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What is the difference between groups ##SU(n)## and ##SO(n,\mathbb{C})##? They look completely the same.
Unitary matrices satisfy ##A\cdot \bar A^\tau=1## whereas orthogonal matrices satisfy ##A\cdot A^\tau =1##. There is no complex conjugation in the definition of orthogonal groups.What is the difference between groups ##SU(n)## and ##SO(n,\mathbb{C})##? They look completely the same.
Your personal opinion doesn't change the definition.I disagree.
A definition can't be incorrect, and as ##\mathbb{R}\subseteq\mathbb{C}## we get ##\operatorname{SO}(n,\mathbb{C}) \neq \{1\}.## Whether this is reasonable is a different question and the reason why we introduced unitary matrices in the complex case. However, some theorems in linear algebra require an algebraically closed field, so it makes sense to allow any field, not just the real numbers.If you have complex entries in some matrix then condition ##A \cdot A^T=1## is incorrect.
It seems like you have made a mistake in your diagonalization. The matrix you start with does satisfy ##A \cdot A^T=1##, and that property should be invariant under diagonalization, so the matrix you end with should satisfy it too.[tex]
\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix} [/tex]
if you diagonalise this matrix you will get
[tex]\begin{bmatrix}
e^{i \theta} & 0\\
0 & e^{-i\theta}
\end{bmatrix}[/tex]
end then try to emply condition ##A \cdot A^T=1##. You will see that is not valid.
It looks fine to me.It seems like you have made a mistake in your diagonalization.
The property is not invariant under replacing a matrix with its complex diagonalization.The matrix you start with does satisfy ##A \cdot A^T=1##, and that property should be invariant under diagonalization, so the matrix you end with should satisfy it too.
Ah, that's right; it's only invariant for a real diagonalization (which doesn't exist for the rotation matrix given since it has no real eigenvalues).The property is not invariant under replacing a matrix with its complex diagonalization.
$$\frac{1}{\sqrt{3}}\left(I disagree. If you have complex entries in some matrix then condition ##A \cdot A^T=1## is incorrect.
Yes. But you still need to have a group. Show me that group. An infinite group of matrices like this.
Please see this text.Your personal opinion doesn't change the definition.
A definition can't be incorrect, and as ##\mathbb{R}\subseteq\mathbb{C}## we get ##\operatorname{SO}(n,\mathbb{C}) \neq \{1\}.## Whether this is reasonable is a different question and the reason why we introduced unitary matrices in the complex case. However, some theorems in linear algebra require an algebraically closed field, so it makes sense to allow any field, not just the real numbers.
Have a look at https://www.amazon.com/s?k=Gantmacher+Matrix+theory&ref=nb_sb_noss and argue with Gantmacher whether this is "incorrect". Minor difficulty: he passed away in 1964.
Not a problem. I just ask why some people make a difference between ##SU(n)## and ##SO(n,\mathbb{C})##? Could you please show me the two dimensional representation of group ##SO(n,\mathbb{C})##?It looks fine to me.
The property is not invariant under replacing a matrix with its complex diagonalization.
@LagrangeEuler Why is it a problem that the diagonalized form is no longer in ##SO(n,\mathbb{C})##?
Because they're different groups. The matrix @mfb gave is an element of ##SO(2,\mathbb{C})## but not of ##SU(2)##.Not a problem. I just ask why some people make a difference between ##SU(n)## and ##SO(n,\mathbb{C})##?
I don't know what you're asking for here. You've been given the definition and a non-real example. You can take ##\begin{pmatrix}z & -w\\ w & z\end{pmatrix}## where ##z^2+w^2=1## to get the whole group.Could you please show me the two dimensional representation of group ##SO(n,\mathbb{C})##?
##\operatorname{SO}(n,\mathbb{C})## is a group.Yes. But you still need to have a group. Show me that group. An infinite group of matrices like this.
So? Myself has listed some isomorphisms here:Please see this text.
##\varphi\, : \,\operatorname{SO}(2,\mathbb{C})\longrightarrow \operatorname{GL}(2,\mathbb{C})## with ##\varphi(A)(\mathbf{v}):=A\cdot \mathbf{v}.##Not a problem. I just ask why some people make a difference between ##SU(n)## and ##SO(n,\mathbb{C})##? Could you please show me the two dimensional representation of group ##SO(n,\mathbb{C})##?