The special orthogonal group for $n=2$ is defined as: $$SO(2)=\big\{A\in O(2):\det A=1\big\}$$ I am trying to prove that if $A\in SO(2)$ then: $$A=\left(\begin{array}{cc} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{array}\right)$$ My idea is show that $\Phi:S^1\to SO(2)$ defined as: $$z=e^{\theta i}\mapsto \Phi(z)=\left(\begin{array}{cc} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{array}\right)$$ is an isomorphism of Lie groups. It is easy prove that is an monomorphism of Lie groups. How can I prove that is also surjective?

  • 2
    $\begingroup$ In this case, you can use directly the definition of orthogonal group $A^T A=1$ to get equations for coefficients of the matrix. $\endgroup$ Dec 23, 2016 at 20:09

3 Answers 3


Let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in \mathrm{SO}(2)$. Then, $$\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$ and $$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc=1.$$ Thus, $\mathrm{SO}(2)$ is the subset of $\mathbb{R}^4$ satisfying the following four equations: $$ \begin{align*} a^2+c^2 &= 1 \\ b^2+d^2 &= 1\\ ad-bc &= 1\\ ab+cd &= 0. \end{align*} $$ The first two equations imply that $(a,c)$ and $(b,d)$ lie on a circle, so $$a=\cos\alpha,\quad c=\sin\alpha,\quad b=\cos\beta,\quad d=\sin\beta$$ for some angles $\alpha,\beta\in\Bbb R$. Inserting in the last two equations, we get $$ \begin{align*} \cos\alpha\sin\beta-\cos\beta\sin\alpha &= 1 \\ \cos\alpha\cos\beta+\sin\alpha\sin\beta &= 0. \end{align*} $$ Using the angle sum trigonometric identities, these equations are $$ \begin{align*} \sin(\beta-\alpha) &= 1 \\ \cos(\beta-\alpha) &= 0. \end{align*} $$ Hence, $\beta-\alpha\in \pi/2+2\pi\Bbb Z$ and we get $$\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}.$$

  • $\begingroup$ Maybe the $-$ should be on lower left at the end, if it follows this computation. $\endgroup$ Dec 23, 2016 at 20:50
  • $\begingroup$ @PeterFranek Not according to my derivation. Anyway changing $\alpha$ to $-\alpha$ gives the form with the '$-$' on the bottom left corner. So the two forms are equivalent. $\endgroup$
    – Spenser
    Dec 23, 2016 at 20:55


If you want develop your idea, note that the field of complex numbers is isomorphic the the subring of $M(2,\mathbb{R})$ of the matrices of the form $$ \begin{pmatrix} a&-b\\ b&a \end{pmatrix} $$

( that are commuting matrices, so that they are a field).

So there is a bijection from the complex numbers of the form $e^{i\theta}=\cos \theta + i \sin \theta$ and the matrix of the form:

$$ $$ \begin{pmatrix} \cos \theta&-\sin \theta\\ \sin \theta&\cos \theta \end{pmatrix} $$


Maybe one can argue as follows: The above first three equations determine the group variety and show it is connected and one dimensional. The suggested matrix realization satisfies them and form a norm torus, which therefore must be the all group.


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