Find the Lie algebra corresponding to this Lie group

[Kara386]

Homework Statement

The group $G = \{ a\in M_n (C) | aSa^{\dagger} =S\}$ is a Lie group where $S\in M_n (C)$. Find the corresponding Lie algebra.

Homework Equations

The Attempt at a Solution

As far as I've been told the way to find these things is to set $a = exp(tA)$, so:
$exp(tA)exp(tS)exp(tA^{\dagger}) = exp(tA)exp(tA^{\dagger})exp(tS)$ Use series expansion:

$= (I+tA+ \frac{t^2A^2}{2}+O(t^3))(I+tA^{\dagger}+\frac{t^2A^{\dagger 2}}{2}+O(t^3))e^{tS}$
$=[I+t(A+A^{\dagger})+t^2(\frac{A^2}{2}+\frac{A^{\dagger 2}}{2}+AA^{\dagger}) + O(t^3)]e^{tS} = e^{tS}$

So require $A+A^{\dagger}=0$ therefore $A = -A^{\dagger}$. This also means the second bracket is zero and the equation is satisfied. Does that mean the Lie algebra of G is $g= \{a\in M_n(C)| A+A^{\dagger}=0\}$? If not, where did I go wrong? Thanks for any help!

 

[fresh_42]

What is the condition on $aSa^\dagger$? The way you wrote it, it's simply $\mathfrak{gl}_n(\mathbb{C})$.

 

[Kara386]

What is the condition on $aSa^\dagger$? The way you wrote it, it's simply $\mathfrak{gl}_n(\mathbb{C})$.

Sorry, that was really silly. Added it, it should read $aSa^\dagger=S$.

 

[fresh_42]

So to be sure: I assume you don't mean the additive group of all matrices with this property, but the multiplicative group of regular matrices of $GL_n(\mathbb{C})$ instead. However, I'm not sure and you didn't qualify $S$. It is a given fixed matrix? Is $S=0$ a possibility? Do we know something about $S\,$? Is it regular?

A remark on the exponential function. $e^0=1\,.$ This is one way to remember that $exp\, : \,\mathfrak{g} \longrightarrow G$ is a function from the algebra into the group, not vice versa. The Lie algebra of a Lie group is basically the tangent space at $1.$ So one way to calculate it would be to calculate the tangent vectors of curves in $G$ at $a=1$. Or you do it the easy way: multiplication becomes addition, inversion a factor $-1$ and $1$ becomes $0$. Or you compute the left invariant vector fields of $G$.

 

[Kara386]

So to be sure: I assume you don't mean the additive group of all matrices with this property, but the multiplicative group of regular matrices of $GL_n(\mathbb{C})$ instead. However, I'm not sure and you didn't qualify $S$. It is a given fixed matrix? Is $S=0$ a possibility? Do we know something about $S\,$? Is it regular?

A remark on the exponential function. $e^0=1\,.$ This is one way to remember that $exp\, : \,\mathfrak{g} \longrightarrow G$ is a function from the algebra into the group, not vice versa. The Lie algebra of a Lie group is basically the tangent space at $1.$ So one way to calculate it would be to calculate the tangent vectors of curves in $G$ at $a=1$. Or you do it the easy way: multiplication becomes addition, inversion a factor $-1$ and $1$ becomes $0$. Or you compute the left invariant vector fields of $G$.

All I know about S is the aforementioned property $S \in M_n(C)$. We don't go into tangent spaces in the lecture series I'm doing; I'm in the second week of only five, and we've been told that tangent spaces would take too long, since none of us have ever touched group theory before. So I think we have to use the easy way to do stuff. Does that mean using the exponential to go from group to algebra is wrong then? Substituting the exponential into the group conditions?

 

[fresh_42]

All I know about S is the aforementioned property $S \in M_n(C)$. We don't go into tangent spaces in the lecture series I'm doing; I'm in the second week of only five, and we've been told that tangent spaces would take too long, since none of us have ever touched group theory before. So I think we have to use the easy way to do stuff. Does that mean using the exponential to go from group to algebra is wrong then? Substituting the exponential into the group conditions?

Yes, because $0$ is the neutral element of the additive group of the vector space, where the algebra structure is defined on, and $1$ the neutral element of the group, i.e. $exp(0)=1$ means from vector space (Lie algebra) to (Lie) group.

So very likely your group is multiplicative and $G = \{A\in \mathbb{M}_n(\mathbb{C})\,\vert \,ASA^\dagger = S \wedge det(A) \neq 0\}$ and $ASA^\dagger =S$, which is $SA^\dagger=A^{-1}S$ translates to $s+a^\dagger=-a+s$ or $a^\dagger = -a$ in the Lie algebra, i.e. the skew-Hermitian matrices. But I'm still not sure about the role of $S$ in this example. If, e.g. $S=0$, then $G=GL_n(\mathbb{C})$ and the Lie algebra of it is $\mathfrak{g}=\mathfrak{gl}_n(\mathbb{C})=\mathbb{M}_n(\mathbb{C})$, i.e. all matrices.

 

[Kara386]

Yes, because $0$ is the neutral element of the additive group of the vector space, where the algebra structure is defined on, and $1$ the neutral element of the group, i.e. $exp(0)=1$ means from vector space (Lie algebra) to (Lie) group.

So very likely your group is multiplicative and $G = \{A\in \mathbb{M}_n(\mathbb{C})\,\vert \,ASA^\dagger = S \wedge det(A) \neq 0\}$ and $ASA^\dagger =S$, which is $SA^\dagger=A^{-1}S$ translates to $s+a^\dagger=-a+s$ or $a^\dagger = -a$ in the Lie algebra, i.e. the skew-Hermitian matrices. But I'm still not sure about the role of $S$ in this example. If, e.g. $S=0$, then $G=GL_n(\mathbb{C})$ and the Lie algebra of it is $\mathfrak{g}=\mathfrak{gl}_n(\mathbb{C})=\mathbb{M}_n(\mathbb{C})$, i.e. all matrices.

Ok thanks, I'll ask for some clarification from the professor I think. :)