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
$begingroup$
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.
ct.category-theory lo.logic gn.general-topology ultrafilters
$endgroup$
|
show 20 more comments
$begingroup$
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.
ct.category-theory lo.logic gn.general-topology ultrafilters
$endgroup$
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
|
show 20 more comments
$begingroup$
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.
ct.category-theory lo.logic gn.general-topology ultrafilters
$endgroup$
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
ct.category-theory lo.logic gn.general-topology ultrafilters
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
|
show 20 more comments
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
|
show 20 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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]
$endgroup$
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
|
show 2 more comments
$begingroup$
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 $textPro(textFinVect)$.
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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]
$endgroup$
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
|
show 2 more comments
$begingroup$
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]
$endgroup$
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
|
show 2 more comments
$begingroup$
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]
$endgroup$
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]
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
|
show 2 more comments
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
|
show 2 more comments
$begingroup$
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 $textPro(textFinVect)$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $textPro(textFinVect)$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $textPro(textFinVect)$.
$endgroup$
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 $textPro(textFinVect)$.
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
add a comment |
$begingroup$
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
add a comment |
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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.
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– Todd Trimble♦
Mar 7 at 15:43
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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.
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– YCor
Mar 7 at 16:17
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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.
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– Asaf Karagila
Mar 7 at 16:30
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(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.)
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– Asaf Karagila
Mar 8 at 8:57
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@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?
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– მამუკა ჯიბლაძე
Mar 8 at 11:55