Low (Jan 30 2022 at 01:56):

I don't think it is helpful to think of monoids as one-object categories in this case, but if you insist, then these are one-object categories enriched over cocomplete join semilatticse.

David Egolf (Jan 30 2022 at 01:57):

Zhen Lin Low said:

I don't think it is helpful to think of monoids as one-object categories in this case, but if you insist, then these are one-object categories enriched over cocomplete join semilatticse.

What might be a more helpful categorical perspective (if there is one)?
At any rate, thanks for the clarification!

David Egolf (Jan 30 2022 at 02:04):

Matteo Capucci (he/him) said:

The 'functoriality' is in the order-preserving property, and then you ask them to also preserve the quantale operation (set union on the algebra, + on the positive reals)

I'm not quite following this. What order is being preserved here? I am guessing we order the sets of the algebra by inclusion, but how is this being preserved?
Maybe the idea is if $A \subseteq B$, then $\mu(A) \leq \mu(B)$ for any measure $\mu$.

David Egolf (Jan 30 2022 at 02:09):

Let me see if I can figure out what the monoids mentioned here are.
The sets of the algebra could form a monoid as follows:
-We have a category with a single object, and
-... where each morphism is a set of the algebra
-...where composition of morphisms corresponds to taking the union of sets

David Egolf (Jan 30 2022 at 02:10):

The real numbers form a monoid, where the morphisms are the real numbers and composition corresponds to addition.

Zhen Lin Low (Jan 30 2022 at 02:11):

At the risk of getting a scolding from certain people, let me put forward the suggestion that there is no such thing as a "helpful categorical perspective" without qualification. Whether a perspective is helpful or not depends entirely on who needs it. "A quantale is a one-object category enriched over CSL" is not going to be helpful if you don't know enriched category theory, and maybe only marginally helpful if you are not familiar with the special features of categories enriched over CSL. "A quantale is a monoid in the monoidal category CSL" is not going to be helpful if you don't know about monoids in monoidal categories, and certainly not helpful if you have not worked with non-cartesian monoidal categories. etc.

David Egolf (Jan 30 2022 at 02:16):

I think we can make the algebra into an ordered monoid like this:
-The morphisms are the sets of the algebra, with composition corresponding to taking unions
-The ordering corresponds to inclusion, so $A \subseteq B \iff A \leq B$
If $A \leq B$ then I believe $A \cup C \leq B \cup C$. So it seems that composition preserves the ordering.

David Egolf (Jan 30 2022 at 02:18):

Similarly for the real numbers:
-The morphisms are the real numbers, with composition corresponding to addition
-The ordering is the usual $\leq$ ordering on the real numbers
If $a \leq b$ for $a,b \in \mathbb{R}$, then $a + c \leq b + c$ for any $c \in \mathbb{R}$. So composition again preserves the ordering.

Zhen Lin Low (Jan 30 2022 at 02:19):

What about _nullary_ unions?

David Egolf (Jan 30 2022 at 02:20):

I thought the empty set corresponded to the identity morphism in the monoid corresponding to the algebra. So, $A \cup \emptyset = A$.

Zhen Lin Low (Jan 30 2022 at 02:20):

For a quantale the multiplication preserves sups. That includes the empty sup. So you can't have a quantale that way.

David Egolf (Jan 30 2022 at 02:22):

I'm not really following you, unfortunately. Although, I can try to figure out what you just said, it will probably take me a while.

David Egolf (Jan 30 2022 at 02:23):

What does it mean for multiplication to "preserve sups"?

Zhen Lin Low (Jan 30 2022 at 02:24):

You can say "preserves colimits" if you prefer category theory language.

David Egolf (Jan 30 2022 at 02:26):

That helps some, but let me try to make an example to understand better. Say $a$ and $b$ are morphisms in an ordered monoid, and we have their supremum $a \vee b$. Now we want to perform some kind of multiplication and ask if this supremum is "preserved" in some sense.

David Egolf (Jan 30 2022 at 02:27):

Is the idea to require $(c a) \vee b = c( a \vee b)$?

Zhen Lin Low (Jan 30 2022 at 02:27):

No. I mean $c (a \vee b) = c a \vee c b$ (among many other equations).

David Egolf (Jan 30 2022 at 02:27):

Oh, I see!

David Egolf (Jan 30 2022 at 02:29):

The empty supremum would correspond to the colimit of an empty diagram, which should be the smallest morphism of the ordered monoid (I think).

Zhen Lin Low (Jan 30 2022 at 02:29):

Yes.

David Egolf (Jan 30 2022 at 02:30):

The smallest morphism for the ordered monoid on the sets of the algebra is the empty set, I believe. So this should be the empty supremum.

Zhen Lin Low (Jan 30 2022 at 02:30):

Yes.

David Egolf (Jan 30 2022 at 02:31):

Excellent!
Now, to understand what it would mean to preserve the empty supremum under multiplication.

David Egolf (Jan 30 2022 at 02:34):

This is a little different than the case above $c(a \vee b) = ca \vee cb$. That described how multiplication was preserved with respect to the supremum of two morphisms.

David Egolf (Jan 30 2022 at 02:38):

We have above $c CL(a,b) = CL(ca, cb)$ where $CL$ stands for "colimit".
Maybe we want now: $c CL() = CL()$.
So, we would need $c \cup \emptyset = \emptyset$ for any set $c$, working in the ordered monoid described above

David Egolf (Jan 30 2022 at 02:38):

But, the above is not a true statement. Maybe this is what @Zhen Lin Low meant about us not getting a quantale.

David Egolf (Jan 30 2022 at 02:44):

Hmmm. I would be interested in any hints on how to describe an algebra of sets as a quantale.

Zhen Lin Low (Jan 30 2022 at 03:04):

Well, as you see, union doesn't work as multiplication. There is something else that does.

David Egolf (Jan 30 2022 at 03:50):

Maybe we can try intersection as multiplication.
Consider an algebra of sets, we want to make this in to an ordered monoid, and hopefully eventually in to a quantale.
Let the morphisms of the monoid be the sets, and let composition of two morphisms (sets) yield their intersection.
We can order sets by inclusion, so $A \leq B \iff A \subseteq B$.
Let $c, a, b$ be sets. Do we have $c(a \vee b) = ca \vee cb$?
That would correspond to $c \cap (a \cup b) = (c \cap a) \cup (c \cap b)$, which I believe is true.
We also need $c CL() = CL()$, which corresponds to $c \cap \emptyset = \emptyset$, which is true.
This seems like it might be working (?).

Zhen Lin Low (Jan 30 2022 at 03:55):

Yes, that works. Incidentally, a quantale in which the multiplication is intersection is called a frame.

Matteo Capucci (he/him) (Jan 30 2022 at 13:11):

John Baez said:

It preserves countable directed colimits.

I've been working too much with the quotient under null sets and I forgot the basics :face_palm:

David Egolf (Jan 30 2022 at 17:12):

Another thought on directed monoids... In the directed monoid described above, the sets sort of live in two categories at once. In one category (the monoid), they are the morphisms. In the other category (the ordering), they are the objects. It seems like we have a kind of ladder going on:
Level 1 [monoid]: object = *, morphisms $M$ (sets)
Level 2 [ordering]: objects = $M$, ordering morphisms

It might be interesting to consider higher order "ladders" like this 3-level ladder:
Level 1: objects $O_1$, morphisms $M_1$
Level 2: objects $M_1$, morphisms $M_2$
Level 3: objects $M_2$, morphisms $M_3$

Does this sort of nested "morphisms become objects" structure have a name?

Zhen Lin Low (Jan 31 2022 at 03:18):

That's the essence of higher categories.

Patrick Nicodemus (Feb 07 2022 at 22:40):

Categories, functors and natural transformations fit into a similar ascending pattern - functors are morphisms between categories in $\mathbf{Cat}$ but functors are themselves the objects in a category whose morphisms are natural transformations