$Z(\mathfrak{g})$ equals Lie algebra of $Z(G)$? I have tried to prove the following problem: 


Suppose $G$ is a connected matrix Lie group. Then the center of $\mathfrak{g}$ which is $\ker \textrm{ad}$ is equal to the Lie algebra of $Z(G)$.



Recall that $\textrm{ad} : \mathfrak{g} \to \text{gl}(\mathfrak{g})$ is the map that sends $X \in \mathfrak{g}$ to $\textrm{ad}_X$, where $\textrm{ad}_X$ is defined by
$$\textrm{ad}_X(Y) = [X,Y].$$
Similarly $\textrm{Ad}_A$ is the map such that for all $X \in \mathfrak{g}$, we have $\textrm{Ad}_A(X) = AXA^{-1}$. Now I have proven the problem above as follows: If $Y \in \ker \textrm{ad}$, then certainly for all $t \in \Bbb{R}$ we have $\textrm{ad}_{tY} = 0$. Now we have given any $x \in G$ that 
$$\begin{eqnarray*} e^{tY}xe^{-tY} &=& \textrm{Ad}_{e^{tY}}(x) \\
&=& e^{\textrm{ad}_{tY}}(x) \\
&=& x \end{eqnarray*}$$
by assumption and since this holds for all $x$, we conclude that $e^{tY} \in Z(G)$ for all $t$ and so $Y$ is in the Lie algebra of $Z(G)$. 
For the converse, if $Y$ is in the Lie algebra of $Z(G)$, then $e^{tY} \in Z(G)$ for all $t$ and so given any $x \in G$, $e^{tY}xe^{-tY} = x$. We now have $e^{\textrm{ad}_{tY}}x =x$ for all $x$, which implies that $e^{\textrm{ad}_{tY}} = I \in \textrm{GL}(\mathfrak{g})$. Taking the derivative at $t = 0$ implies that $\textrm{ad}_Y = 0$ and so $Y \in \ker \textrm{ad}$.


However: I have nowhere used the assumption of connectedness of $G$. What have I done wrong?


 A: Since $\operatorname{ad}_{tY}$ acts on the Lie algebra of $G$ (only), you cannot expect $e^{\operatorname{ad}_{tY}}=\operatorname{Ad}_{e^{tY}}$ to hold as identity of functions defined beyond the connected component $G_0$ of the identity of $G$. Indeed if you think of $G=O(2,\mathbf R)$ where $G_0$ is the commutative group of $SO(2,\mathbf R)$, you can see that $\operatorname{ad}_{tY}=0$ for any $Y$, but $\operatorname{Ad}_{e^{tY}}$ is conjugation by a rotation, which does not fix reflections (unless $e^{tY}=\pm \mathrm{Id}$).
In fact now that I reread your question carefully, I see that you defined $\operatorname{Ad}_A$ to be a function defined on the Lie algebra $\mathfrak{g}$, not on $G$, so this makes your argument that applies it to $x\in G$ invalid in the first place. But my point is that although you can define $\operatorname{Ad}_g$ (maybe with a different notation) to be a conjugation by $g$ in $G$, you cannot expect $\operatorname{ad}_{tY}$ to tell you what any such $\operatorname{Ad}_g$ does beyond $G_0$.
A: As pointed out by Marc, I applied a certain identity that held valid only for elements of the Lie algebra to all of the Lie group. I have now reproduced a new proof that relies on the following lemma:


$\textbf{Lemma:}$ Suppose $G$ is a connected matrix Lie group. Let $\textrm{Ad}$, $\textrm{ad}$ be defined as above and let $\mathfrak{g}$ be the Lie algebra of $G$. Then $Z(G) = \ker \textrm{Ad}$.


I will write lower case letters for elements of $G$ and capitals for elements of the Lie algebra.
Proof: Suppose $x \in \ker \textrm{Ad}$. Then connectedness of $G$ implies that I can write any $y \in G$ as $e^{Y_1}\ldots e^{Y_n}$ for some $Y_1,\ldots,Y_n \in \mathfrak{g}$. Then
$$\begin{eqnarray*} 
xyx^{-1} &=& xe^{Y_1}\ldots e^{Y_n}x^{-1} \\
&=& xe^{Y_1}x^{-1}\ldots xe^{Y_n}x^{-1} \\
&=& e^{xY_1x^{-1}}\ldots e^{xY_nx^{-1}}\\
&=& e^{Y_1}\ldots e^{Y_n}\\
&=& y \end{eqnarray*}$$
proving that $x \in Z(G)$. Conversely, suppose that $y \in Z(G)$. Then I can define a group homomorphism
$$\begin{eqnarray*} \psi_y :& G& \to G \\ 
&x&\mapsto yxy^{-1}. \end{eqnarray*} $$
Now if $\phi_y$ is the induced map on Lie algebras, we see given any $X \in \mathfrak{g}$ that
$$\begin{eqnarray*} e^{t\phi_y(X)} &=& e^{\phi_y(tx)} \\
&=& \psi_y(e^{tX}) \\
&=& ye^{tX}y^{-1} \\
&=& e^{t(yXy^{-1})}. \end{eqnarray*}$$
Now if I differentiate both sides at $t = 0$, I see that $\phi_y(X) = yXy^{-1}$, so that $\phi_y$ is actually $\textrm{Ad}_y$. However from the second last step, $ye^{tX}y^{-1} = e^{tX}$ by definition of $y \in Z(G)$. It follows that $\phi_y(X) =\textrm{Ad}_y(X) = X$ for all $X \in \mathfrak{g}$, from which it follows that $\textrm{Ad}_y = I \in \textrm{GL}(\mathfrak{g})$. It now follows that $y \in \ker \textrm{Ad}$, completing the proof of the lemma.
$$\hspace{6in} \square$$
Having completed the proof of the lemma above, we can now prove our main result:


The Lie algebra of $Z(G)$ is equal to $\ker \textrm{ad}$.


If $Y $ is in the Lie algebra of $Z(G)$, then $e^{tY} \in Z(G)$ for all $t$. By the lemma, this means that given any $X \in \mathfrak{g}$,
$$e^{t \textrm{ad}_Y}= e^{\textrm{ad}_{tY}} (X) = \textrm{Ad}_{e^{tY}}(X) = X.$$
Differentiating at $t = 0$ now gives that $\textrm{ad}_Y = 0$, proving that $Y \in \ker \textrm{ad}$. 
Reverse inclusion: If $Y \in \ker \textrm{ad}$, by the lemma we just need to prove that $e^{tY} \in \ker \textrm{Ad}$ for all $t$. Now for all $X \in \mathfrak{g}$, we have
$$\begin{eqnarray*} X &=& e^{\textrm{ad}_{tY}}(X) \\
&=& \textrm{Ad}_{e^{tY}}(X) \end{eqnarray*}$$
implying that $\textrm{Ad}_{e^{tY}} = I \in \textrm{GL}(\mathfrak{g})$ for all $t$, from which it follows that $e^{tY}$ is in $\ker \textrm{Ad}$.
$$\hspace{6in} \square$$
