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For a while I had been thinking that the path algebra of a quiver $Q$ over a commutative ring $R$ is the same as the "category ring" $R[P]$ (analogous to "group ring", "monoid ring", "semigroup ring", and the like), where $P$ consists of paths in $Q$, and multiplication in $P$ is composition (or zero when the domain doesn't match the codomain).

However, as I have been writing something in more detail, I find that it's not always possible to find a corresponding quiver for some given category. For example, when $C$ is the thin category representing the partial order of $\mathbb R$ (objects are real numbers, and morphisms are pairs $(x, y)$ with $x \le y$), I cannot find the corresponding quiver.

My questions are

  1. Am I just not aware of the quiver that will give rise to the category in question?
  2. If it is really the case that there are no corresponding quivers for some categories, then this "category ring" is a more general object that the path algebra. Should it still be called the path algebra?
  3. Why do people usually start with a quiver, then make it into a category to define the path algebra? Why not start with a category?

Edit: Thank you for the answers below by Aaron and Julian. So the answer to my question number 1 is that I was not aware of the quiver algebra with relations. Now that I am, I have a follow-up question. Is the path algebra with relations the same as the category ring when $R$ is a field? (Why would one want to consider the quotient over a two-sided ideal that does not come from identifying paths anyway?)

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Could you explain your question in brackets a bit more? What else would you want to consider? –  Julian Kuelshammer May 8 '13 at 11:00

2 Answers 2

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  1. I think (but might be false because I'm not that well-informed when it comes to infinite-dimensional algebras, it is certainly true if you replace $\mathbb{R}$ by an appropriate finite set) that in this case there is no quiver $Q$ such that $R[(\mathbb{R},\leq)]\cong RQ$. Instead you will have that $R[(\mathbb{R},\leq)]\cong RQ/I$, where $Q$ is the quiver with arrows $a\to b$ whenever $a\leq b$ and $I$ is the ideal spanned by all $\alpha-\gamma\beta$, where $\alpha$ is an arrow representing $a\leq c$, $\beta$ is an arrow representing $a\leq b$ and $\gamma$ is an arrow representing $b\leq c$.
    In the finite-dimensional world over algebraically-closed fields the path algebras are in fact only the hereditary algebras (i.e. the algebras such that $\operatorname{Ext}^2$ vanishes) up to Morita equivalence.
  2. I think that this should indeed then not be called a path algebra but maybe a path algebra with relations if you don't want to use category ring.
  3. To give a quiver (with relations) is giving much more information than giving just a category. In some sense the quivers satisfy some minimality assumptions. An example: The number of arrows between two vertices in a given quiver is the $\operatorname{Ext}^1$ between the corresponding simple modules. The $\operatorname{Ext}^2$ describes the number of relations needed in a minimal set of relations on the quiver.
  4. For your question whether a category algebra is the same as a quiver with relations (I'm only talking about the case of a finite quiver.) That depends on what equivalence relation you are imposing. If you want to use "up to isomorphism" then the answer is "No". For example consider the category $\mathcal{M}_2$ with two isomorphic objects (and no other morphisms except the two isomorphisms and the two identities). Then it is easy to check $k[\mathcal{M}_2]\cong M_2(k)$ (the ring of $2\times 2$-matices over $k$.) This is not isomorphic to any path algebra of a quiver since its module category is semisimple (all representations are finite sums of simple ones) but the only path algebras which are semisimple are $k^n$ (as the quiver take $n$ vertices and no arrows). And these two algebras are never isomorphic. That's why I wrote in 1. that you have to consider them up to Morita equivalence. Two algebras are called Morita equivalent if their module categories are equivalent. By the theorem of Gabriel there is exactly one path algebra in each equivalence class of finite dimensional algebras. This is exactly the basic algebra (meaning that the finite dimensional simple modules are $1$-dimensional).
  5. One other reason to consider quivers first is maybe educational. It is quite easy to speak about representations of quivers without mentioning a category (although implicitely you are dealing with one). So in principle a second year student can take such a course with only knowledge of linear algebra (and not abstract algebra or category theory).
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(1) I wasn't aware of quiver with relations. Thank you. (2) So you think "category ring" is quite a reasonable term? I'm glad :) I might actually use it. (3) That does not quite answer my question. It's like when we study groups, we define "free groups" after we know what groups are. The situation for quivers and categories is a bit reversed, and I don't know of any good reasons except probably historical ones (which I also don't know of right now). –  Tunococ May 8 '13 at 10:48
    
@Tunococ Tried to answer your expanded question. –  Julian Kuelshammer May 8 '13 at 11:12
    
I fail to understand what $k[\mathcal M_2]$ is. From my understanding, the category you mention has two "generators", say $a:A \to B$ and $b:B \to A$, where $A$ and $B$ are the two objects. The category ring is, as a $k$-module, freely generated by $a, b, A$ and $B$, and $a: A \to B$ and $b: B \to A$ are the two isomorphisms. Allowing $A$ and $B$ to denote their identities, multiplication is defined by $aa = bb = AB = BA = aB = bA = Aa = Bb = 0$, $aA = a = Ba$, $bB = b = Ab$, $ab = B$ and $ba = A$. I believe these equations can be used to define relations for the corresponding path algebra. –  Tunococ May 9 '13 at 19:33
    
Ah, but for a quiver with relations you ask that the relations are of "length" $\geq 2$. –  Julian Kuelshammer May 9 '13 at 21:11
    
Oh. Now I know what you're talking about. I misunderstood the definition of a quiver with relations (again). And now I know that the "category ring" is in fact more general than the path algebra with relations. –  Tunococ May 9 '13 at 21:59

That's a trap I used to fall into as well. This is the example that you should always always always keep in mind:

Let $G$ be a finite group, then construct a category with one object, and all arrows correspond to elements of $G$ with composition of arrows being group element multiplication, then the category algebra is a group algebra. This is the reason why people, such as Peter Webb and Markus Linklemann, are interested in category algebras as they are generalisation of group algebras. On the other hand, given a group, it is extremely difficult to construct the corresponding path algebra (basic algebra).

Note the "corresponding path algebra" always exists in the following sense (a result of Gabriel): Let $A$ be a finite dimensional algebra with direct sum decomposition $ A= \bigoplus_{i=1}^n P_i^{\oplus m_i}$ (i.e. $P_i$ are indecomposable projective) and take $M=\bigoplus_{i=1}^n P_i$, then the path algebra $kQ/I$ corresponding to $A$ (constructed using Ext-quiver) is isomorphic to $End_A(M)$.

So to answer your question (1) and (2): Given a path algebra $A=kQ/I$, you can construct a category which capture homological behaviour of $kQ/I$ is simply given by treating $Q$ as a category, and relations of compositions given by $I$. Here each object (=vertex in $Q$) in the category has ONLY ONE invertible endomorphism corresponding to the primitive idempotent of that vertex. In this way you can say category algebra is a generalisation of path algebra.

Concerning to your third question, I can't give a very clear answer, these are what my supervisor told me years ago, so I hope someone can give some supplementary comments: There are various reason that we go from quiver to category. Sometimes because we do not want to deal with the algebra directly. (for example, V. Mazorchuk (and possibly Y. Drozd) is one of the people who uses this approach very often.) A better reason is that "path algebra of an infinite quiver" is (was?) an unclear concept, as you don't have things like Gabriel's theorem/lemma as mentioned above, and you need to deal with certain issues arising from infinite dimensional algebras. On the other hand, there is no obvious obstruction when we look at a category with infinite objects.

So why do we still favour path algebras? The reason is most techniques that work on path algebras do not carry over to category algebras. When you have a path algebra, most homological work becomes combinatorial. Yet if you have a category algebra, it is still very difficult to see what the projective indecomposables are (remember the group algebra!).

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I like your answer very much, Aaron (+1). I would probably just add that, for me at least (and presumably other people who sometimes do representation theory), the problem of studying the representations of a naturally occurring algebra is considered "solved" once one has written down its AR quiver. Sometimes, it is even considered "solved" once one has written down its ordinary quiver with relations, but as you write, what this has really done is "reduced" everything to combinatorics (which may still be quite interesting and non-trivial). –  S123 May 17 '13 at 17:08

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