Let $G$ be a matrix Lie group and $(Π, V )$ be a finite dimensional unitary representation of $G$ in $V$ . Let $(π, V )$ be the induced representation on the Lie algebra $g$. Show that for each $X ∈ g, π(X)^∗ = −π(X)$.
I know that $π^*(X)=-π(X)^{tr}$ as $(Π, V )$ is unitary so I have $(Π(X)^{tr}=\overline{Π(X)}$. But here what is given to deduce is $π(X)^∗$.
On
What you have is representation $\Pi\colon G\to U(n)$, so $\pi\colon \mathfrak g\to \mathfrak u(n)$ and Lie algebra $\mathfrak u(n)$ consists of skew-Hermitian matrices.
One way to see this is to use exponential map. In particular, $$\mathfrak u(n) = \{ X\in\mathfrak{gl}(n,\mathbb C)\, | \, e^{tX}\in U(n), \,\forall t\in\mathbb R \}$$ so consider function $$t\mapsto \langle\, e^{tX}v, e^{tX}w\,\rangle \tag{1}$$ for $X\in\mathfrak {gl}(n,\mathbb C)$ and some vectors $v,w$.
Assume that $X\in\mathfrak u(n)$. Since $e^{tX}$ is unitary for all $t$ by definition, the function in $(1)$ is constant, so taking it's derivative at $0$ gives you
$$\langle Xv,w \rangle + \langle v, Xw \rangle = 0.$$
For the other direction, take $X$ skew-Hermitian. Taking the derivative of the function in $(1)$ gives you $0$, so it must be constant. Evaluating it at $0$ gives you $\langle\, e^{tX}v, e^{tX}w\,\rangle = \langle v, w\rangle$, so $e^{tX}\in U(n)$ for all $t$, and thus $X\in\mathfrak u(n).$
On
This answer is someway related to @Enner answer, it is given that $\Pi$ is unitary.
Now we know that $\pi({X})^*\pi({X})=1\\ \Rightarrow\frac{d}{dt}\Pi(e^{tX})^*\Pi(e^{tX})|_{t=0}=0\\ \Rightarrow\frac{d}{dt}(e^{t\pi(X)})^*(e^{t\pi(X)})|_{t=0}=0\\ \Rightarrow\frac{d}{dt}e^{t\pi(X)^*}(e^{t\pi(X)})|_{t=0}=0\\ \Rightarrow(\pi(X)^*e^{t\pi(X)^*}(e^{t\pi(X)})+e^{t\pi(X)^*}\pi(X)(e^{t\pi(X)}))|_{t=0}=0\\ \Rightarrow \pi(X)^*=-\pi(X)$
Hence proved.
Hint: If $\Pi$ is orthogonal, the image of $\pi$ is contained in the Lie algebra of infinitesimal orthogonal matrices, so you have: $\langle \pi(X)(u),v\rangle+\langle u,\pi(X)(v)\rangle=0$.