Theorem The simple finite dimensional sl 2 modules Lpdq are indexed by

The Lie algebra sl 2 pcq has basis x, y, and h, with relations rh, xs 2x, rh, ys 2y, and rx, ys h. Theorem The simple finite dimensional sl 2 modules Lpdq are indexed by d P Z ě0 with basis tv`, yv`, y 2 v`,..., y d v`u and action hpy l v`q pd 2lqpy l v`q, xpy l v`q lpd ` 1 lqpy l 1 v`q, with xv` 0 and ypy l v`q y l`1 v`, with y d`1 v` 0. 0 y h y d v` -d y x h y d 1 v` -d+2 y x y x h y 2 v` d-4 y x h yv` d-2 y x h v` d x 0 Then for any (not nec. simple) sl 2 -module M, we can compute the decomposition of M by computing the dimensions of the eigenspaces for h. Namely, as a Ch-module, M à µpc M µ where M µ tm P M hm µmu.

A Hopf algebra is an algebra A (over F ) with three maps : A Ñ A b A, ε : A Ñ F, 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, is a A-module. [This is called Sweedler notation] (2) The vector space F, with action xα εpxqα is an A-module (called the trivial module). (3) If M is a A-module then M HompM, F q with action is an A-module. pxϕqpmq ϕpspxqmq (4) The maps Y and X are A-module homomorphisms. In general, we call a bilinear form on A invariant if xxm, ny xm, Spxqny.

Example: For any finite group G, the group algebra F G is a Hopf algebra with : F G Ñ F G b F G defined by g ÞÑ g b g ε : F G Ñ F defined by g ÞÑ 1, and S : F G Ñ F G defined by g ÞÑ g 1. These define the following for us... (1) If M and N are F G-modules, then M b N with action g pm b nq pg b gqpm b nq gm b gnn. (2) The vector space F, with action gα εpgqα 1α α is a F G-module (called the trivial module).. (3) If M is a F G-module then the dual M HompM, F q is also a module with action pgϕqpmq ϕpg 1 mq. (4) With the dual and trivial modules defined as above, the following maps are F G-module homomorphisms. X : M b M Ñ F pϕ, mq ÞÑ ϕpmq and We call a bilinear form invariant if xgm, ny xm, g 1 ny. Y : F Ñ M b M 1 ÞÑ ř i m i b m i

Example: For any Lie algebra g, the enveloping algebra Ug is a Hopf algebra with : Ug Ñ Ug b Ug defined by x ÞÑ x b 1 ` 1 b x ε : Ug Ñ F defined by x ÞÑ 0, and S : Ug Ñ Ug defined by x ÞÑ x. These define the following for us... (1) If M and N are Ug-modules, then M b N with action x pm b nq px b 1 ` 1 b xqpm b nq xm b n ` m b xn. (2) The vector space F, with action xα εpxqα 0α 0 is a Ug-module. (3) If M is a Ug-module then the dual M HompM, F q is also a module with action pxϕqpmq ϕp xmq. (4) With the dual and trivial modules defined as above, the following maps are U g-module homomorphisms. X : V b V Ñ F pϕ, vq ÞÑ ϕpvq and Y : F Ñ V b V 1 ÞÑ ř i v i b v i We call a bilinear form invariant if xxm, ny xm, xny.

Example: Let M and N be finite-dimensional sl 2 -modules. Again, we can always pick weight bases for M and N with respect to H: with M Ctm 1,..., m r u and N Ctn 1,..., n s u, hm i h pm i qm i and hn j h pn j qn j with h pm i q, h pn j q P C. So h pm i b n j q hm i b n j ` m i b hn j ph pm i q ` h pn j qqpm i b n j q. So M b N has weight basis tm i b n j i 1,..., r, j 1,..., su, where h pm i b n j q h pm i q ` h pn j q. Exercise: Let M Lp2q and N Lp3q, so that the weight spaces of M and N are M M 2 M 0 M 2 and N N 3 N 1 N 1 N 3. Compute the decomposition of M b N.

Example: Let M and N be finite-dimensional sl 2 -modules. Again, we can always pick weight bases for M and N with respect to H: with M Ctm 1,..., m r u and N Ctn 1,..., n s u, hm i h pm i qm i and hn j h pn j qn j with h pm i q, h pn j q P C. So h pm i b n j q hm i b n j ` m i b hn j ph pm i q ` h pn j qqpm i b n j q. So M b N has weight basis tm i b n j i 1,..., r, j 1,..., su, where h pm i b n j q h pm i q ` h pn j q. In general, if the weight space decompositions of M and N are M à M µ and N à N ν, µ ν then the weight space decomposition of M b N is M b N à pm µ b N ν q, µ,ν where pm b Nq λ à µ,ν pm µ b N ν q. µ`ν λ

Example On the homework, you will compute the s 2 pcq modules Lp1q b Lp1q and Lp1q b3.

Example On the homework, you will compute the s 2 pcq modules Lp1q b Lp1q and Lp1q b3. In general, you can show that for any d ą 0, Lpdq b Lp1q Lpd ` 1q Lpd 1q.

Example On the homework, you will compute the s 2 pcq modules Lp1q b Lp1q and Lp1q b3. In general, you can show that for any d ą 0, Lpdq b Lp1q Lpd ` 1q Lpd 1q. So the dimension of Lpaq in Lp1q bk is given by 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..

Forms Let U be a Hopf algebra (over C) with module M. A bilinear form is a map x, y : M b M Ñ C. A bilinear form is symmetric if xm, ny xn, my for all x, y P M. A symmetric bilinear form is invariant if xxm, ny xm, Spxqny for all x P U. A form is nondegenerate if xx, Uy 0 for all x P U. Recall, we defined ad x P Endpgq by ad x pyq rx, ys. The canonical nondegenerate, symmetric, invariant bilinear (NIBS) form on g is the Killing form, given by xx, yy K trpad x ad y q. Good news: If g is simple, then every NIBS form is a constant multiple of the Killing form.

A Cartan subalgebra of g is a maximal abelian subalgebra h Ă g consisting of semisimple (diagonalizeable) elements. (Analogous to Jucys-Murphy elements!!)

A Cartan subalgebra of g is a maximal abelian subalgebra h Ă g consisting of semisimple (diagonalizeable) elements. (Analogous to Jucys-Murphy elements!!) Example: sl n has basis tx ij E ij, y ij E ji, h l E ll E l`1,l`1 1 ď i ă j ď n, 1 ď l ď n 1u. Our favorite Cartan subalgebra is h Cth l l 1,..., n 1u. Some facts about Cartans: 1. Cartan subalgebras are generated by 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.

The weights of a Cartan h is the dual set h tµ : h Ñ Cu.

The weights of a Cartan h is the dual set h tµ : h Ñ Cu. Example: In sl n, our fav. Cartan is h Cth l E ll E l`1,l`1 l 1,..., n 1u.

The weights of a Cartan h is the dual set h tµ : h Ñ Cu. Example: In sl n, our fav. Cartan is h Cth l E ll E l`1,l`1 l 1,..., n 1u. For i 1,..., n, define ε i : h Ñ C by h ÞÑ trpe ii hq, (i.e. it picks the ith diagonal element). So ε i ph l q δ i,l δ i,l`1. Note ε 1 ` ` ε n 0. So we have h Ctε l ε l`1 l 1,..., n 1u. Let x, y be a NIBS form on g. Then the map h ÝÑ h h ÞÑ xh, y h µ ÞÑ µ is an isomorphism, where h µ is the unique element of h such that xh µ, hy µphq for all h P h.

Let g act on itself via ad, and define g α tg P g ad h pxq αphqxu. The set of weights R tα P h α 0, g α 0u is called the set of roots of g (i.e. these are the non-zero simultaneous eigenspaces inside the left-regular (adjoint) module). Example: sl 3 has basis x 1 E 1,2, x 2 E 2,3, x 3 E 1,3 h 1 E 1,1 E 2,2, h 2 E 2,2 E 3,3, y 1 E 2,1, y 2 E 3,2, y 3 E 3,1 You try: Compute rh l, x j s and rh l, y j s for j 1, 2, 3, and compute R in terms of ε i s (again, where ε i ph l q trpe ii h l q δ i,l δ i,l`1 q and g α for each α P R. Note: As a vector space, Usl 3 pcq Ctpy λ 1 1 yλ 2 2 yλ 3 3 qphλ 4 1 hλ 5 2 qpxλ 6 1 xλ 7 2 xλ 8 3 qˇˇλ i P Z ě0 u U U 0 U `, where U is lower triangular, U 0 is diagonal, and U ` is upper triangular. This is called a triangular decomposition.

In general, for sl n pcq, letting α i,j ε i ε j, we have R t α i,j 1 ď i ă j ď nu and g αij CtE ij u for i j, and g 0 h.

In general, for sl n pcq, letting α i,j ε i ε j, we have R t α i,j 1 ď i ă j ď nu and g αij CtE ij u for i j, and g 0 h. In sl 3, we had R t pε i ε j q 1 ď i ă j ď 3u

In general, for sl n pcq, letting α i,j ε i ε j, we have R t α i,j 1 ď i ă j ď nu and g αij CtE ij u for i j, and g 0 h. In sl 3, we had R t pε i ε j q 1 ď i ă j ď 3u t β 1, β 2, pβ 1 ` β 2 qu where β 1 ε 1 ε 2 and β 2 ε 2 ε 3.

In general, for sl n pcq, letting α i,j ε i ε j, we have R t α i,j 1 ď i ă j ď nu and g αij CtE ij u for i j, and g 0 h. In sl 3, we had R t pε i ε j q 1 ď i ă j ď 3u t β 1, β 2, pβ 1 ` β 2 qu where β 1 ε 1 ε 2 and β 2 ε 2 ε 3. We call subset of B Ă R a base of R is every element of R can be written as (positive integral combination of elements of B). In our example, B tβ 1, β 2 u is a base.

In sl 3, we had R t pε i ε j q 1 ď i ă j ď 3u t β 1, β 2, pβ 1 ` β 2 qu where β 1 ε 1 ε 2 and β 2 ε 2 ε 3. We call subset of B Ă R a base of R is every element of R can be written as (positive integral combination of elements of B). In our example, B tβ 1, β 2 u is a base. In particular, since R sits all within a 2-dimensional space, with xε 1 ε 2, ε 2 ε 3 y 1 β 1 =β 2 arccos a arccos xε1 ε 2, ε 1 ε 2 yxε 2 ε 3, ε 2 ε 3 y 2 2π 3, we can plot R as β 2 β 1 ` β 2 -β 1 β 1 pβ 1 ` β 2 q β 2