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Stream: learning: questions

Topic: Is taking the ultrapower a monad?

Oscar Cunningham (Oct 09 2022 at 09:15):

Suppose we fix a nonprincipal ultrafilter on some set $N$. Then we can denote the ultrapower of $A$ with respect to this ultrafilter as ${}^*A$. Then ${}^*$ is a functor and we have a canonical inclusion $A\mapsto{}^*A$. Is ${}^*$ a monad? In particular do we also have a canonical map ${}^{**}A\mapsto{}^*A$? For example, can we turn a hyperhyperreal into a hyperreal?

My first thought was that we can think of elements of ${}^{**}A$ as equivalence classes of maps $N\times N\to A$, and composing this with the identity would give us a map $N\to A$ and hence an element of ${}^*A$. But this operation isn't well defined because it maps different elements of the same equivalence class to different places.

John Baez (Oct 09 2022 at 11:48):

I don't understand this stuff at all, but this category-theoretic description of the ultrapower functor associated to an ultrafilter may help us see whether or not we should expect it to be a monad.

Fabrizio Genovese (Oct 11 2022 at 14:07):

Oscar Cunningham said:

Suppose we fix a nonprincipal ultrafilter on some set $N$. Then we can denote the ultrapower of $A$ with respect to this ultrafilter as ${}^*A$. Then ${}^*$ is a functor and we have a canonical inclusion $A\mapsto{}^*A$. Is ${}^*$ a monad? In particular do we also have a canonical map ${}^{**}A\mapsto{}^*A$? For example, can we turn a hyperhyperreal into a hyperreal?

My first thought was that we can think of elements of ${}^{**}A$ as equivalence classes of maps $N\times N\to A$, and composing this with the identity would give us a map $N\to A$ and hence an element of ${}^*A$. But this operation isn't well defined because it maps different elements of the same equivalence class to different places.

I'm going to say something very wrong: basically an ultrapower is just a product of $N$ copies of $A$ quotiented by the relation generated by your ultrafilter. So you could see this as 'lists of length $N$ quotiented in some way. If $N$ is finite, a list of length $N$ of lists of length $N$ has length $N^2$. So you can 'flatten the list' basically. I think the problem arises where $N$ is infinite, because mapping indexes around becomes very messy

Fabrizio Genovese (Oct 11 2022 at 14:08):

I was thinking about doing the mapping itself in the hyperreal world but I don't think that will help you either, most likely the unquotiented sequences will be non-transferable for some reason and you'll end up with the same problem

Fabrizio Genovese (Oct 11 2022 at 14:09):

Fabrizio Genovese said:

Oscar Cunningham said:

Suppose we fix a nonprincipal ultrafilter on some set $N$. Then we can denote the ultrapower of $A$ with respect to this ultrafilter as ${}^*A$. Then ${}^*$ is a functor and we have a canonical inclusion $A\mapsto{}^*A$. Is ${}^*$ a monad? In particular do we also have a canonical map ${}^{**}A\mapsto{}^*A$? For example, can we turn a hyperhyperreal into a hyperreal?

My first thought was that we can think of elements of ${}^{**}A$ as equivalence classes of maps $N\times N\to A$, and composing this with the identity would give us a map $N\to A$ and hence an element of ${}^*A$. But this operation isn't well defined because it maps different elements of the same equivalence class to different places.

I'm going to say something very wrong: basically an ultrapower is just a product of $N$ copies of $A$ quotiented by the relation generated by your ultrafilter. So you could see this as 'lists of length $N$ quotiented in some way. If $N$ is finite, a list of length $N$ of lists of length $N$ has length $N^2$. So you can 'flatten the list' basically. I think the problem arises where $N$ is infinite, because mapping indexes around becomes very messy

BTW as you said the problem here is the equivalence relation, if you don't have that you could flatten the infinite lists Hilbert's hotel style, but that will most likely mess your equivalence relation completely