diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -1815,47 +1815,47 @@ consequence of something known as \emph{simultaneous diagonalization}, which is
the primary tool we'll use to generalize the results of the previous section.
What is simultaneous diagonalization all about then?
+\begin{definition}\label{def:sim-diag}
+ A set of square matrices \(\{X_i\}_i\) is called \emph{simultaneously
+ diagonalizable} if there some invertible matrix \(P\) such that \(P X_i
+ P^{-1}\) is diagonal for every \(i\).
+\end{definition}
+
\begin{proposition}
+ A set of diagonalible matrices is simultaneously diagonalizable if, and only
+ if all of its elements commute with one another.
+\end{proposition}
+
+\begin{corollary}
Let \(\mathfrak{g}\) be a Lie algebra, \(\mathfrak{h} \subset \mathfrak{g}\)
be an Abelian subalgebra and \(V\) be any finite-dimensional representation
of \(\mathfrak{g}\). Then there is a basis \(\{v_1, \ldots, v_n\}\) of \(V\)
so that each \(v_i\) is simultaneously an eigenvector of all elements of
\(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as a diagonal
- matrix in this basis. In other words, there is some linear functional
- \(\lambda \in \mathfrak{h}^*\) so that
+ matrix in this basis. In other words, there are linear functionals
+ \(\lambda_i \in \mathfrak{h}^*\) so that
\[
- H v_i = \lambda(H) \cdot v_i
+ H v_i = \lambda_i(H) \cdot v_i
\]
- for all \(H \in \mathfrak{h}\) and all \(i\).
-\end{proposition}
+ for all \(H \in \mathfrak{h}\).
+\end{corollary}
-% TODOOO: h is not semisimple. Fix this proof
+% TODOO: Prove that the operators are diagonalizable
\begin{proof}
- We claim \(\mathfrak{h}\) is semisimple. Indeed, if \(\{H_1, \ldots, H_m\}\)
- is basis of \(\mathfrak{h}\) then
- \[
- \mathfrak{h} \cong \bigoplus_i K H_i
- \]
- as vector spaces. Usually this is simply a linear isomorphism, but since
- \(\mathfrak{h}\) is Abelian this is an isomorphism of Lie algebras -- here
- \(K H_i\) represents the 1-dimensional subalgebra spanned by \(H_i\), which
- is isomorphic to the trivial Lie algebra \(K\). Each \(K H_i\) is simple,
- so \(\mathfrak{h}\) is isomorphic to a direct sum of simple algebras -- i.e.
- \(\mathfrak{h}\) is semisimple.
-
- Hence
- \[
- V
- = \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} V
- \cong \bigoplus V_i,
- \]
- as representations of \(\mathfrak{h}\), where each \(V_i\) is an irreducible
- representation of \(\mathfrak{h}\). Since \(\mathfrak{h}\) is Abelian, it
- follows from Schur's lemma that each \(V_i\) is 1-dimensional. Say \(V_i =
- k v_i\) and consider the basis \(\{v_1, \ldots, v_n\}\) of \(V\). Now the
- assertion that each \(v_i\) is an eigenvector of all elements of
- \(\mathfrak{h}\) is equivalent to the statement that each \(K v_i\) is
- stable under the action of \(\mathfrak{h}\).
+ It suffices to show that \(H : V \to V\) is a diagonalizable operator for
+ each \(H \in \mathfrak{h}\). Indeed, if this is the case it follows from the
+ fact that \(\mathfrak{h}\) is Abelian that the set
+ \(\{[H\!\restriction_V]_{\mathcal{B}} : H \in \mathfrak{h}\}\) is
+ simultaneously diagonalible for any basis \(\mathcal{B}\) for \(V\).
+
+ If \(P\) is as in definition~\ref{def:sim-diag} and \(T\) is the operator \(V
+ \to V\) such that \([T]_{\mathcal{B}}\), then \(\mathcal{C} =
+ T(\mathcal{B})\) is a basis for \(V\) such that
+ \([H\!\restriction_V]_{\mathcal{C}} = P [H\!\restriction_V]_{\mathcal{B}}
+ P^{-1}\) is diagonal for each \(H \in \mathfrak{h}\). If we then take
+ \(\{v_1, \ldots, v_n\} = \mathcal{C}\) and denote by \(\lambda_i(H)\) the
+ coefficient of \(E_{i i}\) in \([H\!\restriction_V]_{\mathcal{C}}\), we find
+ \(H v_i = \lambda_i(H) \cdot v_i\) as intended.
\end{proof}
As promised, this implies\dots
@@ -2159,7 +2159,7 @@ theorem~\ref{thm:weak-dominant-weight} we used for the case when \(\mathfrak{g}
knowledge of the roots of \(\mathfrak{sl}_3(K)\). Instead, we need a new
strategy for the general setting.
-% TODOOO: Add further details. Turn this into a proper proof?
+% TODOO: Add further details. Turn this into a proper proof?
Alternatively, one could construct a potentially infinite-dimensional
representation of \(\mathfrak{g}\) whose highest weight is some fixed dominant
integral weight \(\lambda\) by taking the induced representation