Uniqueness of Hermitian inner product Let V be an irreducible representation of a finite group G.How to show that up to scalars,there is a unique Hermitian inner product on V preserved by G. i know of how to get an inner product. but i have no idea on the uniqueness part. i think i have to use schur's lemma in some way
 A: This is basically the idea of the answer above (which I learned from), but I want to make it easier to read for less experienced people (such as myself, several hours ago) and also correct what I think is a typo in it.


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*For any complex vector space $V$, there is a complex vector space $\bar{V}$, the conjugate vector space. We construct $\overline{V}$ as follows: 


a. as a set $\overline{V}$ consists of symbols $\overline{v}$ for $v \in V$.
b. $\bar{v} + \overline{w} = \overline{v + w}$.
c. For $\alpha \in \mathbb{C}$, $\alpha \overline{v} = \overline{\overline{\alpha} v}$. (The scaling is conjugated.)


*The significance of the construction $\overline{V}$ is that (like tensor products), it takes a nearly linear problem into a linear problem. In particular...


Def: Let $V$ and $W$ be two vector spaces over $\mathbb{C}$. An additive map $\phi : V \to W$ is anti-linear if $\phi(\alpha v) = \overline{\alpha} \phi(v)$.
Proposition: Anti-linear maps from $V$ to $W$ are in canonical isomorphism with linear maps from $\bar{V}$  to $W$. (The isomorphism is as a $\mathbb{C}$vector space.)
Pf: For $\phi : V \to W$, define $\overline{\phi} : \bar{V} \to W$ by $\overline{\phi}(\overline{v}) = \phi(v)$. Then $\overline{\phi}( \alpha \overline{v}) = \overline{\phi}(\overline{\overline{\alpha} v} ) = \phi (\bar{\alpha} v) = \alpha \phi(v) = \alpha \overline{\phi}(\overline{v})$. The correspondence $\phi$ to $\overline{\phi}$ has an obvious inverse (if we identify the spaces of $V$ and $\overline{V}$, which I don't recommend doing because it is confusing, then $\phi = \overline{\phi}$), and moreover is $\mathbb{C}$ linear.


*Now we can give a very succient definition of a sesquilinear form. Recall that a sesquilinear form is a bi-additive map $V \times V \to \mathbb{C}$ which is linear in the first entry, and anti-linear in the second entry. Given our construction of $\overline{V}$, we can now define a sesquilinear form as a bilinear map $V \times \overline{V} \to \mathbb{C}$, or as a linear map $V \otimes \overline{V} \to \mathbb{C}$.

*Generalities from tensor products tell us : $(\overline{V} \otimes V)^* \cong  (\overline{V})^* \otimes V^* \cong Hom_{\mathbb{C}-linear} (\overline{V}, V^*)$. So each sesquilinear form induces a $\mathbb{C}$ linear map from $\overline{V}$ to $V^*$. If you unpack this, the correspondence is : $\langle, \rangle$ to $\psi : \overline{V} \to V^*$ with $\psi(\overline{v}) = \langle \_ , v \rangle$.

*Now we work on representation theory. If $V$ is a $G$-module, then $\overline{V}$ and $V^*$ are both naturally $G$-modules.
a. $g \times \overline{v} = \overline{g \times v}$ defines the action on the conjugate vector space.
b. $(g \times f)(\_) = f( g^{-1} \times \_)$ defines the action on functionals.


*Thus, $Hom(\overline{V}, V^*)$ is a $G$ module. We can interpret the fixed points in two different ways:


a. For any two $G$ representations $V$,$W$, it is a general fact the $G$ fixed points of $Hom_{\mathbb{C} linear} (V,W)$ are the $G$-homomorphisms / intertwiners.
b. Identifying $Hom_{\mathbb{C} linear} (V,W)$ with sesquilinear forms, the G-invariant elements are precisely the $G$ invariant forms. (It was an invariant functional on $\bar{V} \otimes V$, and the representation on $\bar{V} \otimes V$ is $g \times (\overline{v} \otimes w) = (g \times \overline{v}) \otimes (g \times w)$.)


*Moreover, $V$ is an irreducible $G$ representation implies that $\bar{V}$ and $V^*$ are irreducible $G$ representations. Indeed, the map $v \to \overline{v}$ is a $\mathbb{C}$ anti -linear intertwiner,and hence will preserve invariant subspaces. Also, if you pick any (non-degenerate) hermitian form on $V$, we get a $\mathbb{C}$ anti-linear intertwiner from $V$ to $V^*$, which will preserve the invariant subspace lattice. (A $\mathbb{C}$ invariant set is also invariant under the conjugate action. Note that these are definitely not linear isomorphims, and in particular the representations are not isomorphic.)

*Now we can put $6$ and $7$ together. (7) If $V$ is irreducible, then so are $\overline{V}$ and $V^*$, hence by Schur's lemma any two $G$isomoprhisms $\psi, \psi'$ from $\overline{V}$ to $V^*$ are equal up to a scalar multiplication, i.e. $\psi = \lambda \psi'$. (6) A G invariant non-degenerate sesquilinear form correspond bijectively to isomoprhisms from $\overline{V}$ to $V^*$, and the bijection commutes with scaling.
Thus, any two $G$ invariant non-degenerate sesquilinear forms are equal up to a scalar.
Something curious -- this seems to imply that if $V$ is an irreducible representation of $G$, and there is a $G$ invariant Hermitian form $\beta$ on $V$, then any $G$ invariant sesquilinear form is automatically Hermitian, since it must be a multiple of $\beta$. This is odd, so perhaps I made a mistake somewhere above (?). On the other hand, $G$ invariance on an irreducible representation is a strong condition, so perhaps not.
A: An easier way to see it is this, let $\phi:G\rightarrow GL(\mathbb{C}^n)$ be a representation that is unitary with respect to the standard inner product and another inner product given by $H$, a positive definite matrix. Then $\phi(g)^\ast H\phi(g)=H$ for all $g\in G$. However $\phi(g)$ is unitary for each $g$, we have $H\phi(g)=\phi(g)H$ for all $g$, so $H=\lambda I$ by Schur's lemma.
A: An inner product is the same as a map from $V \to \bar{V}$: $\langle -, - \rangle$ corresponds to $v \to \langle -, v \rangle$. $G$-invariant inner product corresponds to $G$-invariant maps $Hom_G(V, \bar{V})$. What can you say about this space by Schur's lemma?
