Suppose we have a smooth curve $\alpha$ in U$(n)$, the Lie group of unitary matrices, with $\alpha(0) = I$. Then we have $\alpha(t)^*\alpha(t) = I$. Let $\dot{\alpha}(0) = X$. I want to show that $\frac{d}{d t}\big{\rvert}_{t = 0}\alpha(t)^*\alpha(t) = X^* + X = 0$.
Now my approach was the following: For general Lie groups with multiplication map $m$, inversion $i$ and $\Delta(g) = (g,g)$ I know that $dm_{(p,q)}(X,Y) = dR_q X + dL_p Y$, $di_e(X) = -X$ and $d\Delta_p (X) = (X,X)$. Because conjugate transpose is inversion in this example we have $\alpha(t)^*\alpha(t) = I = (m \circ (i,id) \circ \Delta )(\alpha(t))$. I think that applying the chain rule gives
$$ \frac{d}{d t}\bigg{\rvert}_{t = 0}\alpha(t)^*\alpha(t) = 0 = dm_{(I,I)}(di_I X,X) = X- X$$
Where is my mistake or why does this approach not work? (I know that I could use the product rule instead)
Just use the product rule: $$ \frac{d}{dt}\alpha(t)^*\alpha(t)=\dot{\alpha}(t)^*\alpha(t)+\alpha(t)^*\dot{\alpha}(t). $$ Then plug in $t=0$ on both sides to get $$ \left.\frac{d}{dt}\right|_{t=0}\alpha(t)^*\alpha(t)=X^*+X. $$