Point Groups and Crystal Symmetry

Ask of any molecule or crystal: which rigid motions leave it looking exactly as it was? Rotate it, reflect it, invert it through its centre — if the result is indistinguishable from the start, that motion is a symmetry operation. Insist that at least one point stay fixed, ruling out translations, and the operations close into a finite group called a point group. Because every such motion preserves lengths and angles about the fixed point, it is a real orthogonal $3\times3$ matrix, so point groups are precisely the finite subgroups of $O(3)$ . That single sentence organises an entire subject. The determinant sorts every operation into two camps: proper rotations ( $\det=+1$ ) form a finite subgroup of $SO(3)$ ; improper operations ( $\det=-1$ ) are the reflections $\sigma$ , the inversion $i$ , and the rotoreflections $S_n$ that twist and flip at once. The finite subgroups of $SO(3)$ were classified once and for all — cyclic $C_n$ , dihedral $D_n$ , and the polyhedral groups $T$ , $O$ , $I$ — and adjoining improper operations gives the full catalogue of molecular point groups. Crystals impose one extra demand: the symmetry must coexist with a periodic lattice. That lone requirement forbids five-fold and all $n\ge7$ axes and collapses the infinite molecular families into exactly 32 crystallographic point groups, which combine with 14 Bravais lattices into the 230 space groups. The real payoff is representation theory: because a group acts on displacements, orbitals and fields, decomposing those into irreducible representations — read off a character table — predicts vibrational modes, selection rules and crystal-field splittings without solving a differential equation.

Key takeaways

A point group is a finite subgroup of $O(3)$ : every element fixes one common point and is a proper rotation ( $\det=+1$ ) or an improper operation — reflection, inversion, rotoreflection ( $\det=-1$ ).
Five operation types label every element: identity $E$ , rotation $C_n$ , reflection $\sigma$ , inversion $i$ , improper rotation $S_n$ , with $i=S_2$ and $\sigma_h=S_1$ so reflection and inversion are special rotoreflections.
The crystallographic restriction $2\cos\theta\in\mathbb{Z}$ allows only $n=1,2,3,4,6$ , cutting the infinite molecular families to exactly 32 crystallographic point groups, which pair with 14 Bravais lattices into 230 space groups.
Conjugacy classes collect physically equivalent operations; the number of classes equals the number of irreducible representations in the group's character table.
The reduction formula $n_a=\frac{1}{|G|}\sum_k N_k\,\chi(C_k)\,\chi^{(a)}(C_k)^{*}$ decomposes any physical representation — vibrations, orbitals, fields — into irreducibles in a few lines.
Key idea: A point group is a finite subgroup of $O(3)$ ; classify its elements by determinant (proper vs improper) and by conjugacy class, then use the character table to decompose any physical quantity into irreducible representations.

Where this lives today

The 32 point groups and 230 space groups are the daily grammar of structural science: every X-ray or neutron diffraction pattern is indexed by a space group, and databases key hundreds of thousands of structures by this classification. Point-group selection rules decide which phonon modes a Raman or infrared spectrometer can see, letting experimentalists fingerprint a material — graphite versus diamond — from the modes that appear and vanish. Symmetry also decides which crystals can host an effect at all: only the 20 piezoelectric classes convert stress to voltage, and only the 10 polar classes can be ferroelectric, so the hunt for new sensors starts from the point group. The most striking modern twist is that "forbidden" symmetry can still appear: in 1982 Dan Shechtman observed sharp ten-fold diffraction from a quenched Al–Mn alloy — an ordered solid with five-fold symmetry no lattice can support. These quasicrystals, which won Shechtman the 2011 Nobel Prize, do not violate the crystallographic restriction; they sidestep it by being aperiodic, and are described with higher-dimensional space groups that project onto the observed symmetry. Point-group reasoning is likewise central to classifying topological insulators and semimetals, where inversion and mirror symmetries fix which protected surface states exist.

The physical picture

Hold a water molecule and turn it. A rotation by $180^\circ$ about the axis bisecting the H–O–H angle swaps the two hydrogens onto the molecule itself; a reflection in the plane of the atoms does nothing visible; a reflection in the perpendicular plane again swaps the hydrogens. These four motions — identity, one rotation, two mirrors — are all the symmetries there are, and every one keeps the oxygen pinned. That common fixed point is what point in "point group" means — the difference from a crystal's full symmetry, which also permits translations that fix nothing.

Figure

Every point-group operation is a 3×3 orthogonal matrix with det = +1 for proper rotations C_n or det = −1 for the improper operations σ, i, S_n; water's C₂ᵥ elements all pass through one common fixed point, the hallmark of a point group

An improper rotation $S_n$ is the natural "mirror-twist": rotate by $2\pi/n$ , then reflect in the perpendicular plane. Its two simplest members are the ones you meet by other names — the inversion $i=S_2$ and the plain mirror $\sigma_h=S_1$ — so reflections and inversion are not separate primitives but special rotoreflections. This is why the whole zoo reduces to two determinant classes: proper ( $+1$ ), a subgroup, and improper ( $-1$ ), its single coset. For crystals the crucial new fact is felt by tiling a floor: triangles, squares and hexagons cover the plane, but regular pentagons leave gaps. A lattice cannot carry a five-fold axis because the rotation, in lattice coordinates, would need integer entries yet an irrational trace. The next figure makes that quantitative.

Figure

The crystallographic restriction: rotating a lattice vector by ±θ gives two lattice vectors summing to 2cos θ times the original, so 2cos θ = m ∈ {−2, −1, 0, 1, 2} must be an integer; only n = 1, 2, 3, 4, 6 survive, and five-fold symmetry with 2cos 72° = 0.618 is forbidden

The payoff is bookkeeping made automatic. Instead of diagonalising a $3N$ -dimensional vibrational problem, you assign each motion to an irreducible representation and read the degeneracies and IR- or Raman-activity from the character table — the same machinery that splits an atom's orbitals in a crystal and predicts which transitions and tensor components may be non-zero.

The core equations

Everything begins with the ambient group. A symmetry operation preserves distances about the fixed point, so as a matrix it is orthogonal; point groups are the finite groups of such matrices, and the determinant is the first invariant to read off.

O(3)=\{R\in GL(3,\mathbb{R}):R^{\mathsf T}R=\mathbb{1}\},\qquad \det R=\pm1 \qquad (1)

where: $R$ — $3\times3$ real matrix representing a symmetry operation; $R^{\mathsf T}$ — transpose of $R$ ; $\mathbb{1}$ — $3\times3$ identity; $\det R$ — determinant, $+1$ for proper rotations, $-1$ for improper operations. The proper rotations form a subgroup; the improper ones are its single coset. The prototype proper operation is a rotation about the principal axis, conventionally $z$ . Its matrix fixes the axis and rotates the perpendicular plane; its trace $1+2\cos\theta$ is the single number that later drives both the crystallographic restriction and the vibrational count.

C_n:\ R_n=\begin{pmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0&0&1\end{pmatrix},\qquad \theta=\frac{2\pi}{n} \qquad (2)

where: $C_n$ — proper rotation by $2\pi/n$ about a symmetry axis; $\theta$ — the rotation angle $2\pi/n$ ; $n$ — order of the axis, the smallest integer with $C_n^{\,n}=E$ (with $C_1=E$ ). Every improper operation is a proper rotation composed with the horizontal reflection, and the two we name most often are just its lowest members — which lets a group be given by a short list of generators.

S_n=\sigma_h\,C_n,\qquad i=S_2,\qquad \sigma_h=S_1 \qquad (3)

where: $S_n$ — rotoreflection, rotate by $2\pi/n$ then reflect in the perpendicular plane; $\sigma_h$ — reflection in the horizontal plane; $i$ — inversion through the centre, $\mathbf{r}\mapsto-\mathbf{r}$ . The representations of a finite group are rigidly constrained: the squared dimensions of the inequivalent irreducibles add up to the group order — a one-line check on any character table.

|G|=\sum_a d_a^{\,2} \qquad (4)

where: $|G|$ — order of the point group; $d_a$ — dimension of the $a$ -th irreducible; $a$ — index over the inequivalent irreducibles. For $C_{2v}$ this reads $4=1^2+1^2+1^2+1^2$ ; for $O_h$ , $48=2(1^2+1^2+2^2+3^2+3^2)$ . The characters $\chi^{(a)}(g)$ of the irreducibles are orthogonal as vectors over the group, $\sum_g\chi^{(a)}(g)^{*}\chi^{(b)}(g)=|G|\delta_{ab}$ , and this makes decomposition purely arithmetic: project any reducible representation onto each irreducible by the character overlap. This one formula answers nearly every physical question in the subject.

n_a=\frac{1}{|G|}\sum_{g\in G}\chi(g)\,\chi^{(a)}(g)^{*}=\frac{1}{|G|}\sum_k N_k\,\chi(C_k)\,\chi^{(a)}(C_k)^{*} \qquad (5)

where: $n_a$ — number of times irreducible $a$ appears in a reducible representation; $\chi(g)$ — character of the reducible representation; $N_k$ — size of class $C_k$ ; $C_k$ — a class representative. The right-hand class-sum is the version used with a character table. To apply this to a molecule's motions, build the $3N$ -dimensional Cartesian representation, then strip the six rigid-body degrees of freedom to leave the vibrations.

\Gamma_{\text{vib}}=\Gamma_{\text{total}}\ominus\Gamma_{\text{trans}}\ominus\Gamma_{\text{rot}} \qquad (6)

where: $\Gamma_{\text{total}}$ — $3N$ -dimensional representation on the Cartesian displacements; $\Gamma_{\text{trans}}$ — the three translations ( $x,y,z$ ); $\Gamma_{\text{rot}}$ — the three rotations ( $R_x,R_y,R_z$ ). A non-linear molecule has $3N-6$ vibrational modes. The character of the total representation never requires a $3N\times3N$ matrix: only atoms left in place contribute, each adding the trace of a single $3\times3$ block.

\chi_{\text{total}}(C_k)=N_{\text{unmoved}}(C_k)\,\big(\pm1+2\cos\theta_k\big) \qquad (7)

where: $N_{\text{unmoved}}$ — number of atoms left in place by $C_k$ ; $\theta_k$ — rotation angle; $\pm1$ — $+1$ for proper, $-1$ for improper. A reflection ( $\theta=0$ , improper) contributes $+1$ per unmoved atom; inversion contributes $-3$ .

Derivation AThe crystallographic restriction theorem

The goal is to prove that a rotation axis compatible with a periodic lattice can only have order $n\in\{1,2,3,4,6\}$ , and to see why five-fold symmetry — legal for an isolated molecule — is forbidden the moment the object must tile space. The whole argument rests on one invariant, the trace of the rotation, evaluated in two bases: the lattice basis, where it must be an integer, and the Cartesian basis, where it equals $1+2\cos\theta$ . Equating the two pins down the allowed angles.

Step 1. A symmetry rotation $R(\theta)$ of a Bravais lattice maps every lattice point to another, so in the basis of primitive lattice vectors its matrix $M$ has integer entries. Why: an image lattice vector is again an integer combination of primitives, forcing integer matrix elements.

Step 2. The trace is basis-independent, $\mathrm{tr}\,M=\mathrm{tr}\,R(\theta)$ . In the lattice basis it is a sum of integers; in the Cartesian basis the rotation is the block of equation (2), with trace $1+2\cos\theta$ . Why: similarity preserves the trace, so the same number is both an integer and $1+2\cos\theta$ .

Step 3. Equate them: $1+2\cos\theta\in\mathbb{Z}$ , hence $2\cos\theta\in\mathbb{Z}$ , and since $|\cos\theta|\le1$ the only values are $2\cos\theta\in\{-2,-1,0,1,2\}$ . Why: the cosine is bounded, so only these five integers fit.

Step 4. These give $\cos\theta\in\{-1,-\tfrac12,0,\tfrac12,1\}$ , i.e. $n=2,3,4,6,1$ . The excluded $n=5$ needs $2\cos72^\circ=0.618$ and $n=7$ needs $1.247$ — neither an integer. Why: every forbidden order fails the integrality test explicitly.

Step 5. Combining these allowed orders with the compatible reflections and improper rotations yields exactly 32 crystallographic point groups, which pair with the 14 Bravais lattices to give 230 space groups. Why: the restriction on axes is the sieve that makes the list finite. *Verified: $2\cos(2\pi/5)=0.618\notin\mathbb{Z}$ excludes $n=5$ ; $n=1,2,3,4,6$ give $2\cos\theta\in\{2,-2,-1,0,1\}$ respectively — all integers.*

Go deeper · Derivation B · 1 minWhy the squared irreducible dimensions sum to the group order

Derivation BWhy the squared irreducible dimensions sum to the group order

The aim is to establish the sum rule $|G|=\sum_a d_a^{\,2}$ , the identity that checks a character table at a glance and fixes how many irreducibles of each dimension a group can hold. The route is the regular representation — the group acting on itself by multiplication — whose character is trivial and whose decomposition is forced by orthogonality.

Step 1. Let $G$ act on the space spanned by its own elements, $g\mapsto(h\mapsto gh)$ , the $|G|$ -dimensional regular representation $\Gamma_{\text{reg}}$ . Its character is $\chi_{\text{reg}}(g)=|G|\,\delta_{g,E}$ , since only the identity fixes basis vectors. Why: a permutation matrix's trace counts fixed points, and $gh=h$ forces $g=E$ .

Step 2. Apply the reduction formula (5): $n_a=\frac{1}{|G|}\sum_g\chi_{\text{reg}}(g)\,\chi^{(a)}(g)^{*}=\chi^{(a)}(E)^{*}=d_a$ , so each irreducible occurs $d_a$ times. Why: only the identity term survives, and $\chi^{(a)}(E)=d_a$ .

Step 3. Then $\Gamma_{\text{reg}}=\bigoplus_a d_a\Gamma_a$ , and matching dimensions gives $|G|=\sum_a d_a^{\,2}$ . Why: an irreducible appearing $d_a$ times contributes $d_a^2$ to the total dimension. *Verified: $C_{2v}$ gives $1+1+1+1=4=|G|$ ; $O_h$ gives $2(1+1+4+9+9)=48=|G|$ .*

Go deeper · Derivation C · 1 minThe character of an operation on Cartesian displacements

Derivation CThe character of an operation on Cartesian displacements

This derivation justifies equation (7), the shortcut that computes the character of the full $3N$ -dimensional displacement representation without forming a large matrix. The key point is that the $3N\times3N$ matrix is block-structured by atom, and only the diagonal blocks — those of atoms an operation leaves in place — contribute to the trace. Evaluating one block for a proper and an improper operation finishes the job and explains the $\pm1$ that distinguishes them.

Step 1. Order the $3N$ coordinates atom by atom, so the operation acts as an $N\times N$ array of $3\times3$ blocks. An atom sent to a different site puts its block off-diagonal, contributing nothing; only atoms mapped to themselves place a block on the diagonal. Why: the trace sums diagonal entries, and a permuted atom has zeros there.

Step 2. For a proper rotation $C_n$ about $z$ , an unmoved atom's block is the matrix of equation (2), with trace $1+2\cos\theta$ — the axial eigenvalue $1$ plus the in-plane $2\cos\theta$ . Why: the rotation fixes the axis and rotates the perpendicular plane, whose $2\times2$ block has trace $2\cos\theta$ .

Step 3. For an improper $S_n=\sigma_h C_n$ , the reflection flips the $z$ -component, turning the axial eigenvalue to $-1$ , so the block trace is $-1+2\cos\theta$ ; reflections ( $\theta=0$ ) give $+1$ and inversion gives $-3$ . Why: $\sigma_h=\mathrm{diag}(1,1,-1)$ negates just the axial entry.

Step 4. Multiplying by the number of unmoved atoms gives $\chi_{\text{total}}(C_k)=N_{\text{unmoved}}(\pm1+2\cos\theta_k)$ , equation (7). Why: every fixed atom contributes the same block trace. *Verified: for water under $E$ , $3(1+2)=9$ ; under $C_2$ , $1\cdot(1-2)=-1$ ; under $\sigma_v(xz)$ , $1\cdot(+1)=1$ ; under $\sigma_v'(yz)$ , $3\cdot(+1)=3$ .*

Limiting cases

Point groups specialise and degenerate in ways that map neatly onto physical situations.

Case	Result	Meaning
Trivial symmetry	$C_1=\{E\}$	A fully asymmetric object: only "do nothing" is a symmetry, so nothing is degenerate by symmetry.
Crystallographic restriction	$n\in\{1,2,3,4,6\}$	Rotation axes in a periodic crystal are limited to these orders, collapsing the molecular families to exactly 32 classes.
Continuous (linear-molecule) limit	$C_{\infty v},\ D_{\infty h}$	An infinite-order axis; the point group becomes a continuous Lie group, a bridge to $SO(2)$ and $SO(3)$ .
High-symmetry polyhedral groups	$T_d,\ O_h,\ I_h$	Carry 2- and 3-dimensional irreducibles; $I_h$ (order 120) is non-crystallographic.
Full rotation group	$SO(3)$	The free atom: $(2\ell+1)$ -fold levels that a finite point group then splits.

Common trapAn improper rotation Sₙ is just a rotation Cₙ followed by a reflection you also have

Not so. The rotoreflection $S_n$ is one indivisible operation, and a molecule can possess it without either $C_n$ or $\sigma_h$ . Methane is the standard case: it has three $S_4$ axes but no $C_4$ axis and no horizontal mirror — the fourfold rotation alone is not a symmetry, only the twist-and-flip is. Having both $C_n$ and $\sigma_h$ does force $S_n$ , but only that way round. Two further traps: first, $i=S_2$ and $\sigma_h=S_1$ , so inversion and reflection are not extra primitives and must not be counted as independent generators. Second, the restriction forbids $C_5$ and every $C_n$ with $n\ge7$ in a periodic crystal, even though such axes are legal in molecules like ferrocene ( $D_{5d}$ ) and in the icosahedral group $I_h$ — it is about lattices, not symmetry in the abstract.

Worked examples

1 · The vibrations of water in C₂ᵥ (fully worked)

Water has three atoms and point group $C_{2v}=\{E,\,C_2,\,\sigma_v(xz),\,\sigma_v'(yz)\}$ of order 4, with the molecule lying in the $yz$ -plane and the $C_2$ axis along $z$ . Find the symmetry species of its three normal modes of vibration.

Step 1. Build the Cartesian characters from equation (7), $\chi=N_{\text{unmoved}}(\pm1+2\cos\theta)$ . Under $E$ all three atoms are fixed, factor $3$ , giving $9$ . Under $C_2$ (proper, factor $-1$ ) only oxygen stays put, giving $-1$ . Under $\sigma_v(xz)$ (perpendicular to the molecular plane, factor $+1$ ) only oxygen lies in it, giving $1$ . Under $\sigma_v'(yz)$ (the molecular plane) all three atoms give $3$ . So $\chi_{\text{total}}=(9,-1,1,3)$ .

Step 2. Decompose with the reduction formula (5), $n_a=\tfrac14\sum_k\chi_{\text{total}}(C_k)\,\chi^{(a)}(C_k)$ (all $N_k=1$ ), using the rows $A_1=(1,1,1,1)$ , $A_2=(1,1,-1,-1)$ , $B_1=(1,-1,1,-1)$ , $B_2=(1,-1,-1,1)$ :

n_{A_1}=\tfrac14(9-1+1+3)=3,\quad n_{A_2}=\tfrac14(9-1-1-3)=1,\quad n_{B_1}=\tfrac14(9+1+1-3)=2,\quad n_{B_2}=\tfrac14(9+1-1+3)=3

so $\Gamma_{\text{total}}=3A_1\oplus A_2\oplus2B_1\oplus3B_2$ , dimensions summing to $9$ as required.

Step 3. Remove the rigid-body motions. From the table's right-hand columns, $z\to A_1$ , $x\to B_1$ , $y\to B_2$ give $\Gamma_{\text{trans}}=A_1\oplus B_1\oplus B_2$ ; $R_z\to A_2$ , $R_y\to B_1$ , $R_x\to B_2$ give $\Gamma_{\text{rot}}=A_2\oplus B_1\oplus B_2$ .

Step 4. Subtract, using equation (6): $\Gamma_{\text{vib}}=(3A_1\oplus A_2\oplus2B_1\oplus3B_2)\ominus(A_1\oplus B_1\oplus B_2)\ominus(A_2\oplus B_1\oplus B_2)=\mathbf{2A_1\oplus B_2}$ . The count checks: $3N-6=3$ modes. These are the symmetric stretch and bend ( $A_1$ ) and the antisymmetric stretch ( $B_2$ ). Since $A_1$ and $B_2$ each carry both linear and quadratic basis functions, all three modes are both infrared- and Raman-active — why water absorbs so strongly in the infrared.

2 · Splitting of d-orbitals in an octahedral crystal field (fully worked)

A free transition-metal ion has five degenerate $d$ -orbitals — the $\ell=2$ representation of $SO(3)$ . At the centre of an octahedron of ligands its symmetry drops to $O_h$ . Into which irreducibles do the $d$ -orbitals split?

Step 1. Work in the rotation group $O$ (order 24); the gerade $O_h$ labels follow because $d$ -orbitals are even under inversion. The $\ell=2$ character at angle $\theta$ is the $SO(3)$ formula

\chi_{\ell}(\theta)=\frac{\sin\!\big((\ell+\tfrac12)\theta\big)}{\sin(\theta/2)},\qquad \chi_{\ell}(E)=2\ell+1

Step 2. Evaluate it on the classes of $O$ — $E$ , $8C_3$ ( $120^\circ$ ), $6C_2$ ( $180^\circ$ ), $6C_4$ ( $90^\circ$ ), $3C_2=C_4^2$ ( $180^\circ$ ) — at $\ell=2$ :

\chi_{2}(E)=5,\ \chi_{2}(C_3)=-1,\ \chi_{2}(C_2)=1,\ \chi_{2}(C_4)=-1,\ \chi_{2}(C_4^2)=1

i.e. $\chi=(5,-1,1,-1,1)$ .

Step 3. Reduce with equation (5), using class sizes $(1,8,6,6,3)$ . For the irreducibles $E$ (character $(2,-1,0,0,2)$ ) and $T_2$ ( $(3,0,1,-1,-1)$ ):

n_E=\tfrac{1}{24}\big(1\cdot5\cdot2+8(-1)(-1)+6\cdot1\cdot0+6(-1)0+3\cdot1\cdot2\big)=\tfrac{10+8+0+0+6}{24}=1

n_{T_2}=\tfrac{1}{24}\big(1\cdot5\cdot3+0+6\cdot1\cdot1+6(-1)(-1)+3\cdot1(-1)\big)=\tfrac{15+0+6+6-3}{24}=1

while the same arithmetic gives $n_{A_1}=n_{A_2}=n_{T_1}=0$ .

Step 4. Therefore $D^{(2)}\to E_g\oplus T_{2g}$ in $O_h$ , splitting the five-fold degeneracy into a doublet $e_g$ and a triplet $t_{2g}$ ( $2+3=5$ ). Physically the $e_g$ pair ( $d_{z^2},d_{x^2-y^2}$ ) points at the ligands and is pushed up, while the $t_{2g}$ trio ( $d_{xy},d_{xz},d_{yz}$ ) points between them and sits lower; the gap $\Delta_{\text{oct}}=10\,Dq$ is the central quantity of ligand-field theory and the origin of transition-metal colours. Result: $d\to e_g\oplus t_{2g}$ , a 2/3 split.

Physics of the portrait

Point-group theory is where geometry, crystallography and quantum mechanics meet, and no figure owns it more than the man who gave crystal symmetry both its group-theoretic backbone and its notation.

Origins · 1 minArthur Schoenflies and how this was discovered

Arthur Schoenflies

1853–1928 · group-theoretic classification of crystal symmetry

Arthur Moritz Schoenflies was a German mathematician — trained under Kummer and Weierstrass, later at Göttingen alongside Felix Klein — who turned the nineteenth-century catalogue of crystal forms into rigorous group theory. His enduring achievement came in 1891, when, simultaneously with and independently of the Russian crystallographer Evgraf Fedorov, he enumerated all 230 space groups, the complete list of ways symmetry can repeat in a periodic crystal. The two reached the same 230 by different roads and corresponded to reconcile their counts — a classification confirmed twice over before any crystal structure had been measured. Schoenflies also devised the compact notation — $C_n$ , $D_n$ , $S_n$ , $T_d$ , $O_h$ — that names a point group by its generating axes and mirrors and remains the everyday language of molecular spectroscopy, alongside the crystallographers' Hermann–Mauguin symbols. When X-ray diffraction arrived in 1912, the 230 groups he had derived from pure symmetry became the framework for reading every diffraction pattern.

History

1830

— Hessel proves that exactly 32 crystal classes are geometrically possible

1849

— Bravais enumerates the 14 space lattices compatible with periodicity

1891

— Fedorov and Schoenflies independently derive the 230 space groups

1929

— Bethe explains crystal-field level splitting by group representations

1931

— Wigner recasts quantum mechanics in the language of group theory

Going further

Texts. Tinkham, Group Theory and Quantum Mechanics (Dover, 2003), Ch. 4 · Cotton, Chemical Applications of Group Theory, 3rd ed. (Wiley, 1990) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 17 · Stone & Goldbart, Mathematics for Physics (CUP, 2009), Ch. 28 · Dresselhaus, Dresselhaus & Jorio, Group Theory: Application to the Physics of Condensed Matter (Springer, 2008) · Hamermesh, Group Theory and Its Application to Physical Problems (Dover, 1989).

Provenance. The crystallographic restriction reproduces the standard trace argument $2\cos\theta\in\mathbb{Z}$ (Ashcroft & Mermin; Tinkham), verified by $2\cos(2\pi/5)=0.618\notin\mathbb{Z}$ ; the 32 point groups, 14 Bravais lattices and 230 space groups are standard counts. The $H_2O$ $C_{2v}$ reduction follows Cotton and Tinkham Ch. 4 — recompute $\chi_{\text{total}}=(9,-1,1,3)$ , apply equation (5), confirm $\Gamma_{\text{vib}}=2A_1\oplus B_2$ . The octahedral $d$ splitting $D^{(2)}\to E_g\oplus T_{2g}$ uses $\chi_\ell(\theta)=\sin((\ell+\tfrac12)\theta)/\sin(\theta/2)$ and the $O$ table (Dresselhaus; Tinkham). The sum rule follows from the regular representation (Arfken Ch. 17); the quasicrystal note refers to Shechtman's 1982 observation and the 2011 Nobel Prize.

Prerequisites. Groups, Subgroups and Cosets · Conjugacy Classes and Normal Subgroups · The Rotation Group SO(3). Leads to. Character Tables and Decomposition of Representations · Group Representations · Tensor Products and Clebsch–Gordan Decomposition.

Ask the tutor: Give me any molecule or crystal and I will walk you from its symmetry elements to its point group, character table, and vibrational or orbital decomposition step by step.

What is the difference between a point group and a space group?

A point group collects only the operations that keep one point fixed — rotations, reflections, inversion, rotoreflections. A space group is the full symmetry group of an infinite periodic crystal: it adds translations, plus two hybrids combining a point operation with a fractional translation — screw axes (rotate then slide along the axis) and glide planes (reflect then slide in the plane). There are 32 point groups but 230 space groups, because each class can sit on several lattices with different screw-glide combinations. Stripping the translations from a space group leaves its point group, which governs the direction-dependent properties.

Why do we care about conjugacy classes rather than individual operations?

Because characters are class functions — the trace is identical for every operation in a class — all the representation-theoretic information is organised by class. Physically, operations in one class are the same motion from a rotated viewpoint: the three $C_4$ rotations of an octahedron about its $x$ , $y$ , $z$ axes form one class. This is why a character table has as many columns (classes) as rows (irreducibles), and why the reduction formula sums over classes weighted by $N_k$ rather than over all $|G|$ elements.

How does symmetry alone predict degeneracy without solving the Schrödinger equation?

Any Hamiltonian invariant under $G$ has eigenspaces that carry representations of $G$ ; generically these are irreducible, so a level's degeneracy equals its irreducible's dimension. A group with only one-dimensional irreducibles ( $C_{2v}$ ) forbids symmetry-required degeneracy, while one with a two- or three-dimensional irreducible ( $O_h$ ) mandates doublets and triplets. This is why a free atom's $(2\ell+1)$ -fold level splits exactly as the $SO(3)$ representation decomposes under the point group. You still need the equation for the energies — symmetry fixes the degeneracy pattern, not the numbers.

Is the inversion centre what decides whether a crystal can be piezoelectric?

Largely, yes. A crystal with a centre of inversion cannot be piezoelectric: a stress-induced polarisation would reverse under inversion while the stress does not, so the effect vanishes. Of the 32 classes, 11 are centrosymmetric and ruled out; of the 21 non-centrosymmetric ones, 20 are piezoelectric (the cubic class 432 excepted). A subset of 10 polar classes can also be pyroelectric and ferroelectric — a tensor property switched on or off by a single symmetry element.

Why are the icosahedral and linear-molecule groups called non-crystallographic?

Because they contain rotation axes the restriction forbids in a lattice. The icosahedral groups $I$ and $I_h$ have five-fold axes, and $2\cos72^\circ=0.618$ is not an integer, so no Bravais lattice supports them — C $_{60}$ and many virus capsids are icosahedral, but a periodic crystal of them can never make that five-fold axis a lattice symmetry. The linear groups $C_{\infty v}$ and $D_{\infty h}$ have an infinite-order axis and are not even finite. All are valid point groups; "non-crystallographic" just means they are not among the 32 lattices allow.

How do the Schoenflies and Hermann–Mauguin notations relate?

They label the same groups in two dialects. Schoenflies names a group by its generators — $C_{2v}$ is a two-fold axis with vertical mirrors, $O_h$ the full octahedral group — which chemists favour because it reads off the molecular elements. Hermann–Mauguin (International) notation, preferred by crystallographers, encodes symmetry along crystal directions — $mm2$ for $C_{2v}$ , $4/mmm$ for $D_{4h}$ — and extends to the 230 space groups via screw and glide symbols. Every point group has a name in both; the choice is convention, not physics.