Confusion regarding Matrix representation of Lorentz transformation in Jackson In Jackson (3rd edition p. 545) there are the following equations:
$$A = e^L \tag{11.87}$$
$$\det A = \det(e^L) = e^{Tr L}$$
$$g\widetilde{A}g = A^{-1} \tag{11.88}$$
$$ A = e^L  , g\widetilde{A}g = e^{{g\widetilde{L}g}} , A^{-1} = e^{-L}$$
$$ g\widetilde{L}g = -L $$
$\widetilde{A} $ is the transpose of $A$. I have several doubts:


*

*How is it derived the equation $\det(e^L) = e^{TrL}$? Are we assuming a special type of $L$? 

*In $ g\widetilde{A}g = e^{{g\widetilde{L}g}}$, how is it possible to have $g = e^{g}$?
 A: *

*The property $\det(e^L) = e^{Tr L}$ follows from the matrix identity \begin{equation}
\log \det (M) = Tr \log(M) \to \det(M) = e^{\text{Tr}\, log (M)}
\end{equation}
Choosing $M=e^L$ yields $\,\det(e^L) = e^{\text{Tr}\, L}$.

*The second property can be shown using a Taylor expansion for the matrix $L$ if we assume $g^2=1$. Suppose we want to show that \begin{equation}g \,e^\tilde{L} g = e^{g \tilde{L} g} \end{equation}
By expanding the l.h.s. one has 
\begin{equation}
g \,e^\tilde{L} g = g\left( 1 + \tilde{L} + \frac{1}{2}\tilde{L}^2+...
 \right) g.
\end{equation}
Expanding the r.h.s. we have
\begin{equation}
e^{g \tilde{L} g} = 1 + g \tilde{L} g+\frac{1}{2} g \tilde{L} g \cdot g \tilde{L} g + ...
\end{equation}
Assuming $g^2=\mathbb{1}$ the two expressions agree.

A: Hint: Eq. (11.88) is derived from
$$ \widetilde{A}g A~=~g, \tag{11.86} $$
which reveals that it should really read
$$ g^{-1}\widetilde{A}g ~=~A^{-1}. \tag{11.88'} $$
Note that in many books the shorthand notation for the inverse metric $(g^{-1})^{\alpha\beta}$ is $g^{\alpha\beta}$. Also Jackson is talking about the Minkowski metric $g_{\alpha\beta}={\rm diag}(1,-1,-1,-1)$ in SR where the metric and inverse metric have the same component matrix.
