1. Linear algebra is fussy but super useful. 2. Instead of studying an algebra A all at once, study its representations/modules!

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1 BIG LESSONS SO FAR 1. Linear algebra is fuss but super useful. 2. Instead of studing an algebra A all at once, stud its representations/modules! (If ou re luck, A is semisimple, and is thus the direct product of endomorphism rings corresponding to simple A-modules.) 3. Lots of modules can often be found inside of A b considering the left regular action. 4. Decomposing modules, even when possible, can be hard. But computing eigenvectors is eas. So ou re reall happ when ou can use eigenvectors to compute our decompositions for ou!! Find 1-dimensional submodule b finding vectors v that satisf Av Fv. Find a big commutative subalgebra C Ñ A whose action determined the module for ou. A (complex) Lie algebra is a vector space g over C with a bracket r, s : g b g Ñ g satisfing (a) (skew smmetr) rx, s r, xs, and (b) (Jacobi identit) rx, r, zss ` r, rz,xss ` rz,rx, ss 0, for all x,, z P g. Ver favorite example: sl n pcq. We have gl n pcq M n pcq with bracket rx, s x x, and sl n pcq tx P gl n pcq trpxq 0u. In particular, sl 2 pcq has basis ˆ0 1 ˆ0 0 x with relations and h ˆ rh, xs 2x, rh, s 2, and rx, s h. Abstractl, we can instead define sl 2 as the Lie algebra generated b x,, and h, with these relations. Specificall, we call ( ) the standard representation. ( )

2 Standard and adjoint representations A representation of a Lie algebra is a vector space V together with a Lie algebra homomorphism : g Ñ EndpV q satisfing prx, sq pxq pq pq pxq. Our definitions of sl n, so n,andsp n we actuall the standard representations (the re faithful, though, so that s ok). The adjoint representation of a Lie algebra g is ad : g Ñ Endpgq x fiñ ad x rx, s, i.e. ad x pq rx, s. Algebras Ø Lie algebras Let A be an algebra over F. Then let LpAq be the Lie algebra with Vector space: A Bracket: rx, s x x. Let g be a complex Lie algebra. Then let Ug be the algebra with Vector space: Ct finite words in g-basisu Multiplication: satisfies relation x x rx, s U g is called the universal enveloping algebra of g. Example: Compute Usl 2. Thm. A representation of a Lie algebra g is equivalent to a representation of the associated universal enveloping algebra U g. (Analogous to the relationship between a group and its group algebra.) Fact: If g is the finite direct sum of simple Lie algebras, then it is semisimple, i.e. all g-modules are completel decomposable (Artin-Wedderburn).

3 Representations of sl 2 pcq Let M be a finite-dimensional simple sl 2 pcq-module, and let be the associated representation. Step 1: phq has at least one eigenvector v P M, i.e.hv some P C. Step 2: Use hx xh `rh, xs to show that x`v is also an eigenvector for h with weight ` 2` for each ` P Z 0. Step 3: Since the eigenvalues of h on the x`v s are distinct, the x`v s are distinct. Since M is finite dimensional, there must be a non-zero v` P M with xv` 0 and hv` µv` for some µ P C. v for Step 4: Similarl as with x, useh h `rh, s to show that `v` is also an eigenvector for h with weight µ 2` for each ` P Z 0. Step 5: Again, since M is finite-dimensional, there must be some d P Z 0 with d v` 0 and d`1 v` 0. 0 x h v` h v` 2 v` h d 1 v` h d v` 0 In summar, the sl 2 -module has basis `v`, for` 0,...,d;and with respect to that basis, phq is a diagonal matrix with µ, µ 2, µ 4,...,µ 2d on the diagonal, pq has 1 s on the sub-diagonal and zeros elsewhere, and pxq has the weights µ, 2µ 2, 3µ 6,...,dpµ pd 1qq on the super-diagonal µ h µ 2 µ 4... µ 2d 0 µ 0 2µ 2 x 0 3µ 6... dpµ pd 1qq 0 Further, using the relation h rx, s x x, we get µ 2d 0 dpµ pd 1qq. So µ d dimpmq 1.

4 Theorem The simple finite dimensional sl 2 modules Lpdq are indexed b d P Z 0 with basis tv`,v`, 2 v`,..., d v`u and action hp`v`q pd 2`qp`v`q, xp`v`q `pd ` 1 `qp` 1 v`q, p`v`q ``1 v`, with d`1 v` 0. with xv` 0 and Example: Let M be a (not nec. simple) sl 2 -module. As a Ch-module, M à µpc M µ where M µ tm P M hm µmu, is the µ weight space. Suppose h has weight spaces of the following dimensions: µ i i dimpm µ q More modules Trivial modules. What is our favorite trivial module? Group algebras: The trivial module of FG is V F, with action given b g. Lie algebras: The trivial module of Ug is V F, with action given b x 0. Tensor products. For an algebra A and A-modules V and W, when is V b W naturall an A-module? Group algebras: FG acts on V b W b g pv b wq pg vqbpg wq. Lie algebras: Ug acts on V b W b x pv b wq px vqbw ` v bpx wq. (You ll show this is well-defined on the homework.)

5 More modules Duals. Recall, the dual avectorspacev is V HompV,Fq. If V is an A-module, when is V also an A-module? Group algebras: FG acts on V b Check: pg 'qpvq 'pg 1 vq for all g P G, v P V. pg 1 pg 2 'qqpvq pg 2 'qpg 1 1 vq 'pg 1 2 g 1 1 vq 'ppg 1 g q 1 2 vq ppg 1 g 2 q 'qpvq.x Lie algebras: Ug acts on V b px 'qpvq 'p xvq for all x P g,v P V, and extended canonicall, e.g. Check: ppxq 'qpvq px p 'qqpvq p 'qp xvq 'ppxqvq. ppx xq 'qpvq ppxq 'qpvq ppxq 'qpvq 'ppxqvq 'ppxqvq 'ppx xqvq 'p rx, svq prx, s 'qpvq.x More modules Duals. Recall, the dual avectorspacev is V HompV,Fq. Group algebras: FG acts on V b pg 'qpvq 'pg 1 vq for all g P G, v P V. Lie algebras: Ug acts on V b px 'qpvq 'p xvq for all x P g,v P V, and extended canonicall, e.g. ppxq 'qpvq px p 'qqpvq p 'qp xvq 'ppxqvq. If V is an A-module, there are two canonical momorphisms: X : V b V Ñ F Y : F Ñ V b V and p',vq fiñ 'pvq 1 fiñ i v i b v i where the sum is over an basis tv i u of V and tv i u is the dual basis in V.

6 A Hopf algebra is an algebra A (over F )withthreemaps : A Ñ A b A, " : A Ñ C, and S : A Ñ A such that (1) If M and N are A-modules, then M b N with action xpm b nq ÿ x p1q m b x p2q n x where pxq ÿ x x p1q b x p2q,isaa-module. [This is called Sweedler notation] (2) The vector space F, with actions x "pxq is an A-module. (3) If M is a A-module then M HompM,Fq with action is an A-module. px'qpmq 'pspxqmq (4) The maps Y and X are A-module homomorphisms. In general, we call a bilinear form invariant if xxm, n xm, Spxqn. Example On the homework, ou will compute the s 2 pcq modules Lp1qbLp1q and Lp1q b3. In general, ou can show that for an d 0, LpdqbLp1q Lpd ` 1q Lpd 1q. So the dimension of Lpaq in Lp1q bk is given b the number of downward-moving paths from Lp1q on level on, to Lpaq on level k in the lattice k 1: Lp1q k 2: Lp0q Lp2q k 3: Lp1q Lp3q k 4: Lp0q Lp2q Lp4q k 5:. Lp1q Lp3q Lp5q..

7 Forms Let U be a Hopf algebra (over C) withmodulem. A bilinear form is a map x, : M b M Ñ C. Abilinearformissmmetric if xm, n xn, m for all x, P M. Asmmetricbilinearformisinvariant if xxm, n xm, Spxqn for all x P U. A form is nondegenerate if xx, U 0 for all x P U. Recall, we defined ad x P Endpgq b ad x pq rx, s. The canonical nondegenerate, smmetric, invariant bilinear (NIBS) form on g is the Killing form, given b xx, K trpad x ad q. Good news: If g is simple, then ever NIBS form is a constant multiple of the Killing form. In particular, the trace form is often easier to compute: let : g Ñ M n pcq be the standard representation, and set xx, trp pxq pqq. A Cartan subalgebra of g is a maximal abelian subalgebra h Ä g consisting of semisimple (diagonalizeable) elements. (Analogous to Jucs-Murph elements!!) Some facts about Cartans: 1. Cartan subalgebras are generated b taking a (nice) semisimple element h and setting h tg P g ad h pgq 0u 2. Cartan subalgebras exist and are unique up to inner automorphisms. 3. The centralizer of h is h. 4. All elements of h are simultaneousl diagonalizeable. 5. The restriction of the Killing form to h is non-degenerate. The rank of a semisimple Lie algebra is defined b rankpgq dimphq.

8 The weights of a Cartan h is the dual set h tµ : h Ñ Cu. Let g tg P g ad h pxq phqxu. The set of weights R t P h 0,g 0u is called the set of roots of g (i.e. these are the non-zero simultaneous eigenspaces inside the left-regular (adjoint) module). Let x, be a NIBS form on g. Thenthemap h Ñ h h fiñ xh, h µ fiñ µ is an isomorphism, where h µ is the unique element of h such that xh µ,h µphq for all h P h. Define x, : h b h Ñ C b xµ, xh µ,h.