About the classification of continuous group morphisms from $(\mathbb{S},\times)$ the circle of elements of module $1$ in $\mathbb{C}$ to $\text{GL}_n (\mathbb{R})$ which is a well known problem, I prove it without reference and had a completely different proof than the usual proof using (I believe) classification of morphisms from $\mathbb{R}$ to $\text{GL}_n ( \mathbb{R})$ with exponentials.
EDIT: I realized this is a well known result for student that aim the ENS (French very hard to get math school), I have the original reference, but it is in French (sorry guys) here are all the references that I know about
- S. Francinou, H. Gianella, S. Nicolas, Exercices de mathématiques, Oraux X-ENS, Algèbre 2, 2e édition, Cassini. Exercise 4.29 page 251.
- A PDF written by a fellow student that does the exact same proof (it follows from the book).
However, I did not find any reference that use my proof or something analogous.
My question is (since it has very probably been done) do you have one reference?
The proof uses various mechanisms and I find it interesting.
If my question is off-topic, please tell me, I'll close the subject.
Here is the theorem:
Theorem: Let $\varphi: \mathbb{S} \longrightarrow \text{GL}_n (\mathbb{R})$ be a continuous group morphism, then there exist $r \leqslant \frac{n}{2}$, $m_1, \dots m_r \in \mathbb{Z}$, $P \in \text{GL}_n (\mathbb{R})$ such that:
$$\forall \theta \in \mathbb{R}, \qquad \varphi(e^{i\theta}) = P \begin{pmatrix} R(m_1 \theta) \\ & \ddots \\ && R(m_r \theta) \\ &&&I_{n-2r} \end{pmatrix} P^{-1}$$ With $R(\theta)$ the rotation matrix $\begin{pmatrix} \cos(\theta) &\sin(\theta)\\ -\sin(\theta) & \cos (\theta) \end{pmatrix}$
Some intuition about the theorem in itself: $\varphi$ stabilizes planes in $\mathbb{R}^n$ and rotates them as you would rotate $\mathbb{R}^2$, the more you move on the circle, the more the planes rotate. For the rest of the space, it has to be trivial on it.
Here is the proof in multiple steps:
Prove that every continuous group endomorphism $\varphi$ of $\mathbb{S}$ is of the form $$z \longmapsto z^m$$ for some $m \in \mathbb{Z}$. To do so, I apply the lifting path theorem to $t \mapsto \varphi(e^{it})$ and get a continuous endomorphism of $\mathbb{R}$, $\phi$. These morphisms are well known and of the form $t \mapsto at$. A quick study gives $a \in \mathbb{Z}$ which finishes the proof.
We get into the deep and fix a goal: use codiagonalisation on the image. For that we already have commutativity since the source group is abelian, we just need to show diagonalisation. Let us take $z \in \mathbb{S}$ of finite order, i.e. $$z \in \bigcup_{n \in \mathbb{N}^*} U_n = U_\infty .$$ Then there exist $n$ for which $z^n = 1$, so $\varphi(z)^n = I_n$ is idempotent, thus diagonalisable over $\mathbb{C}$ with spectrum in $\mathbb{S}$. So the image of $U_\infty$ by $\varphi$ is codiagonalisable, which means there exists $ Q \in \text{GL}_n ( \mathbb{C} ) $ such that
$$ \varphi_{ | U_\infty } (z) = Q \begin{pmatrix} \tilde{\lambda_1} (z) \\ & \ddots \\ && \tilde{\lambda_n} (z) \end{pmatrix} Q^{-1}. $$ with $\tilde{\lambda_i}$ continuous (hence uniformly continuous) functions from $ U_\infty $ to $ \mathbb{S} $.
Thanks to the theorem of extension for uniformly continuous mappings, we extend each $\lambda$ and $\varphi_{|\mathbb{U}_\infty}$ and by unicity, we obtain: $$\varphi (z) = Q \begin{pmatrix} \lambda_1 (z) \\ & \ddots \\ && \lambda_n (z) \end{pmatrix} Q^{-1}.$$ Obviously, the $\lambda_i$ are continuous endormorphism of the circle, thus there exists $k_1, \dots , k_n \in \mathbb{Z}$ such that $$\varphi (z) = Q \begin{pmatrix} z^{k_1} \\ & \ddots \\ && z^{k_n} \end{pmatrix} Q^{-1}.$$
By a quick study of the characteristic polynomial of $\varphi \left( e^{i \frac{1}{ \max_{k_i \neq 0} k_i}} \right) $ which does not depend on the field extension thus it has real coefficients, we can reorganize the $k_i$ in $(m_1, -m_1, m_2, -m_2,\dots, m_r,-m_r, 0, \dots 0)$. It is common knowledge that these two matrices are $\mathbb{C}$-conjuguate $$\begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i \theta} \end{pmatrix} \sim R(\theta)$$
We get then $$ \varphi(e^{i\theta}) = P' \begin{pmatrix} R(m_1 \theta) \\ & \ddots \\ && R(m_r \theta) \\ &&&I_{n-2r} \end{pmatrix} P'^{-1}$$ with $P' \in \text{GL}_n ( \mathbb{C} ) $, but two real $\mathbb{C}$-conjuguate matrices are $\mathbb{R}$-conjuguate, which proves the result.
In more "traditional" representation-theoretic language, the first three steps is the classification of $\mathbb C$-representation of $S^1$:
Proposition. Let $V$ be a smooth finite-dimensional representation of $S^1$ with coefficients in $\mathbb C$. Then $V$ decomposes as the direct sum of irreducible representations. Moreover, every irreducible representation of $S^1$ is of the form $V_n=\mathbb C$, where $z\in S^1$ acts as $z^n$, for some $n\in\mathbb Z$. In other words, every finite-dimensional representation of $S^1$ is of the form $\bigoplus_{i\in\mathbb Z}V_i^{m_i}$, for some integers $m_i\ge0$.
The usual proof relies on integration over $S^1$, and mimics the classical proof for finite groups. Your proof looks new (!), and it would be interesting to see if the argument generalizes to arbitrary compact Lie groups. In general I believe the finite order elements still are a dense subset, but they no longer commute with each other, so the argument is more complicated (so the integration technique may be simpler there).
The next part is a "Galois-descent" for such representations:
Proposition. A representation $\bigoplus_{n\in I}V_n$ descends to a representation over $\mathbb R$ if and only if $m_i=m_{-i}$ for each $i\in\mathbb Z$.
For this part, your proof looks pretty much like the standard argument. The generalization to arbitrary compact Lie groups would involve a calculation of Frobenius-Schur indicators. For the theory for finite groups (which extends readily to compact groups), see Conrad's notes.