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\begin{center}
Introduction to Lie Theory\\
Homework \#7\\
\end{center}


\begin{enumerate}
\item
Let $A$ be a finite-dimensional algebra (associative or Lie or
whatever) over an algebraically closed field $\k$. Let $d:A\rightarrow A$ be a derivation. Show that the
semisimple and nilpotent parts $d_s$ and $d_n$ of $d \in \End_\k(A)$ are also
derivations of $A$.

({\em Hint.} It suffices to prove that $d_s \in
\operatorname{Der}(A)$. Decompose $A$ into generalized
eigenspaces
$A = \bigoplus_{\lambda \in \k} A_\lambda$ for $d$, so that $d_s$ acts
on $A_\lambda$ by multiplication by $\lambda$, then
show that $A_\lambda A_\mu \subseteq A_{\lambda+\mu}$.
)

\item
Recall for a Lie algebra $\mathfrak{g}$ that
$\operatorname{Der}(\mathfrak{g})$ is a Lie subalgebra of
$\mathfrak{gl}(\mathfrak{g})$.
\begin{enumerate}
\item
Show that $\operatorname{ad}(\mathfrak{g})$ is an ideal of
$\operatorname{Der}(\mathfrak{g})$, the so-called {\em inner
  derivations}.
\end{enumerate}
Now assume $\mathfrak{g}$ is a finite-dimensional semisimple Lie
algebra over $\C$.
\begin{itemize}
\item[(b)]
Given a derivation $d:\mathfrak{g}\rightarrow \mathfrak{g}$,
show that the vector space $\mathfrak{g}\oplus\C$ is a well-defined
$\mathfrak{g}$-module with action $x(y,\lambda) = ([x,y] + \lambda
d(x), 0)$.
\item[(c)]
Use Weyl's theorem on complete reducibility to prove that
every derivation of $\mathfrak{g}$ is inner, i.e.,
$\operatorname{Der}(\mathfrak{g}) = \operatorname{ad} (\mathfrak{g})$.
\end{itemize}
\item
For a finite-dimensional semisimple Lie algebra $\mathfrak{g}$
and $x \in \mathfrak{g}$, we showed in L6-3 that there are unique
elements $x_s, x_n \in \mathfrak{g}$ such that $x = x_s+x_n, [x_s,x_n] = 0$,
$\operatorname{ad} x_s:\mathfrak{g}\rightarrow\mathfrak{g}$ is
semisimple and $\operatorname{ad} x_n:\mathfrak{g}\rightarrow \mathfrak{g}$ is nilpotent.
Use Q1 and Q2 to give another proof of this.
\item
This question is the beginning of
{\em Lie algebra
cohomology}. 
Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over
$\C$. 
\begin{enumerate}
\item
Suppose that there is a Lie algebra extension
$$
0 \rightarrow \mathfrak{g} \rightarrow \widehat{\mathfrak{g}}
\rightarrow \C \rightarrow 0,
$$ i.e., $\widehat{\mathfrak{g}}$ is a Lie
algebra containing $\mathfrak{g}$ as an ideal of codimension one. 
Prove that $\widehat{\mathfrak{g}} \cong \mathfrak{g} \oplus \C$,
i.e., it is a split extension.
\item
Suppose that there is a Lie algebra extension
$$0 \rightarrow \C \rightarrow \widehat{\mathfrak{g}}
\rightarrow \mathfrak{g} \rightarrow 0.
$$ 
Show that the first map
embeds $\C$ into the center $\mathfrak{z}(\widehat{\mathfrak{g}})$,
i.e., it is a {\em central extension}. 
Then prove that the extension is split.
\end{enumerate}
({\em Hints.} For (a), the data of such an extension is really just
the data of a derivation $d$ of $\mathfrak{g}$; then you can use
Q2. Part (b) is more difficult!)
\item
The infinite-dimensional Lie algebra constructed in the next question is the
{\em affine Lie algebra} $\widehat{\mathfrak{sl}}_2(\C)$.
Let $\mathfrak{g} = \mathfrak{sl}_2(\C)$.
Let $\mathfrak{g}[t,t^{-1}]$ be the 
infinite-dimensional Lie algebra
$\mathfrak{g} \otimes_\C \C[t,t^{-1}]$ with
$[xt^m, y t^n] = [xy] t^{m+n}$ for $x,y \in \mathfrak{g}, m,n \in \Z$.
\begin{enumerate}
\item
Show that there is a non-split Lie algebra extension
$$
0 \rightarrow \C \rightarrow \widehat{\mathfrak{g}}' \rightarrow
\mathfrak{g}[t,t^{-1}] \rightarrow 0
$$
such that $\widehat{\mathfrak{g}}' = \C c \oplus
\mathfrak{g}[t,t^{-1}]$ as a vector space with Lie bracket defined so
that $c$ is central and
$$
[x t^m, y t^n] = [x,y] t^{m+n} + \delta_{m+n,0} \tau(x,y)
mc,
$$
where $\tau$ is the trace form on $\mathfrak{sl}_2(\C)$.
\item
Let $\widehat{\mathfrak{g}}'$ be as in (a). Show that there is a
non-split Lie algebra extension
$$
0 \rightarrow \widehat{\mathfrak{g}}'
\rightarrow\widehat{\mathfrak{g}} \rightarrow \C \rightarrow 0
$$
such that $\widehat{\mathfrak{g}} = \widehat{\mathfrak{g}}'
\oplus \C d$ as a vector space with the Lie bracket defined so that
$\widehat{\mathfrak{g}}'$ is a Lie subalgebra, $[d,c] = 0$ and 
$[d, x t^n] = n x t^n$.
\item Show that $\mathfrak{z}(\widehat{\mathfrak{g}}) = \C c$ and that
  $\widehat{\mathfrak{g}}'$ is the derived subalgebra of $\widehat{\mathfrak{g}}$.
\end{enumerate}
\end{enumerate}
\end{document}
