# Wednesday, September 07 ## 4.3: Cartan's criterion for semisimplicity :::{.remark} For the rest of the course, $V$ is a vector space of finite dimension. Goal: get a criterion for semisimplicity. ::: :::{.theorem title="Cartan's criterion for linear Lie algebras"} Let $L\leq \liegl(V)$ be a linear Lie algebra and suppose $\trace(xz)=0$ for all $x\in [LL]$ and $z\in L$. Then $L$ is solvable. ::: :::{.lemma title="?"} Let $A \subseteq B$ be subspaces of $\Endo(V) = \liegl(V)$ and define \[ M = \ts{w\in \liegl(V) \st [w, B] \subseteq A} .\] Suppose that $w\in M$ satisfies $\trace(wz) = 0$ for all $z\in M$. Then $w$ is nilpotent. ::: :::{.proof title="of Cartan, assuming the lemma"} To show $L$ is solvable, it STS that $[LL]$ is nilpotent since the ideals used to check nilpotency are bigger than those to check solvability. By Engel's theorem, it STS to show each $w\in [LL]$ is ad-nilpotent. Since $L \leq \liegl(V)$, it STS to show each $w\in [LL]$ is a nilpotent endomorphism. As in the setup of the lemma, set $B = L, A = [LL]$, then \[ M \da \ts{z\in \liegl(V) \st [zL] \subseteq [LL] } \contains L \contains [LL] .\] Let $w\in [LL] \subseteq M$, then note that $\trace(wz) = 0$ for all $z\in L$, but we need to know this for all $z\in M$. Letting $z\in M$ be arbitrary; by linearity of the trace it STS $\trace(wz) = 0$ on generators $w = [xy]$ on $[LL]$ for $x,y\in L$. We thus WTS $\trace([xy]z) = 0$: \[ \trace([xy]z) &= \trace(x [yz] ) \\ &=\trace([yz] x) \qquad \in \trace(LMx) \subseteq \trace([LL]L) \\ &= 0 \] by assumption. By the lemma, $w$ is nilpotent. ::: :::{.corollary title="Cartan's criterion for general Lie algebras"} Let $L\in \Lie\Alg$ with $\trace(\ad_x \ad_y) = 0$ for all $x \in [LL]$ and $y\in L$. Then $L$ is solvable. ::: :::{.proof title="of corollary"} Use $\ad: L\to \liegl(V)$, a morphism of Lie algebras. Its image is solvable by Cartan's criterion above, and $\ker \ad = Z(L)$ which is abelian and hence a solvable ideal.[^derived_alg_term] Therefore $L$ is solvable. [^derived_alg_term]: The derived series terminates immediately for an abelian algebra. ::: :::{.proof title="of lemma"} Let $w = s + n$ be the Jordan-Chevalley decomposition of $w$. Choose a basis for $V$ such that this is the JCF of $w$, i.e. $s = \diag(a_1,\cdots, a_n)$ and $n$ is strictly upper triangular. Idea: show $s=0$ by showing $A\da \QQ\gens{a_1,\cdots, a_n} = 0$ by showing $A\dual = 0$, i.e. any $\QQ\dash$linear functional $f: A\to \QQ$ is zero. If $\sum a_i f(a_i) = 0$ then \[ 0 = f(\sum a_i f(a_i)) = \sum f(a_i)^2 \implies f(a_i) = 0 \,\,\forall i ,\] so we'll show this. Let $y = \diag( f(a_1), \cdots, f(a_n) )$, then $\ad_y$ is a polynomial (explicitly constructed using Lagrange interpolation) in $\ad_s$ without a constant term. So do this for $\ad_y$ and $\ad_s$ (see exercise). Since $\ad_s$ is a polynomial in $\ad_w$ with zero constant term, and since $\ad_w: B\to A$, we have $\ad_s(B) \subseteq A$ and the same is thus true for $\ad_y$. So $y\in M$ and $w\in M$, and applying the trace condition in the lemma with $z\da y$ we get \[ 0 = \trace(wy) = \sum a_i f(a_i) ,\] noting that $w$ is upper triangular and $y$ is diagonal. So $s=0$ and $w=n$ is nilpotent. ::: :::{.exercise title="?"} Show $\ad_y$ is a polynomial in $\ad_s$. ::: :::{.remark} Recall that $\Rad L$ is the unique maximal (not necessarily proper) solvable ideal of $L$. This exists, e.g. because sums of solvable ideals are solvable. Note that $L$ is semisimple iff $\Rad L = 0$. ::: :::{.definition title="Killing form"} Let $L\in \Lie\Alg^\fd$ and define the **Killing form** \[ \kappa: L\cross L &\to \FF \\ \kappa(x, y) &= \trace(\ad_x \circ \ad_y) .\] This is an associative[^assoc_invt] bilinear form on $L$. [^assoc_invt]: Associative is $f([xy]z) = f(x [yz])$, sometimes called *invariant*. ::: :::{.example title="?"} Let $L = \CC\gens{x, y}$ with $[xy] = x$. In this ordered basis, \[ \ad_x = \matt 0 1 0 0 \qquad \ad_y = \matt {-1} 0 0 0 ,\] and one can check $\kappa(x,x) = \kappa(x, y) = \kappa(y, x) = 0$ and $\kappa(y,y) = 1$. Moreover $\Rad \kappa = \CC\gens{x}$. > See the text for $\kappa$ defined on $\liesl_2$. ::: :::{.lemma title="?"} Let $I \normal L$. If $\kappa$ is the Killing form of $L$ ad $\kappa_I$ that of $I$, then \[ \kappa_I = \ro{\kappa}{I\times I} .\] ::: :::{.proof title="?"} Let $x\in I$, then $\ad_x(L) \subseteq I$ since $I$ is an ideal. Choosing a basis for $I$ yields a matrix: ![](figures/2022-09-07_09-56-24.png) So if $x,y\in I$, we have \[ \kappa(x,y) &= \trace(\ad_x \circ \ad_y) \\ &= \trace(\ad_{I, x} \circ \ad_{I, y}) \\ &= \kappa_I(x, y) .\] :::