How vector space look like after applying weyl unitarian trick on finite group representation

abstract-algebrafinite-groupslinear algebrarepresentation-theoryunitary-matrices

I have a question about following theorem.

Theorem: Every representation of finite group is equivalent to some unitary representation.

To prove this, we make a new inner product from original one by
$$
\langle v,w \rangle
:= \sum_{g\in G} \bigl(\rho(g)v,\rho(g)w\bigr),
$$
where $(v,w)$ is original inner product,
and then check that this is invariant under linear transformation by representation of group.
because this new inner product is invariant, our representation is considered to be unitary in this new vector space. on the other hand dimension of this new vector space is same as original one, these two vector space are isomorphic and hence our representation is equivalent to some unitary representation.

I was curious how this new vector space looks like.
for example i considered irreducible non-unitary representation of $S_3$
\begin{align}
\Gamma(e)&=\left(\begin{array}{cc} 1&0 \\ 0&1\end{array}\right)\,,&
\Gamma(P_{12})&=\left(\begin{array}{cc} 1&-1 \\ 0 & -1\end{array}\right)\,, &
\Gamma(P_{13})&=\left(\begin{array}{cc} 0&1 \\ 1 & 0\end{array}\right)\,,\\
\Gamma(P_{23})&=\left(\begin{array}{cc} -1&0 \\ -1 & 1\end{array}\right)\,, &
\Gamma(P_{123})&=\left(\begin{array}{cc} 0&-1 \\ 1 & -1\end{array}\right)\,, &
\Gamma(P_{132})&=\left(\begin{array}{cc}-1&1 \\ -1 & 0\end{array}\right)\,,
\end{align}
and then applied weyls unitarian trick and observed how equidistant points transforms.

On original space these equidistant points forms circle and any reversible linear transformations should transform this to some rotated ellipse.
but what I actually observed was ellipse with some dimple.

enter image description here

Am I missing some thing?

Also, how can I obtain linear transformation that leads original vector space to new vector space with this new inner product.

Thanks.

Code I used to make a plot is following.

import matplotlib.pyplot as plt
import numpy as np

#define angle parameter for unit vectors
theta = np.arange(0,2*np.pi,0.01)

#define representation matrix
reps = [np.mat([[1,0],[0,1]]),
        np.mat([[1,-1],[0,-1]]),
        np.mat([[0,1],[1,0]]),
        np.mat([[-1,0],[-1,1]]),
        np.mat([[0,-1],[1,-1]]),
        np.mat([[-1,1],[-1,0]])]

#make polar plot figure
plt.figure()
ax = plt.subplot(polar=True)
#plot original vectors
ax.scatter(theta,np.array([1 for t in theta]),label="original circle")

inners = []
# loop for all unitvectors
for vec in [[np.cos(t),np.sin(t)] for t in theta]:
    inner = 0
    for rep in reps:
        #transform unit vector by representation matrix
        tran = rep*np.mat(vec).T
        # calculate inner product
        prod = np.dot(np.array(tran).T,np.array(tran))
        #add inner products of all represantation transformation
        inner = inner + prod
    # take sqrt to check norm of vector
    inners.append(np.sqrt(inner))

#plot result    
ax.scatter(theta,inners,c="red", label="after appling weyls trick")
ax.legend(bbox_to_anchor=(1.5, 1.2),loc="upper right")

Best Answer

Say $\langle -,-\rangle$ is the Euclidean norm and the $G$-invariant inner product is

$$ \langle u,v\rangle_G:=\frac{1}{|G|}\sum_{g\in G} \langle\rho(g)u,\rho(g)v\rangle=u^T\left(\frac{1}{|G|}\sum_{g\in G}\rho(g)^T\rho(g)\right)v $$

with norm $\|v\|_G^2=\langle v,v\rangle_G$.

Here is how I understand your graph. Parametrizing the unit circle by $v(\theta)=(\cos\theta,\sin\theta)$ ("unit" here with respect to the Euclidean norm), you graph the polar equation $r(\theta)=\|v(\theta)\|_G$. If we think of this curve as the result of applying a transformation pointwise to the unit circle, that transformation would be $T(v)=\|v\|_Gv$, which is not a linear transformation of $v$. Since $T$ is nonlinear, there's no expectation that applying $T$ to a circle should yield an ellipse, as there would be for linear transformations.

What is true is that for any constant $C$, the locus of points $\|v\|_G=C$ is an ellipse. Specifically, in this case I compute that $\|(x,y)\|_G^2=\frac{4}{3}(x^2-xy+y^2)$. Graphing $x^2-xy+y^2=1$ (for instance) on Wolfram|Alpha, it appears to be an ellipse with diagonal axes. Setting it equal to $a(x+y)^2+b(y-x)^2$, we get

$$ \|(x,y)\|_G^2 = \frac{1}{3}(y+x)^2+(y-x)^2. $$

Note the coordinates $y\pm x$ are, up to a scalar multiple of $\frac{1}{\sqrt{2}}$, the scalar projections of the point $(x,y)$ onto the diagonal lines $y=\pm x$, which are the $x,y$ axes rotated $45^\circ$.

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