## May 14, 2014

### Categories vs. Algebras

#### Posted by Tom Leinster

Representation theorists make good use of the “category algebra” construction. This is a way of turning a linear category (one whose hom-sets are vector spaces) into an associative algebra. In this post, I’ll describe what the category algebra is and why it seems to be important.

I’ll also ask two basic questions about the category algebra construction. I hope someone can tell me the answers.

First I’ll describe the category algebra construction. To make life easier for all of us, I’ll always use the word category to mean what category theorists would call a “category enriched in vector spaces”: one in which the hom-sets are vector spaces and composition is bilinear. Similarly, functors will be assumed to preserve this linear structure.

The construction takes as input a category $\mathbf{C}$ (assumed to have just a set of objects, not a proper class) and produces as output an associative algebra $\Sigma \mathbf{C}$, called its category algebra. As a vector space,

$\Sigma\mathbf{C} = \bigoplus_{c, d \in \mathbf{C}} \mathbf{C}(c, d)$

— the direct sum or coproduct of all the hom-sets of $\mathbf{C}$ (which, remember, are vector spaces). To define the multiplication on $\Sigma\mathbf{C}$, it’s enough to define the product $g \cdot f$ whenever $g$ and $f$ are maps in $\mathbf{C}$, and we do this by putting

$g \cdot f = \begin{cases} g \circ f &\text{if }  domain(g) = codomain(f)\\ 0 &\text{otherwise.} \end{cases}$

In other words, multiplication is composition where that makes sense, and zero elsewhere.

(The notation $\Sigma\mathbf{C}$ for the category algebra is something I just made up. Is there standard notation?)

So far I’ve followed the time-honoured tradition of not bothering to say whether “algebras” are supposed to have a multiplicative identity. But actually, the issue of multiplicative identities is crucial to understanding category algebras.

Let me briefly try to explain why. The first observation is that if the category $\mathbf{C}$ has only finitely many objects, then the algebra $\Sigma\mathbf{C}$ does have a multiplicative identity. It’s $\sum_{c \in \mathbf{C}} 1_c$. If $\mathbf{C}$ has infinitely many objects then this sum makes no sense, so $\Sigma \mathbf{C}$ usually doesn’t have a multiplicative identity.

I’ll assume from now that $\mathbf{C}$ has only finitely many objects, so that $\Sigma\mathbf{C}$ is a unital algebra.

We’ll come back to the significance of identities in $\mathbf{C}$ and in $\Sigma\mathbf{C}$, but for now, let’s just observe:

• Taking the category algebra has the effect of concentrating all the identities of $\mathbf{C}$ into a single identity for $\Sigma\mathbf{C}$, with the individual identities $1_c$ of $\mathbf{C}$ being merely idempotents in $\Sigma\mathbf{C}$.

(The sum of all these idempotents is the multiplicative identity of $\Sigma\mathbf{C}$.) Alternatively, looking at it from the point of view of the algebra $\Sigma\mathbf{C}$:

• The multiplicative identity of $\Sigma\mathbf{C}$ is smeared all across the category $\mathbf{C}$, with one summand of the multiplicative identity $\sum_{c \in \mathbf{C}} 1_c$ attached to each object of $\mathbf{C}$.

Time for some examples.

1. Let $S$ be a finite preordered set — that is, a finite set equipped with a reflexive, transitive binary relation $\leq$. We can construct a category $\mathbf{C}$ from it as follows. Abstractly: view $S$ as an ordinary, unenriched category, then let $\mathbf{C}$ be the free linear category on it. Concretely, the objects of $\mathbf{C}$ are the elements of $S$, the hom-set $\mathbf{C}(s, t)$ is the ground field $k$ if $s \leq t$ and zero otherwise, and composition (where it’s nontrivial) is multiplication of scalars.

Now, the category algebra $\Sigma\mathbf{C}$ is a subalgebra of the algebra $M_S(k)$ of all $S \times S$ matrices over $k$. It consists of just those matrices $P$ satisfying the condition that $P(s, t)$ is only allowed to be nonzero when $s \leq t$.

2. A special case of the last example: let $S = \{1, \ldots, n\}$, with the obvious ordering. Then $\mathbf{C}$ is the category that you’d usually draw as $\bullet \to \bullet \to \quad \cdots \quad \to \bullet$ and $\Sigma\mathbf{C}$ is the algebra of $n \times n$ upper-triangular matrices.

3. Another special case: let $S = \{1, \ldots, n\}$ with the discrete ordering: $s \leq t \iff s = t$. Then $\mathbf{C}$ is the discrete category on $n$ objects — the disjoint union of $n$ copies of the ground field $k$ — and $\Sigma\mathbf{C}$ is the algebra of $n \times n$ diagonal matrices. Equivalently, $\Sigma\mathbf{C}$ is the $n$-fold product $k^n$.

4. A final special case: let $S = \{1, \ldots, n\}$ with the other trivial ordering: $s \leq t$ for all $s, t$. Then $\mathbf{C}$ is the codiscrete category on $n$ objects (so that all objects are isomorphic and all hom-sets are $k$), and $\Sigma\mathbf{C}$ is the full matrix algebra $M_n(k)$.

Why are category algebras important? I’m not sure I fully know, but here’s a fundamental fact:

A category and its category algebra are Morita equivalent.

What this means is that for any category $\mathbf{C}$, there’s an equivalence of categories

$[\mathbf{C}, \mathbf{Vect}] \simeq {\Sigma\mathbf{C}}\text{-}\mathbf{Mod}$

where the left-hand side is the category of functors $\mathbf{C} \to \mathbf{Vect}$. If you regard the algebra $\Sigma\mathbf{C}$ as a one-object category, then the right-hand side is the category of functors $\Sigma\mathbf{C} \to \mathbf{Vect}$.

So as far as linear representations are concerned, $\mathbf{C}$ and $\Sigma\mathbf{C}$ are the same thing.

How can we prove this equivalence? It’s one of those follow-your-nose proofs… but in following your nose, you discover the pivotal role of the identities.

In one direction, it’s straightforward: given a functor $F: \mathbf{C} \to \mathbf{Vect}$, put $X = \bigoplus_{c \in \mathbf{C}} F(c)$; then $X$ is a $\Sigma\mathbf{C}$-module in what is, if you think about it, an obvious way.

The other direction isn’t quite so obvious. Starting with a $\Sigma\mathbf{C}$-module $X$, how can we manufacture a functor $F: \mathbf{C} \to \mathbf{Vect}$? Given $X$ and an object $c \in \mathbf{C}$, we have to cook up a vector space $F(c)$. The key here is that, for each $c \in \mathbf{C}$, the element $1_c$ of $\Sigma\mathbf{C}$ is idempotent. It follows that $1_c \cdot - : X \to X$ is idempotent. The image of this map (which is also its set of fixed points) is a vector space; and that’s what we take $F(c)$ to be.

The rest of the details of this equivalence are easy enough, and I won’t bother you with them. Instead, I’ll show you one consequence of the equivalence, then ask you two questions.

First, here’s the consequence. It begins with the observation that equivalent categories don’t usually have isomorphic category algebras. Indeed, suppose we have two equivalent categories, $\mathbf{C}$ and $\mathbf{D}$. Then they’re certainly Morita equivalent. So, by using the result above, we get a chain of equivalences:

$\Sigma\mathbf{C}\text{-}\mathbf{Mod} \simeq [\mathbf{C}, \mathbf{Vect}] \simeq [\mathbf{D}, \mathbf{Vect}] \simeq \Sigma\mathbf{D}\text{-}\mathbf{Mod}$

The end result is that the categories $\Sigma\mathbf{C}$-modules and $\Sigma\mathbf{D}$-modules are equivalent. And, since $\Sigma\mathbf{C}$ and $\Sigma\mathbf{D}$ are not usually isomorphic, this isn’t quite trivial.

The most famous example is due to Morita. Say $\mathbf{C}$ is the codiscrete category on $n \geq 1$ objects (so that all hom-sets are the ground field $k$), and $\mathbf{D}$ is the codiscrete category on a single object. Then, as we saw earlier, $\Sigma\mathbf{C}$ is the full matrix algebra $M_n(k)$, while $\Sigma\mathbf{C} = M_1(k) = k$. But $\mathbf{C}$ and $\mathbf{D}$ are equivalent categories (since all objects of $\mathbf{C}$ are isomorphic), so

$M_n(k)\text{-}\mathbf{Mod} \simeq \mathbf{Vect}$

for any $n \geq 1$.

Now here are my two questions.

First question  Is there a good categorical explanation of the category algebra construction?

The first observation is that the construction isn’t even functorial, or at least, not in the obvious way. A functor $F: \mathbf{C} \to \mathbf{D}$ does induce a linear map $\Sigma\mathbf{C} \to \Sigma\mathbf{D}$, but it doesn’t usually preserve multiplication. For instance, consider the obvious functor from the discrete category $\mathbf{C}$ on two objects to the discrete category $\mathbf{D}$ on one object. The induced linear map $k^2 \to k$ is addition, which is not a homomorphism of algebras.

Second question  Does the category algebra construction suggest that we should study representations of categories rather than representations of algebras?

I need to explain the thinking behind this. The Morita equivalence between a category $\mathbf{C}$ and its category algebra $\Sigma\mathbf{C}$ tells us that from a representation-theoretic viewpoint, it doesn’t much matter which we use. However, if $\mathbf{C}$ has infinitely many objects then $\Sigma\mathbf{C}$ is usually not a unital algebra, and one may view a non-unital algebra as a rather deficient sort of thing. In that case, the thinking goes, it’s better to stick with the original category than pass to the category algebra.

Another way to put it: whenever you see a non-unital algebra (especially an infinite-dimensional one), ask yourself whether it’s the category algebra of some category with infinitely many objects. If it is, you might be better off working with the category rather than the algebra.

I picked up this point of view from a couple of different conversations with algebraists, but I’m not sure I’ve properly understood it. Let me test it out on a couple of examples. One of them kind of “works”, in the sense of corroborating this viewpoint. The other appears not to work at all.

Example  Let $L^1(\mathbb{T})$ be the set of complex-valued integrable functions on the circle $\mathbb{T} = \mathbb{R}/\mathbb{Z}$. It’s a $\mathbb{C}$-algebra under addition and convolution.

This algebra has no multiplicative identity. If it did have one, it would be the Dirac delta function — that is, a function $\delta$ such that $\int_\mathbb{T} f \cdot \delta = f(0)$ for all integrable $f$. But, of course, no such delta function exists. This is what gives Fourier analysis its richness.

So we have before us an infinite-dimensional, non-unital algebra. The viewpoint described above tells us to look for a category of which it’s the category algebra. How can we do this?

Well, if $L^1(\mathbb{T}) = \Sigma\mathbf{C}$ for some category $\mathbf{C}$, then each object of $\mathbf{C}$ gives rise to an idempotent in $L^1(\mathbb{T})$. So we start by looking for the idempotents in $L^1(\mathbb{T})$. Since multiplication in $L^1(\mathbb{T})$ is convolution, this means looking for functions $f: \mathbb{T} \to \mathbb{C}$ such that $f \ast f = f.$ Since taking Fourier coefficients turns convolution into multiplication, this implies that each Fourier coefficient of $f$ is a multiplicative idempotent, that is, $0$ or $1$. Let’s write $e_k: x \mapsto e^{2\pi i k x}$ for the $k$th character of the circle ($k \in \mathbb{Z}$). Then $f = \sum_{k \in S} e_k$ for some $S \subseteq \mathbb{Z}$.

My thinking gets a bit fuzzy around here, but I think one can follow the argument through to show that the objects of $\mathbf{C}$ must be the integers (or if you prefer, the characters of $\mathbb{T}$), and that all the hom-sets are zero apart from an endomorphism ring $\mathbb{C}$ on each object. In other words, $\mathbf{C}$ is the discrete category on $\mathbb{Z}$.

The category algebra of this discrete category $\mathbf{C}$ consists of those double sequences $(c_k)_{k \in \mathbb{Z}}$ that are zero in all but finitely many places. Alternatively, you can think of this as the algebra of all trigonometric polynomials (that is, finite linear combinations of characters $e_k$).

This is not, of course, our original algebra $L^1(\mathbb{T})$. Most integrable functions on $\mathbb{T}$ are not trigonometric polynomials. So, you might say that the viewpoint advocated above has failed. However, perhaps it’s achieved some kind of moral victory. Although not every integrable function is a trigonometric polynomial, the whole theory of Fourier series tells us how, under various hypotheses and in various senses, arbitrary integrable functions can be expressed as limits of trigonometric polynomials. So perhaps it’s a category algebra in some suitably analytic sense.

Here’s another sign that this is a good point of view. When $\mathbf{C}$ is a category with infinitely many objects, we want to say that the identity of $\Sigma\mathbf{C}$ is $\sum_{c \in \mathbf{C}} 1_c$, except that this sum, being infinite, doesn’t exist. In the case of our particular $\mathbf{C}$, this sum is $\sum_{k \in \mathbb{Z}} e_k$, the sum of all the characters of $\mathbb{T}$. On the other hand, if the identity for convolution — the Dirac delta function — existed, then all its Fourier coefficients would be $1$. So this is exactly the nonexistent sum that the Dirac delta wants to be.

Non-example  Here’s another commonly-encountered non-unital algebra, also arising in a soft-analytic context. But this one doesn’t seem to support the viewpoint advocated at all.

Gelfand duality tells us that the commutative not-necessarily-unital $C^\ast$-algebras are dual to the locally compact Hausdorff spaces, with a space $X$ corresponding to the $C^\ast$-algebra $C_0(X)$ of continuous functions $X \to \mathbb{C}$ that vanish at infinity. (The algebra operations are pointwise.) This restricts to a duality between unital commutative $C^\ast$-algebras and compact Hausdorff spaces.

In particular, if $X$ is a Hausdorff space that is locally compact but not compact, then $C_0(X)$ is a commutative algebra without a multiplicative identity. Concretely, the multiplicative identity of $C_0(X)$ would have to be the function with constant value $1$, but this does not vanish at infinity.

Is $C_0(X)$ a category algebra? Apparently not. Again, if it was one, each object of the category would give rise to an idempotent in $C_0(X)$. But in general, $C_0(X)$ has no nontrivial idempotents. For the algebra structure on $C_0(X)$ is pointwise multiplication, so an idempotent in $C_0(X)$ is just a function taking only the values $0$ and $1$; but assuming $X$ is connected, this forces the function to have constant value zero.

Perhaps this failure is somehow due to my ignoring the extra structure on $C_0(X)$. I’ve been treating it as a mere associative algebra, not a $C^\ast$-algebra. But I’m not convinced… can anyone help?

What I’d most like is for someone to explain a bit more the viewpoint that non-unital algebras are often categories in disguise. And some compelling examples would be even better.

Posted at May 14, 2014 5:24 PM UTC

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### Re: Categories vs. Algebras

Concerning C*-algebras: it might be worthwhile to look into groupoid C*-algebras, which are C*-algebras associated to topological groupoids equipped with a suitable system of measures. The duality between C*-algebras and locally compact Hausdorff spaces is then the analogue of your example 3. The simplest non-commutative non-unital C*-algebra is the algebra of compact operators on a Hilbert space $\mathcal{H}$. There is a chance that it arises from the C*-algebra version of your example 4 with the cardinality of $S$ equal to the dimension of $\mathcal{H}$. I can’t say for sure off the top of my head.

One might also want to try and generalize the definition of groupoid C*-algebra to a definition of category C*-algebra by replacing topological groupoids with topological categories. But this is likely not to work, since taking inverses is a necessary ingredient for defining the involution on the groupoid C*-algebra! One could try to associate a category C*-algebra to any topological dagger category, but my feeling is that the C*-condition $||a^\ast a||=||a||^2$ is only going to hold if the dagger actually maps every morphisms to its inverse.

Admittedly, all of this lies in a somewhat different realm: the groupoid or category that one starts with is not assumed to be enriched over vector spaces.

Posted by: Tobias Fritz on May 14, 2014 6:01 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

[As an aside, I would use the terminology slightly differently. I would start with an ordinary category (not a linear category) and say that the category algebra is the algebra with a basis consisting of the morphisms in $C$. In your terminology it would be the category algebra of the free linear category on my category. My use chimes better with the usual notion of group algebra.]

Anyway, let me explain my favourite example of a category algebra (actually a groupoid algebra) that I learnt of from Dan Freed. Take a finite group $G$ and form the action groupoid of $G$ acting on itself by conjugation. This means that the objects are the elements of $G$ and for each pair of elements $g,x\in G$ there is a morphism $g\xrightarrow{x} x g x^{-1}$. We take the category algebra of this and obtain the Drinfeld double of $G$.

My contribution to this was to explicitly describe the twisted Drinfeld double as a twisted groupoid algebra. Once you have that interpretation the representation theory of the twisted Drinfeld double becomes easy to describe.

Posted by: Simon Willerton on May 14, 2014 8:10 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Thanks for the example.

On the terminological point, we’re both being consistent. Most of the time, I (like you) use “category” to mean “unenriched category”. Then, it’s logical to use “category algebra” to mean the algebra constructed from an unenriched category.

In this post, I’m following certain algebraists in routinely using “category” to mean “linear category” (just as certain people routinely use “group” to mean “topological group”). Then, it’s logical to use “category algebra” to mean the algebra constructed from a linear category.

So everyone’s using “category algebra” to mean the algebra constructed from a category. It’s just the meaning of “category” that varies.

Posted by: Tom Leinster on May 14, 2014 11:34 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Tom, I guess I meant a little more than what I actually said about starting with an ordinary category rather than a linear category. In that case you will have much nicer properties because your algebra will have a canonical basis with particularly nice structure constants. I think that all of your examples were of this form weren’t they? Did you have in mind particular examples that weren’t of this form?

Posted by: Simon Willerton on May 20, 2014 6:23 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Yes, I did have in mind some examples of linear categories that aren’t freely generated by ordinary categories. Of course, a trivial example is an associative algebra that isn’t freely generated by a monoid. But there are some other examples I was thinking of, principally the category of indecomposable projective modules over an arbitrary finite-dimensional algebra. I’ll go into this if I get round to writing a sequel to this post.

Posted by: Tom Leinster on May 20, 2014 11:00 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Emily Riehl’s recent series of posts, “An Emerging Pattern in Algebra and Topology” (sorry I don’t know how to include URLs without markup mishaps) explain how linear functors out of the category of finite sets and injections have been studied in topology and homological algebra. But I’ve never seen the corresponding category algebra appear.

Elsewhere, a nice result is that the category of finite sets and partially defined maps is Morita equivalent to the category of finite sets and surjections. You can find this in “Dold-Kan type theorem for $\Gamma$-groups”. The proof involves construction of idempotents, so may be related to the category algebra construction.

Many similar examples turn up when studying the homological algebra of linear functors.

Posted by: James Griffin on May 14, 2014 10:24 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

James wrote: “sorry I don’t know how to include URLs without markup mishaps”.

Posted by: Tom Leinster on May 14, 2014 11:16 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

a nice result is that the category of finite sets and partially defined maps is Morita equivalent to the category of finite sets and surjections.

Let me clarify what I think you mean by that.

Two ordinary, unenriched categories $\mathbf{C}$ and $\mathbf{D}$ are said to be Morita equivalent if the functor categories $[\mathbf{C}, \mathbf{Set}]$ and $[\mathbf{D}, \mathbf{Set}]$ are equivalent.

Let $k$ be a commutative ring. Two $k$-linear, i.e. $k\text{-}\mathbf{Mod}$-enriched, categories $\mathbf{C}$ and $\mathbf{D}$ are said to be Morita equivalent if the linear functor categories $[\mathbf{C}, k\text{-}\mathbf{Mod}]$ and $[\mathbf{D}, k\text{-}\mathbf{Mod}]$ are equivalent.

But I don’t think you mean either of those. The former seems implausible to me. (It would imply that the Cauchy completion of the category of finite sets and partial maps was equivalent to the category of finite sets and surjections. This sounds unlikely, though I’d be interested to be proved wrong.) The latter doesn’t make sense, because the two categories you mention aren’t linear.

What I think you mean is that, writing $\Gamma$ for the category of finite sets and partial maps and $\Omega$ for the category of finite sets and surjections, the ordinary functor categories $[\Gamma, k\text{-}\mathbf{Mod}]$ and $[\Omega, k\text{-}\mathbf{Mod}]$ are equivalent. This follows from Theorem 3.1 and Remark 3.2 of the paper you mention (Teimuraz Pirashvili, Dold-Kan type theorem for $\Gamma$-groups). Right?

I don’t know what the connection to category algebras is, if any. I’m not sure the involvement of idempotents is any indication that category algebras are involved, since idempotents are so intimately connected to Morita equivalence anyway. But it’s definitely an interesting example of Morita equivalence to think about.

Posted by: Tom Leinster on May 14, 2014 11:28 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Apologies for being vague, your guess at what I meant is correct. I would normally use the term (right) $\mathcal{C}$-module for a contravariant functor from $\mathcal{C}$ to an abelian category. Then the result I was looking for is that the category of $\Gamma$-modules is equivalent to the category of $\Omega$-modules. Specifying left instead of right means using covariant functors.

In light of Qiaochu Yuan’s comment below, the process of going from an algebra to a category can be thought of as supposing the algebra is an endomorphism algebra of the generator of some category and then reconstructing the category around it, by finding idempotents. Going from $\Gamma$-modules to $\Omega$-modules also involves finding idempotents in endomorphism algebras.

But I agree fully with your comment that in the context of Morita equivalence, this isn’t much of a coincidence.

Posted by: James Griffin on May 15, 2014 9:48 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

If I’m understanding right, the category algebra of the free category on a directed graph is the very commonly studied path algebra on the graph. (Just to add to your list of standard examples.)

Posted by: Allen Knutson on May 14, 2014 11:17 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Yes, absolutely. My understanding is that anyone who’s anyone these days is generating algebras from quivers all over the place. And of course, you can quotient the path algebra by relations too. I should have mentioned the “path algebra” terminology in the post — that’s important, as you say.

Posted by: Tom Leinster on May 14, 2014 11:30 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

I think the category algebra construction is best understood in the context of Morita equivalence for linear categories. The short story is that 1) formally adjoining finite biproducts to a linear category is a functorial construction, 2) it produces a Morita equivalent linear category, and 3) the category algebra occurs as an endomorphism algebra of a generator of this category.

Posted by: Qiaochu Yuan on May 15, 2014 2:29 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

I also want to point out that we already study representations of categories all the time, only we call them presheaves.

Posted by: Qiaochu Yuan on May 15, 2014 2:30 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Thanks. In (3), do you mean “generator” in the sense of an object $X$ such that $Hom(X, -)$ is faithful? If so, this still doesn’t answer my question 1 (seeking a good categorical account of the category algebra construction), as generators are highly non-unique.

Re your other comment, yes, agreed, there’s a bunch of words meaning more or less the same thing, but with subtly different emphases: representation, module, action, presheaf, ….

Posted by: Tom Leinster on May 15, 2014 2:17 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Here’s my take on the first question. I believe I can explain how category algebras are functorial and I will speculate about the universal property. At the moment I don’t have time to verify this carefully, but if this is true, then it’s only a matter of having enough patience to go through the calculations.

I will follow your convention and say “category” for “$k$-linear category” over a fixed field $k$. The Morita theory suggests that we should be looking at the bicategory of categories. Hence I will denote by $\mathbf{Cat}$ the bicategory of categories, (linear) profunctors and (linear) transformations of profunctors and by $\mathbf{Alg}$ its full subbicategory spanned by the algebras.

The category algebra construction can be enhanced to a bifunctor $\Sigma \colon \mathbf{Cat} \to \mathbf{Alg}$ as follows. If $X \colon \mathbf{C}^{op} \otimes \mathbf{D} \to \mathbf{Vect}$ is a profunctor, then we set

(1)$\Sigma X = \bigoplus_{c \in \mathbf{C}, d \in \mathbf{D}} X(c,d)$

which becomes a $\Sigma\mathbf{C}^{op} \otimes \Sigma\mathbf{D}$-module where we let the elements of $\Sigma\mathbf{C}$ and $\Sigma\mathbf{D}$ act as the elements of $\mathbf{C}$ and $\mathbf{D}$ when this makes sense and by $0$ otherwise. The action on 2-morphisms is clear and this makes $\Sigma$ into a bifunctor, but I’m not sure whether this is an honest bifunctor or only an (op)lax one. If $Y \colon \mathbf{D}^{op} \otimes \mathbf{E} \to \mathbf{Vect}$ is another bifunctor, then I think I can write down natural maps $\Sigma(X \otimes_{\mathbf{D}} Y) \to \Sigma X \otimes_{\Sigma \mathbf{D}} \Sigma Y \to \Sigma(X \otimes_{\mathbf{D}} Y)$ that make $\Sigma(X \otimes_{\mathbf{D}} Y)$ into a retract of $\Sigma X \otimes_{\Sigma \mathbf{D}} \Sigma Y$ but I don’t see whether they are isomorphisms at the moment.

This handles the functoriality and my speculation about the universal property is that $\Sigma$ is (either left or right) adjoint of the inclusion $\mathbf{Alg} \to \mathbf{Cat}$ where the profunctor realizing the Morita equivalence between $\mathbf{C}$ and $\Sigma \mathbf{C}$ should become either the unit or the counit. But I admit that I haven’t tried to see whether this is the case.

Posted by: Karol Szumiło on May 15, 2014 9:49 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Thanks, Karol, that looks very promising. My instinct is that your $\Sigma: \mathbf{Cat} \to \mathbf{Alg}$ does preserve composition (up to coherent isomorphism), but so far I’m no further in the detailed calculations than you.

Posted by: Tom Leinster on May 15, 2014 2:40 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

I think this all works out: we have an adjoint biequivalence of bicategories given in one direction by the inclusion $\mathbf{Alg} \hookrightarrow \mathbf{Cat}$, and in the other by the category algebra construction $\Sigma: \mathbf{Cat} \to \mathbf{Alg}$. (For anyone coming into this conversation partway through, that’s $\mathbf{Cat}$ and $\mathbf{Alg}$ in the sense that Karol defined them, i.e. with bimodules as 1-cells.) The unit and counit, as you say, are the bimodules realizing the Morita equivalence between a category and its category algebra.

The functor $\Sigma$ preserves composition up to isomorphism, just because its quasi-inverse $\mathbf{Alg} \hookrightarrow \mathbf{Cat}$ does. To see that they really are quasi-inverse to each other, you have to check that given any bimodule $X: \mathbf{C} \to \mathbf{D}$ (that should really be an arrow with a cross through it, but I’m too lazy to typeset it), the composite

$\Sigma\mathbf{C} \stackrel{\varepsilon}{\to} \mathbf{C} \stackrel{X}{\to} \mathbf{D} \stackrel{\eta}{\to} \Sigma\mathbf{D}$

is isomorphic to the bimodule $\Sigma X$ that you defined. (Here $\varepsilon$ and $\eta$ are the bimodules realizing the Morita equivalence between a category and its category algebra.) This is easy, I think.

This is all very satisfying, and definitely advances my understanding of the situation. However, it does leave something to be desired. Saying that $\Sigma: \mathbf{Cat} \to \mathbf{Alg}$ is adjoint quasi-inverse to the inclusion $\mathbf{Alg} \hookrightarrow \mathbf{Cat}$ only characterizes $\Sigma$ up to equivalence, which (because of the definition of $\mathbf{Cat}$) is Morita equivalence. So, given a category $\mathbf{C}$, we’ve only characterized $\Sigma\mathbf{C}$ up to Morita equivalence. (And of course, we already knew its Morita equivalence class.) A characterization up to isomorphism would be ideal.

Posted by: Tom Leinster on May 16, 2014 5:27 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

A characterization up to isomorphism would be ideal.

I’m a bit out of my depth here, but I guess that this can be done if you consider $\mathbf{Cat}$ and $\mathbf{Alg}$ as 2-categories with proarrows, where the biequivalence has to preserve the proarrow equipment in a suitable sense. The “ordinary” arrows in $\mathbf{Cat}$ and $\mathbf{Alg}$ should be functors and algebra homomorphisms, respectively, while the proarrows should be profunctors and bimodules.

Posted by: Tobias Fritz on May 16, 2014 5:56 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

But then you’re back to the problem that $\Sigma$ is not functorial on functors.

Posted by: Mike Shulman on May 17, 2014 12:04 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Perhaps this hasn’t been communicated widely so well by algebraists, but the main thrust (as I see it) of categorification/decategorification is that you can often say more about the category than about the algebra. The most common example of this is when the category involved is a category of (finite dimensional) modules over some other algebra. Then things like the fact that dimensions are positive integers have consequences for the corresponding elements in the algebra. Or extensions might correspond to relations. Indeed properties of the “other” algebra can also come into play.

That’s why algebraists are not always satisfied by having some category giving their favourite algebra; they want a realisation of the category in some more concrete terms. In some sense, this says “no” to your Question 2!

But a more “yes”-oriented take on Question 2 is some work pushing out in this direction is the series of papers by Miemietz and Mazorchuk - starting with arXiv:1011.3322 and most recently arXiv:1304.4698.

Finally, a definitely non-trivial example of a non-unital algebra with lots of idempotents instead is Lusztig’s algebra $\dot{\mathbf{U}}$, a modification of the usual quantized enveloping algebra - see Lusztig’s Introduction to Quantum Groups, Part 4.

Posted by: Jan Grabowski on May 15, 2014 12:14 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Thanks for this.

Perhaps this hasn’t been communicated widely so well by algebraists, but the main thrust (as I see it) of categorification/decategorification is that you can often say more about the category than about the algebra.

I don’t know how well that’s been communicated by algebraists, but I would have thought that anyone acquainted with any kind of categorification (in algebra or any other field) would be familiar with the principle that the categorified object is more informative than the decategorified object. That’s why there are usually multiple ways of categorifying the same object: e.g. the category of finite sets, the category of finite sets and bijections, and the category of finite-dimensional vector spaces are all reasonable categorifications of the set of natural numbers.

I take your point about the need for a concrete description of the category. And thanks for the references.

Posted by: Tom Leinster on May 15, 2014 2:27 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Oops - I was thinking ahead, in that I was referring to the particular point about it being desirable for the objects of the categorification to be e.g. modules with e.g. dimensions. Of course Cafe visitors would surely recognise the rather mild claim in my first sentence! That said, they would probably not find anything so profound in the rest either.

Incidentally, what is the category of finite-dimensional vector spaces..? Or less facetiously, “should” one work up to isomorphism or not?

This is actually a live issue in my own work: the habit of casually working up to isomorphism is not necessarily helpful and on occasion can actually obstruct one’s view, since those isomorphisms can contain significant information. But it seems not to exercise the “algebraist on the Clapham omnibus” very much.

Posted by: Jan Grabowski on May 16, 2014 10:21 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Sorry about the technical problems over the last 24 hours or so. All fixed now, I believe, by our benevolent host.

Posted by: Tom Leinster on May 16, 2014 5:28 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Whenever I see something whose behavior depends on the fact that you can only add up finitely many things, I want to try replacing abelian groups with suplattices, in which you can “add up” (i.e. take the join of) infinitely many things. (You have to give up subtraction, but that’s the price you pay.)

So suppose $C$ is a suplattice-enriched category, not necessarily finite. Then $\Sigma C$ is a quantale which as a suplattice is defined by the same formula

$\Sigma C = \bigoplus_{a,b\in C} C(a,b)$

except that now $\bigoplus$ means the suplattice coproduct, which is also the product (even in the infinitary case). It is always a unital quantale, with unit $\bigvee_a 1_a$. And conversely, given a quantale, we can wonder whether it is the category algebra of some suplattice-enriched category.

An interesting example of this is the frame of open sets of a topological space (or a locale), regarded as a quantale with tensor being meet. Every element is idempotent, so the category we get has objects being the opens. I think you can show that it’s the bicategory which Walters used to construct sheaves as enriched categories. Since sheaves can also be regarded as quantale modules, perhaps there is some kind of Morita equivalence going on here too? Although Walters’ construction is not really “modules”.

Posted by: Mike Shulman on May 17, 2014 12:15 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Interestingly, I think the suplattice version is functorial in a certain sense – if $F:C\to D$ is a functor between suplattice-enriched categories, then $\Sigma F : \Sigma C \to \Sigma D$ is an oplax morphism of quantales.

Posted by: Mike Shulman on May 17, 2014 9:30 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Let k be an algebraically closed field. A finite dimensional algebra A over k is basic if its semisimple quotient is a product of copies of k. Every finite dimensional k-algebra is Morita equivalent to a unique basic algebra up to isomorphism. Representation theorist normally restrict to the case A is basic. In this case A has what is called a quiver presentation. This is nothing more than representing A as a category algebra and then presenting the category by generators and relations. Namely you choose a complete orthogonal set E of primitive idempotents and then let C be the full subcategory of the Cauchy completion of A on the objects E. In other words objects of C are primitive idempotents from E. Hom from e to f is the set fAe.

Posted by: Benjamin Steinberg on May 18, 2014 10:00 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Thanks. I think you’re anticipating what’s going to be in the sequel to this post… assuming I get round to writing it.

Posted by: Tom Leinster on May 18, 2014 10:32 PM | Permalink | Reply to this

### Re: Categories vs. Algebras

Hi Tom,

In the late 70s Ross Street had the idea that sheaves were something like categories enriched in a monoidal category, but that the enrichment was seemingly one “without units” (=identities). Bob Walters subsequently came to the realisation that the reason for this is that really one should be enriching in bicategories, rather than monoidal categories.

The situation you are describing is, I think, analogous. Given a linear category $\mathbf{C}$, the algebra $\Sigma \mathbf{C}$ is more than an algebra. Consider the free vector space $F X$ on the set $X$ of objects of $\mathbf{C}$. Like any free vector space, this has a canonical structure of (cocommutative) coalgebra defined on basis vectors by $\Delta(e_x) = e_x \otimes e_x$.

Now $\Sigma \mathbf{C}$ is a left-$F X$-, right-$F X$-bicomodule; the coactions defined on $v \in \mathbf{C}(c,d)$ by $v \mapsto e_c \otimes v \otimes e_d$. Moreover, the units and identities of $\mathbf{C}$ make $\Sigma \mathbf{C}$ into a monoid in the category of $F X$-$F X$-bimodules. The “cotensor product” of comodules is such that the multiplication need only be defined at a pair of composable maps; while the unit is a map not from $k$ but from $F X$. So one may think of the algebra structure of $\Sigma \mathbf{C}$ as being “spread out” over the objects by dint of the bicomodule structure.

These observations are apparently due to Steve Chase. They are explained very well in Marcelo Aguiar’s PhD thesis, in particular Section 2.4 and Chapter 9.

It seems that the interesting point is that, in the finite dimensional case, one can extend the algebra structure in bicomodules to a genuine algebra structure back in vector spaces, and moreover with the Morita equivalence you mention.

Apropos of this, and following up on some of the previous posts, Steve and Ross have been thinking about such things recently; here’s a bibliography of some of the papers they have turned up dealing with these kinds of Morita equivalences:

Teimuraz Pirashvili, Dold-Kan type theorem for $\Gamma$-groups

J. Słomińska, Dold-Kan type theorems and Morita equivalences of functor categories

Randall Helmstutler, Model category extensions of the Pirashvili-Słomińska theorems

Steve Lack and Ross Street, Combinatorial categorical equivalences

Nicholas Kuhn, Generic representation theory of finite fields in nondescribing characteristic

Posted by: Richard Garner on May 24, 2014 6:55 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Also, I don’t know if you have this, but an original reference for the ring bizzo seems to be:

Barry Mitchell, Rings with several objects, Advances in Mathematics, Volume 8, Issue 1, February 1972, Pages 1-161

Section 7 defines the ring associated to a linear category, and Theorem 7.1 states the Morita equivalence you are interested in.

Also, there’s this in TAC, dealing with the Sup enriched case:

Bachuki Mesablishvili, Every small Sl-enriched category is Morita equivalent to an Sl-monoid, Theory and Applications of Categories, Vol. 13, 2004, No. 11, pp 169-171.

and there’s an exegesis of the construction right at the end of

Heymans and Stubbe, “Grothendieck quantales for allegories of enriched categories”, Bulletin of the Belgian Mathematical Society 19, 861-890.

Posted by: Richard Garner on May 24, 2014 7:28 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Another thought –

Given any base for enrichment $\mathcal{V}$, and any small $\mathcal{V}$-category $\mathcal{C}$, you can look at the coproduct $X$ of all the representables in $[\mathcal{C}^{\mathrm{op}}, \mathcal{V}]$. The hom-object $M_{\mathcal{C}} = [\mathcal{C}^{\mathrm{op}}, \mathcal{V}](X,X)$ is, of course, a monoid in $\mathcal{V}$. In the case of a finite linear category, this is the “category algebra” you describe. For an ordinary $\mathbf{Set}$-category, this is the “monoid of bisections”, an element of which assigns to every object of $\mathcal{C}$ an arrow out of that object (which is something that Lie groupoid people think about.)

In any case, this monoid $M_{\mathcal{C}}$ is a one object sub-$\mathcal{V}$-category of $[\mathcal{C}^{\mathrm{op}}, \mathcal{V}]$. So applying Kan’s construction, one obtains an adjunction between $[\mathcal{C}^{\mathrm{op}}, \mathcal{V}]$ and $[M_{\mathcal{C}}^{\mathrm{op}}, \mathcal{V}]$. In the case of a finite linear category, this is an equivalence. In general, it seems that it is far from an equivalence, but it seems like it might be an interesting object to study in any case.

Posted by: Richard Garner on May 25, 2014 1:39 AM | Permalink | Reply to this

### Re: Categories vs. Algebras

Tom,

Here is an answer to your second question. A ring $R$ is said to have local units if it is the directed union of unital subrings $eRe$ with $e$ an idempotent of $R$.

A right $R$-module $M$ is called unitary if $MR=M$. Equivalently, $M$ is unitary iff $M$ is the directed union of the additive subgroups $Me$ (which are in fact $eRe$-modules).

If $C$ is a category (enriched over vector spaces), then $\Sigma C$ has local units. Indeed, $C$ is the filtered colimit of its full subcategories on finite subsets of the object set and so $\Sigma C$ is the directed union of the corresponding unital subrings.

Now I claim that the proof you outlined in the unital case shows that the category of unitary $\Sigma C$-modules is equivalent to the category of representations of $C$.

There is a satisfactory theory of Morita equivalence for rings with local units and analogues of the usual theorems hold. See the papers of Abrams and of Anh and Marki.

In particular, it follows that $C$ and $D$ are Morita equivalent iff $\Sigma C$ and $\Sigma D$ are Morita equivalent (in the local units sense). Also the functoriality on the level of bicategories should work here since cocontinuous functors can be represented by bimodules in the local units context.

Posted by: Benjamin Steinberg on May 31, 2014 2:13 PM | Permalink | Reply to this

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