Understanding the Relationship between Orthogonal and Unitary Groups

[LagrangeEuler]

I'm a little bit confused. Matrices
$$\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$$
$\theta \in [0,2\pi]$
form a group. This is special orthogonal group $SO(2)$. However it is possible to diagonalize this matrices and get
$$\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i \theta} \end{bmatrix}=e^{i \theta}\oplus e^{-i\theta}.$$
It looks like that $e^{i\theta}$ is irreducible representation of $SO(2)$. However in $e^{i\theta}$ we have complex parameter $i$ and this is unitary group $U(1)$. Where am I making the mistake?

 

[fresh_42]

There is no mistake. It depends on the field, i.e. whether you allow complex scalars or not. The angle $\theta$ represents likewise a rotation or a complex number on the unit circle. Your choice. Or a real number if you forget about the group structure.

 

[martinbn]

The two groups are isomorphic.

 

[LagrangeEuler]

Thanks. I am a bit confused because if I want to speak about rotation in $\mathbb{R}^2$. Is then
\begin{bmatrix}
\cos \theta & \sin \theta\\
-\sin \theta & \cos \theta
\end{bmatrix}
reducible or irreducible representation of $\textrm{SO}(2)$? Because every element I can but in block diagonal form that includes complex numbers in the diagonals.

 

[fresh_42]

Thanks. I am a bit confused because if I want to speak about rotation in $\mathbb{R}^2$. Is then
\begin{bmatrix}
\cos \theta & \sin \theta\\
-\sin \theta & \cos \theta
\end{bmatrix}
reducible or irreducible representation of $\textrm{SO}(2)$? Because every element I can but in block diagonal form that includes complex numbers in the diagonals.

Which dimension does the representation space have in this example?

 

[LagrangeEuler]

It is two-dimensional representation. If I understand the question correctly.

 

[Infrared]

Is then
\begin{bmatrix}
\cos \theta & \sin \theta\\
-\sin \theta & \cos \theta
\end{bmatrix}
reducible or irreducible representation of $\textrm{SO}(2)$?

This is reducible as a complex representation and irreducible as a real representation.

 

[LagrangeEuler]

Thanks. But how to know if I have a complex or real representation of $\mathrm{SO}(2)$? If the vector space is real is then representation real?

 

[Infrared]

Yes, a representation of a group $G$ here means a homomorphism $\rho:G\to GL(V)$ where $V$ is a vector space. You say that the representation is real or complex when $V$ is a real or complex vector space.

 

[fresh_42]

It is two-dimensional representation. If I understand the question correctly.

As real rotation we have only one coordinate $\theta$ hence it is one dimensional and therefore cannot be reducible. No dimensions available.

This is reducible as a complex representation and irreducible as a real representation.

Why that? If it is a complex vector space we still have only one complex dimension: $U(1)$.

 

[Infrared]

Why that? If it is a complex vector space we still have only one complex dimension: $U(1)$.

I mean that the standard representation $SO(2;\mathbb{R})\to GL_2(\mathbb{R})$ is irreducible, but the standard representation $SO(2)\to GL_2(\mathbb{C})$ given by viewing elements of $SO(2;\mathbb{R})$ as complex matrices is reducible. The second representation is the complexification of the first.

As real rotation we have only one coordinate hence it is one dimensional and therefore cannot be reducible. No dimensions available.

The group $SO(2;\mathbb{R})$ is 1-dimensional. That does not mean that all its real irreducible representations are 1-dimensional.

 

[fresh_42]

I mean that the standard representation $SO(2;\mathbb{R})\to GL_2(\mathbb{R})$ is irreducible, but the standard representation $SO(2)\to GL_2(\mathbb{C})$ given by viewing elements of $SO(2;\mathbb{R})$ as complex matrices is reducible. The second representation is the complexification of the first.The group $SO(2;\mathbb{R})$ is 1-dimensional. That does not mean that all its real irreducible representations are 1-dimensional.

Sure, but he asked about a rotation matrix, and then the angle is all we have.

 

[Infrared]

@fresh_42 I don't understand your comment. The OP question as I interpret it is whether the standard (real) representation of $SO(2)$ is irreducible and why it isn't a problem that these matrices are diagonalizable over $\mathbb{C}.$

@LagrangeEuler I'll also add that you have to do a little bit more work to see why the complex standard representation is reducible- what you have written down isn't enough because the change of basis matrix you use depends on which matrix in $SO(2;\mathbb{R})$ you're using. Given a representation $\rho:G\to GL_n(\mathbb{C}),$ it is definitely possible for every matrix $\rho(g)$ to be diagonalizable without the representation itself being reducible.

 

[fresh_42]

@fresh_42 I don't understand your comment.

I thought he meant the one parameter group operating on itself, not as rotation of $\mathbb{R}^2$. My mistake.