Ultrafilters as a double dual The Next CEO of Stack OverflowWhat is the matter with Hecke operators?What is the universal property of the Weyl group?Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?Unique Existence and the Axiom of ChoiceWhat are the algebras for the double dualization monad?Characterization of Stone-Cech compactificationsFor a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X to k$?Ultrafilters and diagonal argumentsIs there a general way to turn a 2-monad into a lax-idempotent (a.k.a. KZ) 2-monad?What, mathematically speaking, does it mean to say that the continuation monad can simulate all monads?Relation between monads, operads and algebraic theories (Again)A monad that unions sets

Ultrafilters as a double dual



The Next CEO of Stack OverflowWhat is the matter with Hecke operators?What is the universal property of the Weyl group?Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?Unique Existence and the Axiom of ChoiceWhat are the algebras for the double dualization monad?Characterization of Stone-Cech compactificationsFor a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X to k$?Ultrafilters and diagonal argumentsIs there a general way to turn a 2-monad into a lax-idempotent (a.k.a. KZ) 2-monad?What, mathematically speaking, does it mean to say that the continuation monad can simulate all monads?Relation between monads, operads and algebraic theories (Again)A monad that unions sets










39












$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:




  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^2^X$.

These are reminiscent of similar claims that can be made about vector spaces and double duals:




  • $V$ canonically embeds into $V^star star$;

  • If $V$ is finite-dimensional, then we have $V = V^star star$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^star star) = 2^2^dim(V)$.

This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?



  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^X$.

Apart from the tempting analogy between $beta X$ and $V^star star$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.










share|cite|improve this question











$endgroup$







  • 28




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble
    Mar 7 at 15:43







  • 3




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    Mar 7 at 16:17







  • 3




    $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    Mar 7 at 16:30










  • $begingroup$
    (My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
    $endgroup$
    – Asaf Karagila
    Mar 8 at 8:57











  • $begingroup$
    @AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
    $endgroup$
    – მამუკა ჯიბლაძე
    Mar 8 at 11:55















39












$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:




  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^2^X$.

These are reminiscent of similar claims that can be made about vector spaces and double duals:




  • $V$ canonically embeds into $V^star star$;

  • If $V$ is finite-dimensional, then we have $V = V^star star$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^star star) = 2^2^dim(V)$.

This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?



  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^X$.

Apart from the tempting analogy between $beta X$ and $V^star star$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.










share|cite|improve this question











$endgroup$







  • 28




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble
    Mar 7 at 15:43







  • 3




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    Mar 7 at 16:17







  • 3




    $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    Mar 7 at 16:30










  • $begingroup$
    (My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
    $endgroup$
    – Asaf Karagila
    Mar 8 at 8:57











  • $begingroup$
    @AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
    $endgroup$
    – მამუკა ჯიბლაძე
    Mar 8 at 11:55













39












39








39


15



$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:




  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^2^X$.

These are reminiscent of similar claims that can be made about vector spaces and double duals:




  • $V$ canonically embeds into $V^star star$;

  • If $V$ is finite-dimensional, then we have $V = V^star star$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^star star) = 2^2^dim(V)$.

This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?



  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^X$.

Apart from the tempting analogy between $beta X$ and $V^star star$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.










share|cite|improve this question











$endgroup$




Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:




  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^2^X$.

These are reminiscent of similar claims that can be made about vector spaces and double duals:




  • $V$ canonically embeds into $V^star star$;

  • If $V$ is finite-dimensional, then we have $V = V^star star$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^star star) = 2^2^dim(V)$.

This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?



  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^X$.

Apart from the tempting analogy between $beta X$ and $V^star star$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.







ct.category-theory lo.logic gn.general-topology ultrafilters






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 7 at 16:19









YCor

28.6k484139




28.6k484139










asked Mar 7 at 15:35









Adam P. GoucherAdam P. Goucher

6,84522958




6,84522958







  • 28




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble
    Mar 7 at 15:43







  • 3




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    Mar 7 at 16:17







  • 3




    $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    Mar 7 at 16:30










  • $begingroup$
    (My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
    $endgroup$
    – Asaf Karagila
    Mar 8 at 8:57











  • $begingroup$
    @AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
    $endgroup$
    – მამუკა ჯიბლაძე
    Mar 8 at 11:55












  • 28




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble
    Mar 7 at 15:43







  • 3




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    Mar 7 at 16:17







  • 3




    $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    Mar 7 at 16:30










  • $begingroup$
    (My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
    $endgroup$
    – Asaf Karagila
    Mar 8 at 8:57











  • $begingroup$
    @AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
    $endgroup$
    – მამუკა ჯიბლაძე
    Mar 8 at 11:55







28




28




$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble
Mar 7 at 15:43





$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_textBool(hom_textSet(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble
Mar 7 at 15:43





3




3




$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
Mar 7 at 16:17





$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrmhom_mathrmBool(mathrmhom_mathrmTop(X,mathbfZ/2mathbfZ))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
Mar 7 at 16:17





3




3




$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
Mar 7 at 16:30




$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^**$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
Mar 7 at 16:30












$begingroup$
(My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
$endgroup$
– Asaf Karagila
Mar 8 at 8:57





$begingroup$
(My point above, is that the canonical embedding of $V$ into $V^**$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $beta X$ does not.)
$endgroup$
– Asaf Karagila
Mar 8 at 8:57













$begingroup$
@AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
$endgroup$
– მამუკა ჯიბლაძე
Mar 8 at 11:55




$begingroup$
@AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $Vto V^**$ without choice? Could you recommend a text about those things?
$endgroup$
– მამუკა ჯიბლაძე
Mar 8 at 11:55










2 Answers
2






active

oldest

votes


















32












$begingroup$

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbbC$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbbC$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbbC$, whose spectrum is naturally identified with $beta X$.



[deleted an additional comment which wasn't accurate]






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
    $endgroup$
    – Alex Kruckman
    Mar 7 at 19:28







  • 10




    $begingroup$
    For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:36







  • 7




    $begingroup$
    Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:38







  • 2




    $begingroup$
    @NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
    $endgroup$
    – Adam P. Goucher
    Mar 8 at 14:57






  • 2




    $begingroup$
    Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
    $endgroup$
    – Nik Weaver
    Mar 8 at 16:25


















14












$begingroup$

This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.



Suppose you have a functor $F : C to D$. A general question to ask about it is this:




What additional structure, beyond being objects in $D$, do the objects $F(c) in D$ canonically have, by virtue of having been spit out by $F$?




A simple construction is that the objects $F(c)$ canonically admit an action by the automorphism group $textAut(F)$ of $F$ as a functor, more or less by definition, and more generally by the endomorphism monoid of $F$. This observation can already be used to motivate Weyl groups and Hecke algebras.



A more elaborate construction is that if $F$ admits a left adjoint $G : D to C$, then the objects $F(c)$ canonically admit an action by the monad $T = FG : D to D$, by which I mean they are canonically algebras over this monad. In nice cases (see monadic adjunction and monadicity theorem) this completely characterizes $C$ in terms of $D$ and $T$, for example if $D = textSet$ and $C$ is a typical algebraic category such as groups, rings, modules. A more unusual example here is that $C$ can be compact Hausdorff spaces, and then $T$ is the ultrafilter monad.



But there's an even more general construction than this, which can be motivated in several ways. Here's one. Suppose a monoidal category $M$ acts by endomorphisms on a category $E$, meaning we have a monoidal functor $M to [E, E]$, where $[E, E]$ is the monoidal category of endofunctors $E to E$. This is the minimal setup we need to talk about a monoid $m in M$ acting on an object $e in E$; see this blog post where I use this setup to motivate the definition of a monad.



Now, given an object $e in E$, we can ask for the universal monoid in $M$ which acts on $e$, which is an "$M$-internal" notion of the endomorphism monoid of $e$. This monoid $m in M$, if it exists, is defined by the universal property that maps $n to m$ of monoids are in natural bijection with actions of $n$ on $e$. If $M = [E, E]$, then this construction, when it exists, recovers the endomorphism monad of $e$. If $E = M$ acting on itself by left multiplication, then this construction, when it exists, recovers the internal endomorphism object of $e$.



In our setting we want to apply this construction to $E = [C, D]$ and $M = [D, D]$, where $[D, D]$ acts on $[C, D]$ by postcomposition. That is, we want a monad $T : D to D$ which universally acts on a functor $F : C to D$ in the sense that maps of monads to $T$ are in natural bijection with actions of monads on $F$.




Claim: This monad, if it exists, is the codensity monad of $F$.




(I don't have a reference for this, although it's closely related to the definition of the codensity monad as the right Kan extension of $F$ along itself; I remember convincing myself of this a few years ago, around the time I wrote this blog post on monads, and then I never wrote up the details. Welp.)



Now the really fun fact, which Todd Trimble alludes to above, is:




The codensity monad of the inclusion $textFinSet to textSet$ is the ultrafilter monad, and the codensity monad of the inclusion $textFinVect to textVect$ is the double dual monad.




This sets up a lovely analogy between compact Hausdorff spaces (algebras over the ultrafilter monad) and whatever algebras over the double dual monad are; Tom and Todd call them "linearly compact vector spaces" but my preferred terminology here is just "profinite vector spaces," in that the category is precisely $textPro(textFinVect)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:39











  • $begingroup$
    (Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:42











Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324867%2fultrafilters-as-a-double-dual%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









32












$begingroup$

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbbC$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbbC$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbbC$, whose spectrum is naturally identified with $beta X$.



[deleted an additional comment which wasn't accurate]






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
    $endgroup$
    – Alex Kruckman
    Mar 7 at 19:28







  • 10




    $begingroup$
    For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:36







  • 7




    $begingroup$
    Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:38







  • 2




    $begingroup$
    @NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
    $endgroup$
    – Adam P. Goucher
    Mar 8 at 14:57






  • 2




    $begingroup$
    Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
    $endgroup$
    – Nik Weaver
    Mar 8 at 16:25















32












$begingroup$

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbbC$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbbC$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbbC$, whose spectrum is naturally identified with $beta X$.



[deleted an additional comment which wasn't accurate]






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
    $endgroup$
    – Alex Kruckman
    Mar 7 at 19:28







  • 10




    $begingroup$
    For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:36







  • 7




    $begingroup$
    Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:38







  • 2




    $begingroup$
    @NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
    $endgroup$
    – Adam P. Goucher
    Mar 8 at 14:57






  • 2




    $begingroup$
    Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
    $endgroup$
    – Nik Weaver
    Mar 8 at 16:25













32












32








32





$begingroup$

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbbC$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbbC$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbbC$, whose spectrum is naturally identified with $beta X$.



[deleted an additional comment which wasn't accurate]






share|cite|improve this answer











$endgroup$



This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbbC$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbbC$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbbC$, whose spectrum is naturally identified with $beta X$.



[deleted an additional comment which wasn't accurate]







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 16 at 21:13

























answered Mar 7 at 16:59









Nik WeaverNik Weaver

22k150131




22k150131







  • 4




    $begingroup$
    What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
    $endgroup$
    – Alex Kruckman
    Mar 7 at 19:28







  • 10




    $begingroup$
    For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:36







  • 7




    $begingroup$
    Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:38







  • 2




    $begingroup$
    @NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
    $endgroup$
    – Adam P. Goucher
    Mar 8 at 14:57






  • 2




    $begingroup$
    Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
    $endgroup$
    – Nik Weaver
    Mar 8 at 16:25












  • 4




    $begingroup$
    What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
    $endgroup$
    – Alex Kruckman
    Mar 7 at 19:28







  • 10




    $begingroup$
    For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:36







  • 7




    $begingroup$
    Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
    $endgroup$
    – Nik Weaver
    Mar 7 at 19:38







  • 2




    $begingroup$
    @NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
    $endgroup$
    – Adam P. Goucher
    Mar 8 at 14:57






  • 2




    $begingroup$
    Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
    $endgroup$
    – Nik Weaver
    Mar 8 at 16:25







4




4




$begingroup$
What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
Mar 7 at 19:28





$begingroup$
What does it mean for a function from a set $X$ to $mathbbC$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
Mar 7 at 19:28





10




10




$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
Mar 7 at 19:36





$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
Mar 7 at 19:36





7




7




$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
Mar 7 at 19:38





$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
Mar 7 at 19:38





2




2




$begingroup$
@NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
$endgroup$
– Adam P. Goucher
Mar 8 at 14:57




$begingroup$
@NikWeaver Elegant! This essentially answers both of my questions, namely why the analogy exists, and also why the 'half-iterate' $delta X$ cannot be defined: when you take the first dual of $c_0(X)$, the result is not a C*-algebra, so you can't take its spectrum. But the double dual $l^infty(X)$ is a C*-algebra, so you can indeed take its spectrum, and you get $beta X$.
$endgroup$
– Adam P. Goucher
Mar 8 at 14:57




2




2




$begingroup$
Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
$endgroup$
– Nik Weaver
Mar 8 at 16:25




$begingroup$
Yes, that's right. The first duals generally don't even have a preferred product, so there's nothing like a spectrum.
$endgroup$
– Nik Weaver
Mar 8 at 16:25











14












$begingroup$

This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.



Suppose you have a functor $F : C to D$. A general question to ask about it is this:




What additional structure, beyond being objects in $D$, do the objects $F(c) in D$ canonically have, by virtue of having been spit out by $F$?




A simple construction is that the objects $F(c)$ canonically admit an action by the automorphism group $textAut(F)$ of $F$ as a functor, more or less by definition, and more generally by the endomorphism monoid of $F$. This observation can already be used to motivate Weyl groups and Hecke algebras.



A more elaborate construction is that if $F$ admits a left adjoint $G : D to C$, then the objects $F(c)$ canonically admit an action by the monad $T = FG : D to D$, by which I mean they are canonically algebras over this monad. In nice cases (see monadic adjunction and monadicity theorem) this completely characterizes $C$ in terms of $D$ and $T$, for example if $D = textSet$ and $C$ is a typical algebraic category such as groups, rings, modules. A more unusual example here is that $C$ can be compact Hausdorff spaces, and then $T$ is the ultrafilter monad.



But there's an even more general construction than this, which can be motivated in several ways. Here's one. Suppose a monoidal category $M$ acts by endomorphisms on a category $E$, meaning we have a monoidal functor $M to [E, E]$, where $[E, E]$ is the monoidal category of endofunctors $E to E$. This is the minimal setup we need to talk about a monoid $m in M$ acting on an object $e in E$; see this blog post where I use this setup to motivate the definition of a monad.



Now, given an object $e in E$, we can ask for the universal monoid in $M$ which acts on $e$, which is an "$M$-internal" notion of the endomorphism monoid of $e$. This monoid $m in M$, if it exists, is defined by the universal property that maps $n to m$ of monoids are in natural bijection with actions of $n$ on $e$. If $M = [E, E]$, then this construction, when it exists, recovers the endomorphism monad of $e$. If $E = M$ acting on itself by left multiplication, then this construction, when it exists, recovers the internal endomorphism object of $e$.



In our setting we want to apply this construction to $E = [C, D]$ and $M = [D, D]$, where $[D, D]$ acts on $[C, D]$ by postcomposition. That is, we want a monad $T : D to D$ which universally acts on a functor $F : C to D$ in the sense that maps of monads to $T$ are in natural bijection with actions of monads on $F$.




Claim: This monad, if it exists, is the codensity monad of $F$.




(I don't have a reference for this, although it's closely related to the definition of the codensity monad as the right Kan extension of $F$ along itself; I remember convincing myself of this a few years ago, around the time I wrote this blog post on monads, and then I never wrote up the details. Welp.)



Now the really fun fact, which Todd Trimble alludes to above, is:




The codensity monad of the inclusion $textFinSet to textSet$ is the ultrafilter monad, and the codensity monad of the inclusion $textFinVect to textVect$ is the double dual monad.




This sets up a lovely analogy between compact Hausdorff spaces (algebras over the ultrafilter monad) and whatever algebras over the double dual monad are; Tom and Todd call them "linearly compact vector spaces" but my preferred terminology here is just "profinite vector spaces," in that the category is precisely $textPro(textFinVect)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:39











  • $begingroup$
    (Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:42















14












$begingroup$

This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.



Suppose you have a functor $F : C to D$. A general question to ask about it is this:




What additional structure, beyond being objects in $D$, do the objects $F(c) in D$ canonically have, by virtue of having been spit out by $F$?




A simple construction is that the objects $F(c)$ canonically admit an action by the automorphism group $textAut(F)$ of $F$ as a functor, more or less by definition, and more generally by the endomorphism monoid of $F$. This observation can already be used to motivate Weyl groups and Hecke algebras.



A more elaborate construction is that if $F$ admits a left adjoint $G : D to C$, then the objects $F(c)$ canonically admit an action by the monad $T = FG : D to D$, by which I mean they are canonically algebras over this monad. In nice cases (see monadic adjunction and monadicity theorem) this completely characterizes $C$ in terms of $D$ and $T$, for example if $D = textSet$ and $C$ is a typical algebraic category such as groups, rings, modules. A more unusual example here is that $C$ can be compact Hausdorff spaces, and then $T$ is the ultrafilter monad.



But there's an even more general construction than this, which can be motivated in several ways. Here's one. Suppose a monoidal category $M$ acts by endomorphisms on a category $E$, meaning we have a monoidal functor $M to [E, E]$, where $[E, E]$ is the monoidal category of endofunctors $E to E$. This is the minimal setup we need to talk about a monoid $m in M$ acting on an object $e in E$; see this blog post where I use this setup to motivate the definition of a monad.



Now, given an object $e in E$, we can ask for the universal monoid in $M$ which acts on $e$, which is an "$M$-internal" notion of the endomorphism monoid of $e$. This monoid $m in M$, if it exists, is defined by the universal property that maps $n to m$ of monoids are in natural bijection with actions of $n$ on $e$. If $M = [E, E]$, then this construction, when it exists, recovers the endomorphism monad of $e$. If $E = M$ acting on itself by left multiplication, then this construction, when it exists, recovers the internal endomorphism object of $e$.



In our setting we want to apply this construction to $E = [C, D]$ and $M = [D, D]$, where $[D, D]$ acts on $[C, D]$ by postcomposition. That is, we want a monad $T : D to D$ which universally acts on a functor $F : C to D$ in the sense that maps of monads to $T$ are in natural bijection with actions of monads on $F$.




Claim: This monad, if it exists, is the codensity monad of $F$.




(I don't have a reference for this, although it's closely related to the definition of the codensity monad as the right Kan extension of $F$ along itself; I remember convincing myself of this a few years ago, around the time I wrote this blog post on monads, and then I never wrote up the details. Welp.)



Now the really fun fact, which Todd Trimble alludes to above, is:




The codensity monad of the inclusion $textFinSet to textSet$ is the ultrafilter monad, and the codensity monad of the inclusion $textFinVect to textVect$ is the double dual monad.




This sets up a lovely analogy between compact Hausdorff spaces (algebras over the ultrafilter monad) and whatever algebras over the double dual monad are; Tom and Todd call them "linearly compact vector spaces" but my preferred terminology here is just "profinite vector spaces," in that the category is precisely $textPro(textFinVect)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:39











  • $begingroup$
    (Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:42













14












14








14





$begingroup$

This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.



Suppose you have a functor $F : C to D$. A general question to ask about it is this:




What additional structure, beyond being objects in $D$, do the objects $F(c) in D$ canonically have, by virtue of having been spit out by $F$?




A simple construction is that the objects $F(c)$ canonically admit an action by the automorphism group $textAut(F)$ of $F$ as a functor, more or less by definition, and more generally by the endomorphism monoid of $F$. This observation can already be used to motivate Weyl groups and Hecke algebras.



A more elaborate construction is that if $F$ admits a left adjoint $G : D to C$, then the objects $F(c)$ canonically admit an action by the monad $T = FG : D to D$, by which I mean they are canonically algebras over this monad. In nice cases (see monadic adjunction and monadicity theorem) this completely characterizes $C$ in terms of $D$ and $T$, for example if $D = textSet$ and $C$ is a typical algebraic category such as groups, rings, modules. A more unusual example here is that $C$ can be compact Hausdorff spaces, and then $T$ is the ultrafilter monad.



But there's an even more general construction than this, which can be motivated in several ways. Here's one. Suppose a monoidal category $M$ acts by endomorphisms on a category $E$, meaning we have a monoidal functor $M to [E, E]$, where $[E, E]$ is the monoidal category of endofunctors $E to E$. This is the minimal setup we need to talk about a monoid $m in M$ acting on an object $e in E$; see this blog post where I use this setup to motivate the definition of a monad.



Now, given an object $e in E$, we can ask for the universal monoid in $M$ which acts on $e$, which is an "$M$-internal" notion of the endomorphism monoid of $e$. This monoid $m in M$, if it exists, is defined by the universal property that maps $n to m$ of monoids are in natural bijection with actions of $n$ on $e$. If $M = [E, E]$, then this construction, when it exists, recovers the endomorphism monad of $e$. If $E = M$ acting on itself by left multiplication, then this construction, when it exists, recovers the internal endomorphism object of $e$.



In our setting we want to apply this construction to $E = [C, D]$ and $M = [D, D]$, where $[D, D]$ acts on $[C, D]$ by postcomposition. That is, we want a monad $T : D to D$ which universally acts on a functor $F : C to D$ in the sense that maps of monads to $T$ are in natural bijection with actions of monads on $F$.




Claim: This monad, if it exists, is the codensity monad of $F$.




(I don't have a reference for this, although it's closely related to the definition of the codensity monad as the right Kan extension of $F$ along itself; I remember convincing myself of this a few years ago, around the time I wrote this blog post on monads, and then I never wrote up the details. Welp.)



Now the really fun fact, which Todd Trimble alludes to above, is:




The codensity monad of the inclusion $textFinSet to textSet$ is the ultrafilter monad, and the codensity monad of the inclusion $textFinVect to textVect$ is the double dual monad.




This sets up a lovely analogy between compact Hausdorff spaces (algebras over the ultrafilter monad) and whatever algebras over the double dual monad are; Tom and Todd call them "linearly compact vector spaces" but my preferred terminology here is just "profinite vector spaces," in that the category is precisely $textPro(textFinVect)$.






share|cite|improve this answer











$endgroup$



This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.



Suppose you have a functor $F : C to D$. A general question to ask about it is this:




What additional structure, beyond being objects in $D$, do the objects $F(c) in D$ canonically have, by virtue of having been spit out by $F$?




A simple construction is that the objects $F(c)$ canonically admit an action by the automorphism group $textAut(F)$ of $F$ as a functor, more or less by definition, and more generally by the endomorphism monoid of $F$. This observation can already be used to motivate Weyl groups and Hecke algebras.



A more elaborate construction is that if $F$ admits a left adjoint $G : D to C$, then the objects $F(c)$ canonically admit an action by the monad $T = FG : D to D$, by which I mean they are canonically algebras over this monad. In nice cases (see monadic adjunction and monadicity theorem) this completely characterizes $C$ in terms of $D$ and $T$, for example if $D = textSet$ and $C$ is a typical algebraic category such as groups, rings, modules. A more unusual example here is that $C$ can be compact Hausdorff spaces, and then $T$ is the ultrafilter monad.



But there's an even more general construction than this, which can be motivated in several ways. Here's one. Suppose a monoidal category $M$ acts by endomorphisms on a category $E$, meaning we have a monoidal functor $M to [E, E]$, where $[E, E]$ is the monoidal category of endofunctors $E to E$. This is the minimal setup we need to talk about a monoid $m in M$ acting on an object $e in E$; see this blog post where I use this setup to motivate the definition of a monad.



Now, given an object $e in E$, we can ask for the universal monoid in $M$ which acts on $e$, which is an "$M$-internal" notion of the endomorphism monoid of $e$. This monoid $m in M$, if it exists, is defined by the universal property that maps $n to m$ of monoids are in natural bijection with actions of $n$ on $e$. If $M = [E, E]$, then this construction, when it exists, recovers the endomorphism monad of $e$. If $E = M$ acting on itself by left multiplication, then this construction, when it exists, recovers the internal endomorphism object of $e$.



In our setting we want to apply this construction to $E = [C, D]$ and $M = [D, D]$, where $[D, D]$ acts on $[C, D]$ by postcomposition. That is, we want a monad $T : D to D$ which universally acts on a functor $F : C to D$ in the sense that maps of monads to $T$ are in natural bijection with actions of monads on $F$.




Claim: This monad, if it exists, is the codensity monad of $F$.




(I don't have a reference for this, although it's closely related to the definition of the codensity monad as the right Kan extension of $F$ along itself; I remember convincing myself of this a few years ago, around the time I wrote this blog post on monads, and then I never wrote up the details. Welp.)



Now the really fun fact, which Todd Trimble alludes to above, is:




The codensity monad of the inclusion $textFinSet to textSet$ is the ultrafilter monad, and the codensity monad of the inclusion $textFinVect to textVect$ is the double dual monad.




This sets up a lovely analogy between compact Hausdorff spaces (algebras over the ultrafilter monad) and whatever algebras over the double dual monad are; Tom and Todd call them "linearly compact vector spaces" but my preferred terminology here is just "profinite vector spaces," in that the category is precisely $textPro(textFinVect)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 12 at 17:26









Will Sawin

68.4k7139285




68.4k7139285










answered Mar 8 at 8:05









Qiaochu YuanQiaochu Yuan

77.6k27318604




77.6k27318604











  • $begingroup$
    There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:39











  • $begingroup$
    (Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:42
















  • $begingroup$
    There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:39











  • $begingroup$
    (Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
    $endgroup$
    – Qiaochu Yuan
    Mar 9 at 20:42















$begingroup$
There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
$endgroup$
– Qiaochu Yuan
Mar 9 at 20:39





$begingroup$
There is a kind of "double dual" in this story, by the way: one way to describe the codensity monad of $F : C to D$, if $C$ is essentially small and $D$ has small limits, is that it's the monad associated to the adjunction between $D$ and $[C, textSet]^op$ whose left adjoint sends $d in D$ to the functor $textHom(d, F(-)) : C to textSet$. When $F$ is the inclusion of finite sets into sets this is a disguised form of $beta X$ and when $F$ is the inclusion of finite-dimensional vector spaces into vector spaces this is a disguised form of taking the dual.
$endgroup$
– Qiaochu Yuan
Mar 9 at 20:39













$begingroup$
(Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
$endgroup$
– Qiaochu Yuan
Mar 9 at 20:42




$begingroup$
(Then the right adjoint is the "second dual," although it's a bit trickier to describe.)
$endgroup$
– Qiaochu Yuan
Mar 9 at 20:42

















draft saved

draft discarded
















































Thanks for contributing an answer to MathOverflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324867%2fultrafilters-as-a-double-dual%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Save data to MySQL database using ExtJS and PHP [closed]2019 Community Moderator ElectionHow can I prevent SQL injection in PHP?Which MySQL data type to use for storing boolean valuesPHP: Delete an element from an arrayHow do I connect to a MySQL Database in Python?Should I use the datetime or timestamp data type in MySQL?How to get a list of MySQL user accountsHow Do You Parse and Process HTML/XML in PHP?Reference — What does this symbol mean in PHP?How does PHP 'foreach' actually work?Why shouldn't I use mysql_* functions in PHP?

Compiling GNU Global with universal-ctags support Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Data science time! April 2019 and salary with experience The Ask Question Wizard is Live!Tags for Emacs: Relationship between etags, ebrowse, cscope, GNU Global and exuberant ctagsVim and Ctags tips and trickscscope or ctags why choose one over the other?scons and ctagsctags cannot open option file “.ctags”Adding tag scopes in universal-ctagsShould I use Universal-ctags?Universal ctags on WindowsHow do I install GNU Global with universal ctags support using Homebrew?Universal ctags with emacsHow to highlight ctags generated by Universal Ctags in Vim?

Add ONERROR event to image from jsp tldHow to add an image to a JPanel?Saving image from PHP URLHTML img scalingCheck if an image is loaded (no errors) with jQueryHow to force an <img> to take up width, even if the image is not loadedHow do I populate hidden form field with a value set in Spring ControllerStyling Raw elements Generated from JSP tagds with Jquery MobileLimit resizing of images with explicitly set width and height attributeserror TLD use in a jsp fileJsp tld files cannot be resolved