How to define a map $V^* \to V$ arising from a non-degenerate Hermitian form on $V$? I am studying Representation of finite groups from the book by Fulton and Harris. There I found the following statement in the form of lemma $3.35$ page no $:$ $40$ which I failed to understand properly.

Let $V$ be an irreducible representation of $G.$ Let $G$ be a finite group and $B$ be a non-degenerate symmetric bilinear form preserved by $G.$ Let $H$ be any non-degenerate $G$-invariant Hermitian form. Then $$V \xrightarrow {B} V^* \xrightarrow {H} V$$ gives a conjugate linear isomorphism $\varphi$ on $V$ such that given $x \in V$ there exists a unique $\varphi (x) \in V$ with $B(x,y) = H(\varphi (x),y).$ Furthermore $\varphi$ commutes with the action of $G$ so that $\varphi^2 = \varphi \circ \varphi$ is a complex linear $G$-module homomorphism.

Question $:$ How does the map $H$ on $V^*$ defined? How do I get hold of $\varphi\ $?
No proper argument has been given in that worthless book. I don't know how such a book gets so famous. Most of non-trivial results are discussed without proofs as if they are so trivial. BTW, Can anybody please help me understanding the lemma?
Thanks for your time.
 A: Note that $H : V\times V \to \mathbb{C}$ has the following properties:

*

*$H$ is sesquilinear (complex antilinear in the first argument and complex linear in the second)

*$H$ is hermitian, i.e. $\overline{H(x, y)} = H(y, x)$, and

*$H$ is non-degenerate, i.e. if $H(x, y) = 0$ for all $y \in V$, then $x = 0$.

There is an induced complex antilinear map $V \to V^*$ given by $x \mapsto H(x, \,\cdot\,)$ which is injective since $H$ is non-degenerate, and is therefore a complex antilinear isomorphism (because $V$ is finite-dimensional). So there is an inverse complex antilinear map $V^* \to V$ - this is the map $H$ they are referring to. It sends a linear functional $L$ to the unique vector $x \in V$ such that $L = H(x, \,\cdot\,)$; more precisely, $x$ is the unique vector for which $L(y) = H(x, y)$ for all $y \in V$.
The non-degenerate bilinear form $B$ gives rise to a complex linear isomorphism $B : V \to V^*$ in a similar way: it is given by $x \mapsto B(x,\,\cdot\,)$. The composition $\varphi := H\circ B : V \to V$ is therefore a complex antilinear isomorphism which sends $x$ to the unique vector $\varphi(x) \in V$ for which $B(x, \,\cdot\,) = H(\varphi(x), \,\cdot\,)$; more precisely, $\varphi(x)$ is the unique vector for which $B(x, y) = H(\varphi(x), y)$ for all $y \in V$.
