Lectures on Representations of Quivers by William Crawley-Boevey ...

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EXAMPLES. (1) If Q consists of one vertex and one loop, then kQ k[T]. If Q has one vertex and r loops, then kQ is the free associative algebra on r letters. (2) If there is at most one path between any two points, then kQ can be identified with the subalgebra {C ¡ M (k) ¢ C =0 if no path from j to i} n ij of M (k). If Q is 1 ¡ 2 ¡ ... ¡ n this is the lower triangular matrices. n (i£ IDEMPOTENTS. Set A=kQ. (1) The e i are orthogonal idempotents, ie e e i j 2 = 0 j), e i = e . i n (2) A has an identity given by 1 = ¤ e . i=1 i (3) The spaces Ae , e A, and e Ae have as bases the paths starting at i i j j i and/or terminating at j. n (4) A = ¥ Ae , so each Ae is a projective left A-module. i=1 i i (5) If X is a left A-module, then Hom (Ae ,X) e X. A i i (6) If 0£ f¡ Ae and 0£ g¡ e A then fg£ 0. i i PROOF. Look at the longest paths x,y involved in f,g. In the product fg the coefficient of xy cannot be zero. (7) The e are primitive idempotents, ie Ae is a indecomposable module. i i 2 PROOF. If End (Ae ) e Ae contains idempotent f, then f =f=fe , so A i i i i f(e -f)=0. Now use (6). i (8) If e ¡ Ae A then i=j. i j PROOF. Ae A has as basis the paths passing through the vertex j. j (9) The e are inequivalent, ie Ae ¦ Ae for i£ j. i i j PROOF. Thanks to (5), inverse isomorphisms give elements f¡ e Ae , g¡ e Ae i j j i with fg=e and gf=e . This contradicts (8). i j 4
PROPERTIES OF PATH ALGEBRAS. These are exercises, but some are rather testing. (1) A is finite dimensional Q has no oriented cycles. (2) A is prime (ie IJ£ 0 for two-sided ideals I,J£ 0) ¡ i,j ¢ path i to j. (3) A is left (right) noetherian if there is an oriented cycle through i, then only one arrow starts (terminates) at i. (4) rad A has basis {paths i to j ¢ there is no path from j to i}. (5) The centre of A is k£ k£ ...£ k[T]£ k[T]£ ..., with one factor for each connected component C of Q, and that factor is k[T] C is an oriented cycle. REPRESENTATIONS. We define a category Rep(Q) of representations of Q as follows. A representation X of Q is given by a vector space X for each i¡ Q and a i 0 linear map X :X ¡ X for each ¢ ¡ Q . ¢ s(¢ ) t(¢ ) 1 A morphism ¤ :X ¡ X¥ is given by linear maps ¤ :X ¡ X¥ for each i¡ Q i i i 0 satisfying X¥¦¤ = ¤ X for each ¢ ¡ Q . ¢ s(¢ ) t(¢ ) ¢ 1 The composition of ¤ with § :X¥ ¡ X¨ is given by (§©¦¤ ) = §©¦¤ . i i i EXAMPLE. Let S(i) be the representation with § k (j=i) S(i) = ¨© j ¢ ¡ ¢ 0 (else) S(i) =0 (all Q ). 1 EXERCISE. It is very easy to compute with representations. For example let Q be the quiver •¥¦ • ¡ •, and let X and Y be the representations 1 1 k¥¦ 1 ¡ k k¥¦ k ¡ k 0. Show that Hom(X,Y) is one-dimensional, and that Hom(Y,X)=0. 5
Page 1 and 2: Lectures on Repres
Page 3: §1. Path algebras Once and for all
Page 7 and 8: THE STANDARD RESOLUTION. ¡ ¥ ¡
Page 9 and 10: §2. Bricks In this section we cons
Page 11 and 12: §3. The variety of representations
Page 13 and 14: 1 LEMMA 1. dim Rep(¤ ) - dim O = d
Page 15 and 16: §4. Dynkin and Euclidean diagrams
Page 17 and 18: PROOF. (2) By inspection the given
Page 19 and 20: §5. Finite representation type In
Page 21 and 22: §6. More homological algebra FROM
Page 23 and 24: LEMMA 3. and - give inverse bijecti
Page 25 and 26: §7. Euclidean case. Preprojectives
Page 27 and 28: LEMMA 4. If ¤ is a positive real r
Page 29 and 30: PROPERTIES. Let X be regular simple
Page 31 and 32: DEFINITION. Given a -orbit of regul
Page 33 and 34: LEMMA 2. If X is regular, X £ 0 th
Page 35 and 36: REMARKS. ~ (1) For the Kronecker qu
Page 37 and 38: (4) Auslander-Reiten theory for wil

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