Definition of the special unitary group

  • #1
redtree
313
13
TL;DR Summary
Defining the special unitary group independent of matrix representation
In matrix representation, the special unitary group is distinguished from the more general unitary group by the sign of the matrix determinant. However, this presupposes that the special unitary group is formulated in matrix representation. For a unitary group action NOT formulated in matrix representation, what differentiates a special unitary group action from a more general unitary group action, such as in the infinite dimensional case?
 
Physics news on Phys.org
  • #2
What are the definitions of the special and general unitary groups which you use?
 
  • #3
Why would you want to do that? I mean, your question literally asks to define a matrix group without matrices. "Wash my hands but don't make me wet!"

Since the result will be an algebraic group, i.e. a subgroup of ##\operatorname{GL}(n,\mathbb{C})## we will have finally matrices. Hence, we need to know where this limit is allowed to be, i.e. what you would accept as an answer that doesn't allow the complaint: "But there are matrices!"

E.g. we could say that a special unitary group is a complex, simple, compact Lie group of type ##A_l.## No matrices were necessary. Otherwise, we would quite quickly speak of rotations, and these are matrices in disguise.
 
  • Like
Likes dextercioby
  • #4
That's really the question: how to define the special unitary group independent of matrix representation. Would it be to define the special unitary group as the subgroup of group actions of the unitary group continuously connected to unity?
 
  • #5
redtree said:
That's really the question: how to define the special unitary group independent of matrix representation. Would it be to define the special unitary group as the subgroup of group actions of the unitary group continuously connected to unity?
As I already mentioned: define it as a Chevalley group by its root system and let it operate on buildings. This is an abstract approach that doesn't need a realization.
 
  • Like
Likes apostolosdt
  • #6
redtree said:
That's really the question: how to define the special unitary group independent of matrix representation. Would it be to define the special unitary group as the subgroup of group actions of the unitary group continuously connected to unity?
Why do uou want to do that? And how is the unitary griup defined?
 
  • #7
If a group action is defined by a particular type of matrix, then why bother to utilize a representation that maps a group action to a matrix? In this case, isn't the mapping provided by the representation already included in the definition of the group action?
 
  • #8
redtree said:
If a group action is defined by a particular type of matrix, then why bother to utilize a representation that maps a group action to a matrix? In this case, isn't the mapping provided by the representation already included in the definition of the group action?
Can you be more specific, may be examples? This is very unclear.
 
  • #9
For a group ##G##, a representation is a mapping ##\pi: g \rightarrow GL(n,\mathbb{C}): g \in G; M(n,\mathbb{C}) \in GL(n,\mathbb{C})##
$$ \pi: g \rightarrow M(n,\mathbb{C})$$

For the unitary group ##U(n)##
$$U(n) = \{u \in \mathbb{C}^n: u^{\dagger} u = \textbf{1} \} $$
such that for ##\pi: u \rightarrow GL(n,\mathbb{C}): u \in U(n); M(n,\mathbb{C}) \in GL(n,\mathbb{C})##
$$ \pi: u \rightarrow M(n,\mathbb{C})$$

Whereas, if one defines the unitary group ##U(n)## as follows:
$$U(n) = \{u \in M(n,\mathbb{C}): u^{\dagger} u = \textbf{1} \} $$
Then the mapping of the representation is already included in the definition.
 
  • #10
And thus ##SU(2)## would be the simply connected subgroup of ##U(2)##?
 
  • #11
redtree said:
For a group ##G##, a representation is a mapping ##\pi: g \rightarrow GL(n,\mathbb{C}): g \in G; M(n,\mathbb{C}) \in GL(n,\mathbb{C})##
$$ \pi: g \rightarrow M(n,\mathbb{C})$$
This does not define a representation. A representation of ##G## (by complex matrices) is a group homomorphism ##\pi : G \to GL(n,\mathbb{C})## for some ##n \ge 1##. So, not only is ##\pi(g)## an invertible ##n\times n## complex matrix for every ##g\in G##, but also ##\pi(g)\pi(h) = \pi(gh)## for all ##g,h\in G##.

It might be a typo, but the notation ##M(n,\mathbb{C}) \in GL(n,\mathbb{C})## does not make sense. The ring ##M(n,\mathbb{C})## of all complex ##n \times n##-matrices contains ##GL(n,\mathbb{C})##.
 
Last edited:
  • Like
Likes dextercioby
  • #12
Euge said:
The group ##M(n,\mathbb{C})## of all complex ##n \times n##-matrices contains ##GL(n,\mathbb{C})## as a subgroup.
Not to be pedantic, but the first one is not a group (under multiplication)
 
  • #13
redtree said:
For the unitary group ##U(n)##
$$U(n) = \{u \in \mathbb{C}^n: u^{\dagger} u = \textbf{1} \} $$
This is not the unitary group. It is not even a group!
 
  • #14
martinbn said:
Not to be pedantic, but the first one is not a group (under multiplication)
Yes, I meant a ring. Fixed.
 
  • Like
Likes martinbn
  • #15
Euge said:
This does not define a representation. A representation of ##G## (by complex matrices) is a group homomorphism ##\pi : G \to GL(n,\mathbb{C})## for some ##n \ge 1##. So, not only is ##\pi(g)## an invertible ##n\times n## complex matrix for every ##g\in G##, but also ##\pi(g)\pi(h) = \pi(gh)## for all ##g,h\in G##.

It might be a typo, but the notation ##M(n,\mathbb{C}) \in GL(n,\mathbb{C})## does not make sense. The ring ##M(n,\mathbb{C})## of all complex ##n \times n##-matrices contains ##GL(n,\mathbb{C})##.
Yes. I should have written a group homomorphism instead of mapping for ##\pi##.

I was using ##M(n,\mathbb{C})## to represent a given nxn matrix in the general linear group (not the matrix ring). Given the confusion, I could have chosen better notation.
 
  • #16
martinbn said:
This is not the unitary group. It is not even a group!
Given a group ##G: G \in \mathbb{C}^n, n\geq 1##, one can define a subgroup ##U \in G##, the unitary group as follows:
$$U = \{u \in U: u^{\dagger} u = 1 \}$$

With a representation ##\pi: U \rightarrow GL(n,\mathbb{C})##
 
  • #17
redtree said:
Given a group ##G: G \in \mathbb{C}^n, n\geq 1##, one can define a subgroup ##U \in G##, the unitary group as follows:
$$U = \{u \in U: u^{\dagger} u = 1 \}$$

With a representation ##\pi: U \rightarrow GL(n,\mathbb{C})##
You can define a group ##H## and any group homomorphism ##\pi: H \rightarrow \operatorname{GL}(n,\mathbb{C})## is an ##n##-dimensional (linear) representation of ##H##.

The unitary group ##U## is the group of complex matrices such that ##U^\dagger \cdot U = I.## These matrices are automatically invertible so they build a multiplicative group.

The embedding ##\operatorname{U}(n) \hookrightarrow \operatorname{GL}(n,\mathbb{C})## is a homomorphism and thus a representation.

The operation ##v \longmapsto U\cdot v## with ##v\in \mathbb{C}^n## and ##U\in \operatorname{U}(n)## defines an operation of the unitary group on the ##n##-dimensional complex vector space. It is therefore also a representation.

Both representations are the same: one written as a homomorphism, the other one as an operation.

##U\in \mathbb{C}^n## as you mentioned earlier does not make much sense, as ##G\in \mathbb{C}^n## doesn't. Notation is important since matrices, vectors, scalars, operations, and homomorphisms must be distinguished and recognized by their notation.

Nevertheless, this is the natural representation, matrices transforming vectors into vectors. This is literally the definition of a matrix. However, you have required a representation independent of matrices. How are linear representations even relevant to your question? If you want (emphasis mine) ...

redtree said:
TL;DR Summary: Defining the special unitary group independent of matrix representation

... you have two possibilities as far as I see it:

a) Keep the matrices in the definition of ##\operatorname{U}(n)##, choose an arbitrary set ##X## and a group homomorphism ##\varphi \, : \,\operatorname{U}(n)\longrightarrow \operatorname{Sym}(X).## This defines a representation on ##X.## It is only a linear representation if ##X## is a vector space. If ##X## is not a vector space, then you get a non-linear representation. Representation is only another word for a group homomorphism in this context, which is the crucial part. ##\operatorname{Sym}(X)## does not have to be a general linear group.

b) Get rid of the matrices that define the unitary group in the first place. Chevalley groups and root systems are such a possibility. Using ##\operatorname{U}(n)## instead of ##\operatorname{SU}(n)## is not really an obstacle since one is a central extension of the other. Adding some diagonal matrices ##(z,z,\ldots,z)## is only inconvenient. Chevalley groups operate on Weyl chambers and buildings IIRC. (Sorry, it has been some time since I looked into that book.) This gives you a representation without matrices. It is the classification of simple Lie groups done backward. Matrices from this perspective are only necessary to assure others that your construction isn't out of thin air and has a realization.
 
  • Like
Likes redtree
  • #18
How would you write the definition of the unitary group under "b", where
$$U(n) = \{ ....\}$$
 
  • #19
redtree said:
TL;DR Summary: Defining the special unitary group independent of matrix representation

In matrix representation, the special unitary group is distinguished from the more general unitary group by the sign of the matrix determinant. However, this presupposes that the special unitary group is formulated in matrix representation. For a unitary group action NOT formulated in matrix representation, what differentiates a special unitary group action from a more general unitary group action, such as in the infinite dimensional case?
Cam you provide more context? What book are you using? What class is this? What are you currently studying?
 
  • #20
redtree said:
How would you write the definition of the unitary group under "b", where
$$U(n) = \{ ....\}$$
Well, this question forced me to have a closer look. I'm afraid I was wrong with Chevalley groups. They seem to be defined only over finite fields. Finite fields yield nice geometric transformations. I thought that would work over any field. So either we give up the complex numbers, or we are sent back to linear algebraic groups.

The more I search for a way to get rid of the matrix definition, the deeper I fall into the rabbit hole. I'm currently at
$$
\operatorname{U}(n,\mathbb{C})=\operatorname{O}(2n,\mathbb{R})\cap \operatorname{GL}(n,\mathbb{C})\cap \operatorname{Sp}(2n,\mathbb{R}).
$$
Wikipedia has an interesting approach
https://en.wikipedia.org/wiki/Unitary_group#Related_groups
which leads to invariant forms and interesting generalizations, but finally still coordinates, i.e. matrix entries.

The concepts that use the Dynkin diagrams all seem to set up the Lie algebra and integrate it (exponentiation) to define the group. And the integrations need coordinates, i.e. matrix entries again.

Remains the topological perspective. But that would either destroy the group property or give us the algebraic group again.
 
  • #21
jbergman said:
Cam you provide more context? What book are you using? What class is this? What are you currently studying?
My primary text is the following: Woit, Peter, Woit, and Bartolini. Quantum theory, groups and representations. Vol. 4. New York, NY, USA:: Springer International Publishing, 2017.
 
  • #22
fresh_42 said:
Well, this question forced me to have a closer look. I'm afraid I was wrong with Chevalley groups. They seem to be defined only over finite fields. Finite fields yield nice geometric transformations. I thought that would work over any field. So either we give up the complex numbers, or we are sent back to linear algebraic groups.

The more I search for a way to get rid of the matrix definition, the deeper I fall into the rabbit hole. I'm currently at
$$
\operatorname{U}(n,\mathbb{C})=\operatorname{O}(2n,\mathbb{R})\cap \operatorname{GL}(n,\mathbb{C})\cap \operatorname{Sp}(2n,\mathbb{R}).
$$
Wikipedia has an interesting approach
https://en.wikipedia.org/wiki/Unitary_group#Related_groups
which leads to invariant forms and interesting generalizations, but finally still coordinates, i.e. matrix entries.

The concepts that use the Dynkin diagrams all seem to set up the Lie algebra and integrate it (exponentiation) to define the group. And the integrations need coordinates, i.e. matrix entries again.

Remains the topological perspective. But that would either destroy the group property or give us the algebraic group again.
Instead of square matrices, why can't these be formulated in terms of tensors?
 
  • #23
redtree said:
Instead of square matrices, why can't these be formulated in terms of tensors?
They can, but this is just another way to write a matrix ##A\in M(n,\mathbb{F})##:
$$
A=\sum_{\rho=1}^R v_\rho^* \otimes w_\rho \Longleftrightarrow A=\sum_{\rho=1}^R v_\rho^*\cdot w_\rho^\tau \Longleftrightarrow A(x) = \sum_{\rho=1}^R v_\rho^*(x) \cdot w_\rho
$$
where ##R## is the matrix rank of ##A## and ##x\in \mathbb{F}^n.## ##\tau## is the transposition turning the column vectors ##w_\rho \in \mathbb{F}^n## into row vectors, and ##v_\rho^* \in \left(\mathbb{F}^n\right)^*## are linear forms, written as columns in the formula in the middle. It is coordinate-free but as soon as you want to use it, you will choose actual vectors ##v_\rho,w_\rho## and thus bases and thus a matrix ##A.##
 
  • #24
I have to say I still don't understand the question and the motivation for it!

Can you choose a different group, let's say ##SL_2## and ask the question about it? And if possible what would be an acceptable answer.
 
  • #25
@redtree Maybe you could quote which part of Woit's book you don't understand.

Woit speaks about linear representations ##(V,\pi)## on complex vector spaces ##V##, i.e. complex-linear mappings
\begin{align*}
\pi\, : \,G&\longrightarrow \operatorname{GL}(V) \\
A&\longmapsto (v\longmapsto A(v)=A\cdot v)
\end{align*}
with ##\pi(A\cdot B)=\pi(A)\cdot\pi(B)##. ##A,B## are group elements and ##\pi(A),\pi(B)## are matrices. If we choose ##G## to be a matrix group itself, e.g. ##U(1)## or ##U(n)## then we already have matrices and we can assume that ##\pi=\operatorname{id}_G,## i.e. ##\pi ## doesn't do anything.

If he says unitary representation, then he requires that the vector space ##V## has an inner product such that
\begin{align*}
\bigl\langle x\, , \,ay+z \bigr\rangle&=a\bigl\langle x\, , \,y \bigr\rangle +\bigl\langle x\, , \,z \bigr\rangle \\
\bigl\langle ax+y\, , \,z \bigr\rangle &=\overline{a}\bigl\langle x\, ,\, z \bigr\rangle + \bigl\langle y\, , \,z \bigr\rangle \tag{1} \\
\bigl\langle x\, , \,y \bigr\rangle &=\overline{\bigl\langle y\, , \,x \bigr\rangle } \\[12pt]
\bigl\langle \pi(A)(v)\, , \,\pi(A)(w) \bigr\rangle &=\bigl\langle v\, , \,w \bigr\rangle
\end{align*}

If ##A\in U(n)## is a unitary matrix, then
$$
\bigl\langle A(v)\, , \,A(w) \bigr\rangle =\bigl\langle A^\dagger A(v)\, , \,w \bigr\rangle =\bigl\langle I(v)\, , \,w \bigr\rangle =\bigl\langle v\, , \,w \bigr\rangle \tag{2},
$$
means the natural representation (of unitary matrices on a complex vector space with a (Hermitian) inner product) is a unitary representation.

We don't have any fancy group homomorphisms, the representations are all linear and finite-dimensional, and the vector spaces are complex with a complex inner product, means a Hermitian inner product because of semi-linearity ##(1).## Unitary matrices lead to unitary representations by simply letting them act on complex vectors.

Woit writes ##\pi ## since the theorems hold for any linear, complex, finite-dimensional representation, some even for infinite-dimensional representations. Still, you can understand the matter by thinking of ##\pi## being the identity:

\begin{align*}
\pi = \operatorname{id}_{U(n)}\, : \,U(n)&\longrightarrow \operatorname{GL}(\mathbb{C}^n) \\
A&\longmapsto (v\longmapsto A(v)=A\cdot v)\\[12pt]
\begin{pmatrix}a_{11}&\ldots & a_{1n}\\ \vdots &&\vdots \\ a_{n1}&\ldots & a_{nn}\end{pmatrix}&\longmapsto \left(\begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}\longmapsto \begin{pmatrix}a_{11}&\ldots & a_{1n}\\ \vdots &&\vdots \\ a_{n1}&\ldots & a_{nn}\end{pmatrix}\cdot \begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}\right)
\end{align*}
If you want to be very picky, then the only difference between elements of ##U(n)## and their representation as a matrix in ##\operatorname{GL}(n,\mathbb{C})=\operatorname{GL}(\mathbb{C}^n)## is, that the matrices ##A## on the LHS, the elements of ##A\in U(n)## are just group elements which happened to be defined as matrices, and their homomorphic images ##\pi(A)=\operatorname{id}_{U(n)}(A)=A \in \operatorname{GL}(\mathbb{C}^n)## on the RHS are explicitly seen as linear transformations that map vectors from ##\mathbb{C}^n## onto vectors from ##\mathbb{C}^n.## But this is a very, very academic way to distinguish them. And such a perspective becomes obsolete anyway the moment we directly apply ##A## on ##v## and write ##A(v)## without referring to a representation like I did (out of laziness) in ##(2).## Formally, it should have been
\begin{align*}
\bigl\langle \pi(A)(v)\, , \,\pi(A)(w) \bigr\rangle &=\bigl\langle \operatorname{id}(A)(v)\, , \,\operatorname{id}(A)(w) \bigr\rangle = \bigl\langle A(v),A(w) \bigr\rangle \\
&=\bigl\langle A^\dagger (A(v))\, , \,w \bigr\rangle =\bigl\langle (A^\dagger A)(v)\, , \,w \bigr\rangle
=\bigl\langle I(v)\, , \,w \bigr\rangle \\
&=\bigl\langle v\, , \,w \bigr\rangle
\end{align*}
 
  • #26
Other examples than the natural representation ##\pi=\operatorname{id}_G## of matrix groups ##G\subseteq \operatorname{GL}(\mathbb{C}^n) ## would be
\begin{align*}
\pi= 1\, : \,G&\longrightarrow \operatorname{GL}(V) \\
\pi\, : \, A&\longmapsto (v\longmapsto v)
\end{align*}
which represents all elements of ##G## by the identity matrix. It is called the trivial representation. Or
\begin{align*}
\pi= \det\, : \,G&\longrightarrow \operatorname{GL}(\mathbb{C}) \\
\pi\, : \, A&\longmapsto (z\longmapsto \det(A)\cdot z),
\end{align*}
or the adjoint representation on the Lie algebra ##\mathfrak{g}## of ##G.## In the case of unitary matrices ##G=U(n)## its Lie algebra ##\mathfrak{u}(n)## consists of all skew-Hermitian matrices and the representation is
\begin{align*}
\pi= \operatorname{Ad}\, : \,G&\longrightarrow \operatorname{GL}(\mathfrak{g}) \\
\pi\, : \, A&\longmapsto (X \longmapsto AXA^{-1})
\end{align*}
Now you have four different representations of Lie groups, but they are all linear representations, i.e. the group elements are represented by matrices.

Note: There are several wordings that all mean the same thing
\begin{align*}
&\underbrace{\pi\quad : \quad G \longrightarrow \operatorname{GL}(V)}_{\text{representation}}\\
&\pi\quad : \quad g \underbrace{\longmapsto }_{\text{group homomorphism}}\left(v\longmapsto \underbrace{g.v = \pi(g)(v)}_{\underbrace{\text{operation or action on}}_{g \text{ acts on } v}} \right)
\end{align*}
 
Last edited:
  • #27
redtree said:
My primary text is the following: Woit, Peter, Woit, and Bartolini. Quantum theory, groups and representations. Vol. 4. New York, NY, USA:: Springer International Publishing, 2017.
Given your initial question where you talked about actions on infinite dimensional spaces, I was thinking that you would want to define your unitary group as the group that preserves Hermitian Inner Products. That gives you something that extends to Hilbert spaces. However, I am not sure what criteria you would use to subset to the special unitary group with this kind of view.
 
  • #28
Woit is an excellent text; it's not so much that I feel I don't understand it (Incidentally, Woit is the reason I used "mapping" instead of group homomorphism for ##\pi##.)--it's more that I was bothered by the mapping/group homomorphism you identify ##\pi =1: G \rightarrow GL(V)##.

In reading Woit, it seemed to me that one should be able to define ##U(n)## in more general ways than formulating ##U(n)## as a matrix group, such as a coordinate-free approach utilizing tensors or an infinite-dimensional approach utilizing integrals. For the finite dimensional case, this would also serve to give the representation a less trivial meaning than ##\pi = 1## and it would allow unitary group to act on tensors. For what it's worth, the infinite dimensional approach has been implemented utilizing a definition for the unitary group very similar to the one I presented above (see https://arxiv.org/pdf/2203.06315).
 
  • #29
jbergman said:
Given your initial question where you talked about actions on infinite dimensional spaces, I was thinking that you would want to define your unitary group as the group that preserves Hermitian Inner Products. That gives you something that extends to Hilbert spaces. However, I am not sure what criteria you would use to subset to the special unitary group with this kind of view.
Path connectedness?
 
  • #30
I will ask again, can you give a matrix free definition of ##SL_2##? So that we know what you expect as an answer.
 
  • #31
redtree said:
Path connectedness?
For finite dimensional groups ##U(n)## is path connected (I think). I'm not familiar enough with the infinite dimensional case to comment.
 
  • #32
martinbn said:
I will ask again, can you give a matrix free definition of ##SL_2##? So that we know what you expect as an answer.
For a finite dimensional, matrix-free formulation of ##GL(n,\mathbb{R})## with a tensor determinate operation similar to that of a matrix (https://hal.science/hal-03298805v1/file/tensors.pdf):

$$GL(n,\mathbb{R}) = \{ T_{i j} \in GL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} \neq 0 \}$$

such that one might define the finite-dimensional, matrix-free ##SL(n,\mathbb{R})## formulation:

$$SL(n,\mathbb{R}) = \{ T_{i j} \in SL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} =1 \}$$
 
  • #33
redtree said:
For a finite dimensional, matrix-free formulation of ##GL(n,\mathbb{R})## with a tensor determinate operation similar to that of a matrix (https://hal.science/hal-03298805v1/file/tensors.pdf):

$$GL(n,\mathbb{R}) = \{ T_{i j} \in GL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} \neq 0 \}$$

such that one might define the finite-dimensional, matrix-free ##SL(n,\mathbb{R})## formulation:

$$SL(n,\mathbb{R}) = \{ T_{i j} \in SL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} =1 \}$$
!!!
 
  • #34
martinbn said:
!!!
What were you hoping to see?
 
  • #35
redtree said:
What were you hoping to see?
What you've written uses matrices, and you wanted a definition without them. More importantly it is written in a way that makes no sense and is just a tautology.
 
  • Like
Likes dextercioby

Similar threads

Replies
13
Views
2K
  • Linear and Abstract Algebra
Replies
5
Views
4K
  • Linear and Abstract Algebra
Replies
1
Views
1K
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
2K
Replies
3
Views
2K
  • Differential Geometry
Replies
11
Views
3K
  • Differential Geometry
Replies
1
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
974
Replies
1
Views
683
Back
Top