Similarity transformation changing the determinant to 1

[spin_100]

In Michael Tinkham's book, Group theory and Quantum Physics, he derives a theorem that any matrix representation can be converted to an equivalent transformation which is unitary. i.e $A$ is converted to $B = S^-1 A S$ such that B is unitary. My question is how is it possible to find such a transformation? We know similarity transformations preserve the determinant, so if we start with a matrix A with $det(A) \neq 1$ then it is not possible to get a matrix B with a determinant equal to 1 since we know the determinant of a Unitary matrix is 1. I may be confusing something. Could someone help me out?

 

[PeroK]

In Michael Tinkham's book, Group theory and Quantum Physics, he derives a theorem that any matrix representation can be converted to an equivalent transformation which is unitary. i.e $A$ is converted to $B = S^-1 A S$ such that B is unitary. My question is how is it possible to find such a transformation? We know similarity transformations preserve the determinant, so if we start with a matrix A with $det(A) \neq 1$ then it is not possible to get a matrix B with a determinant equal to 1 since we know the determinant of a Unitary matrix is 1.

Your analysis is correct. If $B = S^{-1}AS$, then $det(B) = det(A)$.

 

[spin_100]

Your analysis is correct. If $B = S^{-1}AS$, then $det(B) = det(A)$.

So that means all group representations must have det = 1 (should belong to SL(n))?

 

[PeroK]

So that means all group representations must have det = 1 (should belong to SL(n))?

I don't know enough about the subject of group representations. What is the context of this in the textbook?

 

[spin_100]

I don't know enough about the subject of group representations. What is the context of this in the textbook?

It's about representing groups using matrices and their applications. The author hasn't mentioned anything about the form the matrix representation of the group takes.

 

[PeroK]

It's about representing groups using matrices and their applications. The author hasn't mentioned anything about the form the matrix representation of the group takes.

I thought it was GL in general, but perhaps he's limiting himself to some special cases?

 

[martinbn]

Unitary matrices don't have determinat equal to 1, only absolute value 1. Also I am guessing he is talking about finite groups.

 

[spin_100]

Unitary matrices don't have determinat equal to 1, only absolute value 1. Also I am guessing he is talking about finite groups.

But that would still mean that the $|det(A)| = 1$to begin with as similarity transformations on A won't change $|det(A)|$.

 

[martinbn]

But that would still mean that the $|det(A)| = 1$to begin with as similarity transformations on A won't change $|det(A)|$.

Why is that a problem!

 

[spin_100]

Why is that a problem!

So that would mean we can only represent finite groups with matrices whose det have unit modulus. Am i right? I also looked up Zee's group theory in a nutshell. The author states that the matrix representation of group G belongs to the General Linear group and then goes on to prove the orthogonality theorem. Shouldn't it be a Special Linear group? I have taken a course on abstract algebra but representation theory is really new for me. Thanks

 

[martinbn]

So that would mean we can only represent finite groups with matrices whose det have unit modulus. Am i right?

This follows from the fact that the group is finite.

I also looked up Zee's group theory in a nutshell. The author states that the matrix representation of group G belongs to the General Linear group and then goes on to prove the orthogonality theorem. Shouldn't it be a Special Linear group? I have taken a course on abstract algebra but representation theory is really new for me. Thanks

It belongs to the general linear group.

 

[PeroK]

So that would mean we can only represent finite groups with matrices whose det have unit modulus. Am i right? I also looked up Zee's group theory in a nutshell. The author states that the matrix representation of group G belongs to the General Linear group and then goes on to prove the orthogonality theorem. Shouldn't it be a Special Linear group? I have taken a course on abstract algebra but representation theory is really new for me. Thanks

I think in QFT the aim is to look for unitary representations of the Lorentz or Poincare groups. That makes sense. But, I don't see how any matrix representation is automatically similar to a unitary representation. That would be a big theorem! Apparanetly:

Theorem: Any finite dimensional representation of a finite group is equivalent to a unitary representation.

That wasn't hard to find using good old Google!

 

[fresh_42]

So that means all group representations must have det = 1 (should belong to SL(n))?

No. Take $\operatorname{GL}(n)$ as a group and $\mathbb{C}^n$ as representation space.

But if you consider the corresponding Lie algebras which is usually the case in QM, then we have a representation $\mathfrak{g} \stackrel{\varphi }{\rightarrow} \mathfrak{gl}(n)$ of the Lie algebra $\mathfrak{g}.$ Now, we have $\mathfrak{gl}(n)=\mathfrak{sl}(n)\oplus \mathfrak{Z}(\mathfrak{gl}(n))$ where $\mathfrak{Z}(\mathfrak{gl}(n))\cong \mathbb{C}$ is the center of $\mathfrak{gl}(n).$ The commutator is therefore in
$$
[\varphi (X),\varphi (Y)]\in [\mathfrak{gl}(n),\mathfrak{gl}(n)]=[\mathfrak{sl}(n)\oplus \mathfrak{Z}(\mathfrak{gl}(n),\mathfrak{sl}(n)\oplus \mathfrak{Z}(\mathfrak{gl}(n)]=[\mathfrak{sl}(n),\mathfrak{sl}(n)]=\mathfrak{sl}(n)
$$

 

[spin_100]

I think in QFT the aim is to look for unitary representations of the Lorentz or Poincare groups. That makes sense. But, I don't see how any matrix representation is automatically similar to a unitary representation. That would be a big theorem! Apparanetly:

Theorem: Any finite dimensional representation of a finite group is equivalent to a unitary representation.

That wasn't hard to find using good old Google!

Sorry, I didn't mention that I am talking about finite-dimensional representation of finite groups. My confusion is how can I convert a matrix representation of a group element with a modulus of determinant not equal to 1 to a similarity transformation? Or does this mean that we can't do it using a similarity transformation but using a transformation such as by first dividing the representation matrix by its determinant to reduce the matrix to one with unit modulus det and then we can use the similarity transformation as given in the orthogonality theorem?

 

[spin_100]

No. Take $\operatorname{GL}(n)$ as a group and $\mathbb{R}^n$ as representation space.

So in general it is not possible to convert a matrix in $\operatorname{GL}(n)$ to an equivalent unitary matrix by a similarity transformation ?

 

[spin_100]

This follows from the fact that the group is finite.

Could you please explain how it follows from that? Thanks

 

[martinbn]

Could you please explain how it follows from that? Thanks

An element of a finite group is of finite order. Hence, so will be the matrix it is represented by. You have that $A^n=1$, so the determinant will be $n$-th root of unity and of absolute value 1.

 

[fresh_42]

So in general it is not possible to convert a matrix in $\operatorname{GL}(n)$ to an equivalent unitary matrix by a similarity transformation ?

Conjugation ($X\mapsto S^{-1}XS$) is a group isomorphism and $\operatorname{GL}(n,\mathbb{C}) \not\cong \operatorname{U}(n).$

We need therefore the entire context of your claim. A "representation can be converted" does not mean "is"!

 

[spin_100]

An element of a finite group is of finite order. Hence, so will be the matrix it is represented by. You have that $A^n=1$, so the determinant will be $n$-th root of unity and of absolute value 1.

Thank you so much now It all makes sense. Could you recommend a good book on representation theory with applications in physics? I am reading Michael Tinkham's book now.

 

[fresh_42]

Thank you so much now It all makes sense. Could you recommend a good book on representation theory with applications in physics? I am reading Michael Tinkham's book now.

Sorry, I only know math books, but you should emphasize the physics part if you choose a book. E.g. a mathematician probably won't speak of "generators of a group" when he actually means an element of a Lie algebra, i.e. a vector field, and a representation in addition. Physicists have developed a language over the years that is slightly different from the mathematical point of view because many terms have grown historically and have never been changed whereas mathematics has undertaken a major change in how things are represented due to Bourbaki. But it is important that you can understand the "physical slang".

Lecture notes are easy to find and a good alternative. Search for a subject, e.g. "group theory for physicists" or "groups in quantum mechanics" and add "introduction" if necessary, but always add "+ pdf". This grants you to find either public books or lecture notes on one of the really many university servers around the world.

 

[martinbn]

There is a book by van der Waerden "Group Theory and Quantum Mechanics". It may be too advanced depending on your background, but it has a chapter on representations.