Matthias ? - Quivers and Algebras

Abstract

Talk

Outline

$k$ algebraically closed field, $A$ finite dimensional Algebra over $k$. Then there exists a “combinatorial description (kind of)” of $A$ by a quiver.

Given a Quiver, we can define two things:

• A Quiver representation, which can be decomposed into indecomposable set of subrepresentations.
• Path algebra of $Q$: $mod{A}_Q$

Those two things will be isomorphic. This means there is a decomposition oof path algebras as well.

Quiver representations

Definition

A quiver $Q=(Q_1,Q_2,s,t)$. Is a set of vertices, of arrows and a start and end function. Connecting arrows to vertices.

You chose a good example, highlighting every part of the diagram.

Example

We draw a graph

Definition

Let $k$ be a field as in the outline. A representation is a collection of $k$-vector spaces $M_i$ and $k$-linear maps. The dimension vector is the vector containing the dimension of all vector spaces?

What is the connection between a group representation and a quiver representation? Why do they share the same name? Group representations can be transformed into a group representation by taking the Regular representation and putting looping arrows from itself to itself for every group element acting on the space.

Example

We draw the same diagram again, but replacing the vertices with vector spaces. (Let the students see the trivial connection themselves). This is the representation of $Q$

Definition

A morphism is a collection of maps, mapping every vector space to a vector space, such that applying the morphism and then the linear maps commute. A morphism is an isomorphism if it is bijective.

The above gives us the category $\text{rep}Q$ of representations.

Proposition

$repQ$ is a $k$-category, meaning the morphisms give us a $k$-vector space.

Definition

If we have a morphism $f$ between two representation, we define the kernel $kerf$, the image $imf$ and the cokernel $cokerf$.

I don’t quite get the definition, so I just blackbox it.

Definition

$L$ is a subrepresentation if there is a injective morphism $i:L\to M$

Definition

Given two $Q$-representations, we define the direct sum by doing the direct sum on all vector spaces and taking the direct sum linear maps.

Theorem Homomorphism theorem

For a morphism $f$ we have $imf\cong M/kerf$

Proposition

The direct sum makes $repQ$ an abelian $k$-category. (We gave the conditions for this)

This talk is very quick.

The decomposition of An

Definition

$M\in repQ$ is indecomposable if it does not decompose into two non-trivial representations.

This definition is different from a Irreducible quiver. For example the action $\mathbb{Z}\to {\mathbb{R}}^2$ by shearing is indecomposable but not irreducible because the shearing can’t be described by a product but also doesn’t act invariant on two spaces. This means we have indecomposable $⟹$ irreducible. I think I haven’t understood the example. :(

Theorem (Krull-Remak-Schmidt)

Let $M$ be a $repQ$. Then $M$ decomposes uniquely into a finite sum of indecomposable representations $M_i$. (up to order)

Let $A_n$ be a graph having the shape of a line with $n$ vertices. Compare to Luis Paris - Artin groups of type Dn

Theorem

Each indecomposable representation of $A_n$ is isomorphic to $[j,i]:=0\to ...\to 0\to \underset{j}{k}\stackrel{id}{\to }k\to ...\to \underset{i}{k}\to 0\to ...\to 0$

Proof: We look at the longest left sequence of linear maps up to ${\phi }_{{\alpha }_i}$ which is injective. Then we look at the longest right sequence of maps which is surjective. By some induction process, we show that the left and the right must be zero and that the middle is the identity.

Lemma

The morphisms from $[j,i]$ to $[j^{\prime },i^{\prime }]$ are isomorphic to

• $k$ if $j^{\prime }\le j\le i^{\prime }\le i$
• $0$ else

Proof: If the non-trivial parts do not match, then it is impossible to define a morphism.

Definition

A non-zero morphism between two indecomposable represenations is called irreducible if it cannot be written as a sum of compositions where each composition consists of two non-isomorphisms.

Definition

For a given Q-representation, the Auslander-Reiter-Quiver is a quiver with vertices the indecomposable representations and arrows the irreducible arrows. We drew this in a way that the bottom row are exactly the irreducible representation

There is only a small set of quivers with finitely many indecomposables

Path algebras

Definition

Let $Q$ be a quiver. $w$ be a path of length $l$ from some vertex $i$ to another vertex $j$ an $l+2$ tuple $w=(j|{\alpha }_l,...,{\alpha }_1|i)$. The lenght zero path $(i|i)$ is called the identity path and similarly we define a cycle and composition as intuitive.

Definition

We define the path category $kQ$ where the objects are the vertices of $Q$ and the morphisms are the paths. (or the space freely generated by the paths??)

Remark: $repQ$ can be identified with the functor space $Fun(kQ,vect)$. (where we identify the paths with linear maps obtained from concatenating the linear maps along the paths)

Definition (Path algebra)

The path algebra $A_Q$ of $Q$ is the linear space freely generated of all paths. (for each endpoints $i,j$)

We did some examples on the quiver sonsisting of a loop and the quiver $A_n$, which I didn’t really get.

Theorem

Let $Q$ be a finite quiver. Then the isomrorphism $mod{A}_Q\cong repQ$ holds.

• What are the $mod(A_Q)$? Something about “cot left $A_Q$-modules”?

Gabriel's Theorem 1

$A$ finite dim algebra over an algebraically closed field (important!). Then there exists a finite quiver $Q$ and an admissible ideal $I$ s.t. $modA\cong mod{A}_Q/I$ So algebras can be classified by quivers.

• Is this the same construction as Bass-Serre-Theory and of Artin groups? Can we appropriate some concepts?