What does “all irreducible representations” mean?

An irreducible representation is a basic building block only after the group, coefficient field, and representation category have been fixed. For a finite group over $\mathbb C$, all irreducibles form a finite list. For $SU(2)$, the finite-dimensional continuous complex irreducibles form an infinite family indexed by spin. For $SL(2,\mathbb R)$, asking for unitary Hilbert-space representations changes the problem again. A table of small dimensions cannot stand for all three answers.

This lesson assumes linear algebra and the definition of a group. It explains how to formulate the question and how to check representative answers; the representation-category reference provides a compact lookup for subsequent calculations.

Specify what is being represented

A linear representation is a homomorphism $\rho:G\to GL(V)$ over a stated field. It is irreducible if $V\ne0$ and the only invariant linear subspaces are $0$ and $V$. Two representations are equivalent when an invertible linear map intertwines their actions. Changing a matrix basis does not create a new equivalence class. A representation can be irreducible without being faithful: irreducibility concerns invariant subspaces, while faithfulness means that the kernel of $\rho$ is trivial.

For a Lie group, require continuity. For a unitary representation on a complex Hilbert space, require strong continuity and use closed invariant subspaces in the definition of irreducibility. Unitarity means that the action preserves a positive-definite Hermitian inner product; preserving an indefinite form is a different condition.

Before reading an irrep label, record:

Choice	Example of information that changes the answer
Domain	$SU(2)$, $SO(3)$, or their common Lie algebra
Field and characteristic	$\mathbb C$ in characteristic zero, $\mathbb R$, or characteristic $p$
Dimension and topology	Finite-dimensional continuous modules or Hilbert-space representations
Linearity	Ordinary linear representations or projective actions
Equivalence and output	An isomorphism class, its character, or matrices in a specified basis

Finite-dimensional definitions and the role of subrepresentations are developed in Etingof et al., §2.3 and Chapter 4.

A finite list: ordinary complex representations of S₃

For a finite group $G$ over $\mathbb C$, every finite-dimensional representation is a direct sum of irreducibles. One way to see why invariant complements exist is to average an arbitrary positive Hermitian inner product:

$$\langle v,w\rangle_G =\frac1{|G|}\sum_{g\in G}\langle\rho(g)v,\rho(g)w\rangle.$$

The average is positive and $G$-invariant, so the orthogonal complement of an invariant subspace is invariant. The resulting character theory has one irreducible character per conjugacy class, with

$$\langle\chi,\psi\rangle =\frac1{|G|}\sum_C |C|\chi(C)\overline{\psi(C)}, \qquad \sum_{R\in\operatorname{Irr}(G)}(\dim_{\mathbb C}R)^2=|G|.$$

Here $C$ runs over conjugacy classes. The irreducible characters are orthonormal; taking an inner product with one extracts its multiplicity. See Etingof et al., Theorems 4.1.1, 4.2.1, and §4.5.

We can construct the table of the permutation group $S_3$ without guessing matrices. Permuting the coordinates of $\mathbb C^3$ fixes the line spanned by $(1,1,1)$. Its invariant complement

$$W=\{(x_1,x_2,x_3):x_1+x_2+x_3=0\}$$

has dimension two. The permutation character counts fixed coordinates, so subtracting the fixed line gives $\chi_W=(2,0,-1)$ on the identity, transpositions, and three-cycles. Together with the trivial and sign actions, this yields the following original calculation:

Class representative	$e$	$(12)$	$(123)$
Class size	$1$	$3$	$2$
Trivial $\mathbf1$	$1$	$1$	$1$
Sign $\mathrm{sgn}$	$1$	$-1$	$1$
Standard $W$	$2$	$0$	$-1$

For example, $\langle\chi_W,\chi_W\rangle=(4+2)/6=1$, so $W$ is irreducible. Its inner products with the other rows vanish. These are three distinct irreducibles and $1^2+1^2+2^2=6$, exhausting the regular representation. The dimension identity certifies completeness after the representations and their irreducibility have been established; a list of integers satisfying it would not do so. The independently derived table agrees with Etingof et al., §4.8, printed p. 74.

Tensor products multiply characters pointwise. Thus $\chi_{W\otimes W}=(4,0,1)$, whose three multiplicities are

$$\frac{4+2}{6}=1,\qquad \frac{4+2}{6}=1,\qquad \frac{8-2}{6}=1.$$

Consequently

$$W\otimes W\cong\mathbf1\oplus\mathrm{sgn}\oplus W.$$

The dimensions check: $2\cdot2=1+1+2$. In the finite-character workbench, select $S_3$, tensor product, and the standard irrep 2 for both factors. This computes a decomposition, not a normalized basis of invariant tensors.

Changing the field changes the question

A complex irrep of real type still has coefficient field $\mathbb C$. “Real type” says it can be obtained by complexifying a real representation. Quaternionic type is different: an equivariant antilinear structure can be normalized to square to $-1$, rather than $+1$. For finite groups, the Frobenius–Schur indicator separates these possibilities from complex type. The reality lesson develops the test; Etingof et al., Definition 5.1.1 and Problem 5.1.2 give the underlying distinctions.

An elementary field-change example is the real plane rotated through $2\pi/3$. It is irreducible over $\mathbb R$: a real invariant line would supply a real eigenvalue, but its eigenvalues are the nonreal roots $\omega,\omega^2$ of $x^2+x+1$. Over $\mathbb C$ it splits into two one-dimensional eigenspaces. These are the two nontrivial ordinary characters of $C_3$. “Two-dimensional irrep” therefore needs its field.

In characteristic $p$ dividing $|G|$, the averaging factor $1/|G|$ does not exist. The failure is substantive. Let $G=C_p$ and work over a field $k$ of characteristic $p$. A generator can act on $k^2$ by

$$U=\begin{pmatrix}1&1\\0&1\end{pmatrix}=I+N, \qquad N^2=0,\qquad U^p=I.$$

The line $k(1,0)$ is invariant and has no invariant complement: every eigenvector of $U$ lies on that line. This module is reducible but indecomposable. Its two composition factors are trivial, although it is not their direct sum. In fact $C_p$ has only the trivial simple module in this characteristic, since $(U-I)^p=0$ in every representation and a nonzero kernel of $U-I$ is invariant. This is the example in Etingof et al., Example 4.1.3. Ordinary complex character tables do not describe these extensions. GAP likewise distinguishes ordinary tables from modular Brauer tables in its manual, §71.4-1.

An infinite parameterization: SU(2)

Continuous finite-dimensional complex representations of compact Lie groups can also be made unitary, now by Haar averaging. For a simply connected compact semisimple group, the irreducibles are parameterized by dominant integral highest weights. After fixing simple roots, these are nonnegative integer Dynkin labels $(a_1,\ldots,a_r)$; the labels are not bounded by the group’s rank. See Conrad–Landesman, Proposition 8.13 and Theorem 28.2.

For $SU(2)$ there is one label $n\ge0$. A concrete realization is

$$V_n=\operatorname{Sym}^n\mathbb C^2, \qquad \dim_{\mathbb C}V_n=n+1, \qquad j=\frac n2.$$

To read the weights, choose the torus $t(\theta)=\operatorname{diag}(e^{i\theta},e^{-i\theta})$ with $\theta\sim\theta+2\pi$. The symmetric monomials $e_1^{n-k}e_2^k$, for $0\le k\le n$, are eigenvectors with eigenvalues $e^{i(n-2k)\theta}$. There are $n+1$ of them and the torus weights are $n,n-2,\ldots,-n$. The usual angular-momentum eigenvalues are half these integers because a physical rotation angle is $2\theta$ under the spin cover. The identification of these modules with all finite-dimensional complex irreducibles is Conrad–Landesman, Lemma 28.6, printed p. 125.

For $n=3$, the weights are $3,1,-1,-3$ and the dimension is $4$. Enter type $A$, rank $1$, labels 3 in the highest-weight workbench to reproduce the dimension. This is one member of an infinite family; listing the doublet, triplet, and quartet would be a selection, not a classification.

The global group matters even here. Substituting $\theta=\pi$ shows

$$\rho_n(-I)=(-1)^nI.$$

The quotient $SO(3)=SU(2)/\{\pm I\}$ therefore admits precisely the even $n$ modules from this list as ordinary representations. The odd $n$ modules instead give projective actions of $SO(3)$: the sign ambiguity disappears on rays. An ordinary representation of the cover and a projective representation of the quotient are different categories. The covering-group lesson explains the general kernel test; roots and weights explains the labels for higher rank.

A different question: SL(2,ℝ) and unitarity

The real Lie algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb R)$ have the same complexification $\mathfrak{sl}(2,\mathbb C)$. Consequently, their finite-dimensional complex irreducible modules have the same highest-weight labels. For the connected matrix group $SL(2,\mathbb R)$, the modules $V_n=\operatorname{Sym}^n\mathbb C^2$ do integrate: apply its defining real matrices to $\mathbb C^2$ and take symmetric powers. The $\mathfrak{sl}_2$ classification is given in Conrad–Landesman, Appendix J, Theorem J.3.1 and Proposition J.4.1.

These actions have different unitarity properties. The element

$$a(t)=\begin{pmatrix}e^t&0\\0&e^{-t}\end{pmatrix} \in SL(2,\mathbb R)$$

has eigenvalue $e^{nt}$ on $e_1^n$. For $n>0$ and real $t\ne0$, its modulus is not one. A positive-definite unitary operator cannot have that eigenvalue, and changing basis does not change eigenvalues. Thus none of these nontrivial finite-dimensional irreducibles is unitarizable.

The unitary Hilbert-space problem has infinite-dimensional answers. Already the spherical sector contains principal-series representations with an imaginary continuous parameter and a complementary interval of real parameters. In Vogan’s normalization, examples have $\nu=i$ and $\nu=1/2$, respectively; they are not the symmetric-power modules above. See Vogan, Unitary representations of reductive Lie groups, slide 4, “Example: $SL(2,\mathbb R)$ on the upper half plane”. This is an illustration, not a list of the full unitary dual, which also has nonspherical representations. Nor may a classification for this matrix group be transferred unchanged to its universal cover. The website’s compact highest-weight calculator does not classify these noncompact unitary representations.

Read an atlas representation as a precise claim

Open the standard $S_3$ representation record. Its domain is $S_3$, its category is ordinary complex, its coefficient field is $\mathbb C$ in characteristic zero, and its dimension is $2$. Its real-type label adds information about that complex irrep. The named class order and character vector identify it; the record does not select a matrix basis.

Compare the $SU(2)$ doublet record. Its label is a highest weight for the stated simply connected compact group, and its quaternionic type does not turn its coefficient field into $\mathbb H$ or its complex dimension into one. Both records are useful selected objects. Completeness comes from the appropriate character table or highest-weight theorem, not from how many record pages are displayed. The coverage statement separates curated records from implemented calculations and unsupported categories.

Continue through the finite-symmetry path or the Lie-group and topology path. For a quick check, determine whether $V_4$ descends to $SO(3)$ and whether it is unitary as an $SL(2,\mathbb R)$ module. The answers follow directly from the two actions above: its central sign is $+1$, but $a(t)$ has eigenvalue $e^{4t}$.