Suppose we have a finite group $G$ with subgroup $H$, a representation $\rho\_V$ of $H$ on a finite-dimensional vector space $V$, and an $H$-invariant inner product on $V$:


$$\forall x,y\in V, h\in H,\enspace \langle\rho\_V(h)x, \rho\_V(h)y\rangle = \langle x,y\rangle$$


We will write $V\_I$ for the direct sum of $\lvert G:H \rvert$ copies of $V$:
$$V\_I = \oplus\_{i=1}^{\lvert G:H \rvert} V$$
We define a map $L\_i$ that lifts $V$ into the $i$th copy in the direct sum:
$$L\_i: V\to V\_I\\ L\_i v = 0 \oplus 0 \oplus ... \underbrace{v}\_{i\text{th summand}} \oplus 0 + ...$$
We extend the inner product on $V$ to one on $V\_I$:


$$ \langle \sum\_i L\_i x\_i, \sum\_j L\_j y\_j \rangle = \sum\_i \langle x\_i, y\_i \rangle$$
From each left coset $K\_i$ of $H$ in $G$ we pick an element $k\_i$, so $K\_i=k\_i H$. We then have an induced representation $\rho\_I$ of the group $G$ on $V\_I$:
$$\rho\_I(g) \sum\_i L\_i v\_i = \sum\_i L\_{j(g,i)} \rho\_V(k\_{j(g,i)}^{-1} g k\_i) v\_i$$


where $j(g,i)$ is the index of the coset $K\_i$ to which $g k\_i$ belongs.


Now, suppose we have an **irreducible** representation $\rho\_W$ of $G$ on some finite vector space $W$. The Frobenius reciprocity theorem says that $Hom\_H(W,V)$, the space of $H$-intertwiners from $W$ to $V$, i.e. the space of maps $S$ that satisfy:


$$S: W\to V\\
\forall h\in H,\enspace S \rho\_W(h) = \rho\_V(h) S$$
is isomorphic to the space $Hom\_G(W,V\_I)$ of $G$-intertwiners from $W$ to $V\_I$, i.e. the space of maps $T$ that satisfy:


$$T: W\to V\_I\\
\forall g\in G,\enspace T \rho\_W(g) = \rho\_I(g) T$$


Indeed, given an intertwiner $S\in Hom\_H(W,V)$ we can easily construct an intertwiner $T\_S\in Hom\_G(W,V\_I)$:


$$T\_S w = \sum\_i L\_i S \rho\_W(k\_i^{-1}) w$$


As we vary $S$ over any basis of $Hom\_H(W,V)$, the associated map $T\_S$ will map $W$ into distinct subspaces of $V\_I$, each of which is invariant under the action of $\rho\_I$, and each of which has the property that the restriction of $\rho\_I$ to that subspace is equivalent to $\rho\_W$.


My question is: supposing the dimension of $Hom\_H(W,V)$ is greater than 1, does there exist an "easy" strategy to choose a basis $\{S\_1,S\_2,...\}$ of $Hom\_H(W,V)$ such that the subspaces $T\_{S\_1}(W), T\_{S\_2}(W), ...$ of $V\_I$ will be **mutually orthogonal**?


By "easy", I mean something less computationally demanding than simply finding all the subspaces by means of an arbitrary basis for $Hom\_H(W,V)$, and then decomposing their direct sum into orthogonal invariant subspaces. In other words, I am seeking to leverage the fact that $Hom\_H(W,V)$ is a lower-dimensional space than the direct sum of the $T\_{S\_i}(W)$ in order to carry out a less demanding procedure to achieve the same result.


**Edited to add:** One easy way to get a basis of $Hom\_H(W,V)$ is to take a sufficient number of linearly independent maps $S^{(0)}\_i: W\to V$ and average over the subgroup $H$:


$$S\_i = \frac{1}{|H|}\sum\_{h\in H} \rho\_V(h) S^{(0)}\_i \rho\_W(h^{-1})$$


So by starting with a basis of (non-intertwining) maps from $W$ to $V$, you can project as many as required into $Hom\_H(W,V)$ to obtain a basis.


My (possibly naive) hope is that there might be some way of modifying this construction to lead directly to a basis with the property I'm seeking.



Inclusion of copies of an irrep as orthogonal subspaces of an induced representation
I'm going to assume that the field is $\mathbb{C}$ (what I'm about to say will be wrong over $\mathbb{R}$. The space $Hom\_H(W,V)$ has an inner product given by $\langle f,g\rangle=\sum \langle f(e\_i),g(e\_i)\rangle$ for $e\_i$ an orthonormal basis of $W$. If you write $f,g$ as matrices in orthonormal bases of $V$ and $W$, this will be the component-wise dot product of matrices; taking adjoint will identify $ Hom\_H(W,V)\cong Hom\_H(V,W)$ (this map will only be semi-linear) and the inner product is just the trace of the composition $f^\dagger g\colon W\to W$. You can easily check that taking induction will just multiply this by the index $[G:H]$.


By Schur's lemma, $f^\dagger g$ is a scalar matrix, so $\langle f,g\rangle=0$ if and only if $f^\dagger g=0$; this will precisely happen if $f$ and $g$ have orthogonal images (by the definition of adjoint). So, you just want an orthonormal basis of $Hom\_H(W,V)$ or $Hom\_G(W,V\_I)$, you just do Gram-Schmidt.


