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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeFeb 2nd 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am polishing the entry _[[category algebra]]_. In the course of this I noticed that there was old and long-forgotten discussion sitting there, which now first of all I hereby move from there to here: --- [ begin old discussion ] I use $k[S]$ to stand for the free vector space on the set $S$. This is compatible with the notation $k[G]$ for group algebra of $G$. Urs' notation $k[C]$ for the category algebra is also compatible, but in a different way. Why is my notation better? First, because I don't like the clunky notation $span_k(C)$ for the free vector space on the set $S$. Second, because the equation $k[B G] = k[G]$ is inconsistent unless Urs is finally willing to admit that $B G = G$. <img src = "http://math.ucr.edu/home/baez/emoticons/tongue2.gif" alt = ""/> So what would _I_ call the category algebra of $C$? I guess $k[C_1]$ or $k[Mor(C)]$. You might complain that <i>this</i> notation is clunky, and I'd see your point. However, it's a fact that whenever the category algebra is important, its representation on $k[C_0] = k[Ob(C)]$ also tends to be important --- so I think the benefits of a notation that handles both structures outweigh the disadvantages of a slight clunkiness. -- [[John Baez|John]] _[[Urs Schreiber|Urs]] says_: It is good that you said this, because we need to talk about this: I am puzzled by your attitude towards $\mathbf{B}G$ vs $G$. It is not the least a remark in your lecture notes with Mike that it is important to _distinguish_ between a $k$-tuply monoidal structure and the corresponding $k$-tuply degenerate category, even though there is a map identifying them. The issue appears here for instance when discussing the universal $G$-bundle in its groupoid-incarnation. It is $$ G \to \mathbf{E}G \to \mathbf{B}G $$ (where $\mathbf{E}G = G//G$ is the action groupoid of $G$ acting on itself). On the left we crucially have $G$ as a monoidal 0-category, on the right as a once-degenerate 1-category. In your notation you cannot even _write down_ the universal $G$-bundle! ;-) Or take the important difference between group representations and group 2-algebras, the former being functors $\mathbf{B}G \to Vect$, the latter functors $G \to Vect$. This is important all over the place, as you know better than me. Or take an abelian group $A$ and a codomain like $2Vect$. Then there are 3 different things we can sensibly consider, namely 2-functors $$ A \to 2Vect $$ $$ \mathbf{B}A \to 2Vect $$ $$ \mathbf{B}^2A \to 2Vect \,. $$ All of this is different. All of this is needed. The first one is the group 3-algebra of $A$. The second is pseudo-representations of the group $A$. The third is representations of the 2-group $\mathbf{B}A$. We have notation to distinguish this, and we should use it. Finally, writing $\mathbf{B}G$ for the 1-object $n$-groupoid version of an $n$-monoid $G$ makes notation behave nicely with respect to nerves, because then realization bars $|\cdot|$ simply commute with the $B$s in the game: $|\mathbf{B}G| = B|G|$. I think this makes for instance your theorem with Danny appear in a prettier way. This behaviour under nerves shows also that, generally, writing $\mathbf{B}G$ gives the right intuition for what an expression means. For instance, what's the "geometric" reason that a group representation is an arrow $\rho : \mathbf{B}G \to Vect$? It's because this is, literally, equivalently thought of as the corresponding classifying map of the vector bundle on $\mathbf{B}G$ which is $\rho$-associated to the universal $G$-bundle: the $\rho$-associated vector bundle to the universal $G$-bundle is, in its groupoid incarnations, $$ \array{ V \\ \downarrow \\ V//G \\ \downarrow \\ \mathbf{B}G } \,, $$ where $V$ is the vector space that $\rho$ is representing on, and this is classified by the representation $\rho : \mathbf{B}G \to Vect$ in that this is the pullback of the universal $Vect$-bundle $$ \array{ V//G &\to& Vect_* \\ \downarrow && \downarrow \\ \mathbf{B}G &\stackrel{\rho}{\to}& Vect } \,, $$ In summary, I think it is important to make people understand that groups can be identified with one-object groupoids. But next it is important to make clear that not everything that can be identified is actually equal. For instance concerning the crucial difference between the category in which $G$ lives and the 2-category in which $\mathbf{B}G$ lives. [ continued in next message ]

   I am polishing the entry category algebra. In the course of this I noticed that there was old and long-forgotten discussion sitting there, which now first of all I hereby move from there to here:

   ---

   [ begin old discussion ]

   I use $k[S]$ to stand for the free vector space on the set $S$. This is compatible with the notation $k[G]$ for group algebra of $G$. Urs’ notation $k[C]$ for the category algebra is also compatible, but in a different way.

   Why is my notation better? First, because I don’t like the clunky notation $span_k(C)$ for the free vector space on the set $S$. Second, because the equation $k[B G] = k[G]$ is inconsistent unless Urs is finally willing to admit that $B G = G$.

   So what would I call the category algebra of $C$? I guess $k[C_1]$ or $k[Mor(C)]$. You might complain that this notation is clunky, and I’d see your point. However, it’s a fact that whenever the category algebra is important, its representation on $k[C_0] = k[Ob(C)]$ also tends to be important — so I think the benefits of a notation that handles both structures outweigh the disadvantages of a slight clunkiness. – John

   Urs says: It is good that you said this, because we need to talk about this: I am puzzled by your attitude towards $\mathbf{B}G$ vs $G$. It is not the least a remark in your lecture notes with Mike that it is important to distinguish between a $k$-tuply monoidal structure and the corresponding $k$-tuply degenerate category, even though there is a map identifying them. The issue appears here for instance when discussing the universal $G$-bundle in its groupoid-incarnation. It is

   $$G \to \mathbf{E}G \to \mathbf{B}G$$

   (where $\mathbf{E}G = G//G$ is the action groupoid of $G$ acting on itself). On the left we crucially have $G$ as a monoidal 0-category, on the right as a once-degenerate 1-category. In your notation you cannot even write down the universal $G$-bundle! ;-)

   Or take the important difference between group representations and group 2-algebras, the former being functors $\mathbf{B}G \to Vect$, the latter functors $G \to Vect$. This is important all over the place, as you know better than me.

   Or take an abelian group $A$ and a codomain like $2Vect$. Then there are 3 different things we can sensibly consider, namely 2-functors

   $$A \to 2Vect$$ $$\mathbf{B}A \to 2Vect$$ $$\mathbf{B}^2A \to 2Vect \,.$$

   All of this is different. All of this is needed. The first one is the group 3-algebra of $A$. The second is pseudo-representations of the group $A$. The third is representations of the 2-group $\mathbf{B}A$. We have notation to distinguish this, and we should use it.

   Finally, writing $\mathbf{B}G$ for the 1-object $n$-groupoid version of an $n$-monoid $G$ makes notation behave nicely with respect to nerves, because then realization bars $|\cdot|$ simply commute with the $B$s in the game: $|\mathbf{B}G| = B|G|$. I think this makes for instance your theorem with Danny appear in a prettier way.

   This behaviour under nerves shows also that, generally, writing $\mathbf{B}G$ gives the right intuition for what an expression means. For instance, what’s the “geometric” reason that a group representation is an arrow $\rho : \mathbf{B}G \to Vect$? It’s because this is, literally, equivalently thought of as the corresponding classifying map of the vector bundle on $\mathbf{B}G$ which is $\rho$-associated to the universal $G$-bundle:

   the $\rho$-associated vector bundle to the universal $G$-bundle is, in its groupoid incarnations,

   $$\array{ V \\ \downarrow \\ V//G \\ \downarrow \\ \mathbf{B}G } \,,$$

   where $V$ is the vector space that $\rho$ is representing on, and this is classified by the representation $\rho : \mathbf{B}G \to Vect$ in that this is the pullback of the universal $Vect$-bundle

   $$\array{ V//G &\to& Vect_* \\ \downarrow && \downarrow \\ \mathbf{B}G &\stackrel{\rho}{\to}& Vect } \,,$$

   In summary, I think it is important to make people understand that groups can be identified with one-object groupoids. But next it is important to make clear that not everything that can be identified is actually equal.

   For instance concerning the crucial difference between the category in which $G$ lives and the 2-category in which $\mathbf{B}G$ lives.

   [ continued in next message ]

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeFeb 2nd 2013
   • (edited Apr 2nd 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItex[ continuation from previous message ] _Toby says_: John said: >I use $k[S]$ to stand for the free vector space on the set $S$. This is compatible with the notation $k[G]$ for group algebra of $G$. Urs' notation $k[C]$ for the category algebra is also compatible, but in a different way. Wait, are you claiming that $k[S]$ and $k[C]$ are incompatible? I disagree! Just as a [[set]] may be seen as a [[discrete category]], so a vector space (or [[module]]) may be seen as an algebra where all multiplication is zero. (This is well known in the theory of [[Lie n-algebroid]]s, where a vector space is a twice monoidal Lie 2-algebroid, that is a commutative Lie algebra.) Then $k[S] = k[D S]$ (where $D S$ is the discrete category on $S$), just as $k[B G] = k[G]$. _[[Mike Shulman|Mike]] says_: Urs, I'm definitely with you about $G$ and $B G$, for all the reasons that you give and more. (For instance, in classical homotopy theory, it is essential to distinguish between the two, for similar reasons.) I'm not sure exactly what remark you're referring to in "n-categories and cohomology," but it's possible that it can be blamed on me rather than John. In particular, anything in section 5 is my fault. _Toby_: John, I\'m afraid that, despite the compatibility of $k[S]$ and $k[C]$, on the general issue (as at [[action]]), Urs and Mike have convinced me too. _[[John Baez|John]] says_: Okay, fine. I personally find it tiresome to use a notation that distinguishes between groups and one-object groupoids. To me, having 'light notation', with a minimum of symbols, is incredibly important. Every extra symbol makes my work look more complicated, reduces how many people will read it, and distracts attention from the actual ideas. So, I don't want to write $B G$ unless I really need to. For example, I agree with Urs that if I'm simultaneously discussing functors $$ A \to 2Vect $$ $$ \mathbf{B}A \to 2Vect $$ and $$ \mathbf{B}^2A \to 2Vect $$ then I need to carefully distinguish between these. (I actually use $A[n]$ to mean the $\infty$-category or chain complex with the abelian group $A$ as $n$-morphisms; this is pretty standard in homological algebra.) But if in some passage of text I'm only taking about a functor like $$ \mathbf{B}^2A \to 2Vect$$ I prefer to say "think of $A$ as a 2-category with only one object and one morphism", and then write this functor as $A \to 2Vect$, to avoid polluting the page with tons of $B^2$'s. This probably isn't worth arguing about. Some people prefer logically precise notation, while other people (like me) are always focused on maximizing readership. These are different goals. My ideal math paper would be mainly words with just a few equations per page, because that's the most fun to read. But I don't expect everyone else to agree. While we're nitpicking: do we really want the product $f\cdot g$ of morphisms $f$ and $g$ in the category algebra to equal $g \circ f$? This seems designed to trip people up. _[[Mike Shulman|Mike]]_: Here's another argument I just thought of, although it's still along the lines of "logically precise" so given what you just said, I guess it's unlikely to convince you. What we are discussing is, in general, a functor from [[k-tuply monoidal n-category|k-tuply monoidal n-categories]] to $(k-1)$-tuply monoidal $(n+1)$-categories. The delooping hypothesis says that it's an equivalence onto its image (at least as long as "0-tuply monoidal" means "pointed"), so from that point of view it's natural to want to leave it nameless and think of its domain as a subcategory of its codomain. However, there is also another functor from $k$-tuply monoidal $n$-categories to $(k-1)$-tuply monoidal $(n+1)$-categories which adds identity $(n+1)$-cells and forgets one level of monoidal structure. This one is not in general an equivalence onto its image. But in the particular case $k=n=\omega$, in which case the domain and codomain of these functors are _both_ the category of stably monoidal $\omega$-categories, it is the _second_ functor that is the identity functor, not the first one. [ end old discussion ]

   [ continuation from previous message ]

   Toby says: John said:

   I use $k[S]$ to stand for the free vector space on the set $S$. This is compatible with the notation $k[G]$ for group algebra of $G$. Urs’ notation $k[C]$ for the category algebra is also compatible, but in a different way.

   Wait, are you claiming that $k[S]$ and $k[C]$ are incompatible? I disagree! Just as a set may be seen as a discrete category, so a vector space (or module) may be seen as an algebra where all multiplication is zero. (This is well known in the theory of Lie n-algebroids, where a vector space is a twice monoidal Lie 2-algebroid, that is a commutative Lie algebra.) Then $k[S] = k[D S]$ (where $D S$ is the discrete category on $S$), just as $k[B G] = k[G]$.

   Mike says: Urs, I’m definitely with you about $G$ and $B G$, for all the reasons that you give and more. (For instance, in classical homotopy theory, it is essential to distinguish between the two, for similar reasons.) I’m not sure exactly what remark you’re referring to in “n-categories and cohomology,” but it’s possible that it can be blamed on me rather than John. In particular, anything in section 5 is my fault.

   Toby: John, I'm afraid that, despite the compatibility of $k[S]$ and $k[C]$, on the general issue (as at action), Urs and Mike have convinced me too.

   John says: Okay, fine. I personally find it tiresome to use a notation that distinguishes between groups and one-object groupoids. To me, having ’light notation’, with a minimum of symbols, is incredibly important. Every extra symbol makes my work look more complicated, reduces how many people will read it, and distracts attention from the actual ideas. So, I don’t want to write $B G$ unless I really need to. For example, I agree with Urs that if I’m simultaneously discussing functors

   $$A \to 2Vect$$ $$\mathbf{B}A \to 2Vect$$

   and

   $$\mathbf{B}^2A \to 2Vect$$

   then I need to carefully distinguish between these. (I actually use $A[n]$ to mean the $\infty$-category or chain complex with the abelian group $A$ as $n$-morphisms; this is pretty standard in homological algebra.) But if in some passage of text I’m only taking about a functor like

   $$\mathbf{B}^2A \to 2Vect$$

   I prefer to say “think of $A$ as a 2-category with only one object and one morphism”, and then write this functor as $A \to 2Vect$, to avoid polluting the page with tons of $B^2$’s.

   This probably isn’t worth arguing about. Some people prefer logically precise notation, while other people (like me) are always focused on maximizing readership. These are different goals. My ideal math paper would be mainly words with just a few equations per page, because that’s the most fun to read. But I don’t expect everyone else to agree.

   While we’re nitpicking: do we really want the product $f\cdot g$ of morphisms $f$ and $g$ in the category algebra to equal $g \circ f$? This seems designed to trip people up.

   Mike: Here’s another argument I just thought of, although it’s still along the lines of “logically precise” so given what you just said, I guess it’s unlikely to convince you. What we are discussing is, in general, a functor from k-tuply monoidal n-categories to $(k-1)$-tuply monoidal $(n+1)$-categories. The delooping hypothesis says that it’s an equivalence onto its image (at least as long as “0-tuply monoidal” means “pointed”), so from that point of view it’s natural to want to leave it nameless and think of its domain as a subcategory of its codomain.

   However, there is also another functor from $k$-tuply monoidal $n$-categories to $(k-1)$-tuply monoidal $(n+1)$-categories which adds identity $(n+1)$-cells and forgets one level of monoidal structure. This one is not in general an equivalence onto its image. But in the particular case $k=n=\omega$, in which case the domain and codomain of these functors are both the category of stably monoidal $\omega$-categories, it is the second functor that is the identity functor, not the first one.

   [ end old discussion ]

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeFeb 2nd 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have added to _[[category algebra]]_ a [remark](http://ncatlab.org/nlab/show/category+algebra#ForLieGroupoids) about the refinement to convolution algebras of smooth functions on Lie groupoids. In the course of this I have tried to polish the entry a little more. Made the convolution product formula more manifest in the Definition-section.

   I have added to category algebra a remark about the refinement to convolution algebras of smooth functions on Lie groupoids.

   In the course of this I have tried to polish the entry a little more. Made the convolution product formula more manifest in the Definition-section.

4. • CommentRowNumber4.
   • CommentAuthorjim_stasheff
   • CommentTimeFeb 2nd 2013
   • PermaLink
   Author: jim_stasheff
   Format: TextBut k[V] suggest an algebra generated by V - not the vector space spanned??

   But k[V] suggest an algebra generated by V - not the vector space spanned??
5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeFeb 18th 2013
   • (edited Feb 18th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have added to _[[groupoid algebra]]_ more references on groupoid convolution $C^\ast$-algebras: in _[References - For smooth geometry](http://ncatlab.org/nlab/show/category+algebra#ReferencesForSmoothGeometry)_.

   I have added to groupoid algebra more references on groupoid convolution $C^\ast$-algebras: in References - For smooth geometry.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeApr 1st 2013
   • (edited Apr 1st 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI gave _[[groupoid convolution algebra]]_ an [Idea-section](http://ncatlab.org/nlab/show/category+algebra#Idea), mentioning also the relation to noncommutative geometry and to quantization.

   I gave groupoid convolution algebra an Idea-section, mentioning also the relation to noncommutative geometry and to quantization.

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeApr 1st 2013
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexUrs, can you edit your comment #2 so that my text and my quotation of John are not combined? (Simply add a blank line before and after the line beginning with `>`.)

   Urs, can you edit your comment #2 so that my text and my quotation of John are not combined? (Simply add a blank line before and after the line beginning with >.)

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeApr 2nd 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexDone.

   Done.

9. • CommentRowNumber9.
   • CommentAuthorTobyBartels
   • CommentTimeApr 2nd 2013
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexThanks!

   Thanks!

10. • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeOct 3rd 2016
    • PermaLink
    Author: zskoda
    Format: TextI added the following remark: If the category has finitely many objects and finitely many morphisms, then the category algebra is also a coalgebra via the unique comultiplication where every morphism in $C_1\subset R[C_1]$ is group like; the structure of the coalgebra and that of an algebra are compatible in the sense that they form a [[weak Hopf algebra]].

    I added the following remark:
    
    If the category has finitely many objects and finitely many morphisms, then the category algebra is also a coalgebra via the unique comultiplication where every morphism in $C_1\subset R[C_1]$ is group like; the structure of the coalgebra and that of an algebra are compatible in the sense that they form a weak Hopf algebra.