# Sorting out a homological computation concerning Hall algebras

Let me recall some basic definitions concerning Hall algebras given in for example Schiffmann's notes, Hubery's notes or even the excellent master thesis by Sjoerd Beentjes which you can find online.

Troughout let $$\mathcal{A}$$ be a magical abelian category that satisfies a bunch of conditions such that one can actually define the Hall algebra $$\mathbb{H}(\mathcal{A})$$ of this category. As a complex vector space $$\mathbb{H}(\mathcal{A})$$ is the complex vector space whose basis is given by the isomorphism classes of objects in $$\mathcal{A}$$ (denoted by $$\text{Iso}(\mathcal{A})$$ and the class of an object $$X$$ is denoted by $$[X]$$).

Before I can tell you what the product on $$\mathbb{H}(\mathcal{A})$$ is, we need the following definition:

Let $$E,M,N\in \mathcal{A}$$, we denote the set of short exact sequences of the form $$0\to N \to E\to M\to 0$$ by $$\mathcal{P}^E_{M,N}$$. The number of elements in this set is denoted by $$P^E_{M,N}$$. (The magical properties make sure this is a finite number). For whatever reason we are actually interested in the numbers $$F^E_{M,N}=\frac{P^E_{M,N}}{a_Ma_N}$$ where $$a_X:=|\text{Aut}(X)|$$ is again magically finite. Remark that $$M$$ appears first in $$P^E_{M,N}$$ but is a quotient object of $$E$$ and appears on the right in the short exact sequence. This will be important later.

Good now we got that, let's go on to define some useful product on the Hall algebra.

The product on $$\mathbb{H}(\mathcal{A})$$ is defined by $$[M]*[N]=\sum_{[E]\in \text{Iso}(\mathcal{A})} F^E_{M,N}[E].$$

I'm aware this is not the standard product and that usually one twists the multiplication using the Euler form. However it should become apparent that this is not important for this question.

Alright, let's compute an interesting example.

Consider the quiver $$A_2$$ which looks like this $$\require{AMScd} \begin{CD} 1 @>{}>> 2\\ \end{CD}$$ Let $$\mathcal{A}$$ be the magical abelian category of $$k$$-linear quiver representations (here $$k=\mathbb{F}_q$$ is a finite field otherwise the magic doesn't really work). The category $$\mathcal{A}$$ is a Krull-Schmidt category and the only indecomposable $$A_2$$ representations are given by $$S_1,S_2$$ and $$I_{12}$$. Here $$S_i$$ is the simple representation corresponding to the vertex $$i$$ and $$I_{12}$$ looks like $$\require{AMScd} \begin{CD} k @>{1}>> k.\\ \end{CD}$$ The following three short exact sequences are the only short exact sequences beginning and ending on different simple objects: $$0\to S_1\to S_1\oplus S_2\to S_2\to 0,$$ $$0\to S_1\to I_{12}\to S_2\to 0,$$ $$0\to S_2\to S_2\oplus S_1\to S_1\to 0.$$ Thus $$[S_1]* [S_2]= F^{S_2\oplus S_1}_{S_1,S_2} [S_2\oplus S_1]$$ and $$[S_2]*[S_1]=F^{S_1\oplus S_2}_{S_2,S_1}[S_1\oplus S_2]+F^{I_{12}}_{S_2,S_1} [I_{12}].$$

Did I convince you? Every other source I can find tells me it's other way around, i.e. the formulas for $$[S_1]*[S_2]$$ and $$[S_2]*[S_1]$$ are swapped. Where am I wrong?

See for example page 47 here, page 37 and 38 here and page 29 and 30 here.

• I do not know enough to help you, but the title gave me a chuckle so thanks for that. – Nethesis Feb 22 at 14:52
• The question in the title does not appear to be about math within the scope defined in the help center. – Babelfish Feb 22 at 15:00
• Definition: a category $\mathscr{A}$ is completely mental if ... [what he said]. :) – Calum Gilhooley Feb 22 at 15:12
• Please make your title useful and informative. As it stands it is slightly humorous but completely useless to organize the site. Regards, – Pedro Tamaroff Feb 22 at 15:17
• Sure, thanks. I can't count how many times I've done the same sort of thing. – David Hill Feb 22 at 16:46

In the convention used in the linked papers, the module $$S_2$$ is projective and the module $$I_{12}$$ appears in the exact sequence $$0\to S_2\to I_{12}\to S_1\to 0$$ which is opposite yours.