Home GroupMath Sym2Int Susyno Extra data Math & Science

Using GroupMath

0. Getting started

GroupMath is a Mathematica package which can make various calculations related to Lie groups and the permutation group $S_n$. This page shows some examples.

To get started, the program needs to be installed: it can be download here . Decompressing the downloaded zip file will generate a folder GroupMath. This folder should be placed in a directory visible to Mathematica. The best options are to place it in the Applications subfolder of $BaseDirectory or $UserBaseDirectory. Often, the paths are the following:

/usr/share/Mathematica/Applications
Linux

/home/(computer name)/.Mathematica/Applications
Linux



C:\ProgramData\Mathematica\Applications
Windows

C:\Users\(computer name)\AppData\Roaming\Mathematica\Applications
Windows



The GroupMath folder (not just its contents) must be placed in one of these directories. Once this is done the package is installed. To load GroupMath, type in the front end 

<<GroupMath`

For help, there is an easy-to-use built-in documentation for each of the functions described in the following text (and other ones as well); it becomes accessible from within Mathematica once the package is installed. For questions, comments or bug reports, please contact me at

fonseca@ipnp.mff.cuni.cz

1. Working with representations of a Lie group

To indicate a group, just use it's name:


U1, SU2, SU3, SU5, SO10, E6, E8, G2, ...
If a group is the product of $U(1)$'s and/or simple groups, a list of the factor groups should be given. For example, the groups $SU(3)\times SU(2) \times U(1)$, $SU(5)\times SU(5)$ and $SO(10)\times U(1)$ are written as follows: 

{SU3,SU2,U1}
{SU5,SU5}
{SO10,U1}
In fact, even groups with a single factor might have the brackets {} around. For example, one might write $SU(2)$ as 

{SU2}


To indicate a representation of each factor group one needs to write its Dynkin coefficients (unless it is a $U(1)$, in which case the charge [i.e., one number] is enough). These are a list of $n$ non-negative integers, where $n$ is the group's rank. In turn, the group's rank corresponds to the maximal number of elements of the group's algebra which can be made simultaneously diagonal, and in practice this can be obtained with GroupMath by typing 


Length[<group>]
such as 

Length[SU5]

This means that $SU(5)$'s rank is four, and so each of its representations can be uniquely identified by a list {n1,n2,n3,n4} of four non-negative integers. The user can call the RepName function to help identify each representation from the Dynkin coefficients. For example: 


RepName[SU5,{1,0,0,0}]
RepName[SU5,{0,1,0,0}]
RepName[SU5,{0,0,0,1}]
RepName[SU5,{1,0,0,1}]

With the help of the function RepsUpToDimN which lists all the representations of a group up to some size, it is then very easy to get a table of representations. For example, the list of all $SU(3)$ representations up to size 30 can be obtained as follows: 


su3Reps = RepsUpToDimN[SU3, 30];
Grid[Prepend[{#, RepName[SU3, #]}&/@ su3Reps,{"Dynkin coefficients", "Name"}],
Frame -> All, FrameStyle -> LightGray]

For the representations $\boldsymbol{R}$ and $\overline{\boldsymbol{R}}$ with no primes, it is possible to use a simplified input format: R and -R (respectively), where R is just the dimension of the representations. For example, in $SU(3)$ instead of {3,0} and {0,3} one can write 10 and -10. On the other hand, 15 and -15 identify the representations {2,1} and {1,2}, so there is no way to use the simplified input format to identify the representations {4,0} and {0,4}.

In the following, for illustrative purposes I will alternate between both formats (Dynkin coefficients and the simplified input format).

2. Dimension, Casimir, Dynkin index and triangular anomaly

Given a representation $R$, its dimension $d(R)$, Casimir $C(R)$ and Dynkin index $T(R)$ can be calculated with the functions DimR, Casimir and DynkinIndex: 


DimR[SO10, 16]
Casimir[SO10, 16]
DynkinIndex[SO10, 16]

Recall that $C(R)$ and $T(R)$ are defined by the relations $C(R) Id=\sum_{a}T^{a}T^{a}$ and $T(R)\delta^{ab}=\textrm{Tr}\left(T^{a}T^{b}\right)$ where $T^a$ are the representation matrices of the algebra generators. Note also that using the Dynkin coefficients {0,0,0,0,1} above instead of 16 would have produced the same results.
These functions actually accept symbolic Dynkin coefficients. For example, the $SU(2)$ representation $\left\{ d-1\right\}$ has dimension $d$ and its Casimir is $\frac{1}{4} (d-1) (d+1)$: 

DimR[SU2, {d - 1}]
Casimir[SU2, {d - 1}]

Gauge triangular anomalies are known to be associated to the quantity $\textrm{Tr}\left(\left\{ T^{a},T^{b}\right\} T^{c}\right)\equiv\kappa(R) d^{abc}$ where the symmetric tensor $d^{abc}$ can be taken to be fixed (for a given group) while $\kappa(R)$ depends on the representation. It should be noted that the only groups for which there might be anomalies are those with $SU(n>2)$ and/or with $U(1)$ factors, so at least one of the indices $a$, $b$ and $c$ must refer to one of these groups.
The triangular gauge anomalies associated to a representation can be calculated with the TriangularAnomalyValue function: 


TriangularAnomalyValue[{SU3}, {{1, 0}}]
TriangularAnomalyValue[{U1}, {y}]

More complex cases, where the gauge group has more than one factor, are also possible. Consider for example the representation of left-handed quarks $Q$ in the Standard Model: 

Q = {1/6,2,3};
TriangularAnomalyValue[{U1, SU2, SU3}, Q]

There is more than one anomaly in the output, and one case use the Verbose -> True option to identify each of them: 

Q = {1/6,2,3};
TriangularAnomalyValue[{U1, SU2, SU3}, Q, Verbose -> True]

One can use this function, for example, to quickly show that all anomalies cancel between each Standard Model generation of fermions: 

uc = {-2/3, 1, -3};
dc = {1/3, 1, -3};
Q = {1/6, 2, 3};
ec = {1, 1, 1};
L = {-1/2, 2, 1};

fields = {uc, dc, Q, ec, L};
TriangularAnomalyValue[{U1, SU2, SU3}, #] & /@ fields
Total[%]

3. Products of representations

The decomposition of products of representations can be achieved with ReduceRepProduct: 

tripletSU3 = {1, 0};
ReduceRepProduct[SU3, {tripletSU3, tripletSU3, tripletSU3}]

In this case, the output is saying that $\mathbf{3}\times\mathbf{3}\times\mathbf{3}$ in $SU(3)$ (note that {1,0} is the $\mathbf{3}$) contains the representation {3,0} once, {1,1} twice, and {0,0} once. If desired, the option UseName -> True can be used to convert these Dynkin coefficients into the names of the representations: 

ReduceRepProduct[SU3, {tripletSU3, tripletSU3, tripletSU3}, UseName -> True]

The group can have various factors, for example: 

uc = {-2/3 , {0}, {0, 1}};
Q = {1/6 , {1}, {1, 0}};
H = {1/2 , {1}, {0, 0}};
ReduceRepProduct[{U1, SU2, SU3}, {uc, Q, H}, UseName -> True]

Also, there is no limit to the number of representations being multiplied: 

ReduceRepProduct[SO10, {10, 10, 10, 10, 10}, UseName -> True]

Sometimes, the product of representations being calculated contains repeated representations, such as the $\mathbf{3}\times\mathbf{3}\times\mathbf{3}$ and $\mathbf{10}\times\mathbf{10}\times\mathbf{10}\times\mathbf{10}\times\mathbf{10}$ examples above. In those cases, there is a permutation symmetry involved: for instance, it is well known that the singlet in $\mathbf{3}\times\mathbf{3}\times\mathbf{3}$ is completely anti-symmetric. To calculate these symmetries, see section 6.

4. Explicit representation matrices

Consider the generators $T^a$ of some representation $R$. For some invertible matrix $O$ one can make the transformation

$T^{a}\rightarrow{T'}^{a}\equiv O_{ab}T^{b}\,, $

and the ${T'}^{a}$ matrices, just like the original $T^a$, will form a basis of the algebra in the $R$ representation. The same thing happens with the following transformation, as long as $U$ is an invertible matrix:

$T^{a}\rightarrow{T'}^{a}\equiv U^{-1}T^{b}U \,. $

In physics, usually one requires that the group transformations represented by the matrices $\exp\left(i\varepsilon^{a}T^{a}\right)$ are unitary, for some real parameters $\varepsilon^{a}$, which implies that the $T^{a}$ need to be hermitian matrices (${T^{a}}^\dagger=T^{a}$). Furthermore, it is convenient to have $C(R) Id=\sum_{a}T^{a}T^{a}$ and $T(R)\delta^{ab}=\textrm{Tr}\left(T^{a}T^{b}\right)$ (see section 2 above). These requirements constrain somewhat the above transformation freedom: if $T^{a}$ obeys these conditions, then the $O$ and $U$ above must be orthogonal and unitary, respectively, in order to keep having $C(R) Id=\sum_{a}{T'}^{a}{T'}^{a}$ and $T(R)\delta^{ab}=\textrm{Tr}\left({T'}^{a}{T'}^{b}\right)$.

Up to now we have been discussing the computation of quantities which are invariant under these $O$/$U$ basis transformations. However, one often wants to obtain basis-depend quantities (such as the Lagrangian). This inevitably requires taking a particular choice of basis.

Keeping this in mind, GroupMath can make various basis-dependent calculations. The most basic one is to compute the representation matrices $T^a$ with the RepMatrices function. For example, the representation matrices of doublets of $SU(2)$ are half the Pauli matrices: 


MatrixForm /@ RepMatrices[SU2, {1}] (* {1}=doublet representation *)

Just as easily one can calculate, say, the quintuplet representation matrices: 

MatrixForm /@ RepMatrices[SU2, 5]

This function works for any representation of any group: 

MatrixForm /@ RepMatrices[SU3, {1,0}] (* {1,0}=fundamental representation *)

Note carefully that these matrices are similar, but not exactly equal, to half the Gell-Mann matrices. However, both these and $1/2$ the GellMan matrices make a perfectly fine basis for the algebra of $SU(3)$, in the fundamental representation and therefore, one can check that there is an $O$/$U$ basis transformations which converts one set of matrices into the other.

It turns out that the basis used by GroupMath for the $T^a$'s is such that there is a maximal number of diagonal generators (which are always given last by RepMatrices). This is usually a very convenient basis choice, but there is an important consequence for real representations which we will now discuss. A group transformation $V=\exp\left(i\varepsilon^{a}T^{a}\right)$ which is real, $V=V^*$, implies that the $T^{a}$ algebra matrices must be anti-symmetric, $T^{a}=-{T^{a}}^T$, and therefore none of them can be diagonal. This means that GroupMath's basis choice is never the real one. For example, even though the triplet representation of $SU(2)$ is real, the $T^a$'s given by RepMatrices do not lead to a real $V=\exp\left(i\varepsilon^{a}T^{a}\right)$ as can be seen from the fact that not all of them are anti-symmetric: 


MatrixForm /@ RepMatrices[SU2, 3]

However, it is not too complicated to figure out the unitary transformation $T^{a}\rightarrow{T'}^{a}\equiv U^{\dagger}T^{b}U$ which makes the ${T'}^{a}$'s anti-symmetric. In the present case, it turns out that

$U=\left(\begin{array}{ccc} \frac{1}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}}\\ 0 & -i & 0\\ \frac{1}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \end{array}\right)$

will do the trick: 


repMs = RepMatrices[SU2, 3];
U = {{1/Sqrt[2], 0, -(I/Sqrt[2])}, {0, -I, 0}, {1/Sqrt[2], 0, I/Sqrt[2]}};
MatrixForm[ConjugateTranspose[U].#.U] & /@ repMs

5. Group invariant combinations of representations

Often one needs to know how to contract, in a group invariant way, the components of a product of representations. The simplest example would be two $SU(2)$ doublets — let us call them $D=(D_1,D_2)^T$ and $D'=({D'}_1,{D'}_2)^T$: it is well known that the combination $\varepsilon_{ij}D_{i}D_{j}^{\prime}=D_1 {D'}_2-D_2 {D'}_1$ is left invariant under the action of the $SU(2)$ group ($\varepsilon$ being the Levi-Civita symbol). To calculate group invariant combinations with GroupMath, use the Invariants function: 


Invariants[SU2,{2,2}]

The syntax is Invariants[<group>,{<rep1>, <rep2>, ...}], with an arbitrary number of representations. In the output, the program considers that the components of <rep1> are named a[1], a[2], etc., that those of <rep2> are b[1], b[2], etc., and so forth.

Notice that the output above is not just a[2] b[1] - a[1] b[2]; this expression is surrounded by curly brackets. That's because, in general, there might more than one independent way of contracting the representations in an invariant way. To illustrate this point, consider the product of four $SU(2)$ doublets, which is known to have two independent invariants (this statement can be confirmed with the ReduceRepProduct function, by counting the number of singlets in the product of four doublets): 


Invariants[SU2, {2,2,2,2}]

One should keep in mind that if $I_{1}$, $I_{2}$, ... are a list of independent group invariants then any linear combination of them will also be invariant. Therefore instead of the $I_a$ one could take instead $I'_a\equiv P_{ab} I_b$ for some invertible matrix $P$. For example, in the example above of the product of 4 doublets of $SU(2)$, the two invariants can be combined and presented in a different way. In fact, even when there is just one invariant expression, it can be arbitrarily multiplied by some numerical factor.

Another important point is that Invariants assumes that the representations transform in the basis given by RepMatrices. We can test this: if we write an invariant as $I=c_{ijk\cdots}a\left[i\right]b\left[j\right]c\left[k\right]\cdots$, the condition to be verified is

$\left(T^{a}\right)_{ii'}c_{i'jk\cdots}+\left({T'}^{a}\right)_{jj'}c_{ij'k\cdots}+\left({T''}^{a}\right)_{kk'}c_{ijk'\cdots}+\cdots=0$

for all $a$, where $T^{a}$, ${T'}^{a}$, ${T''}^{a}$, ... are the representation matrices of $\left(a[1],a[2],...\right)^{T}$, $\left(b[1],b[2],...\right)^{T}$, $\left(c[1],c[2],...\right)^{T}$, ... respectively. The two invariants which can be formed with the product of four $SU(2)$ above can be tested as follows: 


invariants = Invariants[SU2, {2,2,2,2}] ;

subsRuleA = Table[MapThread[Rule, {{a[1], a[2]}, mat.{a[1], a[2]}}],
{mat, RepMatrices[SU2, {1}]}];
subsRuleB = subsRuleA /. a -> b;
subsRuleC = subsRuleA /. a -> c;
subsRuleD = subsRuleA /. a -> d;

(invariants /. subsRuleA) + (invariants /. 
   subsRuleB) + (invariants /. subsRuleC) + (invariants /. subsRuleD)

The 3 groups of {0,0} in the result show the two invariants do not change under any of the three $SU(2)$ generators ($a=1,2,3$).

6. Permutation symmetries of products of representations

Section 3 explains how one can break a product of representations $R_{1}\times R_{2}\times\cdots\times R_{n}$ in its irreducible parts. However, sometimes, that is not enough: if some of the $R_i$ are the same, then there is a permutation symmetry to consider as well. For example, in $SU(2)$ we have $\mathbf{2}\times\mathbf{2}=\mathbf{1}+\mathbf{3}$ and it is well known that the singlet ($\mathbf{1}$) is an antisymmetric contraction of the two $\mathbf{2}$'s, while the triplet ($\mathbf{3}$) is symmetric. Usually this is indicated as $\mathbf{2}\times\mathbf{2}=\mathbf{1}_A+\mathbf{3}_S$. The function ReduceRepProduct introduced in section 3 does not provide information about what happens under permutations of equal representations in a product $R_{1}\times R_{2}\times\cdots\times R_{n}$, but PermutationSymmetry does.

This function computes such information with full generality, however some users might be surprised to know that at stake is not simply a question of adding S's and A's to each irreducible representation in a product. Perhaps is is then instructive to consider first what happens in a product of four $SU(2)$ doublets: $\mathbf{2}\times\mathbf{2}\times\mathbf{2}\times\mathbf{2}=\mathbf{1}+\mathbf{1}+\mathbf{3}+\mathbf{3}+\mathbf{3}+\mathbf{5}$. Are these irreps symmetric (S) or antisymmetric (A)? It turns out that such question is not well formulated: in the case of $\mathbf{2}\times\mathbf{2}=\mathbf{1}+\mathbf{3}$ there are just two equal factors in the product, so the relevant permutation group is $S_2$, which has two 1-dimensional irreducible representations: the symmetric one (S) and the antisymmetric one (A). In the case of $\mathbf{2}\times\mathbf{2}\times\mathbf{2}\times\mathbf{2}$, one should identify the irreducible $SU(2)$ representations contained in it with $S_4$ irreps. In general then, if there is a product with $n$ repeated representations $R$, one should look to the irreps of the $S_n$ permutation group, which are given by the partitions of $n$ (for example {4}, {3,1}, {2,2}, {2,1,1}, {1,1,1,1} in the case of $n=4$). Further down, in section 8, the permutation groups are discussed in more detail.

Having said this, we are ready to take a look at the output of PermutationSymmetry. Consider first $\mathbf{2}\times\mathbf{2}$: 


PermutationSymmetry[SU2, {2, 2}]

The result contains several pieces of information and many brackets, so let us look at it piece by piece. The first part of the output, {{1,2}}, informs us that representation #1 and #2 are the same (which is quite obvious to the user). The second part says that

  1. there is an $SU(2)$ triplet (={2}) in the product which is in the {2}=symmetric representation of the $S_2$ induced by permutations of the doublets (multiplicity is 1);
  2. there is an $SU(2)$ singlet (={0}) in the product which is in the {1,1}=anti-symmetric representation of the $S_2$ induced by permutations of the doublets (multiplicity is 1).
The output is probably clearer if we convert the Dynkin coefficients of SU(2) representations to names, and the partitions (which indicate irreps of the permutation group) to Young diagrams with boxes. This can be done with the UseName option: 

PermutationSymmetry[SU2, {2, 2}, UseName -> True]

In the case of $\mathbf{2}\times\mathbf{2}\times\mathbf{2}\times\mathbf{2}$, the code would be the following: 


PermutationSymmetry[SU2, {2,2,2,2}, UseName -> True]

It says that, under the $SU(2)_{S_{4}}$ group, the product breaks as $\mathbf{5}_{S}+\mathbf{3}_{\mathbf{3}}+\mathbf{1}_{\mathbf{2}}$ (using the notation $\left\{4\right\}=S$, $\left\{ 3,1\right\}=\mathbf{3}$, $\left\{ 2,2\right\}=\mathbf{2}$ for the $S_4$ representations). So, for example, neither of the two singlets in this product is completely symmetric nor anti-symmetric: the pair of singlets is converted into one-another by permutations of the doublets (the permutation symmetry entangles them), thereby they form a 2-dimensional irreducible representation of $S_4$.

Finally, note that the PermutationSymmetry function is capable of handling more general cases, such as when there are more than one type of repeated representations being multiplied or when the Lie group is not simple: 


PermutationSymmetry[SU2, {2,3,2,3}, UseName -> True]


PermutationSymmetry[{SU3, SU2}, {{3,2}, {3,2}, {3,2}},
 UseName -> True]

7. Symmetry breaking: branching rules and more

Concerning symmetry breaking, one usually wants to know how a given representation of a group breaks into irreducible representations of some subgroup. This can be calculated with DecomposeRep, which requires 4 things from the user:

  1. the group $G$;
  2. the subgroup $H\subset G$;
  3. information on how $H$ is embedded in $G$ — the so-called projection matrix, to be specific;
  4. the representation $R$ of $G$ to be decomposed.
(1), (2) and (4) are trivial to provide, while (3) might require some work, but not much (more on this later). In any case, the user needs to figure out (3) only once for a given symmetry breaking pattern $G\rightarrow H$.

Consider the simple case where $SU(3)$ breaks into $SU(2)\times U(1)$. The projection matrix in this case can be chosen to be

$\left(\begin{array}{cc} 1 & 0\\ 1 & 2 \end{array}\right)$

We may then go ahead and see, for example, how does the triplet of $SU(3)$ decompose: 

group=SU3;
rep=3;
subgroup={SU2,U1};
prjMat = {{1, 0}, {1, 2}};

DecomposeRep[group,rep,subgroup,prjMat,UseName->True]

Likewise, consider $SU(5)\rightarrow SU(3)\times SU(2)\times U(1)$. Once the group, subgroup and projection matrix have been given, it is easy to write a few lines of code which will generate a table with the decomposition of $SU(5)$ representations: 


group=SU5;
subgroup={SU3,SU2,U1};
prjMatrix={{1,0,0,0},{0,1,0,0},{0,0,0,1},{2,4,6,3}};
SU5reps=RepsUpToDimN[SU5,100];
data=Table[{Style[RepName[group,rep],Darker[Red]],
DecomposeRep[group,rep,subgroup,prjMatrix,UseName->True]},{rep,SU5reps}];
Grid[Prepend[data,Style[#,Bold]&/@{"SU(5)","SU(3) x SU(2) x U(1) content"}],
Frame->All,FrameStyle->LightGray]

The initial group does not need to be simple. For example, the projection matrix for the embedding of the diagonal subgroup $SU(2)$ of $SU(2)\times SU(2)\times SU(2)$ is

$\left(\begin{array}{ccc} 1 & 1 & 1\\ \end{array}\right)$

and therefore we could calculate the decomposition of the representation $(\mathbf{2},\mathbf{3},\mathbf{2})$ under $SU(2)$ as follows:


DecomposeRep[{SU2,SU2,SU2}, {2,3,2}, SU2, {{1,1,1}}, UseName->True]

So how can one compute the projection matrix? There are two functions in GroupMath which might be of help for this task. One of them is RegularSubgroupProjectionMatrix, but it requires some knowledge of Dynkin diagrams.

An easier alternative is to use the function Embeddings. For example 


embeddings = Embeddings[SU5, {SU3, SU2}]
returns a list with a single entry of the form {<group>, <projection matrix>}. To see this more clearly we can rewrite the output as follows: 

Grid[{CMtoName[#[[1]]], MatrixForm[#[[2]]]} & /@ embeddings]

There are two important comments to be made. First, the output contains only one element, so there is only one way to embed $SU(3) \times SU(2)$ in $SU5)$. Second, we actually got the embedding of $SU(3) \times SU(2) \times U(1)$, with the extra $U(1)$. That is because Embeddings never destroys $U(1)$'s it finds, so if they are unwanted, the user must get rid of them manually by removing or combining rows of the projection matrices. In this particular example, we would remove the last row from the projection matrix: 


prjMatrix = {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}};
DecomposeRep[SU5, 10, {SU3, SU2}, prjMatrix, UseName -> True]

From the following example, we see that $SU(3)\times SU(3)$ can be embedded in $E(6)$ in three different ways, and the possibility of having extra $U(1)$'s depends on the embedding. 


embeddings =Embeddings[E6, {SU3, SU3}];
Grid[{CMtoName[#[[1]]], MatrixForm[#[[2]]]} & /@ embeddings]

Using this data on the DecomposeRep function it becomes clear that the 3 embeddings are distinct, leading to different branching rules.

How does Embeddings work? Just like humans would do, it starts with a group $G$ and explores alls sequences of maximal subgroups ending in the desired subgroup, removing equivalent embeddings. This stepwise calculation can be done manually be using the function MaximalSubgroups several times. For example, $SU(4)$ has 4 maximal subgroups, and these are the associated projection matrices: 


maximalSubgroups = MaximalSubgroups[SU4];
Grid[{CMtoName[#[[1]]], MatrixForm[#[[2]]]} & /@ maximalSubgroups]

8. Functions related to the $S_n$ permutation group

Up to now, we have been discussing the calculation of quantities related to Lie groups. However, as explained in section 6, there is a connection with the discrete permutation group $S_n$, and for this reason the GroupMath program also contains some $S_n$ related functions which we will now go through.

Recall that the irreducible representations of $S_n$ are given (i.e., can be labeled) by the partitions of $n$: 


(* Example: the representations of S5 *)
IntegerPartitions[5]

(IntegerPartitions is a Mathematica built-in function.) GroupMath includes the function YoungDiagram which draws the Young diagram associated to a partition: 


YoungDiagram[{3, 2}]

Likewise, it is well known that the conjugacy classes of $S_n$ can also be labeled by the partitions of $n$, since this can be used to describe the length of the cycles of the group elements. For example, the identity element of $S_5$ can be written as $(1)(2)(3)(4)(5)$ in the cycle notation, meaning that it belongs to the {1,1,1,1,1} conjugacy class, while the $(143)(52)$ group element belongs to the class {3,2}.

With this reminder, we move on to the function SnIrrepDim which computes the dimension of $S_n$ irreducible representations. For example, {3,3,1} is a 21-dimensional representation of $S_7$: 


SnIrrepDim[{3,3,1}]

It is worth noting that the dimension of the $S_n$ irreducible representation associated to the partition $\lambda$ matches the number of standard Young tableaux with that shape. These can be calculated with the function GenerateStandardTableaux: 


GenerateStandardTableaux[{3, 3, 1}, Draw->True]

As a second example, we may compute the dimensions of the $S_5$ representations: 


S5reps = IntegerPartitions[5]
dimS5reps = SnIrrepDim /@ S5reps
dimS5reps.dimS5reps (*check that the sum of the dimensions squared is 120=5!*)

The function SnClassCharacter is more general, as it calculates the character of a given conjugacy class in some irreducible representation of $S_n$. With it we may for example build the character table of $S_5$: 


n = 5;
characterTable = Reverse /@ Table[SnClassCharacter[i, j],
{i, IntegerPartitions[n]}, {j, IntegerPartitions[n]}];

Print["Character table of ", Subscript[S, n], ":"];
characterTable // MatrixForm

The order of each conjugacy class (i.e., the number of elements in it) can be computed with SnClassOrder: 


SnClassOrder[{1, 1, 1}]
SnClassOrder[{2, 1}]
SnClassOrder[{3}]

With this data, one can easily decompose the product of $S_n$, and indeed that is the purpose of DecomposeSnProduct: 


DecomposeSnProduct[{{2, 1}, {2, 1}}]

The output indicates that $\left\{ 2,1\right\} \times\left\{ 2,1\right\} =\left\{ 3\right\} +\left\{ 2,1\right\} +\left\{ 1,1,1\right\}$ (the number after each partition in the output is its multiplicity in the decomposition of $\left\{ 2,1\right\} \times\left\{ 2,1\right\}$; in this example it is always 1). Here is a slightly more elaborate example with $S_5$, namely $\left\{3,2\right\} \times\left\{3,2\right\}\times\left\{2,2,1\right\}$, where YoungDiagram is also used: 


decomposition = DecomposeSnProduct[{{3, 2}, {3, 2}, {2, 2, 1}}];
{YoungDiagram[#[[1]], ScaleFactor -> 10], #[[2]]} & /@ decomposition

To calculate branching rules there is a function SnBranchingRules. The idea is the following: the permutation group $S_{n1}\times S_{n2}\times\cdots$ is a subgroup of $S_{n_1+n_2+\cdots}$ and we may want to know how does a representation of this larger group behave under the subgroup. For instance the representation {2,1} of $S_3$ decomposes as $\left\{ 2\right\} \times\left\{ 1\right\} +\left\{ 1,1\right\} \times\left\{ 1\right\}$ under the subgroup $S_2 \times S_1$: 


SnBranchingRules[{2, 1}, {2, 1}]

The syntax is SnBranchingRules[<partition of the big group>,<{n1,n2,...}>]. This is the decomposition of the irreducible representation {4,3} of $S_7$ under the subgroup $S_3 \times S_2 \times S_2$: 


result = SnBranchingRules[{4, 3}, {3, 2, 2}];
{YoungDiagram[#,ScaleFactor -> 10] & /@ #[[1]], #[[2]]} & /@ result

There is also the function SnIrrepGenerators which calculates explicitly the representation matrices of the group elements $(12)$ and $(12\cdots)$ of $S_n$. Note that these two elements are enough to generate the whole group, with size $n!$. The matrices returned by this function are always real and orthogonal/unitary. For example: 


MatrixForm /@ SnIrrepGenerators[{2, 1}]
MatrixForm /@ SnIrrepGenerators[{3, 2}]

With successive matrices multiplications, one can recover all group elements: 


gens = SnIrrepGenerators[{2, 1}];
MatrixForm /@ FixedPoint[Sort[DeleteDuplicates[Join[#, 
Dot @@@ Tuples[#, 2]]]] &, gens]

A partition can also identify a representation of an $SU(n)$ group. For example, {3, 3, 1} stands for the representation of $SU(10)$ with Dynkin coefficients {0,2,1,0,0,0,0,0,0}, which is 43560-dimensional. The function ConvertPartitionToDynkinCoef can be used to make this conversion (ConvertToPartitionNotation does the inverse): 


su10representation = ConvertPartitionToDynkinCoef[10, {3, 3, 1}]
DimR[SU10, su10representation]

This last number (43560) is the same as the quantity $\mathcal{S}\left(\left\{ 3,3,1\right\} ,10\right)$ mentioned in the paper arXiv:1907.12584 [hep-ph], which in general can be computed with the function HookContentFormula: 


HookContentFormula[{3, 3, 1}, 10]

The following identity holds: \[ \sum_{\lambda\vdash m}d\left(\lambda\right)\mathcal{S}\left(\lambda,n\right)=n^{m}\,, \] where $\lambda\vdash m$ means that $\lambda$ is a partition of $m$, and $d\left(\lambda\right)$ stands for the dimension of the $S_m$ irreducible representation $\lambda$. This formula can be checked for $m=7$ as follows: 


Simplify[Total[
  SnIrrepDim[#] HookContentFormula[#, n] & /@ IntegerPartitions[7]]]

Author
Renato Fonseca

Emails
renatofonseca@gmail.com
or
renatofonseca@ugr.es

Last updated
05 March 2024