What is the difference between groups SU(n) and SO(n,C)?

[LagrangeEuler]

What is the difference between groups $SU(n)$ and $SO(n,\mathbb{C})$? They look completely the same.

 

[fresh_42]

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.

 

[LagrangeEuler]

I disagree. If you have complex entries in some matrix then condition $A \cdot A^T=1$ is incorrect. Could you please give me a matrix with complex entries that are not just real and show that the condition $A \cdot A^T=1$ is fine? I do not think that this is possible. For instance
$$\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$$
if you diagonalise this matrix you will get
$$\begin{bmatrix} e^{i \theta} & 0\\ 0 & e^{-i\theta} \end{bmatrix}$$
end then try to emply condition $A \cdot A^T=1$. You will see that is not valid.

 

[fresh_42]

I disagree.

Your personal opinion doesn't change the definition.

If you have complex entries in some matrix then condition $A \cdot A^T=1$ is incorrect.

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.

 

[PeterDonis]

$$\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$$
if you diagonalise this matrix you will get
$$\begin{bmatrix} e^{i \theta} & 0\\ 0 & e^{-i\theta} \end{bmatrix}$$
end then try to emply condition $A \cdot A^T=1$. You will see that is not valid.

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.

 

[Infrared]

It seems like you have made a mistake in your diagonalization.

It looks fine to me.

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.

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 $SO(n,\mathbb{C})$?

 

[PeterDonis]

The property is not invariant under replacing a matrix with its complex diagonalization.

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).

 

[mfb]

I disagree. If you have complex entries in some matrix then condition $A \cdot A^T=1$ is incorrect.

$$\frac{1}{\sqrt{3}}\left(
\begin{matrix}
2 & -i \\
i & 2
\end{matrix}\right)$$

At WolframAlpha

 

[LagrangeEuler]

$$\frac{1}{\sqrt{3}}\left(
\begin{matrix}
2 & -i \\
i & 2
\end{matrix}\right)$$

At WolframAlpha

Yes. But you still need to have a group. Show me that group. An infinite group of matrices like this.

 

[LagrangeEuler]

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.

Please see this text.

 

[LagrangeEuler]

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 $SO(n,\mathbb{C})$?

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 $SO(n,\mathbb{C})$?

 

[Infrared]

Not a problem. I just ask why some people make a difference between $SU(n)$ and $SO(n,\mathbb{C})$?

Because they're different groups. The matrix @mfb gave is an element of $SO(2,\mathbb{C})$ but not of $SU(2)$.

Could you please show me the two dimensional representation of group $SO(n,\mathbb{C})$?

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.

 

[fresh_42]

Yes. But you still need to have a group. Show me that group. An infinite group of matrices like this.

$\operatorname{SO}(n,\mathbb{C})$ is a group.

Please see this text.

So? Myself has listed some isomorphisms here:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
That does not change the definition of $\operatorname{SO}(n,\mathbb{K}).$ Whether it is a meaningful definition or not depends on the application. Mathematics only requires contradiction-free.

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 $SO(n,\mathbb{C})$?

$\varphi\, : \,\operatorname{SO}(2,\mathbb{C})\longrightarrow \operatorname{GL}(2,\mathbb{C})$ with $\varphi(A)(\mathbf{v}):=A\cdot \mathbf{v}.$

It is also needed for the complexification of the orthogonal group in Lie theory.

The fact that you do not like it means nothing. This question has been answered and the debate is becoming ridiculous.

Thread closed.