[proofplan] We first verify that each map $\operatorname{ad}_x$ is linear and that $x \mapsto \operatorname{ad}_x$ is itself linear. The main point is to check preservation of the Lie bracket: this follows by evaluating the commutator $[\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}$ on an arbitrary element $z \in \mathfrak g$ and rewriting the result with alternation and the Jacobi identity. Finally, the kernel condition is exactly the defining condition for membership in the center. [/proofplan] [step:Verify that the adjoint maps are linear] Fix $x \in \mathfrak g$. The map \begin{align*} \operatorname{ad}_x: \mathfrak g &\to \mathfrak g \\ z &\mapsto [x,z] \end{align*} is $k$-linear because the Lie bracket is $k$-bilinear in its second argument. Thus $\operatorname{ad}_x \in \mathfrak{gl}(\mathfrak g)$. Now let $a,b \in k$ and let $x,y \in \mathfrak g$. For every $z \in \mathfrak g$, bilinearity of the Lie bracket in its first argument gives \begin{align*} \operatorname{ad}_{a x + b y}(z) &= [a x + b y,z] \\ &= a[x,z] + b[y,z] \\ &= \bigl(a\operatorname{ad}_x + b\operatorname{ad}_y\bigr)(z). \end{align*} Since the two linear maps agree on every $z \in \mathfrak g$, we have \begin{align*} \operatorname{ad}_{a x + b y} = a\operatorname{ad}_x + b\operatorname{ad}_y. \end{align*} Hence \begin{align*} \operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g) \end{align*} is a $k$-[linear map](/page/Linear%20Map). [/step] [step:Rewrite the commutator using the Jacobi identity] Let $x,y,z \in \mathfrak g$. The commutator bracket on $\mathfrak{gl}(\mathfrak g)$ is defined by composition, so \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= (\operatorname{ad}_x \circ \operatorname{ad}_y)(z) -(\operatorname{ad}_y \circ \operatorname{ad}_x)(z) \\ &= \operatorname{ad}_x([y,z])-\operatorname{ad}_y([x,z]) \\ &= [x,[y,z]]-[y,[x,z]]. \end{align*} Since the Lie bracket is alternating and bilinear, for all $u,v \in \mathfrak g$ we have \begin{align*} 0 &= [u+v,u+v] \\ &= [u,u] + [u,v] + [v,u] + [v,v] \\ &= [u,v] + [v,u]. \end{align*} Thus $[v,u] = -[u,v]$ for all $u,v \in \mathfrak g$. Apply this identity with $u=y$ and $v=[x,z]$ to obtain \begin{align*} -[y,[x,z]] = [[x,z],y]. \end{align*} Therefore \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= [x,[y,z]] + [[x,z],y]. \end{align*} The Jacobi identity for $x,y,z \in \mathfrak g$ gives \begin{align*} [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0. \end{align*} Using $[z,x] = -[x,z]$ and $[z,[x,y]] = -[[x,y],z]$, this becomes \begin{align*} [x,[y,z]] - [y,[x,z]] - [[x,y],z] = 0. \end{align*} Equivalently, \begin{align*} [x,[y,z]] - [y,[x,z]] = [[x,y],z]. \end{align*} Hence \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) = [[x,y],z] = \operatorname{ad}_{[x,y]}(z). \end{align*} [guided] We need to prove that $\operatorname{ad}$ preserves the Lie bracket. Since the target Lie algebra $\mathfrak{gl}(\mathfrak g)$ has bracket given by the commutator of linear maps, the expression to compute is \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_x \circ \operatorname{ad}_y -\operatorname{ad}_y \circ \operatorname{ad}_x. \end{align*} To compare this linear map with $\operatorname{ad}_{[x,y]}$, we evaluate both maps on an arbitrary element $z \in \mathfrak g$. Using the definition of $\operatorname{ad}_x$ and $\operatorname{ad}_y$, we get \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) &= (\operatorname{ad}_x \circ \operatorname{ad}_y)(z) -(\operatorname{ad}_y \circ \operatorname{ad}_x)(z) \\ &= \operatorname{ad}_x([y,z])-\operatorname{ad}_y([x,z]) \\ &= [x,[y,z]]-[y,[x,z]]. \end{align*} The goal is to show that this equals $\operatorname{ad}_{[x,y]}(z)$, which by definition is \begin{align*} \operatorname{ad}_{[x,y]}(z) = [[x,y],z]. \end{align*} We now justify the sign changes used in the Jacobi manipulation. The Lie bracket is alternating, so $[w,w]=0$ for every $w \in \mathfrak g$. Applying this to $w=u+v$, where $u,v \in \mathfrak g$, and expanding by bilinearity gives \begin{align*} 0 &= [u+v,u+v] \\ &= [u,u] + [u,v] + [v,u] + [v,v] \\ &= [u,v] + [v,u]. \end{align*} Thus $[v,u] = -[u,v]$ for every pair $u,v \in \mathfrak g$. The Jacobi identity says that \begin{align*} [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0. \end{align*} Using the alternating consequence $[z,x] = -[x,z]$, the middle term becomes \begin{align*} [y,[z,x]] = [y,-[x,z]] = -[y,[x,z]]. \end{align*} Using the same consequence with $u=z$ and $v=[x,y]$, the final term becomes \begin{align*} [z,[x,y]] = -[[x,y],z]. \end{align*} Therefore Jacobi becomes \begin{align*} [x,[y,z]] - [y,[x,z]] - [[x,y],z] = 0. \end{align*} Rearranging gives \begin{align*} [x,[y,z]] - [y,[x,z]] = [[x,y],z]. \end{align*} Substituting this into the commutator computation yields \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) = [[x,y],z] = \operatorname{ad}_{[x,y]}(z). \end{align*} This is exactly the bracket-preservation identity evaluated at the arbitrary element $z \in \mathfrak g$. [/guided] [/step] [step:Conclude that the adjoint map is a representation] The preceding step proves that for every $x,y,z \in \mathfrak g$, \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)}(z) = \operatorname{ad}_{[x,y]}(z). \end{align*} Since two linear maps $\mathfrak g \to \mathfrak g$ are equal if they agree on every element of $\mathfrak g$, it follows that \begin{align*} [\operatorname{ad}_x,\operatorname{ad}_y]_{\mathfrak{gl}(\mathfrak g)} = \operatorname{ad}_{[x,y]}. \end{align*} Together with the $k$-linearity of $\operatorname{ad}$, this says precisely that \begin{align*} \operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g) \end{align*} is a Lie algebra homomorphism. Hence $\operatorname{ad}$ is a representation of $\mathfrak g$ on the [vector space](/page/Vector%20Space) $\mathfrak g$. [/step] [step:Identify the kernel with the center] By definition, \begin{align*} \ker(\operatorname{ad}) = \{x \in \mathfrak g : \operatorname{ad}_x = 0_{\mathfrak{gl}(\mathfrak g)}\}, \end{align*} where $0_{\mathfrak{gl}(\mathfrak g)}: \mathfrak g \to \mathfrak g$ is the zero linear map. For $x \in \mathfrak g$, the equality $\operatorname{ad}_x = 0_{\mathfrak{gl}(\mathfrak g)}$ holds if and only if \begin{align*} \operatorname{ad}_x(y) = 0 \end{align*} for every $y \in \mathfrak g$. By the definition of $\operatorname{ad}_x$, this condition is equivalent to \begin{align*} [x,y] = 0 \end{align*} for every $y \in \mathfrak g$. Hence \begin{align*} \ker(\operatorname{ad}) &= \{x \in \mathfrak g : [x,y] = 0 \text{ for every } y \in \mathfrak g\} \\ &= Z(\mathfrak g). \end{align*} This proves the kernel statement and completes the proof. [/step]