CHEM 548: Materials Chemistry

Lecture 1: Translational Symmetry in 2D

Mon 08/24/2026

Course overview and logistics; translational symmetry in 2D and 3D, lattices, unit cells, and the 14 Bravais lattices

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 1 §1.1, pp. 1–12, and Problem 1.2 (§1.6)

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Crystalline vs. Amorphous Solids

$Left: iron pyrite FeS2, a crystalline solid with sharp cubic faces. Right: obsidian (KAlSi3O8), an amorphous solid with smooth conchoidal fracture.$

This course is about crystalline solids — one repeating unit tells you the whole structure.

A Comprehensive Mathematical Framework for Describing Crystals

One intuitive description: atoms packed as spheres.
FCC packing shows close-packed planes with ABC stacking.
But sphere pictures do not scale to all possible crystals.
We need the minimum mathematical framework for all crystals.

FCC cubic close-packed sphere packing shown as ABC layers with a unit cell outline.

Translational Symmetry in 2D

Why start here

Every crystal in this course is a lattice + motif — one repeating pattern
Master the pattern language and properties follow: which defects can form, how phases arrange, how electrons move
2D is where every idea is visible at a glance

Learning Objectives

Differentiate crystalline vs amorphous solids
Define lattice, vectors, unit cell
Identify shapes that tile space
Recognize allowed rotation axes (2, 3, 4, 6-fold)
See how lattice + motif → crystal (e.g. graphene)

Relationship to Text

Solid State Materials Chemistry, Ch. 1: Structures of Crystalline Materials — assigned §1.1 (pp. 1–12)

1.1.1 Translational Symmetry (pp. 2–3)
1.1.2 Rotational Symmetry (pp. 3–5)
1.1.3 Crystallographic Point Groups and Crystal Systems (p. 5)
1.1.4 Bravais Lattices (pp. 5–8)
1.1.5 Introduction to Space Groups (pp. 8–9)
1.1.6 Symmetry Elements That Combine Rotation and Translation (pp. 9–11)
1.1.7 Space-Group Symbols (pp. 11–12)

We Start From the Beginning

Warning: section 1.1 of the book is very DENSE and difficult to UNDERSTAND.
It is an attempt to condense several chapters of crystallography into a few pages.
We will EXPAND upon this material in the next few lectures.

Other helpful online resources:

West, Solid State Chemistry and its Applications
Hammond, The Basics of Crystallography and Diffraction

Translational Symmetry (in 2D)

$\mathbf a$ and $\mathbf b$ are the lattice vectors, every lattice point can be defined by adding these two vectors (in 2D)
The unit cell is the parallelogram defined by the lattice vectors, all space can be filled by tiling unit cells.

To make the idea concrete we work in two dimensions.

Mark a grid of equivalent points — lattice points
Two arrows: the vectors a and b
The parallelogram they span = the unit cell, your single brick
Copy it by pure translation — no rotations, gaps, or overlaps — and it tiles the whole plane

[advance: add the r = 2a + 3b coordinate path]

All points equivalent → label any one the origin
General point r = m a + n b, m and n integers
Example 2a + 3b: two steps along a, three along b, exactly onto another lattice point — the algebra encodes the whole infinite pattern

So the whole crystal is generated by repeating one small cell, and that raises an obvious question: what shapes can that repeating cell actually be?

Which shapes can we use to tile space?

Eight candidate shapes labeled triangle, square, rectangle, parallelogram, rhombus, pentagon, hexagon, octagon — shown clean, before testing which of them can tile.

Not all shapes can tile space

Regular pentagons packed edge to edge, leaving gaps that cannot be filled.

Regular pentagons leave gaps however they are packed.

Why five-fold can't tile: the 360° test

The rotation rule

A lattice can repeat after a turn by $\theta$ only if $2\cos\theta$ is a whole number.

$\theta$	fold	$2\cos\theta$
60°	6 ✓	1
90°	4 ✓	0
120°	3 ✓	−1
180°	2 ✓	−2
72°	5 ✗	0.618
45°	8 ✗	1.41

72° (5-fold) and 45° (8-fold) don't give whole numbers, so no lattice looks the same after those turns — the pentagon and the octagon fail.

Copies meeting at a point must sum to exactly 360°. Pentagons (108°) give 3×108=324° → a 36° gap; hexagons (120°) give 3×120=360° → perfect. Per Solid State Materials Chemistry, §1.1.3.

Tiles meeting at a point must close the circle: exactly 360 degrees, no gap, no overlap.

[advance: three pentagons meet at a point]

Pentagon interior angle 108°; three around a point give 324°

[advance: reveal the 36 degree gap]

Red wedge = the missing 36°; a fourth pentagon would overlap — pentagons cannot close a point

[advance: compare with hexagons]

Hexagons close exactly: 3 × 120 = 360
Table → the general rule: a lattice repeats after a turn by θ only if 2cosθ is a whole number
2, 3, 4, 6-fold pass; 5- and 8-fold fail — the crystallographic restriction

But pentagons and hexagons are only two examples. Next, let's prove that no other rotation can ever work.

Where a(1 − 2cos θ) comes from

Each rotation moves its endpoint a cos θ along a, so two rotations shorten the gap by 2a cos θ. This is the length; which θ a lattice allows is the next slide.

Before we test any particular angle, let's see where the length in the test actually comes from. The picture is drawn at seventy degrees, but nothing here depends on that choice.

[advance: the two rotations]

Turn B about A by theta, and A about B by minus theta. B prime and A prime land at the same height, so the segment joining them runs parallel to a.

[advance: drop the perpendiculars]

Drop each image straight down to the row. B prime lands a cosine theta to the right of A, because that is the horizontal side of the right triangle with hypotenuse a and angle theta. By the same construction at the other end, A prime lands a cosine theta to the left of B.

[advance: subtract]

So the gap between the two feet is the full a with one a cosine theta eaten off each end: a minus two a cosine theta, which is a times one minus two cosine theta. The red segment up top has that same length, since it sits directly above the gap.

Note that seventy degrees is only a drawing choice. This slide says nothing about whether a lattice can have a seventy-degree axis; it just gives us the length to test. That test is the next slide.

The proof — part 1: what a lattice allows

By definition a lattice is the points m·a + n·b with whole m, n — so along a, separations come only in whole multiples of a.

The angle test told us which rotations to expect. Now let's prove the list is complete.

Go back to the definition of a lattice: every lattice point is m a plus n b, with m and n whole numbers. So along the direction of a, lattice points sit at whole multiples of a: one a, two a, three a, nothing in between.

[advance: suppose a rotation by theta]

Now suppose a rotation by theta is a symmetry of the lattice; here theta is 90 degrees, a fourfold turn. Turn B about A by theta, and A about B by minus theta. Both images, B prime and A prime, must land on lattice points.

[advance: the forced conclusion]

B prime and A prime sit at the same height, so their separation is parallel to a, and the geometry gives its length: a times one minus two cosine theta. At 90 degrees that is exactly one a, a whole multiple, so a fourfold axis is allowed.

In general the separation must be m times a, so two cosine theta must be a whole number. That is the test. Before breaking it, let's check the other allowed turns.

The other allowed turns: 60°, 120°, 180°

Separations: 60° → 0·a, 90° → 1·a, 120° → 2·a, 180° → 3·a — whole multiples, all four.

The proof — part 2: only four angles pass

At θ = 75° the separation is 0.48·a — not a whole multiple, so no lattice has a 75° axis. Only θ = 180°, 120°, 90°, 60° make 2cos θ whole.

Which shapes tile the plane?

The eight candidate shapes shown clean: triangle, square, rectangle, parallelogram, rhombus, pentagon, hexagon, octagon. On advance, the regular pentagon and octagon are crossed out in red and the triangle is shown dashed.

Which of the eight cannot tile the plane alone? Verdict: the pentagon and octagon fail; the triangle needs rotation; the rest tile by translation.

From a hexagon tiling to a lattice

Two translation vectors span a parallelogram, never a hexagon. Where is the lattice?

Mark one equivalent point per repeat: the hexagon centers form the lattice.
Neighboring centers span the unit cell: a rhombus, $|\mathbf a| = |\mathbf b|$, angles $60^\circ/120^\circ$.

“Hexagonal” names the points' six-fold symmetry (six neighbors, $60^\circ$ apart), not the cell shape; each point's territory is one hexagon.

Here is the honeycomb again. The hexagons are the drawn tiling; the lattice is the grid of equivalent points underneath it.

Unit cell = the parallelogram the two translation vectors span
No pair of vectors spans a hexagon → a hexagon cannot be the unit cell; find the grid underneath

[advance: lattice points appear at the hexagon centers]

Mark one equivalent point per repeat → the hexagon centers form the lattice

[advance: connect neighbors into the rhombus unit cell]

Unit cell = rhombus, a = b, angles 60°/120°
"Hexagonal" names the points' six-fold symmetry (six neighbors, 60° apart), not the cell shape
Each point's territory = one hexagon — tiling and lattice, two pictures of the same repeating order

When we list the five Bravais lattices in a moment, the hexagonal one is exactly this.

2D Crystal systems

Four panels: square (a=b, gamma 90, 4-fold), rectangular (a!=b, gamma 90, mirror/glide), hexagonal (a=b, gamma 120, 3/6-fold), oblique (a!=b, arbitrary gamma, no required symmetry).

A crystal system groups patterns by their rotational symmetry: the relations between $a$, $b$, $\gamma$ decide which rotations and mirrors are possible.

Square: $a=b,\ \gamma=90^\circ$ → 4-fold axis.
Rectangular: $a\ne b,\ \gamma=90^\circ$ → mirror / glide.
Hexagonal: $a=b,\ \gamma=120^\circ$ → 3- or 6-fold.
Oblique: $a\ne b,\ \gamma$ arbitrary → none.

We have now met several unit cells: squares, rectangles, general parallelograms, and the hexagonal rhombus.

A crystal system = a family of patterns grouped by their rotational symmetry: a, b, γ decide which rotations and mirrors are possible — exactly four in 2D

[advance: the square system highlighted]

Square: a = b, γ = 90° → requires a fourfold axis

[advance: the rectangular system highlighted]

Rectangular: a ≠ b, γ = 90° → hallmark is a mirror or glide line (reflection + translation along the mirror)

[advance: the hexagonal system highlighted]

Hexagonal (triangular points): a = b, γ = 120° → six- or threefold; graphene reflects this

[advance: the oblique system highlighted]

Oblique: general a, b, γ → nothing required except the free twofold every lattice keeps (every lattice vector has an equal and opposite partner)

Four systems, then. The next question is how many distinct lattices they produce.

2D Bravais Lattices

A Bravais lattice is the pattern of equivalent points before any atoms are placed: every point has the same surroundings.

Exactly five distinct 2D lattices exist: four primitive, plus centered rectangular. Per Solid State Materials Chemistry, Problem 1.2 (§1.6) and its solutions-manual answer.

A Bravais lattice is the pattern of equivalent points before I put atoms on it.

Every point has the same surroundings — the pure translation scaffold
Four primitive choices: square, rectangular, hexagonal (equal vectors at 60/120°), oblique

[advance: the centered rectangular panel appears]

Fifth: centered rectangular — rectangular outline plus an equivalent point at ½a + ½b; a lattice point, not a motif atom
A smaller primitive rhombus describes the same points — but it is oblique and hides the mirror symmetry
The only centered 2D case that cannot reduce to a smaller primitive cell of the same symmetry

That gives exactly five 2D Bravais lattices. Before we look closer at that fifth lattice, let me give you an everyday picture for what "every point has the same surroundings" means.

Which points count as lattice points?

Per Solid State Materials Chemistry, §1.1.4 (Bravais Lattices), p. 5.

Here is an everyday picture for that definition.

Every street corner has the same McDonald's — same building, same golden arches, facing the same way
Blindfold test: driven to any other corner, I cannot tell I moved — every corner-to-corner hop repeats the whole city

[advance: the franchise war — alternate corners become Subways]

Subway buys every other corner — on a map the grid is unchanged
Blindfold off in front of a Subway and you know at once: a one-block hop swaps the chains, so it is not a repeat
The true repeat is the diagonal hop, McDonald's to McDonald's — the bigger, tilted cell, with the Subway inside it as part of the motif, not a lattice point

[advance: the verdict line]

A candidate point must pass the blindfold test: same thing, same surroundings, same orientation
Hold on to the franchise war — a practice pattern at the end of this lecture has a B atom playing the Subway

The next question is different: once the points are settled, which cell do we draw around them?

Choosing a cell: crop the piece that shows the pattern

Per Solid State Materials Chemistry, Problem 1.2 (§1.6) and its solutions-manual answer.

Picking a unit cell is like cropping a scrap of last year's Christmas paper so you could reprint the whole roll.

Crop the upright rectangle: every copy off the press shows a whole tree, star up, with its mirror line down the middle
Check the pattern at a glance — the copies stack edge to edge straight back into the sheet

[advance: the diagonal sliver and its jigsaw reprint]

The thrifty crop: a diagonal sliver, genuinely half the paper, and a legal cell — it holds everything needed to reprint
But every edge slices through trees: no copy shows one whole tree, nothing matches the sheet's mirrors, and the rebuild is a jigsaw of oblique edges

[advance: the verdict line]

Both crops reprint the sheet → both are legal unit cells

We pay double the paper for the cell that shows the symmetry — and that choice is exactly what the next slide makes interactive: a centered rectangle against a smaller rhombus, on this same arrangement of points.

Why keep "centered rectangular"?

Same point set, two cell choices. Toggle to see why the rectangle is preferred over the smaller oblique cell.

The rectangular cell exposes the lattice's mirror symmetry that the smaller oblique primitive cell hides. Per Solid State Materials Chemistry, Problem 1.2 (§1.6) and its solutions-manual answer.

Why aren't there other centered Bravais lattices?

A centered square lattice with a shaded square unit cell.

A "centered square" is not new — a smaller primitive square, rotated $45^\circ$ with half the area, generates the same points.

Drawing the diagonals reveals the smaller primitive square.

Triangles tile too — same hexagonal lattice

A tiling of up- and down-pointing equilateral triangles; later a blue rhombus unit cell is highlighted.

"Up" triangles alone leave gaps; you also need "down" triangles.
Mark the triangle centers → each point has 6 neighbors at $60^\circ$.
Unit cell = rhombus, $|\mathbf a|=|\mathbf b|$, $\gamma=60^\circ$ → same hexagonal Bravais lattice.

Equilateral triangles tile the plane, but there is one extra detail.

All pointing up → gaps; one orientation is not enough

[advance: add the down triangles]

Add the 180°-rotated down triangles → the gaps disappear

[advance: mark the triangle centers]

Centers: 6 neighbors at 60° — the same point pattern as the hexagon tiling: the hexagonal Bravais lattice

[advance: highlight the rhombus unit cell]

Unit cell again a rhombus, a = b, 60°

The two triangle orientations do not create a new crystal system; they are just another way to tile the plane on top of the same hexagonal lattice. The catalogue is complete, and every entry came from choosing a cell on a lattice, so before we put atoms on these patterns let's pin down what counts as a cell at all, and then when a cell is primitive.

What counts as a unit cell?

A unit cell is a parallelogram built on two lattice vectors: copied along its own two edges it tiles the plane, and every copy contains the lattice points identically.

A shape is a unit cell exactly when its two edges are lattice vectors; only then do its copies tile the plane with every copy holding the lattice points the same way.

A unit cell is a parallelogram built on two lattice vectors.

The cell on a and b: a one step across, b one step up
Both are lattice vectors — each carries a lattice point onto another lattice point

[advance: the copies of the cell fill the patch]

Copy it along its own two edges → no gaps, no overlaps
Every copy holds the points the same way: a point at each of its four corners, nothing else
That sameness is what lets me say points per cell at all

[advance: the failing candidate appears]

Edges 1.5a and b — 1.5a is not a lattice vector
Watch the corners: on a point, then between points, alternating
Inside, the points sit in different places copy to copy → not all alike; not a unit cell

The rule to carry forward is simple: the edges have to be lattice vectors. Once they are, the cell tiles the plane and the only question left is how many lattice points a copy holds, so let's count that on a few legal cells drawn on the same square lattice.

Same lattice, many cells

The copies tile with no gaps and every copy is alike, and the lattice carries one point per original square, so a cell covering $k$ squares holds $k$ lattice points. To count them: a corner point is shared by four cells and counts $\tfrac{1}{4}$; an edge point is shared by two and counts $\tfrac{1}{2}$; a point inside counts 1.

Here is a plain square lattice with the obvious cell drawn on it: the square built on a and b.

Touches lattice points only at its four corners — a corner is shared by four cells, so each gives a quarter
Four quarters = one point → a primitive cell

[advance: the sheared cell appears]

Replace a by a + b, keep b — the top edge slides one lattice step sideways
A shear keeps base and height → the area does not change; still four corner quarters, one point
A patch holding a hundred points is covered by a hundred one-point cells → every primitive cell of a lattice has the same area

[advance: the two-by-one cell]

Two a across, b up: covers two squares — four corner quarters, plus a point on the middle of each long edge, each shared by two cells
One + two halves = two points

[advance: the copies of the two-by-one cell fill the patch]

8 copies cover 16 original squares — no gaps, no overlaps
One point per square → 16 points, shared equally at two per copy
No copy can gain a point without a neighbor losing one → points accumulate exactly as fast as area
So a cell covering k squares holds k points

[advance: the tilted cell]

a + b by a − b: also two squares — four corner quarters, plus one whole point inside, this cell's alone
One + one = two points again
The pattern is exact: squares covered = points held

So the job is to work out that number straight from the coefficients that define the two edges, and here is the picture that gives it to us.

Why the cross-multiplied difference is the area

rectangle

$(u_1 + v_1)(u_2 + v_2) = u_1u_2 + u_1v_2 + u_2v_1 + v_1v_2$

minus the two u-triangles

$2 \times \tfrac{1}{2}\, u_1u_2 = u_1u_2$

minus the two v-triangles

$2 \times \tfrac{1}{2}\, v_1v_2 = v_1v_2$

minus the two corner rectangles

$2\, u_2v_1$

parallelogram

$u_1v_2 + u_2v_1 - 2\, u_2v_1 = u_1v_2 - u_2v_1$

For cell edges $m_1\mathbf a + n_1\mathbf b$ and $m_2\mathbf a + n_2\mathbf b$, the same picture gives the area $\lvert m_1n_2 - m_2n_1 \rvert$, counted in original squares.

Here is a cell drawn on the numbers that define it. Edge u goes u one across, u two up; edge v goes v one across, v two up. Boxed into its smallest rectangle, the bottom is u one plus v one and the side is u two plus v two. The parallelogram and six leftover pieces exactly fill that rectangle: two u-triangles, two v-triangles, two corner rectangles. So: take the rectangle, subtract the six pieces.

[advance: the rectangle's area]

The rectangle first: u one plus v one, times u two plus v two. Multiplied out, four terms: u one u two + u one v two + u two v one + v one v two.

[advance: subtract the two u-triangles]

Each u-triangle is half a u one by u two rectangle; the pair comes to exactly u one u two. Subtract it — the first term is gone.

[advance: subtract the two v-triangles]

The v-triangles work the same way; the pair comes to v one v two. Subtract it — the last term is gone, leaving u one v two plus u two v one.

[advance: subtract the corner rectangles, and the answer appears]

The corners are each u two by v one, so the pair is two u two v one. Take that away: u one v two minus u two v one — the two products, crossed and subtracted, are the area. In lattice language, edges m one a plus n one b and m two a plus n two b give m one n two minus m two n one original squares, in absolute value. So the whole test collapses to four whole numbers and one subtraction, and that is worth writing down as a rule.

The shortcut: judge any cell from four numbers

Write each edge on the old ones, with whole-number coefficients:

$\mathbf a' = m_1\mathbf a + n_1\mathbf b$

$\mathbf b' = m_2\mathbf a + n_2\mathbf b$

The count: the cell covers $k = \lvert m_1n_2 - m_2n_1 \rvert$ original squares and holds $k$ lattice points.

The verdict: whole numbers with $k \ge 1$ make a genuine cell of $k$ points, primitive exactly when $k = 1$. An edge with a coefficient that is not whole is not a lattice vector — no cell, no test. $k = 0$ means the edges are parallel — no parallelogram, no cell.

Cell	Coefficients	Cross-multiplied difference	Points	Verdict
$(2\mathbf a,\ \mathbf b)$	$(2,0),\ (0,1)$	$\lvert 2\cdot 1 - 0\cdot 0 \rvert = 2$	2	not primitive
$(\mathbf a + \mathbf b,\ \mathbf a - \mathbf b)$	$(1,1),\ (1,-1)$	$\lvert 1\cdot(-1) - 1\cdot 1 \rvert = 2$	2	not primitive
$(\mathbf a + \mathbf b,\ \mathbf b)$	$(1,1),\ (0,1)$	$\lvert 1\cdot 1 - 0\cdot 1 \rvert = 1$	1	primitive
$(1.5\mathbf a,\ \mathbf b)$	$(1.5, 0),\ (0,1)$	1.5 is not whole — no test	—	not a cell
$(2\mathbf a + \mathbf b,\ 4\mathbf a + 2\mathbf b)$	$(2,1),\ (4,2)$	$\lvert 2\cdot 2 - 4\cdot 1 \rvert = 0$	0	not a cell

Four whole-number coefficients decide everything: cell or not, $k$ points, primitive exactly when $k = 1$. $k$ is the cell's area in primitive-cell units — Solid State Materials Chemistry §1.1.4 counts the hexagonal setting of R this way, three lattice points for triple the volume, and the same count makes C and I two-point cells and F a four-point cell, kept for symmetry.

Here is the rule. Write each edge of the cell you are proposing in terms of the old ones, with whole-number coefficients: a prime is m one a plus n one b, and b prime is m two a plus n two b.

The count: the cell covers m one n two minus m two n one original squares, in absolute value — and holds that many points
Primitive exactly when that number is one — now run it on the three cells we counted by hand

[advance: the two-by-one cell]

Two a and b: pairs (two, zero) and (zero, one) → two times one minus zero times zero = two — two points, not primitive, matches the count

[advance: the tilted cell]

a plus b and a minus b: pairs (one, one) and (one, minus one) → minus two, size two — not primitive, matches again

[advance: the sheared cell]

a plus b and b: pairs (one, one) and (zero, one) → one — one point, primitive
Three cells, three verdicts, no redrawing

[advance: the cell that is not whole]

1.5a and b, the shape from a few slides ago: 1.5 is not whole → not a lattice vector → no cell, no test

[advance: the parallel edges]

2a + b and 4a + 2b: all whole, but 2·2 − 4·1 = 0 → edges parallel, no parallelogram → not a cell
Four numbers, no drawing: cell or not, k points, primitive iff k = 1
Caution: the test counts points, it does not pick your cell — a larger cell that displays the lattice's symmetry, like the centered rectangle, is kept; it just holds two points
Same story next time in three dimensions: standard cells often centered, two, three, or four points; k = area in primitive units — distinct vs redundant lattices via the centered-square move: a smaller primitive cell generating the same points

Next, let's turn these bare point patterns into real crystals by putting atoms on them.

Lattice + Motif = Crystal Structure

A CuO2 motif (one Cu, two O) beside a square lattice of points with one shaded unit cell.

Place the $\text{CuO}_2$ motif on one lattice point (fixed orientation).
Repeat the same motif on every lattice point.
Draw bonds → a $\text{CuO}_2$ square-net layer (the cuprate plane).

A lattice alone is empty scaffolding. To get a crystal structure you associate a motif, one or more atoms, with every lattice point.

Motif here = CuO₂: Cu on the lattice point, two O attached in fixed orientation (one down, one right)

[advance: place the CuO₂ motif on one lattice point]

Place it on one point first

[advance: repeat the motif on every lattice point]

Same group, same orientation, every point — orientation is part of the motif; identical everywhere or translational symmetry breaks

[advance: draw bonds, the CuO₂ square-net emerges]

Bonds: O bridges Cu; each Cu gets 4 O in a square → the CuO₂ square-net layer

That is the same topology as the CuO₂ planes in cuprate superconductors, which in 3D simply stack these layers.

Lattice + Motif (2nd Example): graphene

A two-carbon motif beside an oblique (hexagonal) lattice with one shaded unit cell.

Motif = 2 carbons at fixed offsets from a lattice point (not on it).
Repeat across the hexagonal lattice → carbons line up.
Each C is 3-coordinate at $120^\circ$ → the graphene honeycomb.

Second example: graphene, from a hexagonal lattice, the triangular net of points, plus a two-atom motif.

[advance: the 2-carbon motif at fixed offsets]

Motif = 2 identical carbons inside the cell, not on the lattice points — one above-left, one below-right

[advance: repeat the motif across the hexagonal lattice]

Attach to every point: a fills a row, b fills the plane → edge-sharing hexagons appear though we never drew one

[advance: the graphene honeycomb, each C 3-coordinate at 120°]

Each carbon 3-coordinate at ~120° — the graphene honeycomb, top view of one graphite layer

The key lesson: the motif need not sit on the lattice points; fixed offsets that repeat with the lattice still generate the correct long-range order.

Practice: identify the unit cell & motif

Two checkerboard patterns of light A and dark B atoms with blank fields for Bravais lattice, atom counts, and empirical formula.

Top: square lattice, A at corners + 1 B at center → $4\times\tfrac14$ A $=1$ A, $1$ B → AB (square Bravais, 1-atom motif).
Bottom: B at center is a different atom → not centered-rectangular; it is primitive rectangular with a 2-atom motif → AB.

Here are two 2D checkerboard patterns with two atom types, A light and B dark.

Task: pick a unit cell, name the Bravais lattice, count the atoms in one cell, write the empirical formula — count corner and edge fractions

[advance: reveal the top pattern solution]

Top: square cell, A corners + B center → 4 × ¼ A + 1 B → AB; square lattice, 1-atom counting per site
The B at the center fails the blindfold test — not the same thing as the corners → not a lattice point, and the square cell stands

[advance: reveal the bottom pattern solution]

Bottom: the center is a different atom (B, not A) → not equivalent to the corners → not centered rectangular
That B is the Subway from the franchise war: it sits on the grid, but it fails the blindfold test, so it is not a lattice point

The Bravais lattice is primitive rectangular, the motif has two atoms, and the formula is again AB.

Summary — Lecture 1: Translational Symmetry in 2D

Translational Symmetry in 3D

Goal

Introduce the 3D unit cell and the classification of Bravais lattices as the foundation for describing all crystal structures.

Learning Objectives

Recognize the 7 lattice systems and their defining features
Distinguish primitive, body-, face-, and base-centered lattices
Understand how 14 Bravais lattices arise from translational symmetry

Based on Solid State Materials Chemistry, Ch. 1 §1.1 (pp. 1–12)

The 3D unit cell

A unit cell is the smallest repeating volume that can be translated through space (by lattice vectors) to build the entire crystal lattice.

In 2D, the unit cell is an area (a parallelogram).
In 3D, it’s a volume (a parallelepiped — six faces, opposite faces parallel, each face a parallelogram).

The seven lattice systems

Seven shapes a 3D unit cell can take. Crystals get their own seven-way grouping — the crystal systems — once we can talk about rotations.

Cyan squares: angles forced to 90°. Colored arcs: the free angles each system leaves open.

The most symmetric case is the cubic system: three lattice vectors of equal length and all right angles.

Cubic: a = b = c, all angles 90°
Tetragonal: keep 90°, let the third edge differ — a = b ≠ c (two square faces, four rectangular)
Orthorhombic: all edges different, still orthogonal — all faces rectangles
Rhombohedral: cube stretched along opposite corners — equal edges, one common angle ≠ 90°, rhombus faces
Hexagonal: 2D hexagonal net + perpendicular third axis — a = b ≠ c, γ = 120°
Monoclinic: α = γ = 90° says b ⟂ both a and c — β (the a–c angle) and all three lengths stay free
Triclinic: all edges and all angles different
These seven are the lattice systems, the shapes a unit cell can take; crystals get their own seven-way grouping, the crystal systems, once we can talk about rotations, and we meet those later

Those seven shapes are only half the story. The other half is where the lattice points sit inside the cell, and that is lattice centering.

One lattice, two cells

The rule: choose a cell that keeps the full symmetry first, the smallest such cell second. Reflected, the rectangle lands on a shifted copy of itself; the parallelogram lands on its red mirror twin, and no shift lays the twin back on the drawn cell.

Before the 3D vocabulary, here is the whole cell-choice problem in a lattice you already know.

The 2D centered rectangular lattice again; dashed lines = its mirror planes
Reflect across any dashed line: every point lands on a point — the symmetry belongs to the dots themselves, no cell drawn yet

[advance: the smallest cell appears]

Smallest cell, 1 point — a slanted parallelogram askew to the mirrors; the shape hides the symmetry

[advance: the centered cell appears]

2 points, twice the area — edges run along the mirrors; the rectangle shows the symmetry

[advance: the rule appears]

The rule: full symmetry first, minimal size second
The test: reflect each cell in one of the lattice’s own mirrors
The rectangle lands on a shifted copy of itself, one cell along the row — it keeps the symmetry
The parallelogram lands on its red mirror twin, the opposite slant
The twin would tile the same lattice, but no shift lays it back on the drawn cell — the parallelogram loses the symmetry

In 3D the symmetry-first cell often carries extra points inside; naming them is lattice centering.

Lattice centering

Lattice centering tells us how many lattice points are inside a chosen unit cell (beyond just the corners).

A primitive (P) unit cell has only corner lattice points, which add up to exactly 1 lattice point per cell.
A centered unit cell has additional lattice points located inside the cell (body center or face centers).

The interior points are ordinary points of the same lattice — the gold-circled point of the last slide. A symmetry-first cell is bigger than minimal, so some lattice points land inside it; centering names where.

Why centered cells at all? A primitive cell always exists, but its shape can hide the lattice’s symmetry; the centered cell keeps the system’s symmetry visible in its shape.

The four centering types: P, I, C, F

Four cell drawings: primitive lattice P with no centering; body-centered lattice I with a point at the cell center (another point at half a + half b + half c for every point); base-centered lattice C with points on one pair of opposite faces (another at half a + half b); face-centered lattice F with points on all six faces (three more at half a + half b, half a + half c, half b + half c).

From lattice to Bravais lattice

A Bravais lattice is the set of points that three primitive lattice vectors reach by integer steps:

$\mathbf{R} = n_1\mathbf{a}_1 + n_2\mathbf{a}_2 + n_3\mathbf{a}_3, \quad n_i \in \mathbb{Z}$

Three vectors generate every point, so one unit cell carries the whole infinite lattice. A cell has just two properties: its shape and its extra lattice points. The shape is the lattice system; the extra points are the centering.

Next we sort the combinations of system and centering; fourteen distinct ones survive, the fourteen Bravais lattices.

Rules for a Bravais lattice

The 14 Bravais lattices are the irreducible building blocks of all 3D crystals.

Translational equivalence

Every lattice point must be identical (equivalent environment) when shifted by an integer combination of lattice vectors.

Minimal cell at equal symmetry

Every centered cell contains smaller primitive cells. If one of the primitive cells keeps the system’s full symmetry, the centering is redundant and thrown out: tetragonal C contains a smaller tetragonal P. If none of them does, the centering is kept: the primitive cells inside cubic F are all rhombohedral, so the F cell is the smallest cell with full cubic symmetry.

Consistency with lattice system symmetry

The centering must keep the symmetry of its lattice system; cubic, for example, requires all edges equal and all angles 90°. A centering that breaks that symmetry is not a distinct lattice of the system.

The 14 Bravais lattices are the irreducible building blocks of all 3D crystals. Three rules decide the list.

Translational equivalence: every lattice point identical under integer shifts
Minimal cell: every centered cell contains smaller primitive cells — does any of them keep the system’s symmetry?
Yes → redundant, out (tetragonal C contains a smaller tetragonal P); none does → kept (cubic F: primitive cells all rhombohedral, so F is the smallest cell with full cubic symmetry)
System symmetry: the centering must keep the lattice system's symmetry (cubic: all edges equal, all angles 90°) → a symmetry-breaking centering is not a distinct lattice of the system

Let's apply those rules and watch twenty-eight candidates collapse to fourteen.

From 28 candidates to 14

7 lattice systems × 4 centerings (P, C, I, F) = 28 candidates. Gold dashes: rotation axes. Gold points: the centering.

Triclinic

a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°

no rotation symmetry

With no symmetry to protect, a smaller slanted cell always exists: every centered candidate is just a bigger drawing of P.

Monoclinic

a ≠ b ≠ c, α = γ = 90°, β ≠ 90°

180° turn about b

The turn pins only the b edge; drawing new a and c edges through the same points turns I and F into C.

Monoclinic: the drawn proof

F → base-centered at half the size: C

I → base-centered at the same size: C

The gold dashed edges are new a and c edges drawn through the same points; b never moves, so every redrawn cell is still monoclinic. In both redrawn cells the extra points land on one pair of faces (shaded), not inside: the cells are base-centered, the type written C by convention. Left: the F cell redraws at half the size. Right: the I cell redraws at the same size, its old body center now a face point. So I and F were C all along.

Orthorhombic

a ≠ b ≠ c, all angles 90°

180° turns about a, b, c

The three turns force every edge to stay along a, b, c, and the 45° redraw would need a = b: nothing redraws, so P, C, I, F are four different lattices.

Tetragonal

a = b ≠ c, all angles 90°

90° turn about c

a = b allows the 45° redraw of the square grid: C becomes a smaller P, and F becomes a smaller I.

Tetragonal: the 45° redraw

C → gold points become its corners: a smaller P

F → rings at c/2 become its body center: a smaller I

Looking down c: because a = b, the dashed 45° square is a legal tetragonal cell with half the area. C’s gold points become its corners, a smaller P. F’s remaining points (rings, at height c/2) land at the dashed cell’s body center, a smaller I.

Cubic

a = b = c, all angles 90°

120° turns about all 4 diagonals

C is broken, not redundant: its face points spoil the diagonal turns, leaving a tetragonal lattice. I and F keep every turn.

Cubic C: broken, not redundant

the 120° turn finds no partner points

the cell that fits: base a/√2, height a → tetragonal P

Left: a symmetry of the lattice must land points on points, and the 120° turn about the body diagonal sends the gold face points to empty faces (dashed rings): the diagonal turns are gone, so this point set is not cubic. Right: what it is instead. The gold cell fits the same points exactly — its base is the 45° square through the face points, edge a′ = a/√2, and its height is the full edge c = a. Base and height unequal, one 90° turn kept: primitive tetragonal, a row the table already has.

Hexagonal

a = b ≠ c, α = β = 90°, γ = 120°

60° turn about c

Any added point either spoils the 60° turn or redraws as a smaller hexagonal P.

Rhombohedral

a = b = c, α = β = γ ≠ 90°

120° turn about the diagonal

Every centered candidate redraws as a smaller rhombohedron. The survivor’s letter is R, a centering letter it earns from its hexagonal redraw, not from this cell, which has no interior points.

Count the survivors: 1 + 2 + 4 + 2 + 3 + 1 + 1 = 14 Bravais lattices.

Twenty-eight candidates, and every cut is a symmetry argument: a centering is either redundant (the same points redraw with a smaller cell of equal symmetry) or broken (the added points spoil the system's turns).

[advance: triclinic]

Triclinic: no symmetry to protect → a smaller slanted cell always exists → every centered candidate = a bigger drawing of P

[advance: monoclinic]

Monoclinic: the turn pins only b; new a, c edges through the same points turn I and F into C → P, C

[advance: the monoclinic proof]

Gold edges = the new a, c through the same points; b untouched, every redrawn cell still monoclinic
Both redraws land the points on one face pair (shaded), not inside → base-centered = C by convention
F → C at half the size; I (via c′ = a + c) → C at the same size, the body center now a face point

[advance: orthorhombic]

Orthorhombic: turns force every edge along a, b, c; the 45° redraw would need a = b → nothing redraws → P, C, I, F all different

[advance: tetragonal]

Tetragonal: a = b allows the 45° redraw of the square grid → C = smaller P, F = smaller I → P, I

[advance: the tetragonal proof]

Down c: the dashed 45° square = a legal tetragonal cell, half the area (a = b makes it legal)
C’s gold points → its corners (P); F’s half-height rings → its body center (I)

[advance: cubic]

Cubic: C is broken, not redundant — face points spoil the diagonal turns → tetragonal lattice; I and F keep every turn → P, I, F

[advance: the cubic proof]

Left: a lattice symmetry must land points on points — the 120° diagonal turn sends the gold points to empty faces (dashed rings) → not cubic
Right: the gold cell fits the same points — base = the 45° square through the face points (a′ = a/√2), height = the full edge a → base ≠ height → tetragonal P

[advance: hexagonal]

Hexagonal: any added point spoils the 60° turn or redraws as a smaller hexagonal P → P only

[advance: rhombohedral]

Rhombohedral: every centered candidate redraws as a smaller rhombohedron → R
The letter R comes from the hexagonal redraw of this same lattice, which we draw shortly; the rhombohedron itself has no interior points

[advance: the tally]

Count the survivors: 1+2+4+2+3+1+1 = 14

We derived fourteen lattices from seven cell shapes. Crystals themselves get grouped a different way, by the rotations they support — that is how we sorted the 2D patterns into four systems — and that grouping does not line up one-to-one with the lattices. You have seen the mismatch before, in two dimensions.

Systems and lattices already split in 2D

A crystal system groups patterns by their rotational symmetry. A Bravais lattice is a distinct repeat pattern. 2D has 4 systems but 5 lattices.

2D crystal systemIts Bravais lattice(s)

Obliqueoblique

Rectangularrectangular P or centered rectangular

Squaresquare

Hexagonalhexagonal

One system owning two lattices is not a 3D oddity; you met it in Lecture 1.

In 3D only the trigonal system owns two lattices; the full map is on the next slide.

You have seen a system and a lattice disagree before, in two dimensions.

Last lecture: the 2D patterns sorted into four crystal systems by the rotations they have — oblique, rectangular, square, hexagonal
But five Bravais lattices, one more than there are systems
The extra one is centered rectangular: same rotation and same mirrors as plain rectangular, so it belongs to the rectangular system
A separate lattice: the redraw that made tetragonal C redundant fails — with a ≠ b the cell through the center points is a rhombus, never a rectangle, so its points are a genuinely different repeat pattern
The two nouns: a crystal system is a family of crystals that share their rotational symmetry; a Bravais lattice is a distinct pattern of repeat points
So one system can own more than one lattice, and in two dimensions the rectangular system already does

In three dimensions only the trigonal system owns two lattices; next we map which system uses which of the fourteen.

7 crystal systems, 14 Bravais lattices

Crystal systemDefining rotationIts Bravais lattice(s)

TriclinicnoneP

Monoclinicone 180° turnP, C

Orthorhombicthree 180° turnsP, C, I, F

Tetragonalone 90° turnP, I

Cubicfour 120° turnsP, I, F

Hexagonalone 60° turnhexagonal P

Trigonalone 120° turnhexagonal P or rhombohedral R

Hexagonal P serves two systems, and trigonal is the only system that picks between two lattices. “R-centered hexagonal” (drawn two slides ahead) is the rhombohedral R lattice in a hexagonal-shaped cell; it is a different lattice from hexagonal P, and the count stays 14.

Quartz, P3₂21 — trigonal, on the hexagonal P lattice

Calcite, R3̄c — trigonal, on the rhombohedral R lattice

Graphite, P6₃/mmc — hexagonal, on the hexagonal P lattice

Trigonal names a symmetry, not a cell shape: its crystals turn 3-fold at most and come in two cell shapes, hexagonal P and rhombohedral R. So no trigonal cell appeared among the lattice systems, and there is a rhombohedral lattice but no rhombohedral crystal system.

We have fourteen lattices and seven systems, and the two lists do not line up one to one, so here is which system uses which lattice.

Five systems are simple: triclinic P; monoclinic P, C; orthorhombic P, C, I, F; tetragonal P, I; cubic P, I, F — twelve lattices, each owned by one system
The interesting corner is the bottom two rows: a crystal is hexagonal if it has a six-fold turn and trigonal if its highest turn is three-fold, and both of them can sit on the hexagonal P lattice
Trigonal crystals split: those whose space group starts with P use hexagonal P; those whose space group starts with R use the rhombohedral lattice — that R is the fourteenth
You will also see that lattice written as R-centered hexagonal: the same lattice in a different cell, drawn two slides ahead — an alias, not a fifteenth entry
Quartz is P3₂21: trigonal, on hexagonal P. Calcite is R3̄c: trigonal, on R. Graphite is P6₃/mmc: hexagonal, on hexagonal P
Trigonal is a symmetry class, not a cell shape: a crystal is trigonal when its highest turn is 3-fold, and it sits on either of two cells — the lattice-systems slide listed cell shapes, so a rhombohedron appeared there and no trigonal cell did
Why the rhombohedral lattice belongs to the trigonal crystal system: every rotation a crystal has must also be a rotation of its Bravais lattice; the rhombohedral lattice allows at most a 3-fold turn, so only trigonal crystals fit on it — hexagonal P carries a 6-fold, and a 3-fold fits inside a 6-fold, so it hosts both systems
The vocabulary follows from that: trigonal, tetragonal, and hexagonal name the turns a crystal has; rhombohedral names the lattice whose cell is a rhombohedron

Next we draw one of the fourteen, the FCC lattice, and apply the cell-choice rule to it.

One lattice, two cells: FCC

Both drawings hold the same points. The cube's shape shows the cubic symmetry; the primitive cell's shape does not. Crystallography chooses the cell that shows the full symmetry first, and the smallest such cell second. That primitive cell is a rhombohedron, and the lattice is still cubic F: a rhombohedron-shaped cell does not make a rhombohedral lattice, because symmetry classifies a lattice, not cell shape.

Both drawings show the same points, the FCC lattice; only the choice of cell differs.

Left, conventional cell: corners + gold face centers, 8 × ⅛ + 6 × ½ = 4 points; the cube shape shows the full cubic symmetry
Right, a primitive cell through the same points (a corner + the three nearest face centers, dashed gold): 1 point, ¼ the volume, edges at 60°; legal, but nothing in its shape says cubic
The 2D cell-choice slide, now in 3D: full symmetry first, smallest such cell second → keep the 4-point cube; we accept the extra points to keep the symmetry visible
So which lattice is that rhombohedron-shaped cell? Still cubic F — at exactly α = 60° the rhombohedral shape IS the cubic F lattice; symmetry classifies a lattice, cell shape does not

That rhombohedron takes us to the one lattice that is normally drawn two ways, the rhombohedral one.

R centering: one lattice, two cells

The gold rhombohedron is the same cell in the first two drawings; the third is the conventional R-centered hexagonal cell, ⅓ of the prism. The interior points break the 6-fold the empty prism would have — only the 3-fold survives — so both cells describe the rhombohedral R lattice from the fourteen, not hexagonal P.

R centering is not a new lattice: it is the hexagonal description of the rhombohedral lattice we just counted.

Left, the primitive rhombohedral cell: eight corners × ⅛ = 1 point, primitive vectors a_R, b_R, c_R arrowed, and the 3-fold axis runs down its body diagonal, drawn dashed gold
Middle, the same gold rhombohedron inside a hexagonal prism of the lattice: its body diagonal is the c axis, because the three primitive edges add up to c
Its three lower vertices are the (⅔, ⅓, ⅓)-type points and its three upper the (⅓, ⅔, ⅔)-type: the gold interior points ARE the rhombohedron’s corners
Right, one conventional cell on its own (a = b, γ = 120°): one third of the prism, so three times the rhombohedron’s volume and 3 points — corners plus the two labelled interior points
Why this is the same lattice and not a fifteenth: symmetry belongs to the points, not the box — a 60° turn about c moves (⅔, ⅓, ⅓) to (⅓, ⅔, ⅓), an empty site, so the centering kills the empty prism’s 6-fold and only the 3-fold survives, exactly the rhombohedron’s symmetry
Why we prefer the hexagonal setting: the 3-fold axis lands on c, which matches every other trigonal and hexagonal drawing, and the 120° base is easy to read
Who uses it: the trigonal R space groups — calcite is R3̄c; trigonal P structures like quartz stay on the hexagonal P lattice
Warning: redrawing changes nothing physical → a rhombohedron-shaped cell does NOT make the lattice rhombohedral; at exactly 60° a rhombohedral cell’s lattice IS the cubic F one we just drew; classify by the symmetry the points have

Let’s pull the whole 3D story together.

Summary — Translational Symmetry in 3D

Key points

A 3D crystal is built by repeating a unit cell (a parallelepiped) through translational symmetry.
Seven lattice systems: cubic, tetragonal, orthorhombic, rhombohedral, hexagonal, monoclinic, triclinic.
Centering types: primitive (P), body-centered (I), face-centered (F), base-centered (A, B, C).
Combining systems + centering gives the 14 Bravais lattices, the complete set of 3D lattice types.
This framework underpins all crystallography: every crystal structure belongs to one of these lattices.

Homework

Work the Lecture 1 practice questions before the next class. They cover the 2D half — lattices and unit cells, the allowed rotations, the five 2D Bravais lattices, and lattice + motif — and the 3D half: the 3D unit cell, the seven lattice systems, lattice centering, and the 14 Bravais lattices.

Open the Lecture 1 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.