Understanding the Relationship between Orthogonal and Unitary Groups

  • #1
LagrangeEuler
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I'm a little bit confused. Matrices
[tex]\begin{bmatrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{bmatrix}[/tex]
##\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
[tex]\begin{bmatrix}
e^{i\theta} & 0 \\
0 & e^{-i \theta}
\end{bmatrix}=e^{i \theta}\oplus e^{-i\theta}.[/tex]
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?
 
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  • #2
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.
 
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  • #3
The two groups are isomorphic.
 
  • #4
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.
 
  • #5
LagrangeEuler said:
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?
 
  • #6
It is two-dimensional representation. If I understand the question correctly.
 
  • #7
LagrangeEuler said:
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.
 
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  • #8
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?
 
  • #9
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.
 
  • #10
LagrangeEuler said:
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.
Infrared said:
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)##.
 
  • #11
fresh_42 said:
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.

fresh_42 said:
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.
 
  • #12
Infrared said:
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.
 
  • #13
@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.
 
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  • #14
Infrared said:
@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.
 
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