 Solution

For a Lie group $G$ with identity $e$, its Lie algebra as a set is the tangent space $T_{e}G$:The ambient embedding identifies these tangent vectors with vectors in ${\mathbb{R}}^N$. For a matrix Lie group, the identity is $I$ and the tangent vectors are matrices.

If $\gamma (t)\in {\mathrm{SL}}_n$ with $\gamma (0)=I$, put $A={\gamma }^{\prime }(0)$. The determinant expansion $\det (I+tA+o(t))=1+t\mathrm{tr}A+o(t)$ follows directly from its permutation formula: to first order only the diagonal entries contribute. Since $\det \gamma (t)=1$, differentiation gives $\mathrm{tr}A=0$. ThusIn fact equality holds: if $\mathrm{tr}A=0$, then $\exp (tA)$ has determinant $\exp (t\mathrm{tr}A)=1$ and derivative $A$ at zero.