# Lately in Lie Groups

This came from a lecture given on 11/10/09.

The lecture began with the definition of the Cartan matrix.  Basically you order the simple roots $a_1, ..., a_n$ and the $ij$th entry is $2\frac{(a_i,a_j)}{(a_j,a_j)}$.

Reason to care about the Cartan Matrix.  Closely associated to the Cartan matrix, is the Coxeter matrix whose entries are $(a_i, a_j)$; note it is in general its entries are not all rational.  As it happens the covolume is the determinant of this matrix.

Simply laced Dynkin diagrams are ones where there is at most one edge between any two nodes.  For these diagrams, the Coxeter matrix has 1’s on diagonal and -1/2 on off diagonals.

Now $\det (Cartan(E_8)) = 1$ (example of an even unimodular lattice), even means inner product of every vector with itself is even., the only such one in 8 dimension. (its the first one) they only occur in dimension congruent to zero mod 8 (actually signature has to be congruent to zero mod 8, e.g could have 26 dim with sig = 25 – 1).Above disscussion gives some meaning for the det. of cartan for simply laced Dynkin diagrams.  Cartan Matrix is positive definite (basically because Coxeter matrix is ).

Finding Lie Algebras. Last time talked about classification of Dynkin diagrams.  Now what to realize roots systems as roots systems of some lie algebras.  We’ve know this true already for a lot of irreducible root systems (the Weyl group is also listed):  (at some point I should understand the action of the Weyl group better)

$A_n \leftrightarrow sl(n+1, \mathbb{C})$, $W = S_{n+1}$

$B_n \leftrightarrow so(2n+1, \mathbb{C})$$W = (Z/2)^n * S_n$ ( permuting up to sign) ( because can flip one and leave the others unchanged)

$C_n \leftrightarrow sp(2n, \mathbb{C})$, same Weyl group

$Dn \leftrightarrow so(2n,\mathbb{C})$  $W = (Z/2)^{n-1}*S_n$ rational: there is parody between sign change = parody of the permutation.

Existence of root system associated to each Dynkin diagram is trivial, just take group generated by reflections by simple roots, and take orbits of simple roots.

Getting Dynkin diagrams from $A,D,E$ series

Label the roots as follows

Explicit description (in terms of coordinates) are given of all the roots in the actual root system roughly around pg. 333 of Fulton and Harris.  From E8 one can obtain roots systems associated to other Dynkin diagram.  For example, E6, E7 are natrually subspaces of the root system for E8.

One can get F4 from E6 (the diagram for E6 is the one above except you erase $a7, a8$).  In this case there is an obvious involution: $a1 \mapsto a6$, $a3 \mapsto a5$.  The quotient under this action gives F4.

If you quotient by the dihedral symmetry of $latexD_4$, you get $G_2$.  Also a symmetry of $A_n$ will give you $B_n$, and similarly quotienting by a symmetry of $D_n$ gives $C_n$.  Thus $A_n, D_n, E_8$ are the ones to be concerned about.

Regarding $sp(2n, \mathbb{C})$.  Let $M = \bigl( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix}\bigr)$.  And write $Mat(A,B,C,D)$ for the usual 2×2 block matrix, i.e. $M = Mat(0,I_n, -I_n, 0)$.  Get that

$sp(2n, \mathbb{C}) = \{X | X^\intercal M + M X = 0\}$

consequently, the Cartan subalgebra is

$h = \{H_i = E_ii - E_{n+i, n+i}\}$

etc, etc some details I didn’t write down very well, but here is the rest of it in raw form:

L_i \in h^*,  <L_i, H_j> = d_ij

X_i,j = E_i,j – E_n,jE     , roots L_i – L_j

A     -A^t = D

E_i, n+j + E_j,n+i   L_i +L_j

E_n+i,j + En+j,i   -L_i-L_j

E_n+i,j 2L_i

E_n+i,i -2L_i

latter two are diagonals of B,C

roots \pm L_i \pm L_j \in h^*

———————–

So(2n), M = 0  I

I  0

Is X = (A,B,C,D) as above, see that A = -D^t, and B,C are skey symm.

Eigenvectors:

E_i,j – E_n+i, n+j with root  L_i — L_j

E_i, n+j – E_j,n+i root L_i + L_j

E_n+i, j – E_n+j,i root -L_i – L_j

also have \pm L_i \pm L_j, i \ne j. Above covers An, Cn, Dn, so(2n+1, C) is similar computation.

Last little bit.  Assume for now that can construct $E_8$, If $\mathfrak{g}$ is a simple lie algebra, with dynkin diagram $D$, $D^0$ is a subdiagram obtained from removing some nodes, together will all the lines meeting those nodes.  Then can construct $\mathfrak{g}^0$ sub alg. with $D^0$ as its dynkin diagram.  $\mathfrak{g}^0$ is gen. by $\mathfrak{g}_{\pm a}$ with $a \in D^0$.

Exercise: prove this by verifying positive roots of $\mathfrak{g}^0$ are positive roots $b$ of $\mathfrak{g}$ that are sums of roots in $D_0$ and $\mathfrak{h}^0$ is spanned by $H_b \in \mathfrak{h}$

Basically this shows if two roots systems are abstractly isomorphic they give rise to isomorphic lie algebras. Eventually will construct $E_8$, but now Uniqueness: $g,g'$ have Cartan subalgebras $h,h'$ with choice of positive roots and isomprhic root systems, $\exists$ iso $h \to h'$ taking $H_i \to H'_i$, where $a_1, ..., a_n$ positive roots for $h$

Choose $X_i, X'_i$ arb in $g_{a_i}, g_{a'_i}$, just map $X_i \to X'_i$, claim: there exists a unique iso $g \to g'$ extending $h \to h'$ and mapping $X_i to X'_i$.

proof: (in Fulton and Harris) , map is det. on $Y_i \in g_{a_i}$ by $[Y,X] = H$ and $[Y',X'] = H'$, these generated all of $g, g'$ respectively (i.e. $latexX,H$ uniquely det. $Y$).

Existence: let $g'' \subset g \oplus g'$ be the graph of the map: its generated by

$H''_i = H'_i \oplus H_i$ and $X''_i = X_i \oplus X'_i$

and similarly for $Y_i$.  It suffices to show the two projections are isos from $g'' \to g$ or $g'$ respectively.

Kernel of second projection is $t \oplus 0$, where $t is an ideal of g$.  since $g$ is simple, $t = 0$, or $t = g$.  In the latter case, $g'' = g \oplus g'$. Consider a maximal root $b$, by maximal mean $[X_i, X_b] = 0$ for all $i$ simple roots.

Take $X_b$ and $X'_b$ in corresp. root spaces and set $X''_b = X_b\oplus X'_b$  and $W$ be the subspace $g''$ obtained by applying all the $Y''_i$.  $latex W$ is proper subspace of $g''$.

Claim: $g''$ perserves $W$, this is clear for $H''_i$, just have to show $X''_j$ perserves $W$, inductive argument.

Suppose $[X''_j, [Y''_{j1} [.... [ Y''_{jn}, X''_b ]]]]]$, suppose its in $W$ for n, show it will be in $W$ for $n + 1$

$[X''_j , [Y''_{i, n+1}, Z]]$ = (use jacobi) =  $- [Y''_{k+1}, [Z, X''_j]] - [Z, [X''_j, Y''_{ik_1}]]$

first term is in $W$ by induction, and second term should be 0, or a mutliple of $H''_j$ applied to Z, so by induction perserves $W$, so $W$ is proper invariant subspace.  And $W$ is ideal of $g \oplus g'$ , but must be proper so can’t be everything, so forces $X_b oplus 0 \in W$, so $W_b$ is two dimensional, contradiciton.