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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.

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    $\begingroup$ If you want to solve it on a computer, then it is easy to reformulate the problem to equations for $S$. $$ \forall a,b \; \forall u,v\in W : \sum_i \langle S_a\rho(k_i^{-1})u, S_b\rho(k_i^{-1})v\rangle = 0 $$ $\endgroup$ – Vít Tuček Aug 4 '14 at 13:34
  • $\begingroup$ OK, so that would be a large system of simultaneous quadratics in the components of the $S_a$, with $\frac{d(d-1)n^2}{2}$ equations, where $d$ is the dimension of $Hom_H(W,V)$ and $n$ is the dimension of $W$. I was hoping more for some kind of direct construction. $\endgroup$ – Greg Egan Aug 4 '14 at 14:00
  • $\begingroup$ I was thinking you can build $S_a$ iteratively. If you have a linearly independent set $\{S_1, \ldots, S_a\}$ you can try to solve the system of linear equations: $S_{a+1}$ is an intertwiner that is "orthogonal" to $\{S_1, \ldots, S_a\}$ such that $\{S_1, \ldots, S_{a+1}\}$ is linearly independent. $\endgroup$ – Vít Tuček Aug 4 '14 at 15:48
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    $\begingroup$ Actually, you can see from that equation that if $S_1$ and $S_2$ have orthogonal images, then the same is true for $T_{S_1}$ and $T_{S_2}$. You already have the projection $\mathrm{Hom}(W,V) \to \mathrm{Hom}_H(W,V)$ which should give you all "orthogonal" intertwiners as soon as you introduce appropriate scalar product on $\mathrm{Hom}(W,V)$ that will turn "orthogonal" (as in orthogonal image) to really orthogonal. I guess $\langle A,B \rangle = \mathrm{Tr} A\overline{B}^T$ will do. $\endgroup$ – Vít Tuček Aug 4 '14 at 17:09
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    $\begingroup$ The catch is that the projection $\mathrm{Hom}(W,V)\to \mathrm{Hom}_H(W,V)$ doesn't preserve orthogonality, so you can't start with an orthogonal basis of $\mathrm{Hom}(W,V)$ and then project it. So I suppose it's necessary to project to $\mathrm{Hom}_H(W,V)$ to get a non-orthogonal basis, and then use Gram-Schmidt or Householder to make it orthogonal. $\endgroup$ – Greg Egan Aug 5 '14 at 0:26
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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.

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