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1. • CommentRowNumber1.
   • CommentAuthorIan_Durham
   • CommentTimeJun 29th 2010
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   Author: Ian_Durham
   Format: MarkdownItexIt seems to me that Yoneda embedding requires that a given category, $C$, necessarily have a dual category, $C^{op}$. But it also seems to me that I can imagine all sorts of categories that don't have duals. Question: are both of my above conclusions correct?

   It seems to me that Yoneda embedding requires that a given category, $C$, necessarily have a dual category, $C^{op}$. But it also seems to me that I can imagine all sorts of categories that don’t have duals.

   Question: are both of my above conclusions correct?

2. • CommentRowNumber2.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 29th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexIan, I guess you may be confused here between these two notions: a) $C^{op}$ is the opposite category of $C$ b) $C$ is a category with duals. Yoneda requires a), which is not a property but a definition: $C^{op}$ is the category with the same objects as $C$ and with morphisms defined by $C^{op}(X,Y)=C(Y,X)$. that's it, nothing more: the use of opposite category is just to avoid having the classical distinction between covariant and contravariant functors.

   Ian,

   I guess you may be confused here between these two notions:

   a) $C^{op}$ is the opposite category of $C$

   b) $C$ is a category with duals.

   Yoneda requires a), which is not a property but a definition: $C^{op}$ is the category with the same objects as $C$ and with morphisms defined by $C^{op}(X,Y)=C(Y,X)$. that’s it, nothing more: the use of opposite category is just to avoid having the classical distinction between covariant and contravariant functors.

3. • CommentRowNumber3.
   • CommentAuthorIan_Durham
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
   • PermaLink
   Author: Ian_Durham
   Format: MarkdownItexDomenico, Thanks, although I think there's a confusion of terms (some book I read used "dual" and "opposite" somewhat interchangeably which I take is apparently incorrect). In any case, replace the word "dual" with "opposite" in my question and you've got what I'm aiming at. So, if the opposite of a category is just that category with the morphisms reversed, what if the morphisms can't be reversed? See, I have a very specific physical application I'm working with and the morphisms can't be reversed and yet it meets all the other definitions of a category. To put it another way, while mathematically $C^{op}$ is simply a definition, I've got a category $C$ representing something physical for which $C^{op}$ has no physical meaning (i.e. it doesn't physically exist).

   Domenico,

   Thanks, although I think there’s a confusion of terms (some book I read used “dual” and “opposite” somewhat interchangeably which I take is apparently incorrect).

   In any case, replace the word “dual” with “opposite” in my question and you’ve got what I’m aiming at. So, if the opposite of a category is just that category with the morphisms reversed, what if the morphisms can’t be reversed? See, I have a very specific physical application I’m working with and the morphisms can’t be reversed and yet it meets all the other definitions of a category.

   To put it another way, while mathematically $C^{op}$ is simply a definition, I’ve got a category $C$ representing something physical for which $C^{op}$ has no physical meaning (i.e. it doesn’t physically exist).

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThe opposite category is purely a formal construction. The maps in the opposite category are oftentimes not functions, if that's what you're asking. For instance, a property of the category of sets is that its initial object is strictly initial (some term like that) because the only maps into the empty set are from the empty set (by the definition of a function). In the opposite category, however, we have a bunch of maps into and out of the initial object (the set with one point). It follows that the morphisms in the opposite category can't be functions. With regard to your remark that the opposite category "doesn't exist", note that calling the category of commutative rings the category of commutative rings is just a semantic interpretation for the data of the category (including all of the enrichment etc). The category can also be characterized as the opposite category of affine schemes (this is circular, but it is immaterial to the point I'm making).

   The opposite category is purely a formal construction. The maps in the opposite category are oftentimes not functions, if that’s what you’re asking.

   For instance, a property of the category of sets is that its initial object is strictly initial (some term like that) because the only maps into the empty set are from the empty set (by the definition of a function). In the opposite category, however, we have a bunch of maps into and out of the initial object (the set with one point).

   It follows that the morphisms in the opposite category can’t be functions.

   With regard to your remark that the opposite category “doesn’t exist”, note that calling the category of commutative rings the category of commutative rings is just a semantic interpretation for the data of the category (including all of the enrichment etc). The category can also be characterized as the opposite category of affine schemes (this is circular, but it is immaterial to the point I’m making).

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 29th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexIndeed, thinking of $C^{op}$ may be confusing if you want to see "concretely" the morphisms: by definition a morphism in $C^{op}$ from $X$ to $Y$ is nothing but a morphism in $C$ from $Y$ to $X$, so it does not "go" from $X$ to $Y$ in an intuitive sense. as I was saying in the previous post, you can think of $C^{op}$ as a placeholder for the notion of contravariat functor: a functor $F:C^{op}\to D$ is (by definition of opposite category) precisely what was once (and actually still..) called "a contravariant functor $F:C\to D$"

   Indeed, thinking of $C^{op}$ may be confusing if you want to see “concretely” the morphisms: by definition a morphism in $C^{op}$ from $X$ to $Y$ is nothing but a morphism in $C$ from $Y$ to $X$, so it does not “go” from $X$ to $Y$ in an intuitive sense.

   as I was saying in the previous post, you can think of $C^{op}$ as a placeholder for the notion of contravariat functor: a functor $F:C^{op}\to D$ is (by definition of opposite category) precisely what was once (and actually still..) called “a contravariant functor $F:C\to D$”

6. • CommentRowNumber6.
   • CommentAuthorIan_Durham
   • CommentTimeJun 29th 2010
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   Author: Ian_Durham
   Format: MarkdownItex<blockquote><p>The maps in the opposite category are oftentimes not functions, if that's what you're asking.</p></blockquote> Oh really! That is definitely a misinterpretation I had in the sense that I assumed that if the morphisms in $C$ were functions, then the morphisms in $C^{op}$ must also be functions. So you're saying this is actually _not_ true? If I'm understanding you correctly, then, my big problem is figuring out what the morphisms in $C^{op}$ actually represent _physically_. Hmmm...

   The maps in the opposite category are oftentimes not functions, if that’s what you’re asking.

   Oh really! That is definitely a misinterpretation I had in the sense that I assumed that if the morphisms in $C$ were functions, then the morphisms in $C^{op}$ must also be functions. So you’re saying this is actually not true?

   If I’m understanding you correctly, then, my big problem is figuring out what the morphisms in $C^{op}$ actually represent physically. Hmmm…

7. • CommentRowNumber7.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThey may not be anything. As domenico said, you can think of C^op as being a marker for change of variance.

   They may not be anything.

   As domenico said, you can think of C^op as being a marker for change of variance.

8. • CommentRowNumber8.
   • CommentAuthorIan_Durham
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Ian_Durham
   Format: MarkdownItexWell, maybe I should ask a more general question of the physicists out there then. Are there any references out there to how categories are used to model irreversible physical processes?

   Well, maybe I should ask a more general question of the physicists out there then. Are there any references out there to how categories are used to model irreversible physical processes?

9. • CommentRowNumber9.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
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   Author: Harry Gindi
   Format: MarkdownItexSeems like you'd be better off using partially ordered sets.

   Seems like you’d be better off using partially ordered sets.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJun 29th 2010
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    Author: Urs
    Format: MarkdownItex> Are there any references out there to how categories are used to model irreversible physical processes? Remember the entry [[smooth Lorentzian space]]? It's about time for you to actually read it. ;-)

    Are there any references out there to how categories are used to model irreversible physical processes?

    Remember the entry smooth Lorentzian space? It’s about time for you to actually read it. ;-)

11. • CommentRowNumber11.
    • CommentAuthorIan_Durham
    • CommentTimeJun 30th 2010
    • (edited Jun 30th 2010)
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    Author: Ian_Durham
    Format: MarkdownItex<blockquote><p>Remember the entry smooth Lorentzian space? It's about time for you to actually read it. ;-)</p></blockquote> Indeed, I did read it but it didn't seem like it offered me what I needed (I'm still working on that same problem from the APS March Meeting - I need some way to take a bunch of potentially irreversible things and combine them in such a way as to make them reversible - there is an unproven theorem out there that says it can be done). Or, rather, I could do it if I had a way to unitarily represent End(X).

    Remember the entry smooth Lorentzian space? It’s about time for you to actually read it. ;-)

    Indeed, I did read it but it didn’t seem like it offered me what I needed (I’m still working on that same problem from the APS March Meeting - I need some way to take a bunch of potentially irreversible things and combine them in such a way as to make them reversible - there is an unproven theorem out there that says it can be done).

    Or, rather, I could do it if I had a way to unitarily represent End(X).

12. • CommentRowNumber12.
    • CommentAuthorEric
    • CommentTimeJun 30th 2010
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    Author: Eric
    Format: MarkdownItexHi Ian! Don't give up the fight :) And don't listen to Domenico. At least not on this point :) [Note: Domenico may not know how super highly I think of him, so just for the record, I'm kidding of course :)] The word "duality" is extremely overloaded, so when talking about opposites, maybe I'll call it "oppositality" or maybe "oppotality". In some cases duality and oppotality are the same thing, e.g. in Vect, but not in general. We discussed this recently and I was asking very similar questions, so have a look at that discussion. One of the deep themes of the past 30 years and a deep theme here in the nCommunity is [[space and quantity]]. I don't know who was the first (Gelfand-Naimark? Connes?) but it has become recognized that there is an important oppotality between spaces and algebras. If your category is a category of spaces, its opposite is a category of algebras and vice versa. If your category is a category of algebras, its opposite is a category of spaces. The nature of the space depends on the nature of the algebra. Commutative algebras are opposite commutative spaces. Noncommutative spaces are opposite noncommutative algebras. This is a VERY physical idea and trying to trivialize it by saying the opposite categories are somehow not interesting in themselves to try to understand is kind of an injustice to a beautiful idea. Although everything said here is accurate from a mathematical perspective, it misses the physics in my opinion. It is obvious to you and me and everyone that many physical systems can be encoded into a category. For example, states are objects and processes are morphisms. Simple. We have a category. Forming the opposite category of some familiar category and trying to interpret that physically is a worthy exercise in my opinion. Regarding your original line of questioning... Yeah, opposite morphisms are not always of the same kind of beast as the morphisms you started with. For example, in Set, morphisms are functions. On elements, functions are "many to one" and not "one to many". For example, you can have element $x$ and $y$ with $f(x) = f(y)$, but you can't have a function with $f(x) = a$ and $f(x) = b$ if $a\ne b$ (obviously). But that is precisely the kind of thing that you can get with $f^{op}$ (I think!). With $f^{op}$, I'm not even sure if it makes sense to ask what is $f^{op}(y)$ for some $f:X\to Y$ and $y\in Y$. [Note: Since [[Set]] has both an [[initial object]] and a [[terminal object]], then so does $Set^{op}$. If a category has an initial object, you can "probe" each other object with morphisms from this initial object. In this way we can think of the morphisms as "elements" of the object. For example, in Set, an object is a set $X$. We know $X$ has elements, but how do we get at these objects from a category theoretic perspective when all we have is sets and functions to work with? The answer is that for each element $x\in X$ there is a function $x:\bullet\to X$. So the set of morphisms $Hom(\bullet, X)$ corresponds to the set of elements of $X$. See [[generalized element]]. So maybe it does make sense to talk about $f^{op}(y)$ as long as we consider $y$ to be a generalized element.] About reversible processes... If you have in mind a category where states are objects and processes are morphisms, then if all processes are reversible, then your category is actually a groupoid. The reverse of a process $f$ is not $f^{op}$, it is actually $f^{-1}$. A category in which all morphisms are invertible is a [[groupoid]]. If you have irreversible processes, then your category is not a groupoid. It is likely an $(\infty,1)$-[(infinity,1)-category|category], but I doubt you will need to know or understand that (I don't!) for anything you're working on now. I'm guessing for your purposes, it is sufficient to understand the difference between categories and groupoids and just know that you are working with a category and not a groupoid. It sounds like you might be wanting to construct a groupoid given a category. For every category, there is a full subcategory that is a groupoid. If I understand correctly (which is a stretch), you get this by simply removing all non-invertible morphisms from your set of morphisms. This is one way to get a groupoid from a category, but I don't think this is what you want. It sounds to me like you want to somehow modify your original category so that irreversible processes become reversible somehow. I don't think this is possible in an exact sense, but you might be able to add invertible morphisms to your category in such a way that two states that were not "equivalent" before become equivalent. By the way, if two states are related by a reversible process, category theorists would consider these two states to be equivalent. PS: I'm not as shy about talking about stuff that I'm not 100% sure about as I probably should be, so read the above with a grain of salt. If not 100% correct, however, the message should remain somewhat in tact once an expert straightens things out.

    Hi Ian!

    Don’t give up the fight :)

    And don’t listen to Domenico. At least not on this point :)

    [Note: Domenico may not know how super highly I think of him, so just for the record, I’m kidding of course :)]

    The word “duality” is extremely overloaded, so when talking about opposites, maybe I’ll call it “oppositality” or maybe “oppotality”. In some cases duality and oppotality are the same thing, e.g. in Vect, but not in general.

    We discussed this recently and I was asking very similar questions, so have a look at that discussion.

    One of the deep themes of the past 30 years and a deep theme here in the nCommunity is space and quantity. I don’t know who was the first (Gelfand-Naimark? Connes?) but it has become recognized that there is an important oppotality between spaces and algebras. If your category is a category of spaces, its opposite is a category of algebras and vice versa. If your category is a category of algebras, its opposite is a category of spaces.

    The nature of the space depends on the nature of the algebra. Commutative algebras are opposite commutative spaces. Noncommutative spaces are opposite noncommutative algebras.

    This is a VERY physical idea and trying to trivialize it by saying the opposite categories are somehow not interesting in themselves to try to understand is kind of an injustice to a beautiful idea.

    Although everything said here is accurate from a mathematical perspective, it misses the physics in my opinion. It is obvious to you and me and everyone that many physical systems can be encoded into a category. For example, states are objects and processes are morphisms. Simple. We have a category. Forming the opposite category of some familiar category and trying to interpret that physically is a worthy exercise in my opinion.

    Regarding your original line of questioning…

    Yeah, opposite morphisms are not always of the same kind of beast as the morphisms you started with. For example, in Set, morphisms are functions. On elements, functions are “many to one” and not “one to many”. For example, you can have element $x$ and $y$ with $f(x) = f(y)$, but you can’t have a function with $f(x) = a$ and $f(x) = b$ if $a\ne b$ (obviously). But that is precisely the kind of thing that you can get with $f^{op}$ (I think!). With $f^{op}$, I’m not even sure if it makes sense to ask what is $f^{op}(y)$ for some $f:X\to Y$ and $y\in Y$.

    [Note: Since Set has both an initial object and a terminal object, then so does $Set^{op}$. If a category has an initial object, you can “probe” each other object with morphisms from this initial object. In this way we can think of the morphisms as “elements” of the object. For example, in Set, an object is a set $X$. We know $X$ has elements, but how do we get at these objects from a category theoretic perspective when all we have is sets and functions to work with? The answer is that for each element $x\in X$ there is a function $x:\bullet\to X$. So the set of morphisms $Hom(\bullet, X)$ corresponds to the set of elements of $X$. See generalized element. So maybe it does make sense to talk about $f^{op}(y)$ as long as we consider $y$ to be a generalized element.]

    About reversible processes…

    If you have in mind a category where states are objects and processes are morphisms, then if all processes are reversible, then your category is actually a groupoid. The reverse of a process $f$ is not $f^{op}$, it is actually $f^{-1}$. A category in which all morphisms are invertible is a groupoid.

    If you have irreversible processes, then your category is not a groupoid. It is likely an $(\infty,1)$-[(infinity,1)-category|category], but I doubt you will need to know or understand that (I don’t!) for anything you’re working on now. I’m guessing for your purposes, it is sufficient to understand the difference between categories and groupoids and just know that you are working with a category and not a groupoid.

    It sounds like you might be wanting to construct a groupoid given a category.

    For every category, there is a full subcategory that is a groupoid. If I understand correctly (which is a stretch), you get this by simply removing all non-invertible morphisms from your set of morphisms. This is one way to get a groupoid from a category, but I don’t think this is what you want.

    It sounds to me like you want to somehow modify your original category so that irreversible processes become reversible somehow. I don’t think this is possible in an exact sense, but you might be able to add invertible morphisms to your category in such a way that two states that were not “equivalent” before become equivalent.

    By the way, if two states are related by a reversible process, category theorists would consider these two states to be equivalent.

    PS: I’m not as shy about talking about stuff that I’m not 100% sure about as I probably should be, so read the above with a grain of salt. If not 100% correct, however, the message should remain somewhat in tact once an expert straightens things out.

13. • CommentRowNumber13.
    • CommentAuthorIan_Durham
    • CommentTimeJun 30th 2010
    • PermaLink
    Author: Ian_Durham
    Format: HtmlEric, Thanks for that extremely enlightening response! In particular, I found this quite useful: <blockquote><p>If your category is a category of spaces, its opposite is a category of algebras and vice versa. If your category is a category of algebras, its opposite is a category of spaces.</p></blockquote> I will need to digest much of what you said, but I've been discussing with John Sidles, <a href="http://quantummoxie.wordpress.com/2010/06/24/the-sterilization-of-science/">on my blog</a> (really not related to the original topic anymore), about the fact that there aren't many books that get both the math and the physics "right," i.e. they're either pure math books that skimp on the physics or pure physics books that skimp on the math. This is especially true of category theory (since the closest thing I've seen to a true application was a reference or two in Awodey's book). Thus it is very difficult to simultaneously learn the subject and try to apply it. I'm sure some people will say to learn the subject <em>first</em> and <em>then</em> worry about applying it. Unfortunately my brain just doesn't work that way.

    Eric,
    
    Thanks for that extremely enlightening response! In particular, I found this quite useful:
    

    If your category is a category of spaces, its opposite is a category of algebras and vice versa. If your category is a category of algebras, its opposite is a category of spaces.

    
    I will need to digest much of what you said, but I've been discussing with John Sidles, on my blog (really not related to the original topic anymore), about the fact that there aren't many books that get both the math and the physics "right," i.e. they're either pure math books that skimp on the physics or pure physics books that skimp on the math. This is especially true of category theory (since the closest thing I've seen to a true application was a reference or two in Awodey's book). Thus it is very difficult to simultaneously learn the subject and try to apply it. I'm sure some people will say to learn the subject first and then worry about applying it. Unfortunately my brain just doesn't work that way.
14. • CommentRowNumber14.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 30th 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItex> If you have irreversible processes, then your category is not a groupoid. It is likely an (infinity,1)-category, but I doubt you will need to know or understand that (I don't!) for anything you're working on now. I think that thinking about higher categories at this point would not be worthwhile. Developing the theory of quasicategories, for example, requires all of the machinery of ordinary category theory (maybe not all of it, but a pretty good amount), a lot of the machinery from enriched category theory, and a whole lot of machinery from places like homotopy theory, model category theory, some nontrivial 2-category theory. Developing the theory of algebraic omega-categories requires a deep understanding of other parts of category theory like monads, operads, internal category theory, enriched category theory, monoidal categories, bicategories, etc. There's no easy way to get there (if anyone comes up with one, let me know!), so it's best to keep focused on the relevant stuff.

    If you have irreversible processes, then your category is not a groupoid. It is likely an (infinity,1)-category, but I doubt you will need to know or understand that (I don’t!) for anything you’re working on now.

    I think that thinking about higher categories at this point would not be worthwhile. Developing the theory of quasicategories, for example, requires all of the machinery of ordinary category theory (maybe not all of it, but a pretty good amount), a lot of the machinery from enriched category theory, and a whole lot of machinery from places like homotopy theory, model category theory, some nontrivial 2-category theory.

    Developing the theory of algebraic omega-categories requires a deep understanding of other parts of category theory like monads, operads, internal category theory, enriched category theory, monoidal categories, bicategories, etc.

    There’s no easy way to get there (if anyone comes up with one, let me know!), so it’s best to keep focused on the relevant stuff.

15. • CommentRowNumber15.
    • CommentAuthorIan_Durham
    • CommentTimeJun 30th 2010
    • PermaLink
    Author: Ian_Durham
    Format: MarkdownItex<blockquote><p>It sounds to me like you want to somehow modify your original category so that irreversible processes become reversible somehow. I don't think this is possible in an exact sense, but you might be able to add invertible morphisms to your category in such a way that two states that were not "equivalent" before become equivalent.</p></blockquote> OK, hmmm. Actually, the proposition I'm trying to prove using category theory says roughly the following: given $n$ copies of a single, non-unitary quantum channel, as $n \to \infty$, there is a way to make this combination behave unitarily (it's the quantum extension of Birkhoff's theorem, or, more accurately, I'm _looking_ for a quantum extension of Birkhoff's theorem along with a lot of other people - I just happen to think it can be done with category theory).

    It sounds to me like you want to somehow modify your original category so that irreversible processes become reversible somehow. I don’t think this is possible in an exact sense, but you might be able to add invertible morphisms to your category in such a way that two states that were not “equivalent” before become equivalent.

    OK, hmmm. Actually, the proposition I’m trying to prove using category theory says roughly the following: given $n$ copies of a single, non-unitary quantum channel, as $n \to \infty$, there is a way to make this combination behave unitarily (it’s the quantum extension of Birkhoff’s theorem, or, more accurately, I’m looking for a quantum extension of Birkhoff’s theorem along with a lot of other people - I just happen to think it can be done with category theory).

16. • CommentRowNumber16.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 30th 2010
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    Author: Harry Gindi
    Format: MarkdownItexCould you explain which Birkhoff's theorem this is? Also, could you clarify your question? What do you mean by "combination" and what do you mean by "behave unitarily"?

    Could you explain which Birkhoff’s theorem this is? Also, could you clarify your question? What do you mean by “combination” and what do you mean by “behave unitarily”?

17. • CommentRowNumber17.
    • CommentAuthorEric
    • CommentTimeJun 30th 2010
    • (edited Jun 30th 2010)
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    Author: Eric
    Format: MarkdownItexSome background at [[Birkhoff-von Neumann theorem]] PS: I haven't spent much time following your research Ian, but I can see why you're interested in category theory. This definitely smells category theoretic. I suspect some single sweeping statement about functors or natural transformations will make all these mysteries disappear.

    Some background at Birkhoff-von Neumann theorem

    PS: I haven’t spent much time following your research Ian, but I can see why you’re interested in category theory. This definitely smells category theoretic. I suspect some single sweeping statement about functors or natural transformations will make all these mysteries disappear.

18. • CommentRowNumber18.
    • CommentAuthorIan_Durham
    • CommentTimeJun 30th 2010
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    Author: Ian_Durham
    Format: MarkdownItexOh, right. Too many darned Birkhoff's theorems out there... This is actually the Birkhoff-von Neumann theorem. Classically, it states that doubly stochastic matrices of order _n_ form the convex hull of permutation matrices of the same order where the latter are the vertices (extreme points) of the former, i.e. doubly stochastic matrices are convex combinations of permutation matrices. In the quantum context, doubly stochastic matrices become doubly stochastic channels, i.e. completely positive maps preserving both the trace and the identity. Quantum mechanically the permutations are understood to be the unitarily implemented channels. That is, we expect doubly stochastic quantum channels to be convex combinations of unitary channels. Unfortunately, it is known that some quantum channels can't be written this way. Collectively, a solution to this problem has taken on the nom-de-guerre of "the quantum Birkhoff theorem" or some such thing. One of the many "end-arounds" that has been proposed by Andreas Winter is based on the idea that large tensor powers of a channel may be easier to represent as a convex hull of permutation matrices, because one need not use only product unitaries in the decomposition. From the standpoint of an empiricist/experimentalist, this implies that, should a quantum channel prove to be irreversible, _n_ copies of such a channel (or some rough approximation of it) ought to start to look more and more reversible as the _n_ gets larger. So imagine we have an irreversible channel that is actually like a small wire on a microchip. You could design a chip (if this theory proved true) to come within some arbitrarily determined range of being reversible simply by increasing the number of these wires you use to connect the two things you're connecting.

    Oh, right. Too many darned Birkhoff’s theorems out there… This is actually the Birkhoff-von Neumann theorem. Classically, it states that doubly stochastic matrices of order n form the convex hull of permutation matrices of the same order where the latter are the vertices (extreme points) of the former, i.e. doubly stochastic matrices are convex combinations of permutation matrices.

    In the quantum context, doubly stochastic matrices become doubly stochastic channels, i.e. completely positive maps preserving both the trace and the identity. Quantum mechanically the permutations are understood to be the unitarily implemented channels. That is, we expect doubly stochastic quantum channels to be convex combinations of unitary channels. Unfortunately, it is known that some quantum channels can’t be written this way.

    Collectively, a solution to this problem has taken on the nom-de-guerre of “the quantum Birkhoff theorem” or some such thing. One of the many “end-arounds” that has been proposed by Andreas Winter is based on the idea that large tensor powers of a channel may be easier to represent as a convex hull of permutation matrices, because one need not use only product unitaries in the decomposition.

    From the standpoint of an empiricist/experimentalist, this implies that, should a quantum channel prove to be irreversible, n copies of such a channel (or some rough approximation of it) ought to start to look more and more reversible as the n gets larger. So imagine we have an irreversible channel that is actually like a small wire on a microchip. You could design a chip (if this theory proved true) to come within some arbitrarily determined range of being reversible simply by increasing the number of these wires you use to connect the two things you’re connecting.

19. • CommentRowNumber19.
    • CommentAuthorIan_Durham
    • CommentTimeJun 30th 2010
    • PermaLink
    Author: Ian_Durham
    Format: MarkdownItex<blockquote><p>This definitely smells category theoretic. I suspect some single sweeping statement about functors or natural transformations will make all these mysteries disappear.</p></blockquote> Thank you! I wish more people felt that way. Actually, a couple of people at the APS March Meeting thought it sounded like it had a lot of merit, but they had never heard of category theory before and were going on what I told them. Otherwise, people profess interest but never have the time to help me work it out.

    This definitely smells category theoretic. I suspect some single sweeping statement about functors or natural transformations will make all these mysteries disappear.

    Thank you! I wish more people felt that way. Actually, a couple of people at the APS March Meeting thought it sounded like it had a lot of merit, but they had never heard of category theory before and were going on what I told them. Otherwise, people profess interest but never have the time to help me work it out.

20. • CommentRowNumber20.
    • CommentAuthorEric
    • CommentTimeJun 30th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>Otherwise, people profess interest but never have the time to help me work it out. Unfortunately, I'm in this group :)

    Otherwise, people profess interest but never have the time to help me work it out.

    Unfortunately, I’m in this group :)

21. • CommentRowNumber21.
    • CommentAuthorIan_Durham
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Ian_Durham
    Format: MarkdownItex<blockquote><p>Unfortunately, I'm in this group :)</p></blockquote> Ah, well, c'est la vie.

    Unfortunately, I’m in this group :)

    Ah, well, c’est la vie.

22. • CommentRowNumber22.
    • CommentAuthorEric
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexWhen I hear about something that preserves identities, I think of a functor. So I wonder if it is worth reconsidering the category you're working with. For example, I wonder if what you are thinking of as morphisms are actually functors and you're working with a [[functor category]]. If so, then your categories might be more basic than indicated so far (I presume since I haven't seen what has been indicated :)) Wanna try with a simple example of an irreversible quantum channel or something? How is your category defined?

    When I hear about something that preserves identities, I think of a functor. So I wonder if it is worth reconsidering the category you’re working with.

    For example, I wonder if what you are thinking of as morphisms are actually functors and you’re working with a functor category. If so, then your categories might be more basic than indicated so far (I presume since I haven’t seen what has been indicated :))

    Wanna try with a simple example of an irreversible quantum channel or something? How is your category defined?

23. • CommentRowNumber23.
    • CommentAuthorIan_Durham
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Ian_Durham
    Format: MarkdownItexI'm out the door to go fishing, but to get a basic sense, see the entry on [[quantum channel]]s and the PDF file I put together for the APS meeting (it's linked at the bottom of that page).

    I’m out the door to go fishing, but to get a basic sense, see the entry on quantum channels and the PDF file I put together for the APS meeting (it’s linked at the bottom of that page).

24. • CommentRowNumber24.
    • CommentAuthorIan_Durham
    • CommentTimeJul 2nd 2010
    • PermaLink
    Author: Ian_Durham
    Format: MarkdownItexAs per a recent request, I have altered the aforementioned PDF file. The "Anonymous Coward" is me. I forgot to type my name.

    As per a recent request, I have altered the aforementioned PDF file. The “Anonymous Coward” is me. I forgot to type my name.