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\noindent
\parbox{2.25in}{\bf Math 4010/5530}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Feb. $20^\mathrm{th}$, 2023}

\vspace{0.2in}

%\noindent {\bf\large Name:} \underline{\qquad\sc\large Dustin Presnell \qquad}
%
%\vspace{0.2in}

Unless otherwise specified, $V$ is a vector space and $L$ is a Lie algebra over a field $\mathbb{F}$.

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\begin{description}
\item[\#1 Centrally Direct] Let $L_1$ and $L_2$ be subalgebras of $L$ where $L = L_1 \oplus L_2$ (a Lie algebra direct sum): In other words, for every $x \in L$ there exists unique $x_1 \in L_1$ and $x_2 \in L_2$ such that $x=x_1+x_2$. Moreover, given any $y \in L_1$ and $z \in L_2$, we have $[x.y]=0$. Prove that $Z(L)=Z(L_1)\oplus Z(L_2)$. 

\item[\#2 Directly Central] Let $L$ be a Lie algebra and $I \triangleleft L$. Next, let $J = C_L(I) = \{ x \in L \;|\; [x,y]=0 \mbox{ for all } y \in I \}$ (i.e. $J$ is the centralizer of $I$ in $L$).\\ \force \hfill Note: From a previous homework we know $J = C_L(I)$ is an ideal of $L$ since $I$ is an ideal of $L$. 

\begin{enumerate}[(a)]
\item Suppose that all derivations of $I$ are inner (i.e. if $\partial:I \to I$ is a derivation, then there exists some $x \in I$ such that $\partial =\mathrm{ad}_x$). Show that $L=I+J$.
\item If in addition, $Z(I) = \{0\}$, show that $L=I \oplus J$.
\end{enumerate}

\item[\#3 Derived Question] Recall that $L^{(k+1)} = [L^{(k)},L^{(k)}]$ and $L^{(0)}=L$. Let $\varphi:L_1 \to L_2$ be an {\bf epi}morphism.\\ \force \hfill  Show that $\varphi\left(L_1^{(k)}\right) = L_2^{(k)}$ for all $k$.

\item[\#4 Adjoint Representation] Recall that $\mathrm{ad}:L \to \mathfrak{gl}(L)$ is defined by $\mathrm{ad}(x)=\mathrm{ad}_x$ where $\mathrm{ad}_x(y)=[x,y]$ for all $x,y \in L$. We already know that $\mathrm{ad}_x \in \mathfrak{gl}(L)$ for all $x \in L$ (i.e. $\mathrm{ad}_x$ is a linear operator -- in fact, $\mathrm{ad}_x \in \mathrm{Der}(L)$). Show that $\mathrm{ad}$ is a homomorphism. Then show that $\diagquotient{L}{Z(L)} \cong \mathrm{ad}(L)$.

\end{description}

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