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Place the cube vertices at $(\pm 1,\pm 1,\pm 1)$. The four vertices whose coordinate product is $+1$ form $T$; those with product $-1$ form $T^{\prime }$. Within either set, two distinct vertices differ in exactly two coordinates, so every edge has length $2\sqrt{2}$ and each set is a regular tetrahedron.

A member of the cube rotation group is a signed permutation matrix of determinant one. There are $3!$ permutations of the coordinate axes and four choices of signs for each permutation that give determinant one, so $|G|=24$. Such a matrix maps a coordinate-product class to a coordinate-product class. Moreover, the rotation through a quarter-turn $(x,y,z)\mapsto (-y,x,z)$ interchanges the classes. Thus the orbit of $T$ consists of exactly $T,T^{\prime }$. By the orbit-stabilizer theorem, the tetrahedron stabilizer in the cube rotation group has order

The three known vertices have pairwise squared distances $8$, so the regular tetrahedron must have every squared edge length equal to $8$. Comparing the squared distance from $d$ to $b$ with that from $d$ to $a$ givesEquality forces $x=-1$. Similarly,so $y=-1$. The resulting point has squared distance $8$ from each of the other three vertices. Therefore

For one fixed POVM repeated on independently prepared copies, the Born rule gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit density operator has three independent real parameters,The probability map is affine in $\mathbf{r}$. It can be injective on the full Bloch ball only if $m-1\ge 3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\ge 4$ for an informationally complete POVM.

To attain this bound, take the four directionsThey are the vertices of a regular tetrahedron. The tetrahedral qubit POVM has effects $E_i=(I+{\mathbf{n}}_i\cdot \mathbf{\sigma })/4$, each positive and rank one, and ${\sum }_i{E}_i=I$. The observed probabilities areSince ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I_3/3$, they recover the state byThis is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is informationally complete three-observable qubit tomography, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.

Label the vertices of a regular tetrahedron $1,2,3,4$, and include all its Euclidean isometries, both rotations and reflections. The following list describes every subgroup of cardinality at most six, allowing arbitrary relabelling of the vertices. The numbers count actual subgroups, rather than just their group isomorphism types.

Here is why this is exhaustive. Part b leaves only the trivial group, $C_2,C_3,C_4,C_2\times C_2,S_3$. The disjoint cycle decomposition gives exactly the two listed kinds of involutions, the three-cycles, and the four-cycles, accounting for all the cyclic subgroups. In a Klein four-group, the nonidentity elements commute. If it contains $(12)$, any other nonidentity element must be a commuting involution, hence $(34)$ or $(12)(34)$; either choice yields the displayed opposite-edge subgroup. If it has no transposition, its three nonidentity elements must be all three double transpositions, yielding the rotational subgroup.

Finally, an $S_3$ subgroup has a unique cyclic subgroup of cardinality three. Its two three-cycles fix the same unique vertex. Every element of the $S_3$ subgroup preserves this vertex, since conjugation preserves that unique cyclic subgroup. Thus it is the full stabilizer subgroup of that vertex. The counts in the list are consequently exhaustive, givingThe two kinds of $C_2$ and the two kinds of Klein four-groups have different geometric behavior despite having the same abstract group isomorphism type within each pair. This is the distinction recorded by the small subgroups of tetrahedral symmetry.

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

Realize the cube's vertices as $(\pm 1,\pm 1,\pm 1)$ and take the four with coordinate product $+1$. They form a regular tetrahedron, and their negatives form the other one. Every proper rotation of the first tetrahedron also permutes the negatives, hence is a cube rotation. This embeds $G_1$ as a subgroup of $G_2$.

The embedded group preserves each of these two four-vertex sets and acts transitively on each. Thus the resulting action on all eight cube vertices has two orbits, each of size four.

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

Realize the cube's vertices as $(\pm 1,\pm 1,\pm 1)$ and take the four with coordinate product $+1$. They form a regular tetrahedron, and their negatives form the other one. Every proper rotation of the first tetrahedron also permutes the negatives, hence is a cube rotation. This embeds $G_1$ as a subgroup of $G_2$.

The embedded group preserves each of these two four-vertex sets and acts transitively on each. Thus the resulting action on all eight cube vertices has two orbits, each of size four.

The orientation-preserving symmetries of a regular tetrahedron form the alternating group on its four vertices. There are twelve elements: the identity, eight rotational symmetries of order three, and three half-turns.

The full symmetry group of a regular tetrahedron is the symmetric group $S_4$ acting on its vertices. Its subgroups with at most six elements have eight types up to conjugation, with the following representative subgroups and counts:A transposition is a reflection through the plane containing the two fixed vertices and the midpoint of the exchanged edge. A double transposition is a half-turn about the axis joining the midpoints of the two exchanged edges. A three-cycle is a one-third-turn about a vertex-to-opposite-face axis. A four-cycle is a quarter-turn composed with a reflection perpendicular to the opposite-edge midpoint axis; it is an improper orthogonal transformation.

The unique Klein four-group containing only double transpositions consists of the three opposite-edge half-turns and the identity element. The other three Klein four-groups consist of two perpendicular mirror reflections, their half-turn product, and the identity element. Each $S_3$ is a stabilizer subgroup of a vertex, including the three rotations and three mirror reflections preserving that vertex.

For completeness, a Klein four-group containing $(12)$ must contain the commuting involution $(34)$ or $(12)(34)$, yielding the displayed subgroup; one containing no transpositions must contain all three double transpositions. An $S_3$ subgroup has a unique cyclic subgroup of cardinality three, whose unique fixed vertex must therefore be preserved by the entire $S_3$. These observations and the disjoint cycle decomposition exhaust the types allowed by groups of order at most six. The counts sum to $25$.

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.

In tetrahedral molecular geometry, four neighbouring chemical atoms surround a central chemical atom in the directions of the vertices of a tetrahedron. In the regular idealization, the central atomic nucleus lies at the centre of a regular tetrahedron, and every pair of chemical bonds makes the same bond angle. Methane, ${\mathrm{CH}}_4$, provides the standard example: its carbon has four bonding electron domains and no lone pair. See Chemistry 2e, §7.6: tetrahedral methane.

The exact ideal bond angle follows from four directions proportional to $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Each has squared Euclidean norm $3$, and each distinct pair has dot product $-1$. HenceThis is a statement about a regular tetrahedron, rather than a universal exact value for every molecule labelled tetrahedral. It also explains why a cross-shaped flat drawing with $90^{\circ }$ gaps is misleading: four equal directions can spread out more evenly in three dimensions.

The same vectors sum to zero. Consequently, four equal bond-dipole contributions along those directions cancel. The symmetric methane molecule is nonpolar, but a tetrahedral shape alone does not imply a nonpolar molecule: replacing one neighbour can make the electronic distribution and bond contributions unequal. Chemical polarity depends on both molecular geometry and electron density. See McMurry, Organic Chemistry, §2.2: molecular dipoles and methane.

Choose the four unit Bloch vectors at the vertices of a regular tetrahedron, so ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I/3$. The displayed four positive rank-one effects sum to $I$. On input $(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, their probabilities are $p_i=(1+\mathbf{r}\cdot {\mathbf{n}}_i)/4$, giving the reconstruction $\mathbf{r}=3{\sum }_i{p}_i{\mathbf{n}}_i$. This realizes the smallest fixed qubit informationally complete POVM.