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\begin{document}
\lecture{10 --- October 12, 2004}{Fall 2004}{Prof.\ Victor Ka\v{c}}{Steven
Sivek}
Recall that a \emph{trace form} on a Lie algebra $\g$, given a
representation $\pi$ of $\g$ in a finite dimensional vector space $V$, is
defined as $(a,b)_V = \tr_V(\pi(a)\pi(b))$.
\begin{defn}
The \emph{Killing form} on a finite-dimensional Lie algebra $\g$ is the
trace form $K(a,b) = \tr_{\g}((\ad\ a)(\ad\ b))$ in the adjoint
representation of $\g$.
\end{defn}
\paragraph{Exercise 10.1.} Show that the trace form for the defining
representation of $\gl_n(\F)$, $\sll_n(\F)$, $\so_n(\F)$, $\spl_n(\F)$
is nondegenerate.
\paragraph{Solution.}
Let $e_{xy}$ denote the $n$x$n$ matrix with the element in row $x$, column
$y$ equal to 1 and all other elements 0. We note that $\tr(e_{ij}e_{kl})
= \tr(\delta_{jk}e_{il}) = \delta_{jk}\delta_{il}$. To show that this
form is nondegenerate for a Lie algebra $\g \subset \gl_n(\F)$, it
suffices to show for any $a \in \g$ that $\tr(ab) \not= 0$ for some $b \in
\g$, so this is how we will proceed. (We will assume that $\charr(\F) =
0$ for simplicity, since e.g. the trace form for $\sll_n(\F)$ is
degenerate when $\charr(F)\ |\ n$.)
$\g = \gl_n(\F)$: Take nonzero $a = (a_{ij}) \in \g$, and pick $x,y$ such
that $a_{xy} \not= 0$. Then $\tr(a e_{yx}) = a_{xy} \not= 0$.
$\g = \sll_n(\F)$: Choose nonzero $a = (a_{ij}) \in \g$. If $a$ is not
a diagonal matrix, then we may pick $x \not= y$ such that $a_{xy} \not=
0$, and as before we have $\tr(a e_{yx}) \not= 0$. Otherwise, let $a =
\diag(b_1, \dots, b_n)$. Then $a \not= \alpha I_n$ for any
$\alpha$, since otherwise we would have either $a=0$ or $\tr(a) = n\alpha$
with $n,\alpha \not= 0$. It follows that for some $k < n$ we must have
$b_k \not= b_{k+1}$; we calculate that $\tr(a (e_{kk} - e_{k+1}{k+1})) =
b_k-b_{k+1} \not= 0$.
$\g = \so_n(\F)$: Pick a basis for $\F^n$ such that $\so_n(\F)$ is the
algebra of skew-symmetric matrices, i.e. $\so_n(\F) = \{ a \in \gl_n(\F)\
|\ a^T = -a \}$. Then $\g$ has as a basis $\mathcal{B} = \{
e_{ij}-e_{ji}\ |\ i 0$, so
$\lambda(h) = \frac{-Q}{P} \alpha(h)$, where the ratio $\frac{-Q}{P}$
depends only on $\lambda$ and $\alpha$, as desired.
\end{proof}
We may use this to describe solvable subalgebras of $\gl_V$ in several
ways in terms of trace forms.
\begin{thm}[Cartan's criterion]
Let $\g$ be a subalgebra of $\gl_V$ for $V$ a finite-dimensional vector
space over a an algebraically closed field $\F$ of characteristic 0. Then
the following are equivalent:
\begin{enumerate}
\item $(\g, [\g,\g])_V = 0$.
\item $(a,a)_V = 0$ for any $a \in [\g,\g]$.
\item $\g$ is a solvable Lie algebra.
\end{enumerate}
\begin{proof}
$(1) \Rightarrow (2)$: Take $a \in [\g,\g]$, and write $a=[b,c]$. Then
since $(a,[b,c])_V = 0$, we have $(a,a)_V=0$.
$(2) \Rightarrow (3)$: Suppose that $\g$ is not solvable. Then the
derived series of $\g$ stabilizes to some nonzero subalgebra $\p =
\g^{(N)} = \g^{(N+1)} = \cdots$, so $[\p, \p] = \p$. In particular, since
$\p \subset \g$, we have $(a,a)_V = 0$ for any $a \in [\p,\p] = \p$ by our
assumption. Let $\h$ be a Cartan subalgebra of $\p$, and consider the
root and weight space decompositions
\begin{eqnarray*}
\p = \bigoplus_{\alpha \in \h^*} \p_\alpha,\
V = \bigoplus_{\lambda \in \h^*} V_\lambda.
\end{eqnarray*}
Since $[\p_\alpha, \p_\beta] \subset \p_{\alpha+\beta}$ and $[\p,\p] =
\p$, we conclude that $\h = \p_0 = \sum_\alpha [\p_\alpha,
\p_{-\alpha}]$; that is, $\h$ is a span of elements of the form
$h_{\alpha,i} = [e_{\alpha,i}, f_{\alpha,i}]$, where $e_{\alpha,i} \in
\p_\alpha$ and $f_{\alpha,i} \in \p_{-\alpha}$.
Suppose that $V_\lambda \not= 0$ for some fixed $\lambda \in \h^*$. By
Cartan's lemma, we can write $\lambda(h_{\alpha,i}) = r_{\alpha,\lambda}
\alpha(h_{\alpha,i})$ for all $\alpha$ and $i$, where $r_{\alpha,\lambda}
\in \Q$. Since we are assuming (2), $(h_{\alpha,i}, h_{\alpha,i})_V = 0$.
But we compute that $(h_{\alpha,i}, h_{\alpha,i})_V = \sum_\lambda
\lambda(h_{\alpha,i})^2\dim(V_\lambda)$ by considering the restriction of
$h_{\alpha,i}$ to each subspace $V_\lambda$ as in the proof of Cartan's
lemma; this is then equal to $\sum_\lambda r_{\alpha,\lambda}^2
\alpha(h_{\alpha,i})^2 \dim(V_\lambda)$ by the lemma, so
$\alpha(h_{\alpha,i})^2 \left( \sum_\lambda r_{\alpha,\lambda}^2
\dim(V_\lambda) \right) = 0$. Hence, since all the $r_{\alpha,\lambda}$
are rational numbers, either $\alpha(h_{\alpha,i}) = 0$, or
$r_{\alpha,\lambda} = 0$ whenever $V_\lambda \not= 0$, and in either case
we have $\lambda(h_{\alpha,i}) = r_{\alpha,\lambda} \alpha(h_{\alpha,i}) =
0$ whenever $V_\lambda \not= 0$.
For any $\lambda \in \h^*$ such that $V_\lambda \not= 0$, since the
elements $\{h_{\alpha,i}\}$ span $\h$, and $\lambda = 0$ on all of these
elements, we conclude that $\lambda \equiv 0$ on $\h$. Therefore the
weight space decomposition of $V$ is $V = V_0$. But for any $\alpha \not=
0$ we have $\p_\alpha V_0 \subset V_\alpha = 0$, so $\p_\alpha = 0$; hence
$\p = \p_0 = \h$. But if $\p = \h$ then $\p$ is nilpotent, and so $[\p,
\p]$ is a proper subset of $\p$. This is a contradiction, so $\g$ must in
fact be solvable.
$(3) \Rightarrow (1)$: By a corollary to Lie's theorem, we may pick a
basis of $V$ such that all of the matrices in $\g$ are upper triangular,
hence all matrices in $[\g,\g]$ are strictly upper triangular. Then
$\tr_V(ab) = 0$ for any $a \in \g$ and $b \in [\g,\g]$, since $ab$ is
strictly upper triangular, and so $(\g,[\g,\g])_V = 0$.
\end{proof}
\end{thm}
This theorem leads almost immediately to a characterization of solvable
Lie algebras, as follows:
\begin{cor}
A finite dimensional Lie algebra $\g$ over an algebraically closed
field of characteristic 0 is solvable if and only if $K(\g,[\g,\g]) = 0$.
\begin{proof}
We know that $\g$ is solvable if and only if $\g/\centerr(\g)$ and
$\centerr(\g)$ are both solvable, hence if and only if $\g/\centerr(\g)$
is solvable. Consider the adjoint representation $\ad: \g \rightarrow
\gl_{\g}$. Since $\ker(\ad) = \centerr(\g)$, this proof reduces to
consideration of the subalgebra $\ad\ \g \subset \gl_\g$; we now apply the
fact that conditions (1) and (3) of Cartan's criterion are equivalent, and
we are done.
\end{proof}
\end{cor}
Up until this point, we have restricted many of our results to Lie
algebras over algebraically closed fields $\F$ of characteristic zero,
since root space and weight space decompositions exist for such algebras.
Requiring $\F$ to be algebraically closed is not always necessary, as we
shall see:
\begin{remark}
Let $\F$ be a field of characteristic zero which is not necessarily
algebraically closed, and let $\g$ be a Lie algebra over $\F$. Then the
following are true:
\begin{enumerate}
\item Cartan's criterion and the corollary which followed it are both true
over $\F$.
\item $\h_0^a$ is a Cartan subalgebra of $\g$ if $a \in \g$ is regular.
\item $[\g,\g]$ is nilpotent if $\g$ is finite-dimensional and solvable.
\end{enumerate}
\end{remark}
In order to show this, we will introduce some notation: let
$\overline{\F}$ be the algebraic closure of $\F$, and let $\gb =
\overline{\F} \otimes_\F \g$.
\paragraph{Exercise 10.3.}
(a) Show that $\g$ is solvable (resp. nilpotent, abelian) if and only if
$\gb$ is, and that $\overline{[\g,\g]} = [\gb, \gb]$.
(b) Prove the above remark for $\F$ not algebraically closed.
\paragraph{Solution.}
Given two Lie algebras $\g$ and $\h$ over $\F$, we claim that $[\gb, \hb]
= \overline{[\g,\h]}$. Clearly since $\g \subset \gb$ and $\h \subset
\hb$, we have $\overline{[\g,\h]} \subset \overline{[\gb,\hb]} = [\gb,
\hb]$. Given bases $\{ g_\alpha \}_{\alpha \in I}$ of $\g$ and $\{
h_\beta \}_{\beta \in J}$ of $\h$, these same bases generate $\gb$ and
$\hb$ when combined over $\overline{\F}$ rather than $\F$. Hence the set
$\{[g_\alpha, h_\beta]\}_{\alpha \in I, \beta \in J}$ generates
$[\gb,\hb]$ over $\overline{\F}$. But this set also generates $[\g,\h]$
with coefficients in $\F$, so when we extend this to $\overline{F}$ we get
$[\gb,\hb] \subset \overline{[\g,\h]}$. Thus $[\gb, \hb] =
\overline{[\g,\h]}$.
Part (a) now follows easily, since we note that $\gb^i = \overline{\g^i}$
and $\gb^{(i)} = \overline{\g^{(i)}}$ for all $i$. Then $\g$ is solvable
if and only if $\g^{(n)} = 0$ for some $n$; but this is equivalent to
$\overline{\g^{(n)}} = 0$, or $\gb^{(n)} = 0$, and so $\g$ is solvable iff
$\gb$ is. The same is true of nilpotency, as we see by replacing
$g^{(n)}$ with $g^n$ in this argument, and abelianness, which comes from
the solvability argument with $n = 1$. In particular, letting $\h = \g$
above yields $\overline{[\g,\g]} = [\gb,\gb]$.
For part (b), we first look at Cartan's criterion. Since $\g$ is solvable
if and only if $\gb$ is, we have the equivalent conditions (1) $(\gb,
[\gb,\gb]_V = 0$; (2) $(a,a)_V = 0$ for all $a \in [\gb,\gb]$; and (3)
$\g$ is solvable. But $(\gb, [\gb,\gb])_V = (\gb, \overline{[\g,\g]})_V =
\overline{(\g,[\g,\g])_V}$, so $(\g,[\g,\g])_V = 0$ if and only if
$(\gb,[\gb,\gb])_V = 0$. Furthermore, $(a,a)_V = 0$ for all $a \in
[\gb,\gb] = \overline{[\g,\g]}$ iff it does for all $a \in [\g,\g]$, since
$[\g,\g]$ and $\overline{[\g,\g]}$ share the same basis over different
fields and thus $(a,a)_V = 0$ for all $a$ in one iff it does for all $a$
in the other. Therefore Cartan's criterion holds for $\g$ given that it
does for $\gb$; the corollary's proof over $\g$ is identical to its
original proof, since it only needed algebraic closure to satisfy Cartan's
criterion.
Next, we consider the proof that $\h_0^a$ is a Cartan subalgebra if $a$ is
regular. If $a \in \g$ is regular, it is regular in $\gb$, so we know
that $\overline{\h_0^a}$ is a Cartan subalgebra of $\gb$, or
$\overline{\h_0^a} = N_{\gb}(\overline{\h_0^a})$. Since $[b,
\overline{\h_0^a}] \subset \overline{\h_0^a}$ if and only if $[b, \h_0^a]
\subset \h_0^a$ for any $b \in \g$, we conclude that $N_{\g}(\h_0^a) =
\h_0^a$, and so $\h_0^a$ is a Cartan subalgebra of $\g$.
Finally, assume $\g$ is finite dimensional and solvable. Then $\gb$ is
finite dimensional and solvable, so we know that $[\gb,\gb]$ is nilpotent.
But $[\gb,\gb] = \overline{[\g,\g]}$, so $\overline{[\g,\g]}$ is
nilpotent, hence $[\g,\g]$ is nilpotent.
\paragraph{}
\begin{remark}
The basic properties of a trace form are that it is symmetric (i.e.
$(a,b)_V = (b,a)_V$) and invariant (i.e. $([a,b],c)_V = (a,[b,c])_V$).
The basic results on trace forms (like Cartan's criterion) fail, however,
if we assume only that the bilinear forms involved are symmetric and
invariant.
\end{remark}
We can construct an example of this as follows:
\paragraph{Exercise 10.4.}
Consider the 4-dimensional Lie algebra $\D = \F p + \F q + \F c + \F d$,
where $[p,q] = c$, $c$ is central, $[d,p] = p$, and $[d,q] = d$.
Construct a nondegenerate symmetric invariant bilinear form on $\D$.
\paragraph{Solution.}
Define a bilinear form $B(x,y)$ on $\D$ by the values $B(p,q) = B(q,p) =
1$; $B(c,d) = B(d,c) = 1$; $B = 0$ on all other pairs of basis
elements; and all other values of $B$ follow by symmetry and bilinearity.
Then $B$ is nondegenerate, since it has matrix
{\tiny $\left( \begin{array}{cccc} 0&1&0&0 \\ 1&0&0&0 \\
0&0&0&1 \\ 0&0&1&0 \end{array} \right)$ }
with respect to the basis $\{p,q,c,d\}$ and this matrix is invertible.
Invariance is also easy to check, so we are done.
\paragraph{}
The algebra $\D$ is solvable, since $[\D,\D] = \He_1$ is solvable, so by
Cartan's criterion, any trace form on $\D$ must satisfy $(\D, \He_1)_V =
0$. But $B(d,c) = 1$, so $B$ cannot be a trace form.
\paragraph{}
We conclude the lecture by defining a new class of Lie algebras:
\begin{defn}
A Lie algebra $\g$ is called \emph{semisimple} if it contains no nonzero
solvable ideals. Equivalently, $\g$ is semisimple if it contains no
nonzero abelian ideals.
\end{defn}
Equivalence can be proved as follows: If $\g$ has a nonzero abelian ideal,
then it has a nonzero solvable ideal, since abelian ideals are solvable.
Conversely, if $\g$ has a nonzero solvable ideal $\h$, take $n$ such that
$\h^{(n)} \not= 0$ but $\h^{(n+1)} = 0$; then $\h^{(n)}$ is a nonzero
abelian ideal.
In the next lecture, we'll prove that a finite-dimensional Lie algebra
$\g$ over a field of characteristic zero is semisimple if and only if the
Killing form on $\g$ is nondegenerate.
\end{document}