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Suppose I have a linear representation of a finite group $G$. That is a homomorphism $$ \pi: G \rightarrow GL(\mathbb{R}^n) $$

Meaning a collection of matrices, which under matrix multiplication form some subgroup of $G$. My professor claims it's always the case that these transformations are isometries. That is for any matrix $u$ in our representation and vectors $x,y$ we should have have that

$$||u(x-y)|| = ||x-y||$$

So now I'm attempting to derive this property:

Work So Far:

Naturally the determinant of each of the matrices is $\pm 1$ since they are are real matrices, and the group is finite (if this wasn't the case then repeated powers would yield infinitely many values, and the reals are closed under multiplication/addition).

Now going a step further suppose we pick some matrix $u$, and consider a basis where $u$ is diagonal. Then if any of its eigenvalues are not finite roots of unity, we have that the infinite sequence $u,u^2... $ consists of all unique elements, violating our premise that the representation is finite.

So now we have that the eigenvalues for each element or finite roots of unity, whose product is $\pm1$. But this doesn't seem to enough to show that the elements are isometries.

Being explicit take such an element $u$, consider $(X-Y)$ written as linear combination of eigenvectors of $u$. Then,

$$ X-Y = c_1 e_1 + c_2 e_2 + ... c_n e_n $$

After acting by $u$ we have that

$$ u(X-Y) = \lambda_1 c_1 e_1 + \lambda_2 c_2 e_2 + ... \lambda_n c_n e_n $$

But it's not clear that the norm of this $$ \lambda_1^2 c_2 + ... \lambda_n^2 c_n^2 = c_1^2 + ... c_n^2$$

How to finish?

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    $\begingroup$ Let $\pi$ be a liner representation of a finite group over $\mathbb{R}$. Let $g\in G$ and $v\in \mathbb{R}^n$ a non-zero vector. Then $g(v)$, $g^2(v)$, $\cdots$ are also non-zero. But, $g$, being of finite order, must preserve the length of $v$. What is your exact question? $\endgroup$ – p Groups Mar 21 '16 at 10:42
  • $\begingroup$ Arbitrary invertible matrices are not diagonalizable as far as I know. I mean for finite order matrices over $\mathbb{C}$ it is true but I'm not sure if this is true over $\mathbb{R}$ as well. $\endgroup$ – M.U. Mar 21 '16 at 11:33
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    $\begingroup$ The claim is false. It is true that the the image of $G$ is conjugate to a group of isometries, i.e., there is a change of basis after which the image consists of isometries. $\endgroup$ – Chris Godsil Mar 21 '16 at 11:43
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You need to be a little more precise to get the good property because, as stated the statement is false (see the counter-example below).

Let $G$ be a finite group and $\pi:G\rightarrow GL(\mathbb{R}^n)$ a representation. $\underline{\text{There exists}}$ a scalar product $\langle\cdot,\cdot\rangle_{(G,\pi)}$ on $\mathbb{R}^n$ such that for any $g\in G$, $u,v\in \mathbb{R}^n$ :

$$\langle \pi(g)(u),\pi(g)(v)\rangle_{(G,\pi)}=\langle u,v\rangle_{(G,\pi)}$$

$$\underline{\text{proof :}}$$

Let $\langle\cdot,\cdot\rangle$ be $\underline{\text{any}}$ scalar product on $\mathbb{R}^n$, then for $u,v\in\mathbb{R}^n$ define $\langle u,v\rangle_{(G,\pi)}:=\sum_{g\in G}\langle \pi(g)(u),\pi(g)(v)\rangle$. The bilinear form $\langle\cdot,\cdot\rangle_{(G,\pi)}$ is a $(G,\pi)$-invariant scalar product.

An easy counter example is the following $G:=\mathbb{Z}/2$ and :

$$\pi(1):=\begin{pmatrix}1&-2\\0&-1\end{pmatrix} $$

This defines a representation of $G$ in $\mathbb{R}^2$ (because the matrix is of order $2$). If we take the canonical base $(e_1,e_2)$ of $\mathbb{R}^2$ to be orthonormal for the canonical scalar product $\langle\cdot,\cdot\rangle$ then :

$$\langle \pi(1)(e_1),\pi(1)(e_2)\rangle=\langle e_1,-2e_1-e_2\rangle=-2 $$

whereas :

$$\langle e_1,e_2\rangle=0 $$

An interesting exercise for you would be to explicitely write down a scalar product which is $(G,\pi)$-invariant.

$$\underline{\text{solution :}}$$

Let $\langle \cdot,\cdot\rangle$ be the usual scalar product. By the proof we see that if $u,v\in\mathbb{R}^2$ then the invariant scalar product $\langle \cdot,\cdot\rangle_G$ will be $\langle u,v\rangle_G=\langle u,v\rangle+\langle \pi(1)(u),\pi(1)(v)\rangle$ hence the matrix of $\langle \cdot,\cdot\rangle_G$ is $\begin{pmatrix}1+1&0-2\\0-2&1-1\end{pmatrix}=\begin{pmatrix}2&-2\\-2&0\end{pmatrix}$

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  • $\begingroup$ Ah yes, this is extremely similar to the discussion I had with my professor. But then your counterexample is still very disturbing for me, as it clearly is a faithful representation of $\mathbb{Z}_2$ yet it definitely doesn't preserve Euclidean distances, so it cannot be considered a non-translational euclidean isometry (ie a rotation or reflection or composition of such) $\endgroup$ – frogeyedpeas Mar 22 '16 at 8:12

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