Find all automorphisms of the complex representation of $\mathbb{R}$ by rotations in the euclidean space.

My trial:

I know that the representation of $\mathbb{R}$ by rotations in the space $<\cos x, \sin x>$ is given by the matrix:

\begin{bmatrix} \cos t \qquad - \sin t \\ \sin t \qquad \cos t \end{bmatrix}

But what it will be in Euclidean space and what are the generators of this representation as I understood that I must know the generators to know the whole set of isomorphisms, could anyone help me in this ?

Edit: will Schur's lemma help? I feel that it will help but I do not know why.

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  • $\begingroup$ To clarify, are you asking about the two complex dimensional (i.e. four real dimensional) representation of $\mathbb{R}^+$ where $t$ acts by the matrices you wrote down, or the one complex dimensional (two real dimensional) representation where we identify $\mathbb{C}$ with $\mathbb{R}^2$ and $\mathbb{R}$ acts on it by those matrices as a real representation (and by multiplication by $e^{it}$ as a complex representation)? $\endgroup$ – Nate Feb 21 at 20:40
  • $\begingroup$ the question is from Vinberg section 4 but it is stated without the word complex (that I wrote bold in the question) and my professor said to me solve it in case of complex representation only because in case of real representation the solution will be complicated ...... I think I am in the second case.@Nate $\endgroup$ – Smart Feb 21 at 22:24
  • $\begingroup$ @Nate The answer at the back of the book is all rotations ...... but I do not know why ..... could you please clarify this Nate? $\endgroup$ – hopefully Feb 23 at 14:25
  • $\begingroup$ This question is from a book named "Linear Representations of Groups " by Ernest B. Vinberg $\endgroup$ – hopefully Feb 23 at 14:28
  • $\begingroup$ why the first case is a 2 complex dimensional representation while the second case is a 1 complex dimensional representation? @Nate $\endgroup$ – hopefully Feb 24 at 23:54

Okay let me go through this pretty explicitly:

The additive group of real numbers is acting on $\mathbb{C}$, with $t\in \mathbb{R}$ acting by multiplication by $e^{it}$. This is a one complex-dimensional representation, and maybe it's better to think of it as acting by a $1\times 1$ complex matrix $[e^{it}]$. If we were to think of this as a 2 real-dimensional space then in the basis $\{1, i\}$ the action is given by the $2\times 2$ real matrix you wrote down.

Okay so what are the automorphisms? An automorphism invertible complex linear map $\phi$ from the space (in this case $\mathbb{C}$) to itself which commutes with this action - meaning that for all $t \in \mathbb{R}$ we can either act by $t$ and then apply $\phi$ or we can first apply $\phi$ and then act by $t$ and either way we get the same answer.

In this case we have a one dimensional space so $\phi$ will be given by a $1\times 1$ complex matrix. So the condition is which $1\times 1$ matrices are invertible and commute with $[e^{it}]$ for all $t$? Of course though this becomes a kind of silly question: all $1 \times 1$ matrices commute with each other, so the answer is just all non-zero complex $1\times 1$ matrices.

  • $\begingroup$ why when removing the word complex in the question the solution will be all rotations? $\endgroup$ – hopefully Feb 24 at 1:39
  • $\begingroup$ why you said this "If we were to think of this as a 2 real-dimensional space then in the basis {1,i} the action is given by the 2×2 real matrix you wrote down." how is the transformation from the real representation to the complex representation take place? I do not know the details of this step. $\endgroup$ – hopefully Feb 24 at 2:03
  • $\begingroup$ Could you please look at this problem when you have time math.stackexchange.com/questions/3119824/… $\endgroup$ – hopefully Feb 24 at 16:16

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