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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexOn the nLab page for [[semidirect product]], it says: >If a group $G$ [[action|acts]] on a group $\Gamma$ on the left, then there is a [[semidirect product]] group whose underlying set is $\Gamma \times G$ but whose multiplication is $$(\delta,h)(\gamma,g)= (\delta \, ^h \gamma, h g)$$ >for $\delta, \gamma \in \Gamma,\; h,g \in G$. What is $^h\gamma$?

   On the nLab page for semidirect product, it says:

   If a group $G$ acts on a group $\Gamma$ on the left, then there is a semidirect product group whose underlying set is $\Gamma \times G$ but whose multiplication is

   $$(\delta,h)(\gamma,g)= (\delta \, ^h \gamma, h g)$$

   for $\delta, \gamma \in \Gamma,\; h,g \in G$.

   What is $^h\gamma$?

2. • CommentRowNumber2.
   • CommentAuthorTim_Porter
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Tim_Porter
   Format: MarkdownItexIt is the result of acting with $h$ on the left on $\gamma$.

   It is the result of acting with $h$ on the left on $\gamma$.

3. • CommentRowNumber3.
   • CommentAuthorEric
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexThanks. I added your explanation to the definition.

   Thanks. I added your explanation to the definition.

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJul 31st 2010
   • (edited Jul 31st 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexNope, it means conjugation by h, i.e. ${}^h\gamma:=h\gamma h^{-1}$. If not, $(id_N,id_H)\ast (id_N,g)=(g,g)$, but $(id_N,id_H)$ is the identity of the semidirect product. As far as that though, I'd be interested about why that is the formula. I have some vague notions of why, but I'd be interested to listen.

   Nope, it means conjugation by h, i.e. ${}^h\gamma:=h\gamma h^{-1}$. If not, $(id_N,id_H)\ast (id_N,g)=(g,g)$, but $(id_N,id_H)$ is the identity of the semidirect product.

   As far as that though, I’d be interested about why that is the formula. I have some vague notions of why, but I’d be interested to listen.

5. • CommentRowNumber5.
   • CommentAuthorEric
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexI was wondering about that too. But if it is conjugation, then would $G$ need to act on $\Gamma$ on both the left and right? I rolled back my minor change. In any case, the statement could probably use some clarification on the nLab.

   I was wondering about that too. But if it is conjugation, then would $G$ need to act on $\Gamma$ on both the left and right?

   I rolled back my minor change.

   In any case, the statement could probably use some clarification on the nLab.

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeJul 31st 2010
   • (edited Jul 31st 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexThere is a canonical right action assigned to every left action given by the formula $g^{-1}h:=h^{-1}g$ for $h\in H$, $g\in G$, with a fixed left action $H\to Aut(G)$, if I remember correctly. Categorically, a right action is the same as giving a contravariant left action. Let's test it out: $g^{-1}hj=(hj)^{-1}g=j^{-1}h^{-1}g$. So yes, this is the right one.

   There is a canonical right action assigned to every left action given by the formula $g^{-1}h:=h^{-1}g$ for $h\in H$, $g\in G$, with a fixed left action $H\to Aut(G)$, if I remember correctly. Categorically, a right action is the same as giving a contravariant left action.

   Let’s test it out: $g^{-1}hj=(hj)^{-1}g=j^{-1}h^{-1}g$. So yes, this is the right one.

7. • CommentRowNumber7.
   • CommentAuthorTim_Porter
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Tim_Porter
   Format: MarkdownItex@ Harry. You are both right and wrong! It depends on the interior or exterior viewpoint. If, as stated in the n-Lab, the group $G$ is given, together with the action then what I said was correct. You construct a group $\Gamma \rtimes G$ as described, with the given formula as DEFINING the multiplication. You now have a group and it comes with a neat epimorphism to $G$ with kernel ISOMORPHIC to $\Gamma$. There is thus a copy of $\Gamma$ in there and the epimorphism is split so there is a copy of $G$ in there as well. An element $g$ of $G$ sits in there as $(1,g)$ and an element $\gamma$ of $\Gamma$ as $(\gamma,1)$, now if you look at conjugating the copy of $\gamma$ by the copy of $g$ you get $$(1,g)(\gamma,1)(1,g)^{-1} = (^g\gamma,1)$$ so the semidirect product CONVERTS the EXTERIOR action of $G$ on $\Gamma$ into the INTERIOR action, by conjugation, of the copy of $G$ on the copy of $\Gamma$, hence you are both wrong (exteriorly) and right (after the construction of the semidirect product has been done). This sort of point used to be made in books on group theory (I remember one by Rotman) and was very confusing. Another point that may be worth making is that an action need not be given by a morphism $G\to Aut(\Gamma)$. In fact, in some settings there is no automorphism thing around so you use $$G\times \Gamma \to \Gamma$$ satisfying certain obvious rules. (With profinite groups, the group of automorphisms is not always profinite, although it can be topologised in a sensible way.) Of course, you also have, as you pointed out, the idea of an action as a functor from $G$ as a category, and that is also very useful.

   @ Harry. You are both right and wrong! It depends on the interior or exterior viewpoint. If, as stated in the n-Lab, the group $G$ is given, together with the action then what I said was correct. You construct a group $\Gamma \rtimes G$ as described, with the given formula as DEFINING the multiplication. You now have a group and it comes with a neat epimorphism to $G$ with kernel ISOMORPHIC to $\Gamma$. There is thus a copy of $\Gamma$ in there and the epimorphism is split so there is a copy of $G$ in there as well. An element $g$ of $G$ sits in there as $(1,g)$ and an element $\gamma$ of $\Gamma$ as $(\gamma,1)$, now if you look at conjugating the copy of $\gamma$ by the copy of $g$ you get

   $$(1,g)(\gamma,1)(1,g)^{-1} = (^g\gamma,1)$$

   so the semidirect product CONVERTS the EXTERIOR action of $G$ on $\Gamma$ into the INTERIOR action, by conjugation, of the copy of $G$ on the copy of $\Gamma$, hence you are both wrong (exteriorly) and right (after the construction of the semidirect product has been done).

   This sort of point used to be made in books on group theory (I remember one by Rotman) and was very confusing.

   Another point that may be worth making is that an action need not be given by a morphism $G\to Aut(\Gamma)$. In fact, in some settings there is no automorphism thing around so you use

   $$G\times \Gamma \to \Gamma$$

   satisfying certain obvious rules. (With profinite groups, the group of automorphisms is not always profinite, although it can be topologised in a sensible way.) Of course, you also have, as you pointed out, the idea of an action as a functor from $G$ as a category, and that is also very useful.

8. • CommentRowNumber8.
   • CommentAuthorTim_Porter
   • CommentTimeJul 31st 2010
   • PermaLink
   Author: Tim_Porter
   Format: MarkdownItexAha, I can see possibly where the initial confusion might be. The action is an action on the GROUP $\Gamma$ not on the underlying set of $\Gamma$, i.e. an action by automorphisms. Saying it is on the left does not mean left multiplication in any sense, it does mean left action! If this was not the source of confusion, ignore this comment!

   Aha, I can see possibly where the initial confusion might be. The action is an action on the GROUP $\Gamma$ not on the underlying set of $\Gamma$, i.e. an action by automorphisms. Saying it is on the left does not mean left multiplication in any sense, it does mean left action!

   If this was not the source of confusion, ignore this comment!

9. • CommentRowNumber9.
   • CommentAuthorEric
   • CommentTimeJul 31st 2010
   • (edited Jul 31st 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexFor the record, when I said >I was wondering about that too. it was referring to a softer question Harri asked, but he then edited his comment to a more strongly worded disagreement. Thanks Tim! I'm headed to bed and will have a look at the details tomorrow, but your nice description definitely needs a home on the nLab. Edit: Also, when I said the statement could use some clarification, I was referring to the original statement on the nLab and not Tim's comment :)

   For the record, when I said

   I was wondering about that too.

   it was referring to a softer question Harri asked, but he then edited his comment to a more strongly worded disagreement.

   Thanks Tim! I’m headed to bed and will have a look at the details tomorrow, but your nice description definitely needs a home on the nLab.

   Edit: Also, when I said the statement could use some clarification, I was referring to the original statement on the nLab and not Tim’s comment :)

10. • CommentRowNumber10.
    • CommentAuthorTim_Porter
    • CommentTimeJul 31st 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexDon't worry. I find formulae with semidirect products give me trouble. I remember them correctly (usually) but do not trust my memory. Try working out $(\gamma,g)^{-1}$. The answer is very sensible, but worth doing. (Being told it is not the same.) Sometimes I find an n-lab entry assumes too much too early, and perhaps a double level entry i.e. semi-direct products for `les nuls', with a second entry giving more interpretations etc. for once the dummy stage has been passed!!! :-)

    Don’t worry. I find formulae with semidirect products give me trouble. I remember them correctly (usually) but do not trust my memory. Try working out $(\gamma,g)^{-1}$. The answer is very sensible, but worth doing. (Being told it is not the same.)

    Sometimes I find an n-lab entry assumes too much too early, and perhaps a double level entry i.e. semi-direct products for ‘les nuls’, with a second entry giving more interpretations etc. for once the dummy stage has been passed!!! :-)

11. • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexSurely this entry was originally written with the intention of defining the concept for groupoids, assuming that one already knew it for groups. I have re-reverted the explanation of ${}^h{g}$, added some structure, and briefly described internal and right semidirect products.

    Surely this entry was originally written with the intention of defining the concept for groupoids, assuming that one already knew it for groups.

    I have re-reverted the explanation of ${}^h{g}$, added some structure, and briefly described internal and right semidirect products.

12. • CommentRowNumber12.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 2nd 2010
    • (edited Aug 2nd 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexThe notation ${}^h\gamma$ unambiguously means conjugation. If you want to denote a left action, a proper notation is $h\cdot \gamma$.

    The notation ${}^h\gamma$ unambiguously means conjugation. If you want to denote a left action, a proper notation is $h\cdot \gamma$.

13. • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexActually, the notation $^h \gamma $ is not infrequently used for a left action, especially in the context of semidirect products and crossed modules where the usual actions considered are closely related to conjugation (and become conjugation once you construct the semidirect product group).

    Actually, the notation $^h \gamma$ is not infrequently used for a left action, especially in the context of semidirect products and crossed modules where the usual actions considered are closely related to conjugation (and become conjugation once you construct the semidirect product group).

14. • CommentRowNumber14.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex>a proper notation is... it's too late to rewrite my paper with Urs, then :). Also, better tell Ronnie Brown and his collaborators, from whom I picked up the notation.

    a proper notation is…

    it’s too late to rewrite my paper with Urs, then :). Also, better tell Ronnie Brown and his collaborators, from whom I picked up the notation.

15. • CommentRowNumber15.
    • CommentAuthorEric
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexBy the way, I was reading [[semidirect product]] because I'm hoping to understand the relation between functors and forms, which means I need to know something about 2-groups.

    By the way, I was reading semidirect product because I’m hoping to understand the relation between functors and forms, which means I need to know something about 2-groups.

16. • CommentRowNumber16.
    • CommentAuthorTim_Porter
    • CommentTimeAug 2nd 2010
    • (edited Aug 2nd 2010)
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexRe Sdps and actions (Sdp = semi-direct product) I get a bit worried when someone writes `unambiguously'. ${}^h\gamma$ would parse / type incorrectly if it was only used as conjugation. $h\cdot \gamma$, it should be countered also means multiplication! :-) Some people write $\alpha: G\to Aut(\Gamma)$ and then write $\alpha(h)(\gamma)$. (I note that João Faria Martins does this in his papers especially when looking at crossed modules.) That is a good unambiguous notation especially when there may be two different actions of $G$ on $\Gamma$ being considered, but I am a bit lazy and do not use it! :-) The idea of action as functor gives another good careful notation: F: BG\to Groups$ would yield $F(h)(\gamma)$. That is very useful and, Eric, if you are looking at sdps to understand functors and forms, first rewrite sdps in the functor notation and then the multiplication. (Then look at the Grothendieck construction as Ehresmann thought of it i.e. as a generalisation of sdp. Each step is fairly easy but you can bootstrap your intuition nicely by going around this `loop' rewriting things in another way.) Another useful source is the Magnus Forrester-Barker's nice note at http://www.maths.bangor.ac.uk/research/preprints/00/algtop00.html#00.29 Although you may not need that if you have gone through the connections between 2-groups etc, and crossed modules before. It is anyway a useful thing to have online. (Edit: fixed some typos.)

    Re Sdps and actions (Sdp = semi-direct product)

    I get a bit worried when someone writes ‘unambiguously’. ${}^h\gamma$ would parse / type incorrectly if it was only used as conjugation. $h\cdot \gamma$, it should be countered also means multiplication! :-)

    Some people write $\alpha: G\to Aut(\Gamma)$ and then write $\alpha(h)(\gamma)$. (I note that João Faria Martins does this in his papers especially when looking at crossed modules.) That is a good unambiguous notation especially when there may be two different actions of $G$ on $\Gamma$ being considered, but I am a bit lazy and do not use it! :-)

    The idea of action as functor gives another good careful notation: F: BG\to Groups$would yield$F(h)(\gamma)$. That is very useful and, Eric, if you are looking at sdps to understand functors and forms, first rewrite sdps in the functor notation and then the multiplication. (Then look at the Grothendieck construction as Ehresmann thought of it i.e. as a generalisation of sdp. Each step is fairly easy but you can bootstrap your intuition nicely by going around this ‘loop’ rewriting things in another way.)

    Another useful source is the Magnus Forrester-Barker’s nice note at http://www.maths.bangor.ac.uk/research/preprints/00/algtop00.html#00.29

    Although you may not need that if you have gone through the connections between 2-groups etc, and crossed modules before. It is anyway a useful thing to have online. (Edit: fixed some typos.)

17. • CommentRowNumber17.
    • CommentAuthorTim_Porter
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexI have added to [[action]] a brief section on actions of a [[group object]]. I have not mentioned notation :-)

    I have added to action a brief section on actions of a group object. I have not mentioned notation :-)

18. • CommentRowNumber18.
    • CommentAuthorEric
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>The idea of action as functor gives another good careful notation: $F: BG\to Groups$ would yield $F(h)(\gamma)$. That is very useful and, Eric, if you are looking at sdps to understand functors and forms, first rewrite sdps in the functor notation and then the multiplication. (Then look at the Grothendieck construction as Ehresmann thought of it i.e. as a generalisation of sdp. Each step is fairly easy but you can bootstrap your intuition nicely by going around this `loop' rewriting things in another way.) That sounds like a good homework assignment. Maybe a bit over my head though. I wouldn't mind if someone spoiled the fun and showed me how to rewrite sdps in functor notation :) By the way, the idea of relating functors and forms this way is so pretty. Is it really in Urs and Konrad's 2008 paper it was first written down?

    The idea of action as functor gives another good careful notation: $F: BG\to Groups$ would yield $F(h)(\gamma)$. That is very useful and, Eric, if you are looking at sdps to understand functors and forms, first rewrite sdps in the functor notation and then the multiplication. (Then look at the Grothendieck construction as Ehresmann thought of it i.e. as a generalisation of sdp. Each step is fairly easy but you can bootstrap your intuition nicely by going around this ‘loop’ rewriting things in another way.)

    That sounds like a good homework assignment. Maybe a bit over my head though. I wouldn’t mind if someone spoiled the fun and showed me how to rewrite sdps in functor notation :)

    By the way, the idea of relating functors and forms this way is so pretty. Is it really in Urs and Konrad’s 2008 paper it was first written down?

19. • CommentRowNumber19.
    • CommentAuthorTim_Porter
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexHave a look at the menagerie (http://ncatlab.org/nlab/files/menagerie10a.pdf) and its discussion of the Grothendieck construction (p. 274). That may help. That is really the same thing as your 'homework'. (I hope the train is not too noisy and you have a smooth ride when trying this!) One thought that does not always occur to people. In the semidirect product the underlying set is the product, but you can also think of this as the disjoint union of a family of copies of $\Gamma$ indexed by the elements of $G$. Now look at the 'usual' way of building a coproduct of sets, i.e. look at that disjoint union and record as well as each element the index of the set in which you are 'seeing' it. Sort of $A_0, A_1$ gives elements $(a,i)$ where $a \in A_i$, doh! That looks like the product if each $A_i$ is the same set!. You can think of this as a set of fibres over the set of indices. All the fibres are the same. Now look at the Grothendieck construction. (I keep saying that don't I!) There are links between the objects so you get links between the fibres over them. That is the action / functor coming in.

    Have a look at the menagerie (http://ncatlab.org/nlab/files/menagerie10a.pdf) and its discussion of the Grothendieck construction (p. 274). That may help. That is really the same thing as your ’homework’. (I hope the train is not too noisy and you have a smooth ride when trying this!)

    One thought that does not always occur to people. In the semidirect product the underlying set is the product, but you can also think of this as the disjoint union of a family of copies of $\Gamma$ indexed by the elements of $G$. Now look at the ’usual’ way of building a coproduct of sets, i.e. look at that disjoint union and record as well as each element the index of the set in which you are ’seeing’ it. Sort of $A_0, A_1$ gives elements $(a,i)$ where $a \in A_i$, doh! That looks like the product if each $A_i$ is the same set!. You can think of this as a set of fibres over the set of indices. All the fibres are the same.

    Now look at the Grothendieck construction. (I keep saying that don’t I!) There are links between the objects so you get links between the fibres over them. That is the action / functor coming in.

20. • CommentRowNumber20.
    • CommentAuthorTobyBartels
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>That looks like the product if each $A_i$ is the same set! In other words: a product is a repeated sum; multiplication is repeated addition.

    That looks like the product if each $A_i$ is the same set!

    In other words: a product is a repeated sum; multiplication is repeated addition.

21. • CommentRowNumber21.
    • CommentAuthorTim_Porter
    • CommentTimeAug 2nd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexEGGSACTLY!! John had a nice version of that somewhere. Anyone have a link. I remember an array of 1's. Ronnie used it several times.

    EGGSACTLY!!

    John had a nice version of that somewhere. Anyone have a link. I remember an array of 1’s. Ronnie used it several times.

22. • CommentRowNumber22.
    • CommentAuthorEric
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexStill haven't had a chance to read through this and am running out the door to catch a train to another day of mayhem, but had a thought last night while sleeping. If we think of a group $G$ as a one-object category $\mathbf{B}G$ instead, then we could form a functor category $[\mathbf{B}G,\mathbf{B}G]$. It would probably follow that $F(g^{-1}) = F^{-1}(g)$ so each functor would have an inverse and it has identities by definition and it is associative, so this would make the functors in $[\mathbf{B}G,\mathbf{B}G]$ a group. Or something... Am I close? Gotta run!

    Still haven’t had a chance to read through this and am running out the door to catch a train to another day of mayhem, but had a thought last night while sleeping.

    If we think of a group $G$ as a one-object category $\mathbf{B}G$ instead, then we could form a functor category $[\mathbf{B}G,\mathbf{B}G]$. It would probably follow that $F(g^{-1}) = F^{-1}(g)$ so each functor would have an inverse and it has identities by definition and it is associative, so this would make the functors in $[\mathbf{B}G,\mathbf{B}G]$ a group. Or something…

    Am I close? Gotta run!

23. • CommentRowNumber23.
    • CommentAuthorTobyBartels
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>$F(g^{-1}) = F^{-1}(g)$ No. A functor from $\mathbf{B}G$ to $\mathbf{B}G$ is the same as a homomorphism from $G$ to $G$ (although, interestingly, two such functors can be naturally isomorphic even if the corresponding group homomorphisms are not equal). So as with homomorphisms, we have $F(g^{-1}) = F(g)^{-1}$, but plenty of them are not invertible. If you restrict to the invertible homomorphisms, then you get a group, but that's nothing special, *any* category has [[automorphism groups]].

    $F(g^{-1}) = F^{-1}(g)$

    No.

    A functor from $\mathbf{B}G$ to $\mathbf{B}G$ is the same as a homomorphism from $G$ to $G$ (although, interestingly, two such functors can be naturally isomorphic even if the corresponding group homomorphisms are not equal). So as with homomorphisms, we have $F(g^{-1}) = F(g)^{-1}$, but plenty of them are not invertible.

    If you restrict to the invertible homomorphisms, then you get a group, but that’s nothing special, any category has automorphism groups.

24. • CommentRowNumber24.
    • CommentAuthorTim_Porter
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexIf you restrict to the invertible homomorphisms / functors and the natural transformations between them you get a well known 2-group.

    If you restrict to the invertible homomorphisms / functors and the natural transformations between them you get a well known 2-group.

25. • CommentRowNumber25.
    • CommentAuthorEric
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>No. Ok. But I think I at least deserve some points for recognizing that functors $\mathbf{B}G\to\mathbf{B}G$ are the same as homomorphisms $G\to G$. That is kind of cute :) Still in the vortex. Gotta run!

    No.

    Ok. But I think I at least deserve some points for recognizing that functors $\mathbf{B}G\to\mathbf{B}G$ are the same as homomorphisms $G\to G$. That is kind of cute :)

    Still in the vortex. Gotta run!

26. • CommentRowNumber26.
    • CommentAuthorTim_Porter
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexHappy spinning. :-) Now think about the natural transformations between those homomorphisms. That is even cuter!

    Happy spinning. :-)

    Now think about the natural transformations between those homomorphisms. That is even cuter!

27. • CommentRowNumber27.
    • CommentAuthorEric
    • CommentTimeAug 3rd 2010
    • (edited Aug 3rd 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItex>**Lemma 6** For groupoids $\mathcal{G}$ and $\mathcal{H}$, the functor category, $\mathcal{H}^{\mathcal{G}}$, is a groupoid. That _is_ cute :) So for each element of the group $G$, i.e. morphism in $\mathbf{B}G$, and functors $F_1,F_2:\mathbf{B}G\to\mathbf{B}H$, the commuting square coming from the natural transformation $\eta:F_1\to F_2$ gives conjugation by $h = \eta(\ast)$, i.e. $F_1(g) = h F_2(g) h^{-1}.$ PS: This now makes me think that whenever we see a conjugation, there is secretly some natural transformation hiding somewhere.

    Lemma 6 For groupoids $\mathcal{G}$ and $\mathcal{H}$, the functor category, $\mathcal{H}^{\mathcal{G}}$, is a groupoid.

    That is cute :)

    So for each element of the group $G$, i.e. morphism in $\mathbf{B}G$, and functors $F_1,F_2:\mathbf{B}G\to\mathbf{B}H$, the commuting square coming from the natural transformation $\eta:F_1\to F_2$ gives conjugation by $h = \eta(\ast)$, i.e.

    $F_1(g) = h F_2(g) h^{-1}.$

    PS: This now makes me think that whenever we see a conjugation, there is secretly some natural transformation hiding somewhere.

28. • CommentRowNumber28.
    • CommentAuthorTobyBartels
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex@ Eric Actually, you should ignore my comment #23. When you wrote $F(g^{-1}) = F^{-1}(g)$, I assumed that you wanted $F^{-1}$ to be an inverse functor; $F \circ F^{-1} = \id_{\mathbf{B}G}$ and all that. But now I realise that you were defining some *other* group structure on $[\mathbf{B}G,\mathbf{B}G]$. In fact, you want a group structure on $[\mathbf{B}H,\mathbf{B}G]$. If $G$ is commutative, then this works, defining products pointwise, and we do have $F(g^{-1}) = F^{-1}(g)$ (because $F^{-1}(g) = F(g)^{-1} = F(g^{-1})$). But in general, the pointwise product of two group homomorphisms is not a homomorphism. More generally, if $H$ is any *set*, then $Hom_Set(H,G)$ becomes a group by pointwise multiplication. But (when $H$ is a group) $Hom_Grp(H,G)$ is a subset of $Hom_Set(H,G)$ but not in general a subgroup.

    @ Eric

    Actually, you should ignore my comment #23.

    When you wrote $F(g^{-1}) = F^{-1}(g)$, I assumed that you wanted $F^{-1}$ to be an inverse functor; $F \circ F^{-1} = \id_{\mathbf{B}G}$ and all that.

    But now I realise that you were defining some other group structure on $[\mathbf{B}G,\mathbf{B}G]$. In fact, you want a group structure on $[\mathbf{B}H,\mathbf{B}G]$.

    If $G$ is commutative, then this works, defining products pointwise, and we do have $F(g^{-1}) = F^{-1}(g)$ (because $F^{-1}(g) = F(g)^{-1} = F(g^{-1})$). But in general, the pointwise product of two group homomorphisms is not a homomorphism.

    More generally, if $H$ is any set, then $Hom_Set(H,G)$ becomes a group by pointwise multiplication. But (when $H$ is a group) $Hom_Grp(H,G)$ is a subset of $Hom_Set(H,G)$ but not in general a subgroup.

29. • CommentRowNumber29.
    • CommentAuthorTim_Porter
    • CommentTimeAug 3rd 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItex@Eric: Now even more cuter!!!! $\mathcal{G}^\mathcal{G}$ is a monoid (under composition) and a groupoid and the structures satisfy interchange. Looking just at the automorphisms and the nat trans between them gives you a 2-group. It thus gives you a crossed module. (Easy exercise in group theory to work out what it is.)

    @Eric: Now even more cuter!!!! $\mathcal{G}^\mathcal{G}$ is a monoid (under composition) and a groupoid and the structures satisfy interchange. Looking just at the automorphisms and the nat trans between them gives you a 2-group. It thus gives you a crossed module. (Easy exercise in group theory to work out what it is.)

30. • CommentRowNumber30.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexThanks Tim and Toby. This stuff is pretty cool. I'm basically trying to understand "kindergarten 1-transport" or "kindergarten 2-forms as 2-functors". By "kindergarten", I mean I'm not concerned about a continuum so do not need to worry about [[Lie 2-groups]] and all that. I just want to think of space as a finite diagram, i.e. a finite diagram perhaps representing a discretization or finitary model of some continuum (directed) space. Something like a mesh for computational fluid dynamics, electromagnetics, etc. In this case, we can parallel transport abstract edges across abstract faces and our groups do not need to even be smooth. Any 2-groups would suffice I think. I'm still working on my intuition for semidirect products and crossed modules, but this is the direction I hope to go.

    Thanks Tim and Toby.

    This stuff is pretty cool. I’m basically trying to understand “kindergarten 1-transport” or “kindergarten 2-forms as 2-functors”. By “kindergarten”, I mean I’m not concerned about a continuum so do not need to worry about Lie 2-groups and all that. I just want to think of space as a finite diagram, i.e. a finite diagram perhaps representing a discretization or finitary model of some continuum (directed) space. Something like a mesh for computational fluid dynamics, electromagnetics, etc.

    In this case, we can parallel transport abstract edges across abstract faces and our groups do not need to even be smooth. Any 2-groups would suffice I think. I’m still working on my intuition for semidirect products and crossed modules, but this is the direction I hope to go.

31. • CommentRowNumber31.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 4th 2010
    • (edited Aug 4th 2010)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexYou may appreciate the approach in http://arxiv.org/abs/math-ph/0203056 - it treats AUT(G) 2-transport over a simplicial complex, but not in those terms. The 'group-like category' $\mathcal{B}$ in the article is equivalent to the 2-group arising from the crossed module $G \to Aut(G)$.

    You may appreciate the approach in http://arxiv.org/abs/math-ph/0203056 - it treats AUT(G) 2-transport over a simplicial complex, but not in those terms. The ’group-like category’ $\mathcal{B}$ in the article is equivalent to the 2-group arising from the crossed module $G \to Aut(G)$.

32. • CommentRowNumber32.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexThanks David :) I'm hoping to generalize this material on [lattice Yang-Mills](http://arxiv.org/PS_cache/math-ph/pdf/0407/0407005v1.pdf#page=67) (from some mystery authors) to the case of higher transport so this paper definitely looks helpful.

    Thanks David :)

    I’m hoping to generalize this material on lattice Yang-Mills (from some mystery authors) to the case of higher transport so this paper definitely looks helpful.

33. • CommentRowNumber33.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexOne thing to keep an eye out for is the relation between BF theory and surface transport (John and/or Urs has some stuff on this). I don't know of any other physics-like theory that comes up at this 'easy' level of higher gauge theory - otherwise my PhD would have been on it :). One thing that is important is that curvature for 2-transport is valued in a 3-group (see if you can guess which other paper by one of those mystery authors I'm thinking of). And strict functoriality might have to be relaxed in the assignment of (2-)group elements to links/plaquettes, unlike in those early papers on surface transport, else things can become what is known as 'fake flat'.

    One thing to keep an eye out for is the relation between BF theory and surface transport (John and/or Urs has some stuff on this). I don’t know of any other physics-like theory that comes up at this ’easy’ level of higher gauge theory - otherwise my PhD would have been on it :). One thing that is important is that curvature for 2-transport is valued in a 3-group (see if you can guess which other paper by one of those mystery authors I’m thinking of). And strict functoriality might have to be relaxed in the assignment of (2-)group elements to links/plaquettes, unlike in those early papers on surface transport, else things can become what is known as ’fake flat’.

34. • CommentRowNumber34.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexJust for the record, can we write down the numbering conventions here? Ordinary parallel transport. Is that 0-transport or 1-transport? Curvature arising from $k$-transport. Is that $k$-curvature or $(k+1)$-curvature?

    Just for the record, can we write down the numbering conventions here?

    Ordinary parallel transport. Is that 0-transport or 1-transport?

    Curvature arising from $k$-transport. Is that $k$-curvature or $(k+1)$-curvature?

35. • CommentRowNumber35.
    • CommentAuthorTim_Porter
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexI looked at Attal's stuff when I was trying to interpret Turaev's HQFTs in terms of crossed modules, gerbes etc. It looked nice, but I am not really happy with the forms / curvature stuff so did not go further. I think David and I chatted about this one evening at Ross Street's birthday meeting in Sydney. Another author who may be worth looking at for related stuff is Roger Picken. He keeps things fairly down to earth, although he is looking at the non-discrete case mostly. Joao Faria-Martins has useful stuff on this as well.

    I looked at Attal’s stuff when I was trying to interpret Turaev’s HQFTs in terms of crossed modules, gerbes etc. It looked nice, but I am not really happy with the forms / curvature stuff so did not go further. I think David and I chatted about this one evening at Ross Street’s birthday meeting in Sydney.

    Another author who may be worth looking at for related stuff is Roger Picken. He keeps things fairly down to earth, although he is looking at the non-discrete case mostly. Joao Faria-Martins has useful stuff on this as well.

36. • CommentRowNumber36.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex>Ordinary parallel transport. Is that 0-transport or 1-transport? for me this is 1-transport: transport along 1-dimensional objects. >Curvature arising from k-transport. Is that k-curvature or (k+1)-curvature? $(k+1)$-curvature: bundle gerbes have, for example, surface holonomy and 3-curvature.

    Ordinary parallel transport. Is that 0-transport or 1-transport?

    for me this is 1-transport: transport along 1-dimensional objects.

    Curvature arising from k-transport. Is that k-curvature or (k+1)-curvature?

    $(k+1)$-curvature: bundle gerbes have, for example, surface holonomy and 3-curvature.

37. • CommentRowNumber37.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • (edited Aug 4th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItexOk. Thanks. Just to make sure I can count, regular curvature is 2-curvature, right? (Seems a little unfortunate)

    Ok. Thanks. Just to make sure I can count, regular curvature is 2-curvature, right? (Seems a little unfortunate)

38. • CommentRowNumber38.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexWell it is a 2-form... and actually curvature of 1-transport with values in G has values in the 2-group INN(G) (nice, eh?)

    Well it is a 2-form…

    and actually curvature of 1-transport with values in G has values in the 2-group INN(G) (nice, eh?)

39. • CommentRowNumber39.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>Well it is a 2-form... Ok ok. Good point :) >and actually curvature of 1-transport with values in G has values in the 2-group INN(G) (nice, eh?) It might be if I knew what $INN(G)$ was :) So for those mysterious authors, they turned $G$ into a group algebra. Can $INN(G)$ be turned into a group algebra? Or should it be something "higher"? Regarding "discrete" stuff, I will probably try this with directed cubes. Has anyone done this before? I'm slowly learning to love bigons (or trying anyway). I often stare at this diagram $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$ and wonder if the 2-arrow should go from one path to the parallel path or an initial edge to a final edge. I'm guessing this crowd would suggest the former, but I can't give up the latter for some reason. The arrows represent a "flow of time" to me and a bigon from one path to the other parallel path would not "flow". I'd like my 2-morphisms to represent time as well if possible.

    Well it is a 2-form…

    Ok ok. Good point :)

    and actually curvature of 1-transport with values in G has values in the 2-group INN(G) (nice, eh?)

    It might be if I knew what $INN(G)$ was :)

    So for those mysterious authors, they turned $G$ into a group algebra. Can $INN(G)$ be turned into a group algebra? Or should it be something “higher”?

    Regarding “discrete” stuff, I will probably try this with directed cubes. Has anyone done this before?

    I’m slowly learning to love bigons (or trying anyway).

    I often stare at this diagram

    $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$

    and wonder if the 2-arrow should go from one path to the parallel path or an initial edge to a final edge. I’m guessing this crowd would suggest the former, but I can’t give up the latter for some reason.

    The arrows represent a “flow of time” to me and a bigon from one path to the other parallel path would not “flow”. I’d like my 2-morphisms to represent time as well if possible.

40. • CommentRowNumber40.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 4th 2010
    • (edited Aug 4th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexRule of thumb: 2-arrows live between 1-arrows with the same source and target (and this is still true for n-arrows and n-1 arrows).

    Rule of thumb: 2-arrows live between 1-arrows with the same source and target (and this is still true for n-arrows and n-1 arrows).

41. • CommentRowNumber41.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexI read (not absorbed) most of Attal's paper on the train ride home. One thought I had, which probably applies to most similar papers, was that I think parallel transport should come with a "cost". That cost can be interpreted as some notion of time. So instead of a path groupoid, I'd try to work things out on a path category, i.e. a directed space. For example, in the square $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$ we can think of time as being directed diagonally up and to the right. As an aid, I think of the morphisms as rays of light, i.e. light-like paths. With this interpretation in mind, it is difficult to reconcile with the usual notion of 2-morphism. Maybe there is a way. So I would tend to think that the bottom light-like edge should be 2-transported upward toward the top light-like edge like: $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & \Uparrow & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$ Similarly, the left light-like edge should be 2-transported to the right light-like edge: $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & \Rightarrow & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$ Anything else doesn't seem causal. Is there a way to reconcile this with the usual way to think of 2-morphisms? $$\begin{aligned} & & \bullet & & \\ & \nearrow & & \nwarrow & \\ \bullet & &\Rightarrow && \bullet \\ & \nwarrow & & \nearrow & \\ & & \bullet & & \end{aligned}$$

    I read (not absorbed) most of Attal’s paper on the train ride home.

    One thought I had, which probably applies to most similar papers, was that I think parallel transport should come with a “cost”. That cost can be interpreted as some notion of time.

    So instead of a path groupoid, I’d try to work things out on a path category, i.e. a directed space. For example, in the square

    $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$

    we can think of time as being directed diagonally up and to the right. As an aid, I think of the morphisms as rays of light, i.e. light-like paths. With this interpretation in mind, it is difficult to reconcile with the usual notion of 2-morphism. Maybe there is a way.

    So I would tend to think that the bottom light-like edge should be 2-transported upward toward the top light-like edge like:

    $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & \Uparrow & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$

    Similarly, the left light-like edge should be 2-transported to the right light-like edge:

    $$\begin{aligned} \bullet & \to & \bullet \\ \uparrow & \Rightarrow & \uparrow \\ \bullet & \to & \bullet \end{aligned}$$

    Anything else doesn’t seem causal.

    Is there a way to reconcile this with the usual way to think of 2-morphisms?

    $$\begin{aligned} & & \bullet & & \\ & \nearrow & & \nwarrow & \\ \bullet & &\Rightarrow && \bullet \\ & \nwarrow & & \nearrow & \\ & & \bullet & & \end{aligned}$$
42. • CommentRowNumber42.
    • CommentAuthorEric
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>It might be if I knew what $INN(G)$ was :) Haven't absorbed it yet and I'm headed to bed now, but a very very nice looking reference can be found [here](http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.1741v2.pdf#page=3) (by some mystery authors :))

    It might be if I knew what $INN(G)$ was :)

    Haven’t absorbed it yet and I’m headed to bed now, but a very very nice looking reference can be found here (by some mystery authors :))

43. • CommentRowNumber43.
    • CommentAuthorTobyBartels
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>I'm guessing this crowd would suggest the former, but I can't give up the latter for some reason. That is certainly the usual thing, as Harry's rule of thumb suggests. That is what we have in a $2$-category. It's important there that the square is just a special case; you can take any polygon and divide its boundary into any two contiguous pieces (which are oriented oppositely) so that the composite of each piece is a $1$-arrow from a common source to a common target. Then the $2$-arrow goes from one composite to the other. In fact, it's misleading to think about these polygons (including the original squares) at all; really $2$-arrows in a $2$-category always and only fill *bigons*. (You can break up each side of the bigon into a composite of several $1$-arrows if you like, but that's irrelevant to the $2$-arrow.) However, you can also work in a [[double category]]. There we have two classes of $1$-arrows, one called (and usually drawn) ‘horizontal’ and the other ‘vertical’. Then instead of a bigon, you must use a square, and the $2$-arrow goes between the horizontal $1$-arrows in the square. (Again, you could break up each side of the square into a composite of several $1$-arrows if you like, but that's irrelevant to the $2$-arrow.)

    I’m guessing this crowd would suggest the former, but I can’t give up the latter for some reason.

    That is certainly the usual thing, as Harry’s rule of thumb suggests. That is what we have in a $2$-category. It’s important there that the square is just a special case; you can take any polygon and divide its boundary into any two contiguous pieces (which are oriented oppositely) so that the composite of each piece is a $1$-arrow from a common source to a common target. Then the $2$-arrow goes from one composite to the other. In fact, it’s misleading to think about these polygons (including the original squares) at all; really $2$-arrows in a $2$-category always and only fill bigons. (You can break up each side of the bigon into a composite of several $1$-arrows if you like, but that’s irrelevant to the $2$-arrow.)

    However, you can also work in a double category. There we have two classes of $1$-arrows, one called (and usually drawn) ‘horizontal’ and the other ‘vertical’. Then instead of a bigon, you must use a square, and the $2$-arrow goes between the horizontal $1$-arrows in the square. (Again, you could break up each side of the square into a composite of several $1$-arrows if you like, but that’s irrelevant to the $2$-arrow.)

44. • CommentRowNumber44.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 4th 2010
    • (edited Aug 4th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexSecond rule of thumb: Let bigons be bigons. Meanwhile, apparently there are [a lot of complete idiots on the internet](http://www.google.com/search?client=opera&rls=en&q=%22let+bigons+be+bigons%22&sourceid=opera&ie=utf-8&oe=utf-8).

    Second rule of thumb:

    Let bigons be bigons.

    Meanwhile, apparently there are a lot of complete idiots on the internet.

45. • CommentRowNumber45.
    • CommentAuthorzskoda
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: zskoda
    Format: MarkdownItexHarry 4,6 and 12: The definition should not be circular, right ? The claim in 12 is such, impatient. One starts with a left action of $G$ on $\Gamma$. So at the point of definition there is no conjugation action whatsoever used, nor it makes really sense for elements in two different abstract groups. Then ${}^h\gamma$ is the result of the left action of $h$ on $\gamma$. This helps defining the product in the semidirect product. In this new product, new copies of $G$ and $\Gamma$ sit, and the left action will be, **after the identifications** equal to the conjugation action. The same idea is for the smash product of Hopf algebras, see the entry [[crossed product algebra]].

    Harry 4,6 and 12: The definition should not be circular, right ? The claim in 12 is such, impatient.

    One starts with a left action of $G$ on $\Gamma$. So at the point of definition there is no conjugation action whatsoever used, nor it makes really sense for elements in two different abstract groups. Then ${}^h\gamma$ is the result of the left action of $h$ on $\gamma$. This helps defining the product in the semidirect product. In this new product, new copies of $G$ and $\Gamma$ sit, and the left action will be, after the identifications equal to the conjugation action. The same idea is for the smash product of Hopf algebras, see the entry crossed product algebra.

46. • CommentRowNumber46.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 4th 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItex@Zoran: Regarding #12, I was merely stating that most algebraists use the left exponent to denote conjugation.

    @Zoran: Regarding #12, I was merely stating that most algebraists use the left exponent to denote conjugation.

47. • CommentRowNumber47.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex>One thought I had, which probably applies to most similar papers, was that I think parallel transport should come with a "cost". That cost can be interpreted as some notion of time. and this makes me think of categories enriched in the rig $\mathbb{R}^+$, if you are interested in having 'costs' associated to paths (this has to be done in a certain way, not just any old enriched category will do here). Or perhaps your category wants to live over a monoid (say $(\mathbb{R}^+,+)$) interpreted as a category with one object, so that paths are assigned costs, and concatenating paths adds the costs.

    One thought I had, which probably applies to most similar papers, was that I think parallel transport should come with a “cost”. That cost can be interpreted as some notion of time.

    and this makes me think of categories enriched in the rig $\mathbb{R}^+$, if you are interested in having ’costs’ associated to paths (this has to be done in a certain way, not just any old enriched category will do here). Or perhaps your category wants to live over a monoid (say $(\mathbb{R}^+,+)$) interpreted as a category with one object, so that paths are assigned costs, and concatenating paths adds the costs.

48. • CommentRowNumber48.
    • CommentAuthorEric
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>and this makes me think of categories enriched in the rig $\mathbb{R}^+$ Thanks David. This comes back to my old idea about "extruding" groupoids. You and Mike helped me out a lot on this subject before. I need to collect all that wisdom in one place so I can remember everything. I thought of that when I read Attal's paper and he discusses how he keeps "thin homotopies" as distinct paths. For example $(xyxz)$ is not the same path as $(xz)$. When I read that, I said, "Of course they are not the same path. One is three (time) steps and the other is one (time) step." If you extrude everything, it is pretty obvious (although extruding isn't really necessary, I like to use it as a conceptual crutch).

    and this makes me think of categories enriched in the rig $\mathbb{R}^+$

    Thanks David. This comes back to my old idea about “extruding” groupoids. You and Mike helped me out a lot on this subject before. I need to collect all that wisdom in one place so I can remember everything.

    I thought of that when I read Attal’s paper and he discusses how he keeps “thin homotopies” as distinct paths. For example $(xyxz)$ is not the same path as $(xz)$. When I read that, I said, “Of course they are not the same path. One is three (time) steps and the other is one (time) step.”

    If you extrude everything, it is pretty obvious (although extruding isn’t really necessary, I like to use it as a conceptual crutch).

49. • CommentRowNumber49.
    • CommentAuthorMike Shulman
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> apparently there are a lot of complete idiots on the internet. Who knew? > it's misleading to think about these polygons (including the original squares) at all; really 2-arrows in a 2-category always and only fill bigons. I think that's a matter of opinion, or maybe a choice of presentation for a theory. When you say that, you're probably thinking of a 2-category as structure on an underlying 2-quiver (= 2-globular set). But 2Cat is also monadic over the category of 2-computads, so the "underlying data" of a 2-category can just as well be thought of as consisting of 2-cells from strings of n arrows to strings of m arrows, for all nonnegative integers n and m. Double categories are funny because you have these 2-cells that sit in a square, but it's entirely your choice about what to call their "source" and "target". You can say that they go from one horizontal arrow to the other, with the two vertical arrows being the "frame" or something "along which" they go, or vice versa, or you can say that they go from the "formal composite" of a horizontal and vertical arrow to the formal composite of a vertical and horizontal arrow. I think one way of thinking makes more sense for some double categories, but the other makes more sense for others. For instance, if horizontal arrows are bimodules and vertical arrows are ring homomorphisms, I really want to think of a 2-cell as a map of bimodules that happens to be "equivariant" along a pair of ring homomorphisms. But if horizontal arrows are lax monoidal functors and vertical ones are colax monoidal functors, I think it makes much more sense to think of the source of a 2-cell as a formal composite of monoidal functors of opposite laxity.

    apparently there are a lot of complete idiots on the internet.

    Who knew?

    it’s misleading to think about these polygons (including the original squares) at all; really 2-arrows in a 2-category always and only fill bigons.

    I think that’s a matter of opinion, or maybe a choice of presentation for a theory. When you say that, you’re probably thinking of a 2-category as structure on an underlying 2-quiver (= 2-globular set). But 2Cat is also monadic over the category of 2-computads, so the “underlying data” of a 2-category can just as well be thought of as consisting of 2-cells from strings of n arrows to strings of m arrows, for all nonnegative integers n and m.

    Double categories are funny because you have these 2-cells that sit in a square, but it’s entirely your choice about what to call their “source” and “target”. You can say that they go from one horizontal arrow to the other, with the two vertical arrows being the “frame” or something “along which” they go, or vice versa, or you can say that they go from the “formal composite” of a horizontal and vertical arrow to the formal composite of a vertical and horizontal arrow. I think one way of thinking makes more sense for some double categories, but the other makes more sense for others. For instance, if horizontal arrows are bimodules and vertical arrows are ring homomorphisms, I really want to think of a 2-cell as a map of bimodules that happens to be “equivariant” along a pair of ring homomorphisms. But if horizontal arrows are lax monoidal functors and vertical ones are colax monoidal functors, I think it makes much more sense to think of the source of a 2-cell as a formal composite of monoidal functors of opposite laxity.

50. • CommentRowNumber50.
    • CommentAuthorEric
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>However, you can also work in a [[double category]]. I wonder if there is a way to interpret a 2-form as a smooth functor to a Lie double groupoid instead of a Lie 2-group?

    However, you can also work in a double category.

    I wonder if there is a way to interpret a 2-form as a smooth functor to a Lie double groupoid instead of a Lie 2-group?

51. • CommentRowNumber51.
    • CommentAuthorTim_Porter
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItex@Mike: you say 'Double categories are funny because you have these 2-cells that sit in a square, but it's entirely your choice about what to call their "source" and "target".' Surely the point of double categories is that they do not have a source and target as such. They do have a direction 1 source and a direction 2 source etc for each square. (Horizontal and vertical do not generalise to higher triple categories etc.)

    @Mike: you say ’Double categories are funny because you have these 2-cells that sit in a square, but it’s entirely your choice about what to call their “source” and “target”.’

    Surely the point of double categories is that they do not have a source and target as such. They do have a direction 1 source and a direction 2 source etc for each square. (Horizontal and vertical do not generalise to higher triple categories etc.)

52. • CommentRowNumber52.
    • CommentAuthorMike Shulman
    • CommentTimeAug 5th 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> Horizontal and vertical do not generalise to higher triple categories etc. Of course not, but that doesn't give me any qualms about using those words in the case of double categories. For triple categories I like to speak about "horizontal," "vertical," and "depthwise." > They do have a direction 1 source and a direction 2 source etc for each square. That's one way to think about it, which is perhaps closest to the formal definition, if you define double categories to be internal categories in Cat. But I think the intuitions I described are also valid ways to think about it (and not really very different, either).

    Horizontal and vertical do not generalise to higher triple categories etc.

    Of course not, but that doesn’t give me any qualms about using those words in the case of double categories. For triple categories I like to speak about “horizontal,” “vertical,” and “depthwise.”

    They do have a direction 1 source and a direction 2 source etc for each square.

    That’s one way to think about it, which is perhaps closest to the formal definition, if you define double categories to be internal categories in Cat. But I think the intuitions I described are also valid ways to think about it (and not really very different, either).

53. • CommentRowNumber53.
    • CommentAuthorTim_Porter
    • CommentTimeAug 6th 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItex@ Mike I hope I was not implying that I though your way was `invalid' in some way, no I was trying to say that often with double categories it is useful to keep the two directions very separate for some way in their study as they may have different geometric meaning / content. Then taking the double nerve of the double category and working with that as a bisimplicial set, before applying diag or the Artin-Mazur codiag to get a usable simplicial set leads to something that measures the changes in both directions. I did some work some years ago on ordered groupoids which correspond to inverse semigroups and the two directions are very different in feeling, one is a partial order, the other consists of 'local' transformations / partial bijections.

    @ Mike I hope I was not implying that I though your way was ‘invalid’ in some way, no I was trying to say that often with double categories it is useful to keep the two directions very separate for some way in their study as they may have different geometric meaning / content. Then taking the double nerve of the double category and working with that as a bisimplicial set, before applying diag or the Artin-Mazur codiag to get a usable simplicial set leads to something that measures the changes in both directions. I did some work some years ago on ordered groupoids which correspond to inverse semigroups and the two directions are very different in feeling, one is a partial order, the other consists of ’local’ transformations / partial bijections.

54. • CommentRowNumber54.
    • CommentAuthorMike Shulman
    • CommentTimeAug 6th 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> often with double categories it is useful to keep the two directions very separate for some way in their study as they may have different geometric meaning / content I definitely agree with that. I think there are lots of different "flavors" of double categories which often come with different intuitions.

    often with double categories it is useful to keep the two directions very separate for some way in their study as they may have different geometric meaning / content

    I definitely agree with that. I think there are lots of different “flavors” of double categories which often come with different intuitions.

55. • CommentRowNumber55.
    • CommentAuthorEric
    • CommentTimeAug 10th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItexA beautiful beautiful description of [[semidirect product]] is given on Page 20 in: * [Invitation to Higher Gauge Theory](http://arxiv.org/PS_cache/arxiv/pdf/1003/1003.4485v1.pdf#page=20)

    A beautiful beautiful description of semidirect product is given on Page 20 in:

    • Invitation to Higher Gauge Theory
56. • CommentRowNumber56.
    • CommentAuthorTim_Porter
    • CommentTimeAug 10th 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexYes. That is a good description. It essentially dates back to Ehresmann and Grothendieck, but not so neatly expressed. Have you also glanced and Forrester-Barker's description of the process of going between crossed modules and 2-groups (due to Ronnie in his early work with Spencer)?

    Yes. That is a good description. It essentially dates back to Ehresmann and Grothendieck, but not so neatly expressed. Have you also glanced and Forrester-Barker’s description of the process of going between crossed modules and 2-groups (due to Ronnie in his early work with Spencer)?

57. • CommentRowNumber57.
    • CommentAuthorEric
    • CommentTimeAug 10th 2010
    • (edited Aug 10th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItex>Have you also glanced and Forrester-Barker's description of the process of going between crossed modules and 2-groups (due to Ronnie in his early work with Spencer)? Not yet, but I'll put it on the list of reading material. I still haven't made the conceptual leap to understanding 2-categories yet, but this paper helps a lot. My problem is that I learned from watching John lecture in both meatspace and cyberspace that 2-morphisms are maps between 1-morphisms, but I never bothered looking too carefully at the drawings, so all this time, I've thought a 2-morphism should go between 1-morphisms regardless of whether the 1-morphisms share the same source and target. The picture in my head is like a directed graph. Directed 1-cells go between directed 0-cells. Directed 2-cells go between directed 1-cells, but they are not constrained to go between directed 1-cells sharing the same source and target. Therefore, $(C_1,C_0,s:C_1\to C_0,t:C_1\to C_0)$ is a directed graph in the traditional sense. However, I also expect $(C_n,C_{n-1},s:C_n\to C_{n-1},t:C_n\to C_{n-1})$ to also be a quite general directed graph where $C_{n-1}$ are thought of as vertices. The way 2-categories are usually defined in terms of bigons form a particular boring kind of directed 2-graph. I'm probably just thinking about things the wrong way, but this is a mental stumbling block for me. I suspect that my way of thinking about things is essentially equivalent to the standard way because my 2-morphisms would probably have to come with a pair of component 1-morphisms (similar to a natural transformation) that turn the "square" into a bigon anyway. For example, if we had two 1-morphisms that do not share source and target such as $$\begin{aligned} x & & a \\ \mathllap{f}{\downarrow} & & \mathrlap{\,\,\,\,g}{\downarrow} \\ y & & b \end{aligned}$$ I'd still want a 2-map $\alpha:f\rightrightarrows g$ (using different notation to avoid a clash), but it might need to come with component 1-morphisms $\alpha_x:x\to a$ and $\alpha_y:y\to b$ so it would look like $$\begin{aligned} x &\stackrel{\alpha_x}{\to} & a \\ \mathllap{f}{\downarrow} & \stackrel{\alpha}{\rightrightarrows} & \mathrlap{\,\,\,\,g}{\downarrow} \\ y & \stackrel{\alpha_y}{\to} & b \end{aligned}$$ Then we could probably think of $\alpha:f\rightrightarrows g$ as a typical 2-morphism $$\alpha_y\circ f\Rightarrow g\circ\alpha_x.$$ Then my 2-map would correspond to standard 2-morphisms when the components are identities, but would otherwise be more natural in my opinion (understanding that "natural" is a personal opinion not necessarily shared by others, but that is OK). Ok. Since I am thinking out loud... Now I am tempted to introduce a 1-morphism $$\partial\alpha: b\to b$$ for any 2-map $\alpha:f\rightrightarrows g$ such that $$\partial\alpha\circ\alpha_y\circ f = g\circ\alpha_x.$$ _Note: This is intentionally designed to look like a boundary._ For horizontal composition, this leads to a bunch of interesting looking relations among $\partial \alpha$, $\partial\beta$, and $\partial(\beta\circ\alpha)$ that I'm still working out. For example, with $f_i:x_i\to y_i$ and $g_i:y_i\to z_i$ for $i\in\{1,2\}$ and 2-maps $\alpha:f_1\rightrightarrows f_2$ and $\beta:g_1\rightrightarrows g_2$ we have 1. $g_2\circ \partial\alpha\circ\alpha_y\circ f_1 = g_2\circ f_2\circ\alpha_x$ 1. $\partial(\beta\circ\alpha)\circ\beta_z\circ g_1\circ f_1 = g_2\circ f_2\circ\alpha_x$ (implying a relation between $\partial A$ and $\partial(\beta\circ\alpha)$) 1. $\partial\beta\circ\beta_z\circ g_1\circ f_1 = g_2\circ\beta_y\circ f_1$ Edit: I'm also thinking there should be some kind of "gluing" 2-morphism to glue the two parallel components together such that 1. $\partial g_{\alpha,\beta}\circ \beta_y = \partial\alpha\circ\alpha_y$

    Have you also glanced and Forrester-Barker’s description of the process of going between crossed modules and 2-groups (due to Ronnie in his early work with Spencer)?

    Not yet, but I’ll put it on the list of reading material. I still haven’t made the conceptual leap to understanding 2-categories yet, but this paper helps a lot.

    My problem is that I learned from watching John lecture in both meatspace and cyberspace that 2-morphisms are maps between 1-morphisms, but I never bothered looking too carefully at the drawings, so all this time, I’ve thought a 2-morphism should go between 1-morphisms regardless of whether the 1-morphisms share the same source and target.

    The picture in my head is like a directed graph. Directed 1-cells go between directed 0-cells. Directed 2-cells go between directed 1-cells, but they are not constrained to go between directed 1-cells sharing the same source and target. Therefore, $(C_1,C_0,s:C_1\to C_0,t:C_1\to C_0)$ is a directed graph in the traditional sense. However, I also expect $(C_n,C_{n-1},s:C_n\to C_{n-1},t:C_n\to C_{n-1})$ to also be a quite general directed graph where $C_{n-1}$ are thought of as vertices. The way 2-categories are usually defined in terms of bigons form a particular boring kind of directed 2-graph.

    I’m probably just thinking about things the wrong way, but this is a mental stumbling block for me.

    I suspect that my way of thinking about things is essentially equivalent to the standard way because my 2-morphisms would probably have to come with a pair of component 1-morphisms (similar to a natural transformation) that turn the “square” into a bigon anyway.

    For example, if we had two 1-morphisms that do not share source and target such as

    $$\begin{aligned} x & & a \\ \mathllap{f}{\downarrow} & & \mathrlap{\,\,\,\,g}{\downarrow} \\ y & & b \end{aligned}$$

    I’d still want a 2-map $\alpha:f\rightrightarrows g$ (using different notation to avoid a clash), but it might need to come with component 1-morphisms $\alpha_x:x\to a$ and $\alpha_y:y\to b$ so it would look like

    $$\begin{aligned} x &\stackrel{\alpha_x}{\to} & a \\ \mathllap{f}{\downarrow} & \stackrel{\alpha}{\rightrightarrows} & \mathrlap{\,\,\,\,g}{\downarrow} \\ y & \stackrel{\alpha_y}{\to} & b \end{aligned}$$

    Then we could probably think of $\alpha:f\rightrightarrows g$ as a typical 2-morphism

    $$\alpha_y\circ f\Rightarrow g\circ\alpha_x.$$

    Then my 2-map would correspond to standard 2-morphisms when the components are identities, but would otherwise be more natural in my opinion (understanding that “natural” is a personal opinion not necessarily shared by others, but that is OK).

    Ok. Since I am thinking out loud…

    Now I am tempted to introduce a 1-morphism

    $$\partial\alpha: b\to b$$

    for any 2-map $\alpha:f\rightrightarrows g$ such that

    $$\partial\alpha\circ\alpha_y\circ f = g\circ\alpha_x.$$

    Note: This is intentionally designed to look like a boundary.

    For horizontal composition, this leads to a bunch of interesting looking relations among $\partial \alpha$, $\partial\beta$, and $\partial(\beta\circ\alpha)$ that I’m still working out. For example, with $f_i:x_i\to y_i$ and $g_i:y_i\to z_i$ for $i\in\{1,2\}$ and 2-maps $\alpha:f_1\rightrightarrows f_2$ and $\beta:g_1\rightrightarrows g_2$ we have

    1. $g_2\circ \partial\alpha\circ\alpha_y\circ f_1 = g_2\circ f_2\circ\alpha_x$
    2. $\partial(\beta\circ\alpha)\circ\beta_z\circ g_1\circ f_1 = g_2\circ f_2\circ\alpha_x$ (implying a relation between $\partial A$ and $\partial(\beta\circ\alpha)$)
    3. $\partial\beta\circ\beta_z\circ g_1\circ f_1 = g_2\circ\beta_y\circ f_1$

    Edit: I’m also thinking there should be some kind of “gluing” 2-morphism to glue the two parallel components together such that

    1. $\partial g_{\alpha,\beta}\circ \beta_y = \partial\alpha\circ\alpha_y$
58. • CommentRowNumber58.
    • CommentAuthorTim_Porter
    • CommentTimeAug 10th 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexOne thought is that the 2-category model of spaces using bigons is derived from situations where you are mapping a square into the space but two opposite sides of the square are mapped to points, so a homotopy rel endpoints. The picture you seem to like is the free homotopy. Unfortunately they are more difficult to handle in that spatial situation. You are using a double category rather than a 2-category. These can be useful as you can constrain the sides to have some special property, instead of merely being constant. There seems to be a similarity between what you are thinking of an certain diagrams that Larry Breen uses when describing non-Abelian cohomology classes.

    One thought is that the 2-category model of spaces using bigons is derived from situations where you are mapping a square into the space but two opposite sides of the square are mapped to points, so a homotopy rel endpoints. The picture you seem to like is the free homotopy. Unfortunately they are more difficult to handle in that spatial situation. You are using a double category rather than a 2-category. These can be useful as you can constrain the sides to have some special property, instead of merely being constant.

    There seems to be a similarity between what you are thinking of an certain diagrams that Larry Breen uses when describing non-Abelian cohomology classes.

59. • CommentRowNumber59.
    • CommentAuthorEric
    • CommentTimeAug 10th 2010
    • (edited Aug 10th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItexInteresting. Thanks Tim. That was a lot more encouraging than the "He's nuts!" I was expecting :) By the way, in the case of a 2-groupoid where all component 1-morphisms are identities, i.e. standard 2-groupoids as usually considered, my boundary constraint for horizontal composition becomes $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}.$$ Does that look familiar? I like this idea of taking the boundary of a 2-morphism resulting in a 1-morphism that makes the diagram commute. Edit: Note also that when dealing with typical 2-groupoids the boundary of the 2-morphism is even nicer, i.e. $$\partial\alpha = g\circ f^{-1}.$$ Reminiscent of Stokes theorem, eh? :)

    Interesting. Thanks Tim. That was a lot more encouraging than the “He’s nuts!” I was expecting :)

    By the way, in the case of a 2-groupoid where all component 1-morphisms are identities, i.e. standard 2-groupoids as usually considered, my boundary constraint for horizontal composition becomes

    $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}.$$

    Does that look familiar? I like this idea of taking the boundary of a 2-morphism resulting in a 1-morphism that makes the diagram commute.

    Edit: Note also that when dealing with typical 2-groupoids the boundary of the 2-morphism is even nicer, i.e.

    $$\partial\alpha = g\circ f^{-1}.$$

    Reminiscent of Stokes theorem, eh? :)

60. • CommentRowNumber60.
    • CommentAuthorEric
    • CommentTimeAug 10th 2010
    • (edited Aug 10th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItexI didn't realize it at the time I wrote it, but $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}$$ looks like a semidirect product :) In other words, $$\partial(\beta\circ\alpha) = \partial\beta\circ {}^{g_2}\partial\alpha.$$ Edit: A little time later (not wanting to flood with even more posts) Let $\alpha:f_1\Rightarrow f_2$ and $\beta:g_1\Rightarrow g_2$, then let $(f_1,\alpha)$ denote the fact that $\partial\alpha\circ f_2 = f_1$, then we can define a product $$(g_1,\beta)\circ (f_1,\alpha) = (g_1\circ f_1,\beta\circ\alpha).$$ The term on the right-hand side refers to horizontal composition with $$\beta\circ\alpha: (g_1\circ f_1)\Rightarrow (g_2\circ f_2)$$ such that $$\partial(\beta\circ\alpha)\circ g_2\circ f_2 = g_1\circ f_1 = (\partial\alpha\circ g_2)\circ(\partial\beta\circ f_2)$$ from which it follows that $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}.$$

    I didn’t realize it at the time I wrote it, but

    $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}$$

    looks like a semidirect product :)

    In other words,

    $$\partial(\beta\circ\alpha) = \partial\beta\circ {}^{g_2}\partial\alpha.$$

    Edit: A little time later (not wanting to flood with even more posts)

    Let $\alpha:f_1\Rightarrow f_2$ and $\beta:g_1\Rightarrow g_2$, then let $(f_1,\alpha)$ denote the fact that $\partial\alpha\circ f_2 = f_1$, then we can define a product

    $$(g_1,\beta)\circ (f_1,\alpha) = (g_1\circ f_1,\beta\circ\alpha).$$

    The term on the right-hand side refers to horizontal composition with

    $$\beta\circ\alpha: (g_1\circ f_1)\Rightarrow (g_2\circ f_2)$$

    such that

    $$\partial(\beta\circ\alpha)\circ g_2\circ f_2 = g_1\circ f_1 = (\partial\alpha\circ g_2)\circ(\partial\beta\circ f_2)$$

    from which it follows that

    $$\partial(\beta\circ\alpha) = \partial\beta\circ g_2\circ\partial\alpha\circ g_2^{-1}.$$