Representations

A representation is a way for an abstract group to act by linear transformations on a vector space. It turns symmetry composition into operator composition.

This is the step that makes group theory usable in quantum mechanics. The abstract group says which transformations can be composed. A group action says what is transformed. A representation says that the transformation is realized by linear maps on a vector space, such as a Hilbert space, a spin space, a space of wavefunctions, or a vector space of observables.

Helpful background. Groups and Linear Maps supply the two structures a representation connects.

Representing a group by linear maps

Let $G$ be a group and let $V$ be a vector space over a field $\mathbb F$, usually $\mathbb R$ or $\mathbb C$. A representation of $G$ on $V$ is a group homomorphism

$$\rho:G\to \operatorname{GL}(V),$$

where $\operatorname{GL}(V)$ is the group of invertible linear maps $V\to V$.

Equivalently, each $g\in G$ is assigned an invertible linear map $\rho(g)$ such that

$$\rho(gh) = \rho(g)\rho(h),$$

and

$$\rho(e)=I_V.$$

The inverse condition then follows:

$$\rho(g^{-1}) = \rho(g)^{-1}.$$

The vector space $V$ is called the representation space. Its dimension is the dimension of the representation. If $\dim V=n$ and a basis is chosen, each $\rho(g)$ becomes an invertible $n\times n$ matrix. The representation itself is the basis-independent assignment of linear maps; the matrices are coordinates for that assignment.

Linear Actions

A representation is the same structure as a linear left action of $G$ on $V$:

$$g\cdot v = \rho(g)v.$$

The group-action law

$$(gh)\cdot v = g\cdot(h\cdot v)$$

becomes

$$\rho(gh)v = \rho(g)\rho(h)v.$$

Thus representation theory is the linear part of group-action theory. The distinction matters because many physical actions are not linear actions on the first space one writes down. Rotations act on points in ordinary space, permutations act on labels, and symmetry groups may act on parameter spaces of Hamiltonians. Once the action is lifted to states, wavefunctions, observables, or fields, representation theory often enters.

Basic Examples

The trivial representation sends every group element to the identity map:

$$\rho(g)=I_V$$

for all $g\in G$. It is mathematically simple but physically meaningful: it describes degrees of freedom left unchanged by the symmetry.

The sign representation of $\mathbb Z_2=\{e,a\}$ on a one-dimensional complex vector space is

$$\rho(e)=1, \qquad \rho(a)=-1.$$

Since $a^2=e$, this obeys

$$\rho(a)^2 = (-1)^2 = 1 = \rho(e).$$

The permutation representation of the symmetric group $S_n$ on the vector space with basis $\{e_1,\ldots,e_n\}$ is

$$\rho(\sigma)e_i = e_{\sigma(i)}.$$

This is a representation because composing permutations composes their actions on the basis vectors.

The defining representation of $SO(3)$ on $\mathbb R^3$ sends a rotation $R$ to the linear map $\mathbf v\mapsto R\mathbf v$. The defining representation of $SU(2)$ on $\mathbb C^2$ sends a special unitary matrix $U$ to its natural action on two-component spinors. These examples are called defining because the group has already been presented as a matrix group acting on the vector space used in its definition.

Representation Spaces Are Part of the Data

It is not enough to name the group. One must say which vector space carries the representation and how the group acts on it.

For example, the same group $SO(3)$ can act on:

• ordinary vectors in $\mathbb R^3$;
• scalar wavefunctions by rotating their arguments;
• integer-spin spaces of dimension $2j+1$; half-integer spin uses $SU(2)$ or a projective $SO(3)$ action;
• tensor operators with several spatial indices;
• spaces of spherical harmonics of fixed angular momentum.

These are different representations. Some share the same group, but they have different representation spaces and different physical meanings.

This distinction prevents a common mistake: a group is not the same thing as one favorite collection of matrices. Matrices become attached to a group only after a representation and a basis have been chosen.

Equivalent Representations

Two representations can look different in different bases while describing the same group action.

Let

$$\rho_1:G\to \operatorname{GL}(V_1), \qquad \rho_2:G\to \operatorname{GL}(V_2)$$

be representations. A linear map $S:V_1\to V_2$ is an intertwiner if

$$S\rho_1(g) = \rho_2(g)S$$

for every $g\in G$. If $S$ is invertible, the two representations are equivalent. In that case,

$$\rho_2(g) = S\rho_1(g)S^{-1}.$$

This is the representation-theoretic version of a change of basis. Equivalent representations have the same invariant content even though their matrices may look different.

Not every similarity of individual matrices gives an equivalence of representations. The same single map $S$ must work for every group element at once.

Invariant Subspaces

A subspace $W\subset V$ is invariant under a representation $\rho$ if

$$\rho(g)W\subset W$$

for every $g\in G$.

Equivalently, whenever $w\in W$, the transformed vector $\rho(g)w$ also lies in $W$. Invariance means the symmetry never mixes vectors in $W$ out of that subspace.

If a basis is chosen so that the first basis vectors span $W$, then every representation matrix has block upper-triangular form:

$$\rho(g) = \begin{pmatrix} \rho_W(g) & B(g)\\ 0 & D(g) \end{pmatrix}.$$

The lower-left zero block is the invariant-subspace condition. It says that a vector starting in $W$ has no component outside $W$ after transformation.

Invariant subspaces are central in quantum mechanics. If a symmetry preserves a subspace of states, one can study that subspace separately. If a Hamiltonian commutes with the symmetry action, energy eigenspaces and degeneracy patterns are often organized by invariant representation spaces.

Reducible and Irreducible Representations

A representation on $V$ is reducible if it has a nonzero proper invariant subspace:

$$\{0\}\ne W\ne V.$$

It is irreducible if it has no such subspace.

Irreducible representations, often called irreps, are the elementary symmetry types. They play a role in symmetry analysis similar to prime factors in arithmetic: complicated representations are often built from irreducible pieces.

In finite dimension over $\mathbb C$, every representation of a finite group, and every continuous representation of a compact Lie group, decomposes into a direct sum of irreducibles. One can choose an invariant positive-definite inner product, so the orthogonal complement of each invariant subspace is invariant too; Unitary Representations gives the proof. A representation admitting such a decomposition is called completely reducible. Reducibility alone guarantees only an invariant subspace, not an invariant complement.

Worked example: an invariant line without a complement

Consider the additive group $(\mathbb R,+)$ acting on $\mathbb C^2$ by the dimensionless shear parameter $t$:

$$\rho(t)=\begin{pmatrix}1&t\\0&1\end{pmatrix}, \qquad \rho(t)\rho(s)=\rho(t+s).$$

The line $W=\operatorname{span}\{e_1\}$ is invariant because $\rho(t)e_1=e_1$. Thus the representation is reducible. Any complementary line has a generator $v=a e_1+b e_2$ with $b\ne0$. If that line were invariant, $\rho(t)v$ would be proportional to $v$ for every $t$:

$$\rho(t)v=(a+tb)e_1+b e_2=\lambda(t)(a e_1+b e_2).$$

The second component forces $\lambda(t)=1$; the first then requires $tb=0$ for every $t$, contradicting $b\ne0$. No invariant complementary line exists. A basis change cannot turn the whole family into two one-dimensional blocks. This representation is indecomposable, even though it is reducible.

There is no conflict with quantum unitary symmetries: the shear representation cannot preserve any positive-definite inner product. Otherwise the orthogonal-complement argument would supply exactly the invariant line that is missing. The distinction already occurs in two dimensions; it is not solely an infinite-dimensional domain issue. See Etingof et al., §2.3 and Theorem 4.6.3, for the general representation-theoretic distinction.

Direct Sums

If $\rho_1$ is a representation on $V_1$ and $\rho_2$ is a representation on $V_2$, their direct sum is the representation on $V_1\oplus V_2$ defined by

$$(\rho_1\oplus\rho_2)(g)(v_1,v_2) = (\rho_1(g)v_1,\rho_2(g)v_2).$$

In block-matrix form,

$$(\rho_1\oplus\rho_2)(g) = \begin{pmatrix} \rho_1(g) & 0\\ 0 & \rho_2(g) \end{pmatrix}.$$

When a representation decomposes as a direct sum of irreducible representations, symmetry calculations simplify because each block can be analyzed separately. This is the algebraic reason angular-momentum multiplets, selection rules, and degeneracy labels are so powerful.

Lie Group and Lie Algebra Representations

For a Lie group $G$, a representation is usually required to be smooth or continuous in the group parameters. If $R(g)$ is a smooth representation and $X$ is an element of the Lie algebra $\mathfrak g$, differentiating near the identity gives a Lie algebra representation.

In a one-parameter subgroup,

$$g(t)=\exp(tX),$$

the differentiated operator is

$$dR(X) = \left.\frac{d}{dt}R(\exp(tX))\right|_{t=0}.$$

The Lie algebra representation preserves brackets:

$$dR([X,Y]) = [dR(X),dR(Y)].$$

The converse is local rather than automatic. A Lie algebra representation may fail to integrate to a representation of a chosen global group because of topology or domain questions. The distinction between $SO(3)$ and $SU(2)$ is the standard quantum-mechanical warning sign: closely related Lie algebras can support different global representation behavior, as explained in SU(2) versus SO(3).

Quantum-Mechanical Interpretation

In quantum mechanics, representations answer the question:

How does this symmetry act on the state space?

For a complex Hilbert space $\mathcal H$, an ordinary linear representation assigns operators to symmetry elements:

$$g\mapsto U(g).$$

For a unitary representation, the operators preserve inner products:

$$U(g)^\dagger U(g) = I.$$

A Hamiltonian $H$ is invariant under the representation when

$$U(g)HU(g)^{-1} = H$$

for every $g\in G$. Equivalently, $HU(g)=U(g)H$. In a finite-dimensional unitary representation, this prevents $H$ from mixing inequivalent symmetry types, but it can mix repeated copies of the same irreducible representation. Thus an arbitrary choice of irreducible summands need not also diagonalize $H$. Degeneracy and Multiplets explains the remaining matrix acting on the copy labels and the resulting energy degeneracies.

There are two important quantum caveats. First, physical pure states are rays, so a symmetry action on rays may lift only to a projective representation on vectors. Second, Wigner’s theorem allows antiunitary symmetries as well as unitary ones. Those issues are handled in Quantum Symmetries and Projective Representations. This page gives the ordinary linear representation language used by both.

Common Mistakes

• Treating a group as identical to one matrix realization of it.
• Forgetting that the representation space is part of the representation.
• Calling two representations equivalent because one group element has similar matrices, without checking all group elements.
• Assuming every reducible representation is already written as a direct sum.
• Ignoring the difference between an invariant subspace and a basis-dependent coordinate subspace.
• Using Lie algebra commutators while forgetting global group conditions.
• Treating projective quantum representations as ordinary representations without tracking phases.

Exercises

1. Prove that a representation sends inverses to inverse linear maps.

Solution

Using the homomorphism property,

$$\rho(g)\rho(g^{-1}) = \rho(gg^{-1}) = \rho(e) = I_V.$$

Similarly,

$$\rho(g^{-1})\rho(g) = I_V.$$

Therefore $\rho(g^{-1})=\rho(g)^{-1}$.

2. Check that the sign representation of $\mathbb Z_2$ is a representation.

Solution

Let $\mathbb Z_2=\{e,a\}$ with $a^2=e$. Define $\rho(e)=1$ and $\rho(a)=-1$. The identity condition is immediate. The only nontrivial multiplication check is

$$\rho(a^2) = \rho(e) = 1,$$

while

$$\rho(a)\rho(a) = (-1)(-1) = 1.$$

Thus $\rho(gh)=\rho(g)\rho(h)$ for all pairs of elements.

3. For the permutation representation of $S_3$ on the basis $\{e_1,e_2,e_3\}$, write the matrix for the transposition exchanging $1$ and $2$.

Solution

The transposition $\sigma=(12)$ sends $e_1\mapsto e_2$, $e_2\mapsto e_1$, and $e_3\mapsto e_3$. With columns recording the images of basis vectors, the matrix is

$$\rho(\sigma) = \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$

4. Suppose $W\subset V$ is invariant under $\rho$. Explain why an adapted basis makes every $\rho(g)$ block upper triangular.

Solution

Choose a basis of $W$ and extend it to a basis of $V$. A vector in $W$ has coordinates with zero entries in the complementary directions. Since $W$ is invariant, applying $\rho(g)$ to such a vector again gives a vector in $W$. Therefore the matrix entries that would send a $W$-basis vector into the complementary coordinates must vanish. This gives the lower-left zero block:

$$\rho(g) = \begin{pmatrix} \rho_W(g) & B(g)\\ 0 & D(g) \end{pmatrix}.$$

5. Why does equal dimension not imply equivalence of representations?

Solution

Equivalence requires one invertible map $S$ satisfying

$$S\rho_1(g) = \rho_2(g)S$$

for every $g\in G$. Equal dimension only says that such an invertible linear map could exist as a map of vector spaces. It does not say that the map intertwines the group actions. For example, the trivial one-dimensional representation of $\mathbb Z_2$ and the sign representation are both one-dimensional, but they are not equivalent because the nonidentity element acts as $1$ in one representation and as $-1$ in the other.

6. For the shear representation above, test unitarity directly with a general Hermitian inner-product matrix

$$G=\begin{pmatrix}a&b\\b^*&c\end{pmatrix}, \qquad \langle v\vert w\rangle_G=v^\dagger Gw.$$

Show that $\rho(t)^\dagger G\rho(t)=G$ for every real $t$ is incompatible with positive definiteness. Why would an invariant indefinite Hermitian form not contradict the conclusion?

Solution

Here $a,c$ are real. Direct multiplication gives

$$\rho(t)^\dagger G\rho(t) =\begin{pmatrix} a&at+b\\ at+b^*&at^2+(b+b^*)t+c \end{pmatrix}.$$

Equality for all $t$ requires $a=0$ and $\operatorname{Re}b=0$. But positive definiteness requires $\langle e_1\vert e_1\rangle_G=a>0$, so no positive inner product is invariant. For example, $b=i$ and $c=0$ do give an invariant nondegenerate Hermitian form, but its eigenvalues are $+1$ and $-1$. It is indefinite, so it cannot supply Hilbert-space probabilities or the positive orthogonal splitting used in the unitary theorem. In particular, $e_1$ has zero squared length in this form and lies in its own orthogonal complement.

References

• P. Etingof, O. Golberg, S. Hensel, T. Liu, A. Schwendner, D. Vaintrob, and E. Yudovina, Introduction to Representation Theory, American Mathematical Society, 2011, §2.3 and §4.6.
• W. Fulton and J. Harris, Representation Theory: A First Course, Springer, 1991.
• H. Georgi, Lie Algebras in Particle Physics, 2nd ed., Westview Press, 1999.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015.