In the context of group theory (in my case, applications to physics), I frequently come across the phrase "the ${\bf N}$ of a group", for example "a ${\bf 24}$ of $\mathrm{SU}(5)$" or "the ${\bf 1}$ of $\mathrm{SU}(5)$" (the integer is usually typeset in bold).

My knowledge of group theory is pretty limited. I know the basics, like what properties constitute a group, and I'm familiar with simple cases that occur in physics (e.g. rotation groups $\mathrm{SO}(2)$, $\mathrm{SO}(3)$, the Lorentz group, $\mathrm{SU}(2)$ with the Pauli matrices as a representation), but not much more. I've got a couple of related questions:

  • What is meant by "${\bf N}$ of a group"?
  • Is is just shorthand for an ${\bf N}$ representation? If so, what exactly is an ${\bf N}$ representation of a given group?
  • How can I work out / write down such a representation concretely, like the Pauli matrices for $\mathrm{SU}(2)$? I'd be grateful for a simple example.
  • What does it mean when something "transforms like the ${\bf N}$"?
  • $\begingroup$ The dimension of SU(5) is 24. $\endgroup$
    – MBN
    Commented Oct 22, 2012 at 14:45
  • 2
    $\begingroup$ "Standard thing" depends on who you're talking to. Physicists use jargon for talking about Lie groups that isn't standard among mathematicians and vice versa. $\endgroup$ Commented Oct 22, 2012 at 14:47
  • $\begingroup$ ${\bf N}$ is used to describe the representation of the group where $N$ is the real dimension of said representation. Thus, the ${\bf N}$ of SU(N) is the fundamental rep and the $\bf{ N^2 - 1 }$ is the adjoint rep., etc. If there are two representations that have equal dimension, then some sub/superscripts are used to distinguish them. For instance, the left and right Weyl representation of the double cover of $SO(1,3)$ is denoted by ${\bf 4}_\pm$ or ${\bf 4}$ and ${\bar {\bf 4}}$. The vector/fundamental representation of $SO(8)$ is ${\bf 8}_v$ whereas the MW rep is denoted ${\bf 8}_s$. $\endgroup$
    – Prahar
    Commented Jul 28, 2016 at 3:23

3 Answers 3


OP wrote (v1):

What does "the ${\bf N}$ of a group" mean?

1) Physicists are referring to an irreducible representation (irrep) for whatever group $G$ we are talking about. The number ${\bf N}$ refers to the dimension of the irrep. The point is that irreps are so rare that irreps are often uniquely specified by their dimension (modulo isomorphisms). (This is not quite true in general, and physicists then start to decorate the bold-faced dimension symbol with other ornaments, e.g. ${\bf 3}$ and $\bar{\bf 3}$, or e.g. ${\bf 8}_v$ and ${\bf 8}_s$ and ${\bf 8}_c$, etc, to distinguish.)

2) By the way, concerning a group representation $\rho: G \to GL(V,\mathbb{F})$, where $G$ is a group, where $\mathbb{F}$ is a field (typically $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$), where $V$ is a $\mathbb{F}$-vector space, and where $\rho$ is a group homomorphism; be aware that physicists refer to both the map $\rho$ and the vector space $V$ as "a representation".

  • $\begingroup$ From my observation, physics textbooks refer $GL(V, F)$ as the representation of $G$. Math textbooks refer $\rho$ (the mapping) or, $V$ as the representation. Reference: Wiki, Representation Theory: A First Course - William Fulton, Joe Harris [page 3]. $\endgroup$
    – rainman
    Commented Jul 22, 2016 at 13:13
  • $\begingroup$ @omephy: Which physics textbooks? $\endgroup$
    – Qmechanic
    Commented Jul 22, 2016 at 13:52
  • $\begingroup$ 1. Matthew Robinson - Symmetry and the Standard Model (2011) [page 59] 2. Jakob Schwichtenberg - Physics from Symmetry (2015) [page 50]: "Although we define a representation as a map, most of the time we will call a set of matrices a representation." $\endgroup$
    – rainman
    Commented Jul 22, 2016 at 13:55
  • $\begingroup$ These are textbooks on group theory. Most physics literature identify the ${\bf N}$ of $G$ with the vector space $V$, not the map $\rho$ as they properly should, nor $GL(V,\mathbb{F})$. NB: If we discuss representation $\rho(g)$ of a specific group element $g\in G$, then $\rho(g)$ is typically represented by a $N\times N$ matrix in the physics literature. $\endgroup$
    – Qmechanic
    Commented Jul 22, 2016 at 19:48

What is meant by "${\bf N}$ of a group"?

The ${\bf N}$ of a group is in fact a shorthand for the $N$-dimensional irreducible representation of this group.

Is is just shorthand for an ${\bf N}$ representation? If so, what exactly is an ${\bf N}$ representation of a given group?

The group elements are abstract operations defined by how they act on given objects. For example, the rotation group in three dimensions, $\mathrm{SO(3)}$, is formed by elements that rotate coordinate systems in such a way that the length of any vector is invariant. In order to make things more explicit we assign linear representations to these groups, i.e. we map the group elements into matrices acting upon some vector space $\mathbb V$. If $\mathbb V$ is $N$-dimensional so it is the group representation.

If all $N$-dimensional matrices representing the group elements can - by a similarity transformation - be written in a block diagonal form then the representation is said to be reducible. Otherwise it is called irreducible (or simply an irrep.) and can be labelled by $\bf N$ which denotes its dimension. For example, general rotation in the plane, which constitute the group $\mathrm{SO(2)}$, can be written simply as $e^{i\theta}$ giving a one-dimensional irreducible representation. On the other hand, the two-dimensional representation $$ \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}. $$ is reducible since it can be put in the diagonal form $$ \begin{bmatrix} e^{i\theta}&0\\ 0&e^{-i\theta}\\ \end{bmatrix}, $$ by making use of a similarity transformation generated by $$ \frac{1}{\sqrt 2} \begin{bmatrix} 1&1\\ -i&i\\ \end{bmatrix}. $$ We should label that one-dimensional irreducible representation this by $\bf 1$ and therefore the two-dimensional reducible representation is labeled by $\bf 1\oplus \bf 1$, where the $\bf 1$ refers to each of the one dimensional block that can be written after a similarity transformation. This representation is actually acting upon a direct sum of two vector spaces of dimensions $1$.

  • How can I work out / write down such a representation concretely, like the Pauli matrices for $SU(2)$? I'd be grateful for a simple example.
  • What does it mean when something "transforms like the ${\bf N}$"?

To work out the irreducible representations we need to deal with the algebra instead of the group. Among all elements of a Lie group there are special ones that can be used to generate any other. They are called generators of the group and they satisfy a particular structure, called Lie algebra. For example, the group $\mathrm{SU}(2)$ has a Lie algebra $\mathfrak{su}(2)$ whose generators are $T_a$, $a=1,2,3$, satisfying $$[T_a,T_b]=i\epsilon_{abc}T_c.$$ A representation $R$ of these abstract elements has to preserve this structure, i.e., $$[R(T_a),R(T_b)]=i\epsilon_{abc}R(T_c),$$ where $R(T)$ shall be understood as an $N$-dimensional matrix.

From the Lie algebra one can obtain all possible representations. This is usually done by writing the generators in the so-called Cartan-Weyl basis which decomposes the algebra into the Cartan subalgebra (the maximal set of self-commuting or diagonalizable generators) and the step or ladder operatores. The states of a given irreducible representation are then given by the eigenvectors of the Cartan generators. Clearly these states are $N$-dimensional vectors given that the algebra representation is $N$-dimensional. So when we say that something - a field for instance - transforms like the $\bf N$ of an algebra we mean that this object is mapped to a column matrix with $N$ entries whose basis is given by the eigenvectors mentioned above. For example, the $\mathfrak{su}(2)$ algebra has only one step operator, $T_3$. For a two dimensional irrep. the matrix $R(T_3)$ has two eigenvectors. A field transforming like $\mathbf N$ - or simply as a doublet - is $\phi$, such that $$R\phi= \begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} \begin{bmatrix} \phi_1\\ \phi_2 \end{bmatrix} =\begin{bmatrix} \phi_1'\\ \phi_2' \end{bmatrix}. $$

One can show for instance that the $\mathfrak{su}(2)$ algebra has $N$-dimensional representations for any integer $N$. The classical algebras $\mathfrak{su}(n)$, $\mathfrak{so}(n)$ and $\mathfrak{sp}(n)$ have at least the singlet, the defining and the adjoint representations. The singlet is the one dimensional representation, i.e. they are only numbers. Notice that the only possibility numbers can satisfy a non-trivial algebra is that they are all zero. They are useful in physics when something does not transform at all. The defining representation is the $n$-dimensional, e.g. the three-dimensional representation for a quark transforming under the flavor $\mathfrak{su}(3)$. When the vector field $\mathbb V$ is the algebra itself the representation is called adjoint. In this case, the dimension of the algebra equals the dimension of the representation. The gauge fields transform under this representation of the gauge groups. For example, the algebra $\mathfrak{su}(5)$ has $24$ generators so the $\bf{24}$ is the adjoint representation of $\mathfrak{su}(5)$.

Once we know the representation $R(T)$ for a Lie algebra we can induce it to the group by means of an exponential operation, $$R(g)=\exp\left[i\phi R(T)\right],$$ where $g$ denotes the group element. Notice that if we have a singlet of the algebra, the singlet of the group turns out to be just the number $1$.

There is although some subtleties when going from the algebra to the group. Starting from a given Lie algebra and assigning a given representation one can ends up with different Lie groups. So for adjoint representation of $\mathfrak{su}(2)$ the group generated turns out to be $\mathrm{SO}(3)$ instead of $\mathrm{SU}(2)$.

  • $\begingroup$ SO(2) is abelian and therefore, all its irreducible representations are one-dimensional. The $2\times 2$ rotation matrix that you wrote is not irreducible. @Diracology $\endgroup$
    – SRS
    Commented Aug 7, 2017 at 4:15
  • $\begingroup$ @SRS Please have another look. I corrected what you pointed out. Thanks! $\endgroup$
    – Diracology
    Commented Aug 7, 2017 at 13:41
  • $\begingroup$ The 3-dimensional reducible representation of SO(2) that you've written, is block- diagonal. The $2\times 2$ block is still reducible. I don't know whether we can use the notation '$\textbf{2}$' for this. Your answer that '$\textbf{N}$' is "a shorthand for N-dimensional irreducible representation" then clashes with this. But I might be wrong and this notation may be usual. A minor thing: generators are not special group elements (except for certain discrete groups such as the cyclic groups). Generators of the Lie groups are exponentiated to obtain the group elements. @Diracology $\endgroup$
    – SRS
    Commented Aug 7, 2017 at 13:50
  • $\begingroup$ @SRS I definitely should not use $1+2$ for that representation. It should be $1+1+1$ since the $N$ always means the dimension of the irreducible representations. $\endgroup$
    – Diracology
    Commented Aug 7, 2017 at 13:59

''the $N$ of a group $G$'' refers to an $N$-dimensional irreducible (projective) representation of the (typically semisimple) group $G$. A representation is a homomorphism $U$ from $G$ to the space of linear self-mappings of a vector space $V$ (in the projective case acting on the rays); it is irreducible if there is no basis in which all $U(g)$ are block triangular. The dimension of the representation is the dimension of $V$.

For example, the representation theory of $SO(3)$ implies that there is precisely one irreducible projective representation of every dimension $N$. The 2-dimensional representation is the spinor representation, the three-dimensional one the ordinary vector representation.

If an object $x$ transforms like an $N$ then $x$ is a generic element from an $N$-dimensional space with the representation $N$, and hence trasforms under a group element $g$ by means of $x\to U(g)x$. For example in case of $SO(3)$, if $x$ transforms as a $2$ then it is a spinor, if it transforms like a $3$ then it is a vector, etc.

In many cases, the dimension determines the representation up to isomorphism, hence the jargon. (Otherwise, representations may be called $N$ and $\overline N$, etc., to distinguish them.) For example, the dimension of SU(5) is 24, and the 24 characterizes the adjoint representation (which has dimension 24).

  • 1
    $\begingroup$ Maybe you meant $so(3)$ instead of $SO(3)$. The Lie group $SO(3)$ has irreps of only odd dimension but it's Lie algebra $so(3) = su(2)$ has irreps of every dimension. $\endgroup$
    – Eric
    Commented Oct 22, 2012 at 15:20
  • $\begingroup$ @Eric: Thanks. I edited the answer to make clear that I meant projective representations. These are the relevant ones in quantum mechanics. $\endgroup$ Commented Oct 22, 2012 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.