Showing that Lie($U(n)$)=$\mathfrak{u}(n)$
What is the dimension of $U(n)$ and $\mathfrak{u}(n)$? Is it true that $\dim(U(n))=\dim(\mathfrak{u}(n))$?
If the above is true, to prove that Lie($U(n)$)=$\mathfrak{u}(n)$, it is sufficient to prove that if $X\in T_e(U(n))$ then $X\in\mathfrak{u}(n)$. For this, if $X\in T_e(U(n))$ then $X=\dot{\gamma}(0)$ for some $\gamma:I\to U(n)$, then as $\gamma(t)\in U(n)$ for all $t\in I$, it is true that $\gamma(t)\gamma(t)^*=Id$ and thus $0=\frac{d}{dt}|_{t=0}\gamma(t)\gamma(t)^*$, but I would like to use that $\frac{d}{dt}|_{t=0}\gamma(t)\gamma(t)^*=\frac{d}{dt}|_{t=0}\gamma(t)+\frac{d}{dt}|_{t=0}\gamma(t)^*$, why is this true? If the latter is true, we would have to $0=\frac{d}{dt}|_{t=0}\gamma(t)\gamma(t)^*=\frac{d}{dt}|_{t=0}\gamma(t)+\frac{d}{dt}|_{t=0}\gamma(t)^*=X+X^*$ and thus $X\in\mathfrak{u}(n)$, whereby Lie($U(n)$)=$\mathfrak{u}(n)$.