Irreducible representation

An irreducible group representation is nonzero and has no nonzero proper invariant subspace. These are the pieces that cannot be decomposed further by taking a smaller symmetry-preserved vector space.

The real two-dimensional rotation through $120$ degrees illustrates the role of the coefficient field. A one-dimensional real invariant subspace would supply a real eigenvector, but the characteristic polynomial ${\lambda }^2+\lambda +1$ has no real root. The representation is therefore irreducible over the real numbers. Over the complex numbers, that polynomial has distinct roots of unity $e^{2\pi i/3}$ and $e^{-2\pi i/3}$, giving two invariant eigenlines. Irreducibility over the real numbers does not imply absolute irreducibility.

In a completely reducible group representation $V={⨁}_{\lambda }{m}_{\lambda }{V}_{\lambda }$, the nonnegative integer $m_{\lambda }$ counts the copies of the corresponding irreducible representation. For complex representations of a finite group, character orthogonality gives $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V⟩$.

The one-dimensional trivial representation, in which every symmetry operation acts as $+1$. An integral over a symmetry-invariant domain can be nonzero only when its integrand contains such a component. Molecular character tables often label it $A_1$ or $A_g$.

A group representation over a field is absolutely irreducible if it remains irreducible after every field extension. For a finite-dimensional representation it is enough to check an algebraic closure. Irreducibility over the original field alone is weaker: the two-dimensional real rotation representation of the cyclic group of order three splits into two lines over the complex numbers.

A real group representation is absolutely irreducible if its complexification is irreducible. For a finite group this is equivalent to having only scalar real endomorphisms commuting with the representation. Independent coordinate sign changes force a commuting matrix to be diagonal; a transitive cyclic permutation of the coordinates then forces its entries equal. Alternatively, the coordinate projections and the cyclic permutation directly prove irreducibility over either field.

If conjugation sends an operator to its inverse, the conjugating element sends its $\lambda$-eigenspace to the ${\lambda }^{-1}$-eigenspace.

For $D_{2n}=⟨r,s:r^n=s^2=1,srs=r^{-1}⟩$, every irreducible complex representation has degree at most two. If $v$ is an eigenvector of $r$ with eigenvalue $\lambda$, then $sv$ has eigenvalue ${\lambda }^{-1}$, so irreducibility makes the representation equal to $\mathrm{span}\{v,sv\}$.

For the dihedral group of order $2m$ with odd $m$, let $D_k(r)=\mathrm{diag}({\zeta }^k,{\zeta }^{-k})$ and $D_k(s)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, where $\zeta =e^{2\pi i/m}$ and $1\le k\le (m-1)/2$. An invariant tensor defining a bilinear form satisfies $D_k(g{)}^{T}L{D}_k(g)=L$. The rotation forces $L_{11}=L_{22}=0$ and the reflection forces $L_{12}=L_{21}$. Thus the invariant tensor is a multiple of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and a singlet Yukawa interaction gives two equal physical masses by singular value decomposition. Distinct inequivalent doublets admit no such bilinear form, by Schur lemma and their self-duality.

For $\omega =e^{2\pi i/n}$ and $1\le j<n/2$, defineThese representations are irreducible and pairwise nonisomorphic. They give all two-dimensional irreducibles: $(n-1)/2$ of them for odd $n$, and $n/2-1$ for even $n$.

For $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$, the matricesdefine a faithful irreducible two-dimensional complex representation.