MM 219: Lecture 5 - Symmetry Classes & Platonic Solids

Symmetry Classes

Classes: A set of all elements from a group which are conjugates to one another.

An element cannot be in more than one class.
The number of elements in a class is an integral factor of the order of the point group.
Every element of an Abelian group is in a class by itself.

Classes in Staggered Ethane ($D_{3d}$)

Elements of a class have common symmetry operator properties (e.g., rotations about the same axis, reflection planes).

$\{E\}$

$\{i\}$

$\{C_3, C_3^2\} \rightarrow 2C_3$

$\{S_6, S_6^5\} \rightarrow 2S_6$

$\{C_2, C_2', C_2''\} \rightarrow 3C_2$

$\{\sigma_d, \sigma_d', \sigma_d''\} \rightarrow 3\sigma_d$

Reduced Representation: $E, 2C_3, 3C_2, 3\sigma_d, i, 2S_6$

High Symmetry & Platonic Solids

"A solid which has the same convex, regular polyhedron as each one of its faces..."
— Hypothesised by Plato in Timaeus, circa 360 BC.

Tetrahedral Symmetry ($T_d$)

What happens when we consider multiple $C_3$ axes? A solid having multiple $C_3$ axes is only possible if they are arranged at specific angles (e.g., $109.5^\circ$ as in a tetrahedron). Opening the triangular faces about one vertex reveals the deep symmetry of a tetrahedron.

Identifying Symmetry Elements:

4 Vertices / 4 Faces: Yields $4 \times C_3$ axes (8 operations: $C_3$ and $C_3^2$).
6 Edges: Yields 3 perpendicular $C_2$ axes passing through edge midpoints.
$S_4$ Axes: The $C_2$ axes are also $S_4$ collinear axes ($3 \times 2 = 6$ operations: $S_4, S_4^3$).
6 $\sigma_d$ Planes: Planes passing through an edge and bisecting the opposite edge.

$T$ Point Group: $\{E, 8C_3, 3C_2\}$ → Order = 12

$T_d$ Point Group: $\{E, 8C_3, 3C_2, 6S_4, 6\sigma_d\}$ → Order = 24

$T_h$ Point Group: $\{E, 8C_3, 3C_2, i, 8S_6, 3\sigma_h\}$ → Order = 24

Comprehensive Example

Symmetry of Methane ($T_d$)

Exploring the complete $T_d$ point group symmetry of the Methane ($\mathrm{CH}_4$) molecule.

Drag to Rotate Scene

Animate All 24 Symmetry Operations Present in Methane Molecule

Click any button to view the exact mathematical transformation from the initial state.

Complete Set of Symmetry Elements ($T_d$ Point Group)

$\{ E, 8C_3, 3C_2, 6S_4, 6\sigma_d \}$

The 24 symmetry operations form the full tetrahedral group $T_d$. Observe how operations map the 4 distinctly colored hydrogen atoms to each other while preserving the overall tetrahedral shape.

Beyond Tetrahedrons

Constructing More Platonic Solids

Investigating solids formed by regular polygons meeting at vertices and verifying Euler's Formula.

Using Squares

What if we use squares instead of triangles? A square has interior angles of $90^\circ$. If we place three squares meeting at a single vertex, the sum of the angles is $3 \times 90^\circ = 270^\circ$. Because $270^\circ < 360^\circ$, this arrangement leaves a $90^\circ$ gap (or defect), allowing it to fold into a 3D shape: a Cube (Hexahedron).

3 Squares at each vertex

Using Pentagons

Next, consider regular pentagons. Each interior angle is $108^\circ$. Bringing three pentagons together at a vertex gives a sum of $3 \times 108^\circ = 324^\circ$. This is less than $360^\circ$ (a $36^\circ$ defect), allowing it to form a closed 3D solid: a Dodecahedron.

3 Pentagons at each vertex

Using 4 Triangles

If we place four equilateral triangles meeting at a single vertex, the sum of the angles is $4 \times 60^\circ = 240^\circ$. Because $240^\circ < 360^\circ$, this arrangement leaves a $120^\circ$ gap, folding into an Octahedron.

4 Triangles at each vertex

Using 5 Triangles

Taking it a step further, placing five equilateral triangles at a vertex yields a sum of $5 \times 60^\circ = 300^\circ$. This still leaves a $60^\circ$ defect, which folds into the complex 20-sided Icosahedron.

5 Triangles at each vertex

What about Hexagons?

If we try to construct a Platonic solid using regular hexagons (interior angle $120^\circ$), placing three at a vertex results in $3 \times 120^\circ = 360^\circ$. Because the sum is exactly $360^\circ$, the polygons lie perfectly flat, forming a planar tessellation (like a honeycomb) rather than folding into a three-dimensional solid. Therefore, there are no Platonic solids made of hexagons or any regular polygon with more than 5 sides.

Euler’s Formula: $F + V - E = 2$

Euler's formula describes a fundamental topological property of convex polyhedra, relating the number of Faces ($F$), Vertices ($V$), and Edges ($E$). Let's verify it for all five Platonic solids.

Platonic Solid	No. of faces ($F$)	No. of vertices ($V$)	No. of edges ($E$)	Total ($F + V - E$)
Tetrahedron	4	4	6	2
Cube	6	8	12	2
Octahedron	8	6	12	2
Dodecahedron	12	20	30	2
Icosahedron	20	12	30	2