9a. Groups, examples

[math] \newcommand{\mathds}{\mathbb}[/math]

We have seen so far the basics of linear algebra, with the conclusion that the theory is very useful, and quickly becomes non-trivial. We have seen as well some abstract applications, to questions in analysis and combinatorics, and with some results on the infinite dimensional case as well. All this is of course very useful in physics.

In this second half of this book we discuss a related topic, which is of key interest, namely the matrix groups. The theory here is once again very useful in connection with various questions in physics, the general idea being that any physical system [math]S[/math] has a group of symmetries [math]G(S)[/math], whose study can lead to concrete conclusions about [math]S[/math].

Let us begin with some abstract aspects. A group is something very simple, namely a set, with a composition operation, which must satisfy what we should expect from a “multiplication”. The precise definition of the groups is as follows:

Definition

A group is a set [math]G[/math] endowed with a multiplication operation

[[math]] (g,h)\to gh [[/math]]

which must satisfy the following conditions:

• Associativity: we have, [math](gh)k=g(hk)[/math], for any [math]g,h,k\in G[/math].
• Unit: there is an element [math]1\in G[/math] such that [math]g1=1g=g[/math], for any [math]g\in G[/math].
• Inverses: for any [math]g\in G[/math] there is [math]g^{-1}\in G[/math] such that [math]gg^{-1}=g^{-1}g=1[/math].

The multiplication law is not necessarily commutative. In the case where it is, in the sense that [math]gh=hg[/math], for any [math]g,h\in G[/math], we call [math]G[/math] abelian, en hommage to Abel, and we usually denote its multiplication, unit and inverse operation as follows:

[[math]] (g,h)\to g+h\quad,\quad 0\in G\quad,\quad g\to-g [[/math]]

However, this is not a general rule, and rather the converse is true, in the sense that if a group is denoted as above, this means that the group must be abelian.

At the level of examples, we have for instance the symmetric group [math]S_N[/math]. There are many other examples, with typically the basic systems of numbers that we know being abelian groups, and the basic sets of matrices being non-abelian groups. Once again, this is of course not a general rule. Here are some basic examples and counterexamples:

Proposition

We have the following groups, and non-groups:

• [math](\mathbb Z,+)[/math] is a group.
• [math](\mathbb Q,+)[/math], [math](\mathbb R,+)[/math], [math](\mathbb C,+)[/math] are groups as well.
• [math](\mathbb N,+)[/math] is not a group.
• [math](\mathbb Q^*,\cdot\,)[/math] is a group.
• [math](\mathbb R^*,\cdot\,)[/math], [math](\mathbb C^*,\cdot\,)[/math] are groups as well.
• [math](\mathbb N^*,\cdot\,)[/math], [math](\mathbb Z^*,\cdot\,)[/math] are not groups.

Show Proof

All this is clear from the definition of the groups, as follows:

(1) The group axioms are indeed satisfied for [math]\mathbb Z[/math], with the sum [math]g+h[/math] being the usual sum, 0 being the usual 0, and [math]-g[/math] being the usual [math]-g[/math].

(2) Once again, the axioms are satisfied for [math]\mathbb Q,\mathbb R,\mathbb C[/math], with the remark that for [math]\mathbb Q[/math] we are using here the fact that the sum of two rational numbers is rational, coming from:

[[math]] \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} [[/math]]

(3) In [math]\mathbb N[/math] we do not have inverses, so we do not have a group:

[[math]] -1\notin\mathbb N [[/math]]

(4) The group axioms are indeed satisfied for [math]\mathbb Q^*[/math], with the product [math]gh[/math] being the usual product, 1 being the usual 1, and [math]g^{-1}[/math] being the usual [math]g^{-1}[/math]. Observe that we must remove indeed the element [math]0\in\mathbb Q[/math], because in a group, any element must be invertible.

(5) Once again, the axioms are satisfied for [math]\mathbb R^*,\mathbb C^*[/math], with the remark that for [math]\mathbb C[/math] we are using here the fact that the nonzero complex numbers can be inverted, coming from:

[[math]] z\bar{z}=|z|^2 [[/math]]

(6) Here in [math]\mathbb N^*,\mathbb Z^*[/math] we do not have inverses, so we do not have groups, as claimed.

■

There are many interesting groups coming from linear algebra, as follows:

Theorem

We have the following groups:

• [math](\mathbb R^N,+)[/math] and [math](\mathbb C^N,+)[/math].
• [math](M_N(\mathbb R),+)[/math] and [math](M_N(\mathbb C),+)[/math].
• [math](GL_N(\mathbb R),\cdot\,)[/math] and [math](GL_N(\mathbb C),\cdot\,)[/math], the invertible matrices.
• [math](SL_N(\mathbb R),\cdot\,)[/math] and [math](SL_N(\mathbb C),\cdot\,)[/math], with S standing for “special”, meaning [math]\det=1[/math].
• [math](O_N,\cdot\,)[/math] and [math](U_N,\cdot\,)[/math], the orthogonal and unitary matrices.
• [math](SO_N,\cdot\,)[/math] and [math](SU_N,\cdot\,)[/math], with S standing as above for [math]\det=1[/math].

Show Proof

All this is clear from definitions, and from our linear algebra knowledge:

(1) The axioms are indeed clearly satisfied for [math]\mathbb R^N,\mathbb C^N[/math], with the sum being the usual sum of vectors, [math]-v[/math] being the usual [math]-v[/math], and the null vector [math]0[/math] being the unit.

(2) Once again, the axioms are clearly satisfied for [math]M_N(\mathbb R),M_N(\mathbb C)[/math], with the sum being the usual sum of matrices, [math]-M[/math] being the usual [math]-M[/math], and the null matrix [math]0[/math] being the unit. Observe that what we have here is in fact a particular case of (1), because any [math]N\times N[/math] matrix can be regarded as a [math]N^2\times1[/math] vector, and so at the group level we have:

[[math]] (M_N(\mathbb R),+)\simeq(\mathbb R^{N^2},+) [[/math]]

[[math]] (M_N(\mathbb C),+)\simeq(\mathbb C^{N^2},+) [[/math]]

(3) Regarding now [math]GL_N(\mathbb R),GL_N(\mathbb C)[/math], these are groups because the product of invertible matrices is invertible, according to the following formula:

[[math]] (AB)^{-1}=B^{-1}A^{-1} [[/math]]

Observe that at [math]N=1[/math] we obtain the groups [math](\mathbb R^*,\cdot),(\mathbb C^*,\cdot)[/math]. At [math]N\geq2[/math] the groups [math]GL_N(\mathbb R),GL_N(\mathbb C)[/math] are not abelian, because we do not have [math]AB=BA[/math] in general.

(4) The sets [math]SL_N(\mathbb R),SL_N(\mathbb C)[/math] formed by the real and complex matrices of determinant 1 are subgroups of the groups in (3), because of the following formula, which shows that the matrices satisfying [math]\det A=1[/math] are stable under multiplication:

[[math]] \det(AB)=\det(A)\det(B) [[/math]]

(5) Regarding now [math]O_N,U_N[/math], here the group property is clear too from definitions, and is best seen by using the associated linear maps, because the composition of two isometries is an isometry. Equivalently, assuming [math]U^*=U^{-1}[/math] and [math]V^*=V^{-1}[/math], we have:

[[math]] (UV)^* =V^*U^* =V^{-1}U^{-1} =(UV)^{-1} [[/math]]

(6) The sets of matrices [math]SO_N,SU_N[/math] in the statement are obtained by intersecting the groups in (4) and (5), and so they are groups indeed:

[[math]] SO_N=O_N\cap SL_N(\mathbb R) [[/math]]

[[math]] SU_N=U_N\cap SL_N(\mathbb C) [[/math]]

Thus, all the sets in the statement are indeed groups, as claimed.

■

Summarizing, the notion of group is something extremely wide. Now back to Definition 9.1, because of this, at that level of generality, there is nothing much that can be said. Let us record, however, as our first theorem regarding the arbitrary groups:

Theorem

Given a group [math](G,\cdot)[/math], we have the formula

[[math]] (g^{-1})^{-1}=g [[/math]]

valid for any element [math]g\in G[/math].

Show Proof

This is clear from the definition of the inverses. Assume indeed that:

[[math]] gg^{-1}=g^{-1}g=1 [[/math]]

But this shows that [math]g[/math] is the inverse of [math]g^{-1}[/math], as claimed.

■

As a comment here, the above result, while being something trivial, has led to a lot of controversy among mathematicians and physicists, in recent times. The point indeed is that, for the needs of quantum mechanics, the notion of group must be replaced with something more general, called “quantum group”, and there are two schools here:

(1) Certain people, including that unfriendly mathematics or physics professor whose classes no one understands, believe that God is someone nasty, who created quantum mechanics by using some complicated quantum groups, satisfying [math](g^{-1})^{-1}\neq g[/math].

(2) On the opposite, some other mathematicians and physicists, who are typically more relaxed, and better dressed too, and loving life in general, prefer either to use beautiful quantum groups, satisfying [math](g^{-1})^{-1}=g[/math], or not to use quantum groups at all.

Easy choice you would say, but the problem is that, due to some bizarre reasons, the quantum group theory with [math](g^{-1})^{-1}=g[/math] is quite recent, and relatively obscure. For a brief account of what can be done here, mathematically, have a look at my book .

General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].