What is a group?

A group is a very abstract mathematical concept that I like to thing of as an extension to the notion of a set. That is, a set is quite an abstract mathematical concept. A set is an object that “groups” any elements by its properties, or even using any “logic” we can think of.

One can define a set to be, for example, all people of age 32. Mathematically, one can define a set of all real numbers, a set of all even numbers, a set of even functions, etc… A set can be finite, countably infinite, uncountably infinite, etc… A set can be caracterized by different things, for example the cardinality of the set. The cardinality can be thought of as the size of the set.

Now, what is a group then? What the notion of the group does, is it establishes a relationship between the elements in the set. Namely, the set simply “contains” elements in an abstract manner. The group, however, establishes some “action/connection” between the contained elements inside this group.

That is, one can for example define the addition relationship (e.g. 2 real numbers can be added), multiplication relationship (e.g. 2 matrices can be multiplied), composition relationship (e.g. 2 functions can be composed) etc… That is, the relationship we’re talking about is nothing but a function that takes two elements of the set and spits another. One can remember that the notion of groups is often backed up by examples from symmetries, which are nothing but functions.

An important property of a group is that all elements in this set can be obtained via each other. Mathematically speaking, one calls this property “closure”.

What it means is that for any 2 elements of the group, their product/operation will result in another element of the group. Alternatively, any element of the group can be obtained via a binary operation between two elements of the group.

A group is thus an abstract mathematical concept that is based on 2 notions: a set and a binary function. The function in the context of a group is called an operation. This operation maps two elements of a set to another element of this set. Therefore, a group is a set, with a defined operation (the operation can, e.g. be addition, product, composition,…) that we sometimes note as $(G, \circ)$, where $G$ is the set and the $\circ$ corresponds to the binary operation. The formal mathematical definition of a group is therefore:

The group $(G, \cdot)$ is said to be a group if it obeys 3 conditions:

  • $\exists!;e \in G: \forall g \in G, g\circ e = g = e \circ g \quad$. In other words, there exists a neutral element, such that composed with any element in the group, it will give this element (example for the group $(\mathbb{R^*}, \times)$, the neutral element is 1).
  • $\forall g\in G, ;\exists! g^{-1} : g\circ g^{-1} = g^{-1}\circ g = e$. That is, for any element of the group, one can find its unique inverse that by operating, will yield the neutral element.
  • $\forall g ,h \in G,; g\circ h \in G \text{ and } h \circ g \in G$. In other words, the result of the operation between any 2 elements in $G$ will give an element that also belongs to $G$.

Or more informally, we’ll repeat ourselves - it’s a set (either finite or infinite), that is closed (remember the notion of closure) under an operation, containing an identity element
(it can be shown that can only exist one identity element) and an inverse for every element.

This is quite an abstract concept, but also very common, when brought it with correspondance with real examples. Examples of group in physics involve symmetry transformations (geometrical operation of a molecule that leave them invariant), Lorentz transforms, Galilean transforms, and many more.

Apart from the exact notion of the group, there are multiple concepts related to it. For example, the notion of a conjugacy class of a group.

Intuitively, this can be thought of as a kind of a subset of this group containing elements, that share similar properties. More precisely, mathematically, one defines the conjugacy class $C$ as follows:

Two elements $x$ and $y$ belongs to the conjugacy class $C$: $$ y, x\in C \iff \exists u \in G :; u^{-1}\circ x\circ u=y$$ In other words, elements $x,y$ of the group $G$, belong to the same conjugacy class if there is some kind of other element $u$ of $G$, such that they are related by $u^{-1}\circ x \circ u = y$.

It is not very easy to try to understand the meaning of this notion, since the general concept of groups is quite abstract. We could interpret this as follows; let the group $(G, \circ)$ represent some kind of action. Then, two actions $x$ and $y$ belong to the same class if
performing the action $x$ results to the same action as performing some kind of action $u$, then $y$, then remove the action of the $u$ operation. In other words, the actions $x$ and $y$ are somewhat the same, in a different basis.

We could try to make an analogy with matrices i.e. linear transformations. That is, we know there exists such a concept as change of basis. This concept gives the possibility, to express a matrix $A$ in two different basis. This is done via the change of base matrix $P$, namely, $A’ = P^{-1}AP$. The matrix $A’$ represents the exact same matrix as $A$, but in a different basis/from a different point of view.

Symmetries

We’ve mentioned that the notion of groups is an abstract concept and can be associated to many, even non-purely mathematical conepts. For mathematics-related group examples, one can mention the $(\mathcal{Z}, +)$ group (group of integers with the binary operation of addition), the permutation group (the group consisted of permutation operations). Here, we’ll be interested in symmetries, and describe these symmetries using the notion of groups. Symmetries are geometrical transformations that can form a group.

Example

Let’s provide an example of such transformation. Consider a triangle with equal sides with edges denoted $A,B,C$. What are the transformations that could potentially make a group of symmetries? More simply put - what is the set of transformations that will leave the system invariant and make a group?

  • First, we have the identity element. The identity operation does not change anything.
  • The second operation is the rotation by $120$ degrees. This operation will map $A \mapsto B$, $B \mapsto C$ and $C\mapsto A$. However, the rotation by $120$ degrees will not simply permute the edges of the triangle - it will also rotate the coordinates, that are attached to the individual atoms. Indeed, one can attach a fixed coordinate frame to every edge and make sure that these coordinates do indeed rotate with respect to the “fixed coordinate frame”.
  • The third operation is the rotation of the triangle by -120 degrees. This would be the exact same thing as the rotation by 240 degrees.
  • The fourth, fifth and the sixth operations are mirror operations. That is, if one reflects the triangle with respect to the bissectrice going from every single edge ($A, B$ or $C$).

By defining these 6 operations, one can create a group of transformations. It is indeed a group, since for all elements, there exist an inverse element and an identity element. This group is commonly reffered to as a $C_{3v}$ group.

For different shapes and symmetries and even dimensions (one can indeed consider a triangle in 2D or in 3D), one can come up with different groups.

In order to make the notation more consise, one can make arrange them in tables, often reffered to as the Cayleigh table. The Cayleigh table keeps track of all possible operations in a group. We will create a Cayleigh for this $C_{3v}$ group later. For now, we will restrict ourselves to a more simple group.

For an example, one can consider a simpler group $\mathcal{Z}_2$ commonly named as the inversion group or the permutation group of order 2, or $({0,1},+)$ group. This is nothing but the group of addition modulo $2$. The operations between the two elements of the group can be summarized as $$ 0+0 \stackrel{\text{mod }2}{=} 0\\ 0+1 \stackrel{\text{mod }2}{=} 1\\ 1+0 \stackrel{\text{mod }2}{=} 1\\ 1+1 \stackrel{\text{mod }2}{=} 0 $$ One may easily identify the neutral element $0$ and note that this group is Abelian, meaning that all the elements commute. And the corresponding Cayleigh table for the group. $$ \def\arraystretch{1.5} \begin{array}{|c|c|c|} \hline \circ & 0 & 1 \\ \hline 0 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array} $$

There are 2 conjugacy classes in this group, namely, the neutral element ${0}$ conjugacy class and ${1}$ conjugacy class. One can create the same Cayleigh table for the mentioned $C_{3v}$ group. These tables, are however, not very easy to manipulate for large groups. Indeed, for a group of $6$ elements, the table has $6\times 6 = 36$ entries in it. There are ways, however, to represent these groups in a more compact way. We will further see the notion of the character table, which can summarize its properties in a more compact manner.

Let’s try to write down the Cayleigh table of the $\text{C}_{3v}$ group and analyze it using the properties we’ve defined above. So we’ve identified all the 6 transformatins belonging to this group. It can be shown that there exists only 2 types of such groups. That means that one can come up with only two “different Cayleigh tables” for a group of order 6. Up to notation of course. In order to create the table, one can write down all the transformations we’ve encountered before in a table. During the filling process, one can ask ourselves at each intersection of the table the following question:

If I first do [transformation from top] and then do [transformation from right], what would it be equivalent to?

For example, let’s take one trivial case: first apply $e$ - the identity operation, and the apply $\sigma_A$ (the mirror with respect to the bissectrice from the point $A$)? It is clear the result will be $\sigma_A$, since the $e$ operation doesn’t do anything… What about a more complex case? First apply $C_1$-the rotation by $120$ degrees, and then the $\sigma_A$ mirror operation? This may not be very obvious, but it is quite easy to verify by drawing the 2 transformations. In fact, the 2 transformation will create some kind of permutation, and the final permutation has the mapping: $\sigma_A \circ C_2: A \mapsto C; B \mapsto B; C \mapsto A$. which is nothing but the $\sigma_B$ transformation. Thus, at the intersection of $C_1$ (top - first) and $\sigma_A$ (left - second), the resulting element will be $\sigma_B$. By repeating this procedure $6\times 6$ times, one can obtain the Cayleigh table for the $C_{3v}$ group:
$\circ$ $e$ $C_1$ $C_2$ $\sigma_A$ $\sigma_B$ $\sigma_C$
$e$ $e$ $C_1$ $C_2$ $\sigma_A$ $\sigma_B$ $\sigma_C$
$C_1$ $C_1$ $C_2$ $e$ $\sigma_B$ $\sigma_C$ $\sigma_A$
$C_2$ $C_2$ $e$ $C_1$ $\sigma_C$ $\sigma_A$ $\sigma_B$
$\sigma_A$ $\sigma_A$ $\sigma_C$ $\sigma_B$ $e$ $C_2$ $C_1$
$\sigma_B$ $\sigma_B$ $\sigma_A$ $\sigma_C$ $C_1$ $e$ $C_2$
$\sigma_C$ $\sigma_C$ $\sigma_B$ $\sigma_A$ $C_2$ $C_1$ $e$

So what can we say intuitively about this group? First, this group is not abelian, since performing 2 different operations in different order may not yield the same resulting state. Second thing is that there are 3 (2) types of “intuitive” transformations, namely, the identity transformation, the rotations (2 rotations) and reflexions (3 reflexions). One may notice that on the table, there is some kind of pattern. Namely, the table can be divided into 4 quadrants - top left (TL), TR, bottom right (BR) and BL. These quandrants can be characterized by the fact they only contain a specific type of transformations, i.e. either the $\sigma$ ’s or $C$ ’s (without counting $e$). This can be intuitively interpreted as conjugacy classes, however, this is only a visual interpretation and is not always true and useful for larger groups.

In the $C_{3v}$ symmetry group, there are 3 conjugacy classes - the trivial one ${e}$, the rotations class $C \equiv {C_1, C_2}$ and the reflexions class $\sigma \equiv {\sigma_A, \sigma_B, \sigma_C}$ (Note: there is a special notation for groups and their elements, which, unfortunately, I do not fully follow). What this means is that if we take e.g. one element from the rotation class - some $C_i \in C$, then no matter which transformation of group $u \in C_{3v}$ we’re taking; it can be the trivial identity ($u = e$), the element from the same conjugacy class ($u = C_k \in C$) or from the different conjugacy class ($u = \sigma_j \in \sigma$), we will always get that the result of $u^{-1}\circ C_i \circ u$ will be in the same conjugacy class as $C_i$, namely, in the conjugacy class $C \equiv {C_1, C_2, …}$: $\; \; \; u^{-1}\circ C_i \circ u \in C$.
Note: finding conjugacy classes is not always straightforward. They may obey to the notion of “different types of transformations” as in the example with $C_{3v}$ (identity, rotations, reflexions), but this is not always a matter of intuition. There exist databases and tables with groups and their corresponding partition into conjugacy classes.

Representations

Once again, the notion of groups is very broad, and can be associated to many concepts. We’ve mentioned examples of groups, like $(\mathcal{Z}_2, +)$, particular matrix group like the $\text{SO}3$ group ( $3\times 3$ matrix of determinant $1$) and many others.

Here, the considered groups will be symmetries. That is, groups of geometrical transformations.

Representation theory is in a way a branch of group theory. In representation theory, what we do is we associate groups to matrices.That is, symmetry operations are associated to groups, which then are associated to matrices.

Thus, the idea of representation theory to associate every element of a group to a matrix of any size. In addition to that, we want this association to be a homomorphism.

$$ (G, \circ) \equiv {e, g_1, g_2, …, g_N} \xrightarrow{\text{Mapping (homomorphism)}} {M_e, M_{g_1}, M_{g_2}, …} $$

Before going into the notion of group representation, one would need a couple of notions.

Let $f: G \rightarrow H$, a function that maps an element from the group $(G, \circ)$ to the group $(H,\star)$, i.e. $f: g \mapsto h$, with $g\in G$, $h\in H$. Then the function $f$ is a homomorphism if $\forall g_1, g_2 \in G$, $f(g_1\circ g_2) = f(g_1)\star f(g_2)$.

A homomorphism $f’$, which is one-to-one (bijective) is an isomorphism. Thus, intuitively speaking, a homomorphism is a map from a group to another group such that the operations inside the “target” group is the same as in the “original” one.

Now, one defines the central concept, i.e. the representation of a group.

Let $G$ some group. Let $V$ - some kind of vector space over some field of dimension $n$. A representation of the group $G$ is a function $\Gamma$, sometimes denoted $\Gamma(G)$, defined by $\Gamma(G): G \rightarrow \text{GL}(V)$, that is, $\Gamma(g): g \mapsto \text{gl}$, for some $g\in G$ and $\text{gl}\in \text{GL}(V)$. The representation $\Gamma$ is a homomorphism.

Let’s break down the definition. The $\text{GL}(V)$ is the general linear group over the space V, that is, it is a set of linear transformation in the space $V$ of dimension $n$. Similarly, $\text{GL}(V)$ in nothing but the space of $n\times n$ invertible matrices, that act on the vector space $V$. Note that this vector space can be arbitrary, e.g. $\mathbb{R}^n$, Hilbert space, or something else.

In other words, the representation is a homomorphism (a function) that will map every element $g\in G$ to a matrix $M_g\in \text{GL}(V)$. The space, to which $\Gamma$ maps the elements of $G$ is also a group (the group matrix). The values of these functions must obey the same rules as the elements $g \in G$. Thus, when saying “representation”, one can refer to the most general $\Gamma$ as the set of mappings from $g\in G$ to elements in $\text{GL}(V)$.

Let’s once again emphasize on this: the term representation itself encompasses the mappings from elements $g\in G$ to elements in $\text{GL}(V)$, and NOT one particular matrix $M_g \in \text{GL}(V)$ associated to the element $g\in G$.

One can consider an example of a simple parity group of order 2, that we already discussed. This group (sometimes denoted as $\mathbb{Z}_2$) has 2 elements: ${e, p}$ and they follow the same multiplication rules as ${1, -1} \equiv {e, p}$ or same additions rules as ${0,1}$ $\text{mod} 2$. One can try to create the representation $\Gamma$ for this group. Since the group is small, one can easily come up with a set of $2$ matrices, that will obey to the same multiplication rule as the group elements (see the Cayleigh table above). The $2$ possible matrices are given by:

$$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \equiv e, \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \equiv p $$

which indeed follow the same multiplication rules as the group $\mathbb{Z}_2 \equiv {e, p}$. One must note that the group itself, containing ${e,p}$ is an abstract group, which simply has multiplication rules $e\cdot p = p\cdot e= p$; $p\cdot p = e$ and $e \cdot e = e$. The group has nothing to do with matrices or even $\mathbb{Z}_2$. Indeed, the group is anything that will obey these kind of relations. The representations, however, do indeed have a relation with matrices and $\mathbb{Z}_2$. In other words, one can come up with 2 representations:

$$ \Gamma _{\mathbb{Z} _z}: e \mapsto 1, p\mapsto -1 $$

$$ \Gamma _{\text{GL}(V)}: e \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , p \mapsto \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $$ both representations are valid. Indeed, the two elements of $\mathbb{Z}_2$ are indeed elements of $\text{GL}(V)$ of dimension 1 and the two matrices are elements of $\text{GL}(V)$ of dimension 2. We will then see that it is possible to come up with another one.

In quantum mechanics, as said, we’re working with finite groups, that are groups of symmetry. This group of $2$ elements represents the symmetry of inversion.

Now, one should introduce the very important concept of reductibility. Consider a symmetry group $G$ that describes some system. And let $\Gamma$ be its representation. Let now $\Gamma (g)$ be a representaion, that acts on the vector space $V$. One says that this representation is completely reducible if there exists a non-zero subspace $W \in V$, that is invariant under $\Gamma$. In other, words, it is reducible if there exists a subspace $W \in V$, such that $\forall \ket{w} \in W$, such that $\Gamma (g) \ket{w} \in W$, which is true FOR ALL $g\in G$. In addition to that, the subspace $W^\text{T}$ is orthogonal to $W$. One can visualize it in terms of matrices. That is, $\Gamma (g)$ is reducible if for all $g\in G$, one can put the matrices $\Gamma$ in the form $$ \Gamma (g) \equiv \begin{pmatrix} \Gamma(g) ^{(1)} & 0 \\ 0 & \Gamma(g) ^{(2)} \\ \end{pmatrix} = \Gamma(g) ^{(1)} \oplus \Gamma(g) ^{(2)} $$ Where the two matrices $\Gamma (g) ^{(1),(2)}$ are also matrices and the matrix $\Gamma (g)$ is diagonal by blocks. A reducible representation is a representation that can be block-diagonalized more and more. A representation that is already “maximally reduced”, is called irreducible representation.

Since the representation we’re considering are matrices, there should be a way to go from a reducible to irreducible representation. This could be done via a change of basis matrix. Namely, one can write $$ \Gamma(g) _{\text{red}} = S^{-1} \Gamma(g) _{\text{irred}} S $$

This relation is called an equivalence relation. That is 2 representations are equivalent if there exists a basis that transforms them as given above. This relation is called a similarity transformation. In other words, 2 equivalent representations are related via a similarity transformation.

In representation theory, we’re interested in non-equivalent, but irreducible representations. That is, in representation theory, two representations, are considered to be “the same” if they are irreducible and possibly equivalent. One should, however, note another fact concerning the reducible representations. The general matrix form of the irreducible representation is instead given by $$ \begin{pmatrix} \Gamma(g) _1 ^{(1)} & 0 &0 &0 & 0 & 0\\ 0 & \Gamma(g) _2 ^{(1)} &0 &0 &0 & 0\\ 0 & 0 & \ddots & 0 &0 &0 \\ 0 & 0 & 0 & \Gamma(g) _1 ^{(2)} &0 &0 \\ 0 & 0 & 0 & 0& \Gamma(g) _2 ^{(2)} & 0 \\ 0 & 0 & 0 & 0 & 0 & \ddots \end{pmatrix} $$

That is, the representation, in its most general case, is decomposed as

$$ \Gamma(g) = \bigoplus_{a,x} \Gamma(g)^{(a)}_{x} $$

instead of simply $\Gamma(g) = \bigoplus _{a} \Gamma(g)^{(a)}$. That is, one adds some degenerescence to a given element of the block-representation $\Gamma(g)^{(a)}$. Meaning that some of the irreducible representations are encountered more than once in the final, irreducible representation.

This is a very important formula that describes the notion of (ir)-reducible representations.

Why do we need it? Recap up to now

  • So, for now, what we understand is that there exists some kind of spatial structure (a triangle, cube, molecule, etc…) that has some symmetry associated to it. What symmetry means is that one can apply one of the symmetrical transformation, and we won’t be able to tell the difference between the systems before and after transformations. This is called “invariance under transformations” or simply “symmetries”.

  • These symmetries obey to some mathematical properties between themselves. These mathematical properties are called a group. In other words, the geometrical transformations may obey relations, that are nothing but groups.

  • Instead of working with the group, which can be characterized using a Cayleigh table (containing all possible relations in the group), one can characterize the group using the representation theory. This is equivalent, since a representation is an isomorphism by definition. That means that the relations of all the elements with all the elements is the same as the relations between the matrices that form this representation (recall that a representation is mainly a association between abstract elements of a group to “real”/“tangible” matrices).

  • The set of matrices (the representation) can be different. Indeed, the matrices is a representation as long as it obeys the necessary relation from the group.

But what could we eventually need this for? Suppose the (symmetric) system that we are analyzing is somewhat a physical object (most likely a molecule). We want to retrieve and compute some of its properties. This will be most likely done through a Hamiltonian operator, which can be represented as a matrix. The Hamiltonian will also most likely lead to the solution of an eigenvalue problem, thus requiring diagonalization. The next question is thus “what does the Hamiltonian have to do with the symmetry transformation matrix (representation)?” Well, the answer to this makes sense - they commute. Indeed, the symmetry transformation is by definition a transformation that takes into account symmetry and the system is unvariant under this transformation, meaning the Hamiltonian does not change either. From linear algebra, this commutative property leads to another mathematical property: if 2 operators are commuting, it is possible to find a common basis, such that both of the commuting operators will by (block) diagonal. So, if the symmetry matrix is nice and easy, one can diagonalize the latter and work our way to the diagonalization of the hamiltonian itself.

  • Now, what does the notion of reducible/irreducible has to do with that? When discussing the representations, we mentioned that we’re especially interested in irreducible representations, which, by definition are block-diagonal. This means that irreducible representations are matrices of symmetry transformations (i.e. obeying the group rules), but meant to be easy manipulated.

  • Thus the goal of the representation theory is:

based on symmetry considerations, find a valid representation, and then reduce it in order to obtain the simplest form possible using maths from representation theory. Once this is done, it is possible to infer some properties of the system, and potentially solve the eigenvalue problem for the complex hamiltonian, since the irreducible representation have a simpler structure.

Mathematical description

Before going into the practical examples, trying to understand the role of the notion of group representations and how could that be useful, one should simply mention, without proving, some of important theorems & concepts. These concepts are tools, that are used to accomplish the goal that we’ve written above.

Shur Lemma’s

The Shur’s Lemma’s is the central theorem in the group representation theory. Namely, many results can be simply derived from the definitions and the mentioned Shur’s lemmas. For Shur’s Lemmas, let $\Gamma_1$ and $\Gamma_2$ - two different irreducible representations each belonging to its own vector space $V_1$ and $V_2$ respectively
(for example, the two representations could be matrices of size $3\times 3$ and $2\times 2$ acting on $\mathbb{C}^3$ and $\mathbb{C}^2$). Let in addition $M$ be a operator (matrix) $M: V_1 \mapsto V_2$, such that it commutes with both of the representations $\Gamma_{1,2}$. That is, $M;\Gamma_1(g) = \Gamma_2(g); M ;; \forall g \in G$, then

Shur’s Lemma 1 Shur’s Lemma 2
If $\Gamma_1$ and $\Gamma_2$ are not-equivalent, then $M=0$ If $\Gamma_1 = \Gamma_2 \equiv \Gamma$, then $M$ can only be the multiple of identity, that is, $M$ has the form $M\equiv \lambda I$

These two lemmas state that if some operator commutes with some two representations of a group, then this operator is either zero (the two representations are equivalent) or diagonal (if they are the same).

Note: We mentioned in one of the previous sections how representations are useful. Here, the matrix $M$ in question can be nothing but our Hamiltonian.

Characters

By now, we came across some mathematical objects such as group, subgroup, isomorphism, representation. It is now time to add a new one - the character. We’ve mentioned that in representation theory, we’re interested in representations, up to an equivalence relations (i.e. equivalent representations are considered to be the same). Thus, we may want to somehow characterize a representation (remember the definition of the representation - set of matrices), without having to deal the problem of cheking the equivalence.
The answer to that is the trace of the matrix. The set of traces of a representation is its character. One denotes a character of a matrix $\chi_\Gamma(g)$ of the representation $\Gamma$ of the matrix $\Gamma(g)$. That is, $\chi_\Gamma(g)=\text{Tr}[\Gamma(g)]$. One should recall the properties of the trace - $\text{Tr}(A\oplus B) = \text{Tr}(A)+\text{Tr}(B)$ and $\text{Tr}(A\otimes B) = \text{Tr}(A)\cdot \text{Tr}(B)$.

Orthogonality

We mentioned that should be able to decompose a representation into a direct sum of irreducible representations. That is, we want to decompose the representation (which possibly can be reduced) into the direct sum of irreducible: $\Gamma = \bigoplus_{a,x}\Gamma_{a,x}$, with $a$ - the irreducible representations and $x$ - the corresponding multiplicities or $\Gamma = \bigoplus_{a} b_a \Gamma_{a}$ and $b_a$ - the multiplicities. The resulting representation consists of a direct sum of vector spaces $\bigoplus_{a,x} V_{a,x}$. For every space $V_{a,x}$, one my identify a basis ${\ket{j}}$. One can denote the basis ${\ket{j}}$ belonging to the space $V_{a,x}$ as ${\ket{a,j,x}}$. Thus the basis of a given representation is denoted as ${\ket{a,j,x}}$.

The orthogonality theorem is an important result, that can be shown using the Shur’s lemmas (not done here)
Let $\Gamma_a$ and $\Gamma_b$ - two non-equivalent representations over the space $V_a$, $V_b$ of dimensions $n_a$ and $n_b$, of the group $G$ of order $N$. Then, the great orthogonality theorem states:

$$ \sum_{g \in G}\frac{n_a}{N} [\Gamma_a(g)]_{i,k}^{*} [\Gamma_b (g)] _{m,n} = \delta _{a,b} \delta _{i,m} \delta _{k,n} $$

Where the sum is performed over the group elements $g\in G$. This can be easily rewritten as

$$ \sum_{g \in G} \sqrt{\frac{n_a}{N}} [\Gamma_a (g)]_{i,k}^{*} \sqrt{\frac{n_b}{N}} [\Gamma_b (g)] _{m,n} = \delta _{a,b} \delta _{i,m} \delta _{k,n} $$

and identifying the mentioned elements of the vector space

$$ \ket{a,k,j} \equiv \sqrt{\frac{n_a}{N}} [\Gamma_a (g)]_{j,k} $$

The basis ortogonality relation can be identified as $\braket{b,i,k|a,m,n}=\delta_{a,b}\delta_{i,m}\delta_{k,n}$.

Based on the orthogonality relations, it is possible to derive the Burnside lemma, which gives a restriction to dimensions of the irreducible vector spaces $n_a$: $$ \sum_{a}n_a^2 = N $$ That being said, the sum of squares of the dimensions of the vector spaces into which the representations are operating is equal to the order of the group.
The Burnside Lemma is quite useful for determining the dimensions of representations of groups of small order. That is, suppose there exists a group of order $6$. Then, based on Burnside’s lemma, one state that a possible irreducible representation will be made of $6$ representations of dimension $1$. Indeed, $\sum_{a\in |\Gamma_a|}n_a^2=6\cdot 1^2 = 6 = N$, where $|\cdot|$ stands for cardinality/size and $N$ the size of the group. Or, another possibility is to have $1$ representation of dimension $2$ and $2$ representations of dimension $1$. Indeed, $\sum_{a\in |\Gamma_a|}n_a^2=1^2+1^2+2^2=6=N$. Similarly, for e.g. a group of size $8$, based on only Burnside, there can be either $8$ irreducible representations of dimension 1 ($8\cdot 1^2 = N$), or $1$ representation of dim. $2$ and $4$ of dim. $1$ ($2^2 + 1^2+ 1^2+ 1^2+ 1^2 = N$) or simply $2$ representations of dimension $2$.
If in addition, we know, how many conjugacy classes there is, it is even easier to determine the dimensions of the representations. That is, suppose some kind of group of order $10$ contains $4$ conjugacy classes. Then, there are $4$ irreducible representations. It is straightforward to find out the dimensions of the $4$ irreducible representations - they are $2^2+2^2+1^2+1^2 = N = 10$.

If one takes the trace of the mentioned orthogonality relation, one gets the second orthogonality theorem relation: $$ \sum_{g\in G} \chi_a(g) \chi_b(g) = N\delta_{a,b} $$ or equivalently, one can replace the group elements $g\in G$ by the elements of the conjugacy class, since we remember that the characters are same for all the elements in the conjugacy class. Thus, the equivalent relation is given by

$$ \sum_{\mu \in N_{\text{conj. cl.}}} n_{\mu} \chi_{a}^*(C_\mu)\chi_{b}(C_\mu)=N\delta_{a,b} $$ with $n_\mu$ - the number of elements in the conjugacy class.

The degeneracies $b_a$ mentioned before can be computed using the relation

$$ b_a = \frac{1}{N}\sum_{\mu} n_\mu \chi_{a}^*(C_\mu)\chi_{\Gamma}(C_\mu) $$ with $\chi_{\Gamma}(C_\mu)$ the character of the $C_\mu$ conjugacy class.

One may add the last relation that gives necessary and sufficient conditions for a representation to be irreducible. $$ \sum_{\mu} n_\mu |\chi_\Gamma(C_\mu)|^2 = N $$

Projectors

Projectors are operators that let us decompose and reduce a representation. Recall the goal of problem related to representation theory - to reduce a representation of some symmetric system. Mathematically, given a representation of a symmetry group $\Gamma$, one wants obtain the direct sum decomposition: $$ \Gamma = \bigoplus_{a,x} \Gamma_{a,x} = \bigoplus_{a} b_a \Gamma_a $$

The space that the resulting representation will be acting on will be the exact same direct sum of $V_a$ spaces, each being orthogonal. This is equivalent of finding a representation in the basis that we called ${\ket{a,j,x}}$. Let’s now see how we’ll try to define the projectors. We start by considering a vector $\ket{a,j,x}$ and the representation $\Gamma$. What happens, when we apply the representation to this element $\Gamma (g) \ket{a,j,x} $? Well, we remember that the $a$ in the ket notation represents the number of the non-equivalent irreducible representation, that the initial representation was reduced to. The $x$ represents its multiplicity and only the $j$ represents the “actual” vector inside this $V_{a,x}$ vector space. The usual matrix multiplication definition: $$ \Gamma \ket{a,j,x} = \sum_{k=1} (\Gamma)_{k,j} \ket{a,k,x} $$

But we know that for a given $a$ and $x$, the spaces are orthogonal, so the only non-zero contribution comes from the $\Gamma_{a,x}$ representation:

$$ \Gamma \ket{a,j,x} = \sum_{k=1} (\Gamma)_{k,j} \ket{a,k,x} $$

Using this and the previous orthogonality relations, one can transform it by doing some algebra into: $$ \sum_{g \in G} [ \Gamma_b (g)]^{*}_{k’,j’} ; \Gamma (g) \ket{a,j,x} = \frac{N}{n_a} \delta _{a,b} \delta _{j,j’} \ket{a,k’,x} $$

from which, one defines the projection operator $\hat{\Pi}_{k,j}^{b}$, which is defined as

$$ \hat{\Pi} ^{b} _{k,j} = \frac{n_a}{N} \sum _{g \in G} [\Gamma _b (g)]^{*} _{k,j} \Gamma _{g} $$

which will satisfy the following properties:

$$ \hat{\Pi} ^{a} _{k,j} \ket{a,j,x} = \ket{a,k,x} $$

$$ \hat{\Pi} ^{a} _{k,j} \ket{a,j’,x} = 0 $$

This projector $\hat{\Pi}_{k,j}^{a}$ projects an arbitrary vector onto the vector $\ket{a,k,x}$. If applied on another, orthogonal vector of $\ket{a,j,x}$, the result will be zero.

Idea of how use the representations?

Now, with all that in mind, one can complete the discussion on how are representations useful and used.
Let’s suppose we’re given some kind of symmetric system that represents, for example, some kind of molecular structure or multiple moleculesm, that is unvariant under some kind of transformations. The first thing is to identify the abstract symmetry group. Once identified, one should find (or construct the character table (see below)). The character table gives almost all the information needed for reducing the representation. For example, what if one wants determine the degeneracies of the eigenvalue problem? That is, how many times is a certain energy degenerate. Remember, the number of degeneracy of an operator is equal to the dimension of the eigenspace. Now, we remember that the representation of the system WILL commute with the Hamiltonian. Thus, by identifying the multiplicities $b_a$’s (see identity above), one can find out the degeneracy of the subspace, associated to the same representation.

One often may want to solve the motion of the system (e.g. $x(t)$, $p(t)$ or any general function $\psi(x,t)$). Then, once again, using the fact that the commutator of the Hamiltonian and $\Gamma$ is zero, one finds the seeked solution of the general eigenvalue problem, encountered in quantum mechanics $\hat{H}\ket{\psi} = E\ket{\psi}$. To solve this, one can consider the matrix $\Gamma$ instead and compute the eigenvectors and eigenspaces. The latter is often performed using the projectors.

Worked examples

We’re finally in a position of starting considering concrete examples and apply the mentioned concept for solving quantum mechanical problems. This is the most important part of the article, where we will try to understand the mentioned notions by applying them, since the considered concepts tend to be quite abstract.

Simplest group - reflection group

We start off by considering the simplest, already mentioned group - the reflection group/ parity group/ $C_s$ group. This group has lots of names, since, as we’ve seen a group can be associated to many different notions. The main point is that the $C_s$ group has $2$ elements and there is only one possible group of order $2$. It’s Cayleigh table is given by

$$ \def\arraystretch{1.5} \begin{array}{|c|c|c|} \hline \circ & e & p \\ \hline e & e & p \\ \hline p & p & e \\ \hline \end{array} $$ This is a very simple group and it is quite easy to determine the irreducible representations. Indeed, let’s recall the burnside lemma. From the Burnside lemma, one can determine that there are $2$ irreducible representation of dimension $1$, i.e. $\Gamma = \Gamma_1 \oplus \Gamma_2$, with both $\Gamma_1$ and $\Gamma_2$ of dimension $1$. Thus the representation are simply $2$ numbers. It is quite easy to find that the 2 representations the following:

$$ \def\arraystretch{1.5} \begin{array}{|c|c|c|} \hline & e & p \\ \hline \Gamma_1 & 1 & 1 \\ \hline \Gamma_2 & 1 & -1 \\ \hline \end{array} $$ We can check that this is true - if $e \mapsto 1$, and $p \mapsto 1$, the relations within the group are verified. Similarly, if $e \mapsto 1$ and $p \mapsto -1$, the relation is still verified. Now, what about the most powerful tool available - the character table? What does it look like? Well, in order to find out, one must simply take the trace of the representaions. Here, the representations are simply numbers, thus the trace is equal to the number itself. So the character table:

$$ \def\arraystretch{1.5} \begin{array}{|c|c|c||c|} \hline \chi & e & p & \text{Linear funcs. and rotations } \\ \hline \Gamma_1 & 1 & 1& x,y, R_z \\ \hline \Gamma_2 & 1 & -1& z, R_x, R_y\\ \hline \end{array} $$ Well, this was quite easy… However, let’s suppose we don’t know the irreducible decomposition, and we try to start from the physical situation. We have a simple molecule having $2$ atoms - left and right ones denoted as $L$ and $R$. Their coordinates are given by $L \equiv (0, 0, z=-z_0)$ and $R \equiv (0, 0, z=z_0)$. This can be a 3-dimensional system, but we can always choose a coordinate system so that only one of the components is non-zero; in this case, we take $z$.

Now, we ask ourselves an important question - what will happen when the reflection transformation is applied? We have that the left one becomes the right one and the right one becomes the left one. Thus, the 2 transformations in the matrix form (the ID transformation and the permutation transformation) are given by $$ e \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \;\;\;\;\;\;\;\; p \equiv \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} $$

Now, one considers the character table that we’ve provided above and observe that there are 2 conjugacy classes: $\lbrace e \rbrace$ and $\lbrace p \rbrace$. Therefore, one can apply the different concepts we’ve mentioned before to determine the different vibrations of the molecule.
We know that there are 2 conjugacy classes thus we know that we will have $2$ representations of dimensions $1$. We can verify it mathematically, using the formula for $b_1$ and $b_2$. Let’s start with the formula:

$$ b_a = \frac{1}{N}\sum_{\mu} n_\mu \chi_{a}^*(C_\mu)\chi_{\Gamma}(C_\mu) $$

then, using $N=2$ and $a \equiv 1$ $$ b_1 = \frac{1}{2}\sum_{\mu} n_\mu \chi_{1}^*(C_\mu)\chi_{\Gamma}(C_\mu) = \frac{1}{2} ( 1\cdot 2 \cdot 1 + 1\cdot 0 \cdot 1 ) = \frac{2}{2} = 1 $$ where we have used that $\chi_{1} = 2$ (trace of matrix associated to $e$) and $\chi_2 = 0$, and the character of $\Gamma$ $\chi_{\Gamma}(C_\mu)$ is taken from our character table. Similarly,

$$ b_2 = \frac{1}{2}\sum_{\mu} n_\mu \chi_{2}^*(C_\mu)\chi_{\Gamma}(C_\mu) = \frac{1}{2}(1 \cdot 2 \cdot 1 + 1\cdot 0 \cdot (-1)) = 1 $$ which shows our initial Burnside lemma of $2$ representation of dimensions $1$.

Remember we mentioned that we can come up with different representations. Here, we can add another one - namely, we can consider the problem in terms of coordinates transformations, rather than in terms of element permutation. The transformation will keep $(x,y)$ unchanged and the will flip the $z$ coordinate. The resulting transformation matrix is therefore given by $$ \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \equiv e, \quad \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} \equiv p $$

If using the formulae for $b_a$ - we will find out different things - instead of $\Gamma = \Gamma_1\oplus \Gamma_2$ we find $\Gamma = 2\Gamma_1\oplus \Gamma_2$. This is due to the fact that in the second case, we’ve implicitly converted the problem into 3 dimensions. We should also mention the meaning of the 4th column in the character table of the $C_s$ group. That is, the meaning of those “function basis” and “rotations”. One should read them as follows: the $\Gamma_i$ transforms as <the function in question>. So, the $\Gamma_2$ transforms as $z$, which is very obvious since the $\Gamma_2$ flips the sign of the $z$; also, the $\Gamma_2$ transforms as $R_x$, $R_y$, since flipping the sign would do essentially the same as rotating by $\pi$ over the $x$ or $y$ axis. Similar reasonings can be applied for $\Gamma_1$. So what to do, once the multiplicities have been found?
Once the multiplicities have been found, we want to finally find the common eigenbasis for the obtained representations $\Gamma = \bigoplus_{a,x} \Gamma_{a,x}$ and the Hamiltonian, which is our final goal. This is done through projectors. This is not very complex, but relatively computationally expensive. We will thus not discuss it there, but only the idea:

So the idea is to find the set of eigenvectors that will span every irreducible representation. For that, one should construct the projector operators for each representations $\hat{\Pi}$, and find the eigenbasis for each of them. The number of the eigenvectors, serving as the basis will be equal to the degeneracies of the computed irreducible representations, that we computed.

The D3v group

Now, we can proceed with the next example and the related problem. This problem comes actually from one of my previous year’s exam - on representation group theory. The statement of the problem is as follows:

We consider a molecule of 3 identical atoms disposed on a equilateral triangle in 3D space. The symmetry group of an equilateral triangle in 3D is the $D_{3v}$. We consider the $\hat{z}$ direction to be perpendicular to the triangle plane. The character table of the $D_{3v}$ group is given below. The group consists of the identity operation $E$, two rotations around the $\hat{z}$ axis $C_2$ (as in the $C_{3v}$ group). Three rotations around the 3 possible bissectrices from each vertex $C_3$. One reflection (mirror) operation along the plane, which is parallel to the plane of the triangle. Then two improper rotations that we denote $S_3$, which are given by the composition of $C_3\circ \sigma_h$. Finally, three reflections over the 3 plans along the bissectrices. Every atom has 3 degrees of freedom and can move in 3D.
The task is to compute the characters of the representation $\Gamma$, find the decomposition to determine the potential degeneracies.

Okay, let’s start by providing the character table of the $D_{3v}$ group: $$ \def\arraystretch{1.5} \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \chi_{D_{3v}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_h & &\\ \hline \Gamma_1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, z^2\\ \hline \Gamma_2 & 1 & 1 & -1& 1 & 1 & -1& R_z & \\ \hline \Gamma_3 & 2 &-1 & 0 & 2 &-1 & 0& (x,y) & (x^2-y^2, xy) \\ \hline \Gamma_4 & 1 & 1 &-1 &-1 &-1 & 1& & \\ \hline \Gamma_5 & 1 & 1&-1 &-1 & -1& 1& z & \\ \hline \Gamma_6 & 2 &-1& 0 & -2 & 1 & 0& (R_x, R_y) & (xz, yz) \\ \hline \end{array} $$

Okay, we may ask ourselves a question - what is the difference between the rotation around one of the bissectrice $C_3$ and the reflection operations $\sigma_h$? Indeed, when performing the $C_2$ and $\sigma_v$ operations, the positions of the atoms are the same, as shown on the image:

However, they do differ and we can illustrate it by attaching a local coordinate system to each of the atoms.

Here, we see the difference - the $\sigma_v$ simply changes the red and green atoms (simple permutation operation). The $C_2$ operation, however, performs the full rotation of the local coordinate system. Try to picture the $C_2$ operation - the triangle rotates around the blue bissectrice axis, thus resulting in the flipped $z$ axis. The $\sigma_v$, however, simply flips the $x$ axis (we keep the correct orientation of the coordinate system). We thus see that there is indeed a difference between the 2 rotations.

Okay, so we can get back to the discussion and the problem itself. How can we determine the representation of the problem in question? This will be done via the tensor product of matrices. For that, we can even create a new subsection for that.

Tensor product

So, in order to determine the representations, i.e. $\Gamma(E)$, $\Gamma(C_3)$ etc… we need to determine 2 matrices for each transformation - one $3\times 3$ matrix that will describe the coordinate transformations (e.g. a rotation in 2D will yield a simple 2D rotation matrix) and the second $3\times 3$ permutation matrix that will provide the information on how the atoms are permuted as the result of the transformation. So let the coordinate transformation matrix be $R$ and the permutation matrix $T$; then resulting representation matrix of the transformation will be given by $\Gamma(g) = T\otimes R$ yielding a $9\times 9$ matrix. This does indeed make sense - there are 3 atoms, each having 3 coordinates, thus giving 9 “parameters”.
For that, let’s start by the simplest example - the representation of the identity transformation $\Gamma(E)$. The permutation matrix is easily given by $$ T_E \equiv \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

since no atoms are permuted (changing place). Similarly, for the coordinate matrix, it is given by $$ R_E \equiv \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

Thus giving the representation $\Gamma(E) = I_{9\times 9}$. The next operation to analyze is the $C_3$ operation. We want to find the representation $\Gamma(C_3)$. What will be its permutation matrix? So, one of the elements of the $C_3$ will perform a rotation by $120^{\circ}$. This will make the permutation $R\mapsto B$, $B \mapsto G$ and $G\mapsto R$ (where the letters represnt the colors of the atoms on the image). Therefore, the resulting permutation matrix will be given by $$ P_{C_3} \equiv \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} $$ What about the coordinate transformation matrix for the $C_3$ transformation? For that we ask ourselves - how do the transformations change the coordinates? The $z$ coordinate does not change at all, and the rest is changed as a usual 2D rotation, which is given by $$ R_{C_3} \equiv \begin{pmatrix} \cos(\theta) & -\sin(\theta) & 0\\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ where the angle $\theta = 120^{\circ}$, yielding the rotation matrix $$ R_{C_3} \equiv \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0\\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ The resulting representation matrix of the operation $C_3$ is thus the tensor product of $P_{C_3}$ and $R_{C_3}$. We can proceed further for this procedure for other operations in the list. Let’s quickly go over other example of constructing matrices.
The $C_2$ rotation is the rotation over the bissectrice axis. Let’s, for example, take the blue axis. Then, the rotation will be performed around the $y$ axis, meaning that the coordinate transformation matrix will be in the form of $$ \begin{pmatrix} \cos(\theta) &0 & \sin(\theta) \\ 0 & 1 & 0 \\ -\sin(\theta) &0 & \cos(\theta) \\ \end{pmatrix} $$ with $\theta = 180^{\circ}$ yielding the matrix

$$ \begin{pmatrix} -1 &0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & -1 \\ \end{pmatrix} $$

and the permutations are given by $R\mapsto G$, $G\mapsto R$ and $B \mapsto B$.

We must recall, that we’re mainly interested in characters, rather than the full set of all matrices (the full representation).

We will not write down all the permutations and coordinate transformation matrices. We’ve written down some of them - but it is quite straightforward to go on for other matrices. We will state the resulting character table: $$ \def\arraystretch{1.5} \begin{array}{|c|c|c|c|c|c|c|} \hline \chi & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline \Gamma_{P} & 3 & 0 & 1&3 &0 &1 \\ \hline \Gamma_{R} & 3 & 0 &-1 &1 &-2 &1\\ \hline \Gamma & 9 & 0 & -1 & 3 & 0 & 1\\ \hline \end{array} $$ Finally, we can now use the characters of our representation, that we computed using the tensor product of permutation matrix and coordinate transformation matrix. We can thus find the desired multiplicities using the already very familiar formula $$ b_a = \frac{1}{N} \sum_{\mu}^{N_c}n_{\mu} \chi_a^*(C_{\mu}) \chi(C_{\mu}) $$ where $\chi(C_{\mu})$ - the characters obtained via our procedure (the first table above). The $n_\mu$ - the number of elements inside this conjugacy class. Finally, the $\chi_{a}(C_{\mu})$ - the character of the irreducible representation $a$, which is found in the character table of the $D_{3v}$ group.

Conclusion

This article’s goal was mainly to make a recap of the topic of representation group theory basic applications in (quantum) physics. For that reason, we’ve tried to introduce in the most gentle and intuitive way possible. We’ve introduced the notion of representation in the same way. All the associated notions were introduced in the same way, with examples and analogies.