Representations, Intertwiners, Invariants, and Tensor Decomposition

A representation is a linear action: it assigns each $g\in G$ an invertible linear map $\rho(g)$ on a declared vector space $V$, while preserving the group product. Invariant subspaces reveal smaller representations; intertwiners are linear maps compatible with two actions; and invariant tensors are fixed vectors in tensor constructions. In fact,

$$\operatorname{Hom}_G(V,W) \cong (W\otimes V^*)^G,$$

so intertwiners and invariant tensors are two descriptions of the same symmetry-compatible data.

A decomposition as a direct sum of irreducible pieces is justified only when a splitting theorem applies. Finite-dimensional unitary representations, finite-group representations over $\mathbb R$ or $\mathbb C$, and continuous finite-dimensional representations of compact Lie groups are completely reducible. A general representation need not be: a reducible representation can have no invariant complement. This hypothesis boundary is as important as the decomposition itself.

Required background. Groups, Actions, Quotients, and Covers supplies actions, kernels, and descent; Vector Spaces, Duals, Linear Maps, and Bases supplies typed linear maps, algebraic duals, and basis changes.

Linear actions and invariant structure

Unless stated otherwise, $V$ is finite-dimensional over $\mathbb F=\mathbb R$ or $\mathbb C$, and all duals and tensor products are algebraic. A group may be finite, discrete, or topological; continuity is an extra hypothesis when compact-Lie-group averaging is used. No inner product is assumed until one is specified.

This page does not classify all irreducible representations, develop roots and weights, classify Lorentz or Poincaré representations, or treat infinite-dimensional Hilbert representations. It identifies algebraically permitted invariant couplings but does not decide whether a transformation is an exact QFT symmetry or whether a permitted interaction is local, Hermitian, dynamically present, nonanomalous, or stable under renormalization.

A representation is a homomorphism into linear maps

A representation of a group $G$ on $V$ is a homomorphism

$$\rho:G\longrightarrow GL_{\mathbb F}(V), \qquad \rho(g_1g_2)=\rho(g_1)\rho(g_2).$$

It defines the linear left action $g\cdot v=\rho(g)v$. Conversely, every linear left action defines such a homomorphism. The group, vector space, representation map, and vector being transformed are four different objects.

A basis converts $\rho(g)$ into a matrix, but the matrix is not the representation by itself. If $e'_a=e_bM^b{}_a$, then

$$[\rho(g)]' = M^{-1}[\rho(g)]M.$$

The new matrices describe the same linear maps in a new basis. More generally, representations $(V,\rho)$ and $(W,\sigma)$ are equivalent when an invertible linear map $S:V\to W$ satisfies

$$S\rho(g)=\sigma(g)S \qquad \text{for every }g\in G.$$

Definitions of representations, equivalence, subrepresentations, duals, and tensor products are collected in Etingof 2020, §§4.3 and 11.1, PDF and Kirillov 2008, §§4.1–4.2, PDF.

The action kernel is

$$\ker\rho = \{g\in G:\rho(g)=\mathbf1_V\}.$$

It is normal, and the representation is faithful exactly when this kernel is trivial. The groups-and-actions page gives a faithful representation of $G/\ker\rho$ with exactly the same realized linear transformations. This does not make $G$ and $G/\ker\rho$ interchangeable for other representations.

Invariant subspaces, quotients, and splitting

A subspace $U\subseteq V$ is invariant when

$$\rho(g)U\subseteq U \qquad \text{for every }g\in G.$$

The inclusion is automatically an equality because the same condition applied to $g^{-1}$ gives the reverse inclusion. Restricting $\rho(g)$ to $U$ defines a subrepresentation. There is also a quotient representation

$$\rho_{V/U}(g)(v+U)=\rho(g)v+U.$$

It is well defined precisely because $U$ is invariant.

Several nearby terms answer different questions:

Term	Condition
Fixed vector	$\rho(g)v=v$ for every $g$
Invariant subspace	$\rho(g)U=U$ for every $g$
Irreducible	$V\neq0$ and its only invariant subspaces are $0$ and $V$
Reducible	A nonzero proper invariant subspace exists
Decomposable	$V=U\oplus W$ for two nonzero invariant subspaces
Completely reducible	$V$ is a direct sum of irreducible subrepresentations

The fixed vectors form

$$V^G=\{v\in V:\rho(g)v=v\ \text{for all }g\}.$$

An invariant subspace need not be fixed pointwise, and reducibility does not guarantee decomposability. Equivalently, the exact sequence

$$0\longrightarrow U\longrightarrow V\longrightarrow V/U \longrightarrow0$$

need not admit a $G$-equivariant splitting.

Intertwiners expose invariant structure

For representations $(V,\rho)$ and $(W,\sigma)$ of the same group, an intertwiner is a linear map $A:V\to W$ satisfying

$$A\rho(g)=\sigma(g)A \qquad \text{for every }g\in G.$$

The vector space of such maps is denoted $\operatorname{Hom}_G(V,W)$. A general intertwiner need not be an endomorphism, so writing only a commutator $[A,\rho(g)]$ would lose its domain and codomain.

Two immediate calculations make intertwiners useful. If $v\in\ker A$, then

$$A\rho(g)v=\sigma(g)Av=0,$$

so $\ker A$ is invariant. If $w=Av\in\operatorname{im}A$, then

$$\sigma(g)w=A\rho(g)v\in\operatorname{im}A,$$

so the image is invariant as well.

These observations prove the first form of Schur’s lemma: a nonzero intertwiner between two irreducible representations is an isomorphism. For a finite-dimensional complex irreducible representation, every equivariant endomorphism is scalar,

$$\operatorname{End}_G(V)=\mathbb C\,\mathbf1_V.$$

Indeed, an endomorphism $A$ has a complex eigenvalue $\lambda$. $A-\lambda\mathbf1$ is a noninvertible intertwiner, so the first statement forces it to vanish. The scalar conclusion uses finite dimension and an algebraically closed field. Over $\mathbb R$ it can fail: the standard two-dimensional real representation of $SO(2)$ is irreducible, while its commuting endomorphisms include the complex structure

$$J= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}.$$

The exact field qualifications and proof appear in Etingof 2020, §11.2, Lemma 11.9, PDF and Kirillov 2008, §4.4, Lemma 4.23, PDF.

Dual, tensor, and Hom representations

The dual representation on the algebraic dual $V^*$ is

$$\begin{aligned} \rho_{V^*}(g)\lambda &=\lambda\circ\rho(g^{-1}), \\ \bigl(\rho_{V^*}(g)\lambda\bigr)(v) &=\lambda\bigl(\rho(g^{-1})v\bigr). \end{aligned}$$

The inverse is forced by the homomorphism law. In dual bases, $\rho_{V^*}(g)$ is represented by $\rho(g^{-1})^{\mathsf T}$. For a complex vector space this is the complex-linear dual action; a conjugate representation, Hermitian adjoint, or identification $V\simeq V^*$ requires additional structure.

If $V$ and $W$ are representations of the same group, their tensor product has the diagonal action

$$\rho_{V\otimes W}(g)(v\otimes w) = \rho_V(g)v\otimes\rho_W(g)w.$$

The same $g$ acts on both factors. This differs from the external tensor product representation of $G\times H$, in which two independent group elements act.

On $\operatorname{Hom}(V,W)$, define

$$g\cdot A = \rho_W(g)A\rho_V(g^{-1}).$$

Its fixed vectors are exactly the intertwiners:

$$\operatorname{Hom}(V,W)^G = \operatorname{Hom}_G(V,W).$$

Using the finite-dimensional identification $\operatorname{Hom}(V,W)\cong W\otimes V^*$ gives the opening relation

$$\operatorname{Hom}_G(V,W) \cong (W\otimes V^*)^G.$$

The trivial one-dimensional representation $\mathbf1$ supplies two useful special cases:

$$V^G\cong\operatorname{Hom}_G(\mathbf1,V), \qquad (V^*)^G\cong\operatorname{Hom}_G(V,\mathbf1).$$

An invariant bilinear form is equivalently a fixed element of $V^*\otimes V^*$ or an intertwiner $V\to V^*$. More generally, a scalar $n$-linear coupling on $V_1,\ldots,V_n$ is symmetry-invariant exactly when its coefficient is in

$$\begin{aligned} &(V_1^*\otimes\cdots\otimes V_n^*)^G \\ &\qquad\cong \operatorname{Hom}_G \bigl( V_1\otimes\cdots\otimes V_n,\mathbf1 \bigr). \end{aligned}$$

This is the algebraic origin of singlet tests for couplings and selection rules.

Complete reducibility needs a theorem

The simplest counterexample comes from the noncompact additive group $G=(\mathbb R,+)$. On $V=\mathbb R^2$, let

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

Since $\rho(s+t)=\rho(s)\rho(t)$, this is a representation. The line $U=\mathbb Re_1$ is invariant. Every complementary line is generated by $e_2+ae_1$ for some $a$, but

$$\rho(t)(e_2+ae_1) = e_2+(a+t)e_1$$

does not remain in that line for all $t$. The representation is therefore reducible but indecomposable. Its two quotient factors are trivial, yet it is not their direct sum.

Complete reducibility follows under several distinct hypotheses:

Hypothesis	Mechanism and conclusion
A finite-dimensional orthogonal or unitary representation of any group	Invariant orthogonal complements split every invariant subspace
A finite group acting on a finite-dimensional real or complex space	Average a positive-definite form over the finite group
A compact Lie group acting continuously on a finite-dimensional real or complex space	Average a positive-definite form with normalized Haar measure
A finite-dimensional representation of a semisimple Lie algebra in characteristic zero	Weyl’s complete-reducibility theorem applies
An arbitrary representation of a general group	No complete-reducibility conclusion follows

The first row already supplies an invariant positive-definite form. For the finite-group and compact-Lie-group rows, start with any positive-definite inner product over $\mathbb R$, or Hermitian form over $\mathbb C$, and average it:

$$\langle v,w\rangle_G = \begin{cases} \displaystyle \frac1{|G|} \sum_{g\in G} \langle\rho(g)v,\rho(g)w\rangle_0, &G\ \text{finite}, \\[1.25em] \displaystyle \int_G \langle\rho(g)v,\rho(g)w\rangle_0\,\mathrm d\mu(g), &G\ \text{a compact Lie group}. \end{cases}$$

In the compact-Lie-group case $\rho$ is continuous and $\mu$ is normalized Haar measure. The averaged form remains positive definite and is $G$-invariant. In all three rows, the resulting invariant form supplies the common splitting argument. If $U$ is invariant, then for $w\in U^\perp$, $u\in U$, and $g\in G$,

$$\langle\rho(g)w,u\rangle_G = \langle w,\rho(g^{-1})u\rangle_G =0.$$

Thus $U^\perp$ is invariant and $V=U\oplus U^\perp$. Induction on dimension gives a direct sum of irreducibles. This proof and its precise finite, unitary, and compact hypotheses are given in Etingof 2020, §11.3, Propositions 11.13–11.14 and Corollary 11.15, PDF and Kirillov 2008, §4.5, Theorems 4.30–4.32 and §4.6, Theorem 4.40, PDF.

The compact-Lie-group hypothesis is sufficient here, not necessary; many noncompact-group representations are completely reducible. Conversely, compact-Lie-group averaging does not cover a discontinuous representation.

The same averaging operation extracts fixed vectors:

$$P_G = \begin{cases} \displaystyle\frac1{|G|}\sum_{g\in G}\rho(g), &G\ \text{finite}, \\[1em] \displaystyle\int_G\rho(g)\,\mathrm d\mu(g), &G\ \text{a compact Lie group}. \end{cases}$$

Left invariance of the average gives $\rho(h)P_G=P_G$. Hence $\operatorname{im}P_G=V^G$ and $P_G^2=P_G$. More generally, an equivariant idempotent $P$ yields the invariant splitting

$$V=\operatorname{im}P\oplus\ker P.$$

Tensor decomposition and multiplicity

Assume now that the complex representation $V$ being decomposed is finite-dimensional and completely reducible. Choose one representative $U_\lambda$ of each irreducible isomorphism class. Then

$$V \cong \bigoplus_\lambda M_\lambda\otimes U_\lambda, \qquad M_\lambda=\operatorname{Hom}_G(U_\lambda,V).$$

The group acts trivially on each multiplicity space $M_\lambda$. For a tensor product, additionally assume that the diagonal representation on $V\otimes W$ is finite-dimensional and completely reducible. Then

$$V\otimes W \cong \bigoplus_\lambda M^\lambda_{VW}\otimes U_\lambda, \qquad M^\lambda_{VW} = \operatorname{Hom}_G(U_\lambda,V\otimes W).$$

The Clebsch–Gordan multiplicity is

$$\begin{aligned} N^\lambda_{VW} &=\dim_{\mathbb C}M^\lambda_{VW} \\ &= \dim_{\mathbb C} \bigl( V\otimes W\otimes U_\lambda^* \bigr)^G. \end{aligned}$$

Multiplicities and isotypic components are intrinsic. Individual copies of an irreducible, Clebsch–Gordan maps, coefficient arrays, phases, and bases need not be canonical, especially when a multiplicity exceeds one.

One decomposition exists without any classification theorem. On $V\otimes V$, the flip

$$\tau(v\otimes w)=w\otimes v$$

commutes with the diagonal action. Over $\mathbb R$ or $\mathbb C$,

$$P_\pm=\frac12(\mathbf1\pm\tau)$$

are equivariant projectors and

$$V\otimes V = \operatorname{Sym}^2V \oplus \Lambda^2V.$$

The symmetric and antisymmetric spaces are invariant but need not themselves be irreducible.

The rotation cover inside a tensor product

The $SU(2)$ tensor-product conventions and Clebsch–Gordan decomposition used below can be compared with Hall 2015, Chapter 4 and Appendix C.

Let $V=\mathbb C^2$ carry the defining representation of $SU(2)$. The central element $-\mathbf1$ acts as minus the identity, so this representation does not descend through $SU(2)\to SO(3)$. On $V\otimes V$, however, it acts as $(-\mathbf1)\otimes(-\mathbf1)=+\mathbf1$.

The flip projectors give

$$\mathbb C^2\otimes\mathbb C^2 = \operatorname{Sym}^2\mathbb C^2 \oplus \Lambda^2\mathbb C^2 \cong \mathbf3\oplus\mathbf1.$$

The alternating line is spanned by

$$\epsilon = \frac1{\sqrt2} \left( e_1\otimes e_2-e_2\otimes e_1 \right).$$

It is fixed because the induced action on $\Lambda^2\mathbb C^2$ is multiplication by $\det U=1$. The symmetric three-dimensional component is the irreducible triplet. Both pieces descend to $SO(3)$ because the covering kernel acts trivially on the tensor square. The doublet does not.

This is the representation-level step in the chapter’s recurring $SU(2)$–$SO(3)$ comparison. The standard decomposition is supported by Etingof 2020, §11.4, Theorem 11.18, PDF and Kosmann-Schwarzbach 2022, “Representations of $SU(2)$ and $SO(3)$,” pp. 103–118.

Controlled QFT bridge: an O(N) scalar multiplet

Let $N\geq2$ real scalar fields form the defining representation $V=\mathbb R^N$ of $O(N)$:

$$(R\cdot\phi)^a(x)=R^a{}_b\phi^b(x), \qquad R^{\mathsf T}R=\mathbf1.$$

The Euclidean tensor $\delta_{ab}$ is invariant,

$$\delta_{ab}R^a{}_cR^b{}_d=\delta_{cd},$$

so the Lagrangian

$$\begin{aligned} \mathcal L ={}& \frac12\delta_{ab} \partial_\mu\phi^a\partial^\mu\phi^b -\frac{m^2}{2}\delta_{ab}\phi^a\phi^b \\ &-\frac{\lambda}{4!} \left(\delta_{ab}\phi^a\phi^b\right)^2 \end{aligned}$$

is $O(N)$-invariant. This is the non-Abelian scalar example in Tong 2006, §1.3.4, with the full group correctly identified as $O(N)$ rather than only $SO(N)$.

The invariant form also makes the two-index channels explicit. For $A^{ab}\in V\otimes V$, set

$$\begin{aligned} A_{\mathbf1}^{ab} &= \frac{\delta^{ab}}{N}\, \delta_{cd}A^{cd}, \\ A_{\wedge}^{ab} &= \frac12\left(A^{ab}-A^{ba}\right), \\ A_{S_0}^{ab} &= \frac12\left(A^{ab}+A^{ba}\right) -A_{\mathbf1}^{ab}. \end{aligned}$$

Trace and flip commute with the $O(N)$ action, so

$$V\otimes V = \operatorname{span}\{\delta\} \oplus \Lambda^2V \oplus \operatorname{Sym}^2_0V.$$

This is an invariant-channel decomposition. No claim that every displayed channel is irreducible for every low dimension or changed global group is needed.

The same fixed-tensor test supplies a bounded selection statement. For $N\geq2$, the defining representation has no nonzero $O(N)$-fixed vector, so no nonzero constant coefficient can produce an invariant term linear in $\phi$. The invariant $\delta_{ab}$ permits the quadratic contraction and its powers. By contrast, the Levi-Civita tensor changes by $\det R$ and is not an $O(N)$ invariant; it is invariant only after restricting to $SO(N)$.

These are algebraic permissions and exclusions for the declared group action. They do not prove that the interaction occurs, survives quantum effects, respects every other symmetry, or remains meaningful for a different global group. The exact physical continuation develops those questions.

Common pitfalls

Calling arbitrary matrices a representation. The matrices must be invertible and obey the group multiplication law on a declared vector space. A basis change conjugates every matrix coherently.

Confusing invariant with pointwise fixed. An invariant subspace is carried into itself; its individual vectors can move. Only $V^G$ is fixed pointwise.

Assuming reducible means decomposed. The unipotent $\mathbb R$ example has an invariant line but no invariant complement. Complete reducibility needs a theorem with hypotheses.

Forgetting the inverse in the dual action. The formula $\lambda\mapsto\lambda\circ\rho(g^{-1})$ is what makes the dual action a homomorphism. Transpose, conjugate, and Hermitian adjoint are different operations.

Using Schur’s scalar conclusion over the wrong field. A nonzero intertwiner between irreducibles is an isomorphism over any field. Scalar endomorphisms require the additional finite-dimensional algebraically-closed-field hypothesis.

Making Clebsch–Gordan arrays canonical. Multiplicities survive a basis change, while coefficient arrays depend on bases, phases, normalizations, and choices within multiplicity spaces.

Replacing the group by its Lie algebra or connected component. Global cover kernels and disconnected elements can impose additional invariance conditions. The $SU(2)$ doublet and the $O(N)$ Levi-Civita tensor expose the two failures.

Promoting an invariant tensor to a physical interaction. A singlet test is necessary for symmetry invariance, not sufficient for locality, statistics, Hermiticity, power counting, anomaly freedom, or dynamical generation.

Exercises

1. Decompose the permutation representation

Let $S_3$ act on $\mathbb R^3$ by $\rho(\sigma)e_i=e_{\sigma(i)}$. Verify the representation law and find two nonzero invariant subspaces whose direct sum is $\mathbb R^3$.

Solution

On a basis vector,

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

so the group law holds. The line

$$L=\operatorname{span}\{(1,1,1)\}$$

is fixed pointwise. The plane

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

is invariant because permutations preserve the coordinate sum. Every vector splits uniquely into its mean times $(1,1,1)$ plus a vector of coordinate sum zero, so $\mathbb R^3=L\oplus U$.

2. Prove the first form of Schur’s lemma

Let $A:V\to W$ be a nonzero intertwiner between irreducible representations. Prove that $A$ is an isomorphism without choosing bases.

Solution

The intertwining equation makes $\ker A$ an invariant subspace of $V$. Irreducibility and $A\neq0$ rule out $\ker A=V$, so $\ker A=0$ and $A$ is injective. Its image is a nonzero invariant subspace of $W$, hence equals $W$. Therefore $A$ is also surjective and is an isomorphism.

3. Diagnose a nonsplit representation

For

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

show that $\mathbb Re_1$ is invariant and that the quotient representation on $\mathbb R^2/\mathbb Re_1$ is trivial. Why do these two trivial factors not give a direct sum?

Solution

The matrix fixes $e_1$, so its span is an invariant trivial subrepresentation. Since $\rho(t)e_2-e_2=te_1$, the coset of $e_2$ is fixed in the quotient, which is also trivial. A complementary line would have a generator $e_2+ae_1$, but $\rho(t)$ sends it to $e_2+(a+t)e_1$, outside the same line for $t\neq0$. The exact sequence does not split equivariantly.

4. Test tensor descent through the rotation cover

Let $V=\mathbb C^2$ be the defining $SU(2)$ representation. Determine how $-\mathbf1$ acts on $V$ and $V\otimes V$, and identify the singlet in the tensor square.

Solution

The central element acts as $-\mathbf1_V$ on $V$, so the doublet does not descend to $SO(3)$. On the tensor square it acts as $(-\mathbf1)\otimes(-\mathbf1)=+\mathbf1$, so that representation does descend. The antisymmetric vector

$$\epsilon = \frac1{\sqrt2} \left(e_1\otimes e_2-e_2\otimes e_1\right)$$

spans $\Lambda^2V$. Since $U$ acts there by $\det U=1$, it is the invariant singlet; the complementary symmetric subspace is the triplet.

5. Type an O(N)-invariant source and coupling

Let $\phi\in V=\mathbb R^N$ transform in the defining representation. Explain why $\delta_{ab}\phi^a\phi^b$ is invariant. If a source term $J_a\phi^a$ is to remain invariant while $J$ also transforms, what representation contains $J$? What happens when a nonzero $J$ is instead held fixed?

Solution

Orthogonality gives

$$\delta_{ab}R^a{}_cR^b{}_d=\delta_{cd},$$

so the quadratic contraction is unchanged. The field is in $V$, hence the source must transform in the dual representation $V^*$; then evaluation $J(\phi)$ is invariant. If a chosen nonzero numerical $J$ is held fixed rather than transformed, it selects a direction in $V$ and preserves only its stabilizer. It is an explicit symmetry-breaking source, not an $O(N)$-invariant coefficient.

Where to continue

• Later chapter topics that require this page are Compact Lie Groups, Roots, Weights, and Weyl Structure, Lorentz Field Representations and Poincaré Particle Representations, and Spinors, Conjugations, Bilinears, Chirality, and Fierz Identities. The first two also require the Lie-group leaf; the spinor leaf also requires the Clifford leaf.
• For the first physical application to field multiplets, selection rules, and invariant couplings, continue to Multiplets, Invariants, and Selection Rules. That page also hard-requires the physical global-symmetry page, so this mathematical page alone does not satisfy its preparation.

References

• Pavel Etingof, Lie Groups and Lie Algebras I, PDF, MIT OpenCourseWare 18.745, Fall 2020, §§4.3, 11.1–11.4, and 18.4. These open notes develop representations, intertwiners, invariants, dual and tensor constructions, Schur’s lemma, complete-reducibility hypotheses, and Clebsch–Gordan decomposition.
• Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, second edition, Graduate Texts in Mathematics 222, Springer, 2015, Chapter 4 and Appendix C. These sections cross-check basic representation theory, $SU(2)$ Clebsch–Gordan theory, and convention-dependent coefficient data.
• Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras, Cambridge University Press, 2008, §§4.1–4.6 and 6.3, and Example 8.9; an author-posted preliminary version is available as an Open PDF and supplies exact theorem locators. These sections establish intertwiners, invariant tensors, irreducibility, compact averaging, tensor decomposition, and the stated field and global hypotheses.
• Yvette Kosmann-Schwarzbach, Groups and Symmetries: From Finite Groups to Lie Groups, second edition, Springer, 2022, “Representations of $SU(2)$ and $SO(3)$,” pp. 103–118. This supports continuity with the chapter’s rotation-cover comparison.
• David Tong (2006), Quantum Field Theory, §1.3.4, “Internal Symmetries”, Cambridge Part III lecture notes. This section derives the bounded $O(N)$ scalar multiplet and invariant interactions; developed selection rules remain at the physical continuation.