I'm reading the following post to learn about QCD interactions: Why are 3 colors used in QCD?

However, I can't seem to grasp the conceptual difference between the dimension of a group and the dimension of its representation. The information I found uses representation theory formalism that I don't quite understand yet.

What is the difference in the definition of the dimension of a group and its representation? And in particular, why does $SU(N)$ symmetry corresponds to $N^2-1$ conserved charges?


3 Answers 3


An illustrative example.

The group $SU(2)$ is 3-dimensional. It's generated by the three Pauli matrices. $$\sigma_x = \left(\begin{matrix}0&1\\ 1 & 0\end{matrix}\right) \qquad \sigma_y = \left(\begin{matrix}0&i\\ -i & 0\end{matrix}\right)\qquad \sigma_z = \left(\begin{matrix}1&0\\ 0 & -1\end{matrix}\right)$$

These matrices act by multiplication on a $2$-dimensional vector space, the standard representation of $SU(2)$. The dimension of this representation is $2$.

  • 3
    $\begingroup$ +1 A simple example might be the best way to explain some non trivial concept. I small nitpick is that the group SU(2) is generated by operators satisfying $[T_i,T_j]=i\epsilon_{ijk}T_k$. In a particular representation those operators are given by Pauli matrices. $\endgroup$
    – Diracology
    Mar 17, 2018 at 1:11
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    $\begingroup$ Ah, a pedantry challenge! Such operators generate a group isomorphic to a quotient of $SU(2)$! But the group $SU(2)$ is, by definition, the manifold of $2$ by $2$ unitary matrices of determinant one. The Pauli matrices generate this group. $\endgroup$
    – user1504
    Mar 17, 2018 at 1:24

The dimension of a Lie group is given equivalently by the dimension of its corresponding manifold, the dimension of its associated Lie algebra or the number of group generators.

On the other hand to understand what the dimension of a (linear) representation is we shall understand what a linear representation means. Group elements are abstract in the sense that they are defined by the way they act on certain objects. For example, the rotation group in three dimensions is given by operators that rotate three dimensional vectors. Given a linear representation, the group elements are appropriately mapped into linear operators which act on a linear space. The dimension of this linear space is the dimension of the representation. Recalling that linear operators acting on a linear space can be written as $n$x$n$ matrices, then we see that the dimension of the corresponding representation is $n$. The Lie algebra representation is also defined in the same way and this actually induces the group representation since group elements can be obtained by the exponentiation of the algebra.

In particular, the group $SU(N)$ has dimension $N^2-1$, since it has $N^2-1$ generators, and among others it has the so-called defining and adjoint representations which have dimensions $N$ and $N^2-1$, respectively. This is because the group $SU(N)$ can be defined as the set of $N$x$N$ unitary matrices with unit determinant and the adjoint representation of the algebra uses the algebra itself as the linear space for the representation.

  • $\begingroup$ What do you mean by "the adjoint representation uses the algebra itself..."? $\endgroup$
    – PeaBrane
    Mar 17, 2018 at 0:46
  • $\begingroup$ @PeaBrane Please have another look at the answer. I tried to fill some gaps I had left. $\endgroup$
    – Diracology
    Mar 17, 2018 at 0:55
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    $\begingroup$ @PeaBrane An "adjoint representation" is a representation where the matrix elements are given by the structure constants that describe how the Lie algebra elements commute. I.e. if the Lie algebra elements commute as $[e^a,e^b]=i {f^{ab}}_c\ e^c$ (using the Einstein summation convention), then the adjoint representation consists of the matrices ${(T^a)^b}_c=-i {f^{ab}}_c$. (Use of $i$ and $-i$ differs by author.) $\endgroup$
    – Red Act
    Mar 17, 2018 at 2:29

Diracology's answer is correct in that this is the standard usage of the term "dimension of a representation", but I am guessing that you may be confused simply because there is another plausible meaning the phrase might have, but does not, in the literature.

Recall that a representation $\rho:\mathfrak{G}\to\mathrm{End}(V)$ is a homomorphism between a Lie group $\mathfrak{G}$ and the group $\mathrm{End}(V)$ of linear, homogeneous endomorphisms on a vector space $V$. Example: $\mathfrak{G}$ is often the Lorentz or Poincaré group and $V$ the separable, infinite dimensional Hilbert space of the states of a quantum system and we seek $\rho$ that tells us what unitary transformation our quantum state space undergoes when Poincaré transformations are imparted to the rest frame of the quantum system.

The dimension of this representation is, as discussed in Diracology's answer, $\aleph_0$. Or, in slightly different examples, it could be some finite number: the same as the dimension of the quantum state space in question: the dimension of the vector space $V$.

However, we always demand that $\rho$ should be smooth, so that the group of endomorphisms $\mathfrak{H} = \mathrm{Im}(\rho)\subseteq\mathrm{End}(V)\subseteq GL(V)$ at the pointy end of the homomorphism $\rho$ is also a Lie group. Accordingly, because $\rho$ is a homomorphism, $\mathfrak{H}$'s dimension as a Lie group must be equal to or smaller than that of the original Lie group. This is the second plausible meaning of dimension here, but it is not standard usage to call this number the dimension of the representation. This latter dimension depends on the kernel of the homomorphism. If the kernel is discrete, then $\mathfrak{G}$ is a cover of $\mathfrak{H}$ and the two groups have the same dimension. If the kernel is itself a Lie group, then the $\mathfrak{H}$'s dimension is less than that of $\mathfrak{G}$ such that $\dim(\mathfrak{G})=\dim(\mathfrak{H}) + \dim(\ker(\rho))$.


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