MTH3008 Tensor Analysis - Final Exam Cheat Sheet

Created by William Fayers

Good luck and have fun!! :)) - identify question type via §8, jump to section, apply the recipe, then sanity-check. Ez 100%.

§1. Suffix-notation Foundations

Indices

Free index: appears once per term, labels a component, fixes rank.
Dummy index: appears exactly twice in a term; summed $1, 2, 3$ (Einstein convention).
Rules: same free indices on every term; no index appears $> 2$ times; fresh letters for independent sums.

Kronecker delta $δ_{ij}$

Definition: $1$ if $i = j$ , else $0$ (identity matrix).
Symmetric: $δ_{ij} = δ_{ji}$ .
Isotropic: $δ_{ij}^{'} = L_{ik} L_{jk} = δ_{ij}$ - same in every frame.
Substitution: $δ_{ij} a_{j} = a_{i}$ .
Chain: $δ_{ij} δ_{jk} = δ_{ik}$ .
Trace: $δ_{ii} = 3$ .
Cartesian Jacobian: $\partial x_{j} / \partial x_{i} = δ_{ij}$ .

Alternating tensor $ϵ_{ijk}$

Definition: $+ 1$ even perm of $(1, 2, 3)$ ; $- 1$ odd; $0$ if any two equal.
Cyclic (sign-preserving): $ϵ_{ijk} = ϵ_{jki} = ϵ_{kij}$ .
Swap (sign-flipping): $ϵ_{ijk} = - ϵ_{jik}$ .
Totally antisymmetric.

$ϵ$ – $δ$ identity (shared index in position 1 of both $ϵ$ ; cyclically rotate first)

$ϵ_{ijk} ϵ_{i ℓ m} = δ_{j ℓ} δ_{km} - δ_{jm} δ_{k ℓ}$
Corollaries: $ϵ_{ijk} ϵ_{ij ℓ} = 2 δ_{k ℓ}$ (set $m = j$ ); $ϵ_{ijk} ϵ_{ijk} = 6$ (set $ℓ = k$ ); $δ_{ij} ϵ_{ijk} = 0$ .

Symmetry / antisymmetry

Symmetric: $S_{ij} = S_{ji}$ . Antisymmetric: $A_{ij} = - A_{ji}$ (forces diagonal $= 0$ ).
Decomposition: $T_{ij} = \frac{1}{2} (T_{ij} + T_{ji}) + \frac{1}{2} (T_{ij} - T_{ji})$ .
Frame-independent in Cartesian.
Forced-zero chain: if $T_{ijk}$ is sym in $(j, k)$ AND antisym in $(i, j)$ , then $T_{ijk} = T_{ikj} = - T_{kij} = - T_{kji} = T_{jki} = T_{jik} = - T_{ijk} \Rightarrow T_{ijk} = 0$ .

Kill rule - symmetric × antisymmetric (over contracted pair) $= 0$ .

$ϵ_{ijk} S_{jk} = 0$ if $S_{jk} = S_{kj}$ .
$ϵ_{ijk} a_{j} a_{k} = 0$ .
$ϵ_{ijk} \partial_{j} \partial_{k} (anything) = 0$ (mixed partials commute).

§2. Suffix ↔ Vector

Dictionary ( $\partial_{i} := \partial / \partial x_{i}$ )

Vector	Suffix	Vector	Suffix
$a \cdot b$	$a_{i} b_{i}$	$\nabla f$	$\partial_{i} f$
$(a \times b)_{i}$	$ϵ_{ijk} a_{j} b_{k}$	$\nabla \cdot u$	$\partial_{i} u_{i}$
$a \cdot (b \times c)$	$ϵ_{ijk} a_{i} b_{j} c_{k}$	$(\nabla \times u)_{i}$	$ϵ_{ijk} \partial_{j} u_{k}$
$(A B)_{ij}$	$A_{ik} B_{kj}$	$\nabla^{2} f$	$\partial_{j} \partial_{j} f$
$(A^{T})_{ij} = A_{ji}$	swap	Position $r$	$r_{i} = x_{i}$ , $r = x_{j} x_{j}$ , $\partial_{i} r = x_{i} / r$
$Tr (A) = A_{ii}$		$\nabla f (r)$	$f^{'} (r) x_{i} / r$
$det M$	$ϵ_{ijk} M_{1 i} M_{2 j} M_{3 k}$	$\partial_{k} x_{i}$	$δ_{ki}$ (Cartesian Jacobian)

Computing $a \times b$ and $3 \times 3$ determinant (cofactor expansion along row 1)

$a \times b = det i a_{1} b_{1} j a_{2} b_{2} k a_{3} b_{3} = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1})$ . Cover-row mnemonic: $i$ th component = $2 \times 2$ det of the matrix with row $i$ deleted, with alternating sign $+, -, +$ .
$det M = M_{11} M_{22} M_{32} M_{23} M_{33} - M_{12} M_{21} M_{31} M_{23} M_{33} + M_{13} M_{21} M_{31} M_{22} M_{32}$ where $a c b d = a d - b c$ . Sign pattern $+, -, +$ along the row.
Sarrus (alternative for $3 \times 3$ only): copy first 2 columns to the right; $det M =$ (sum of three down-right diagonals) $-$ (sum of three up-right diagonals).
Scalar triple identity: $a \cdot (b \times c) = det a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3}$ (rows = $a, b, c$ ).

Vector → suffix (in order)

Pick free index $i$ matching the rank.
Attach $i$ to every vector factor (same letter every term).
Fresh dummy for each scalar (dot products, scalar coefficients).
$\times \to ϵ_{ijk}$ ; matrix product $\to A_{ik} B_{kj}$ .

Example. $a \times b + (a \cdot d) c = e \Rightarrow ϵ_{ijk} a_{j} b_{k} + (a_{ℓ} d_{ℓ}) c_{i} = e_{i}$ .

Suffix → vector (in order)

Collapse $δ$ by substitution.
Cyclically rotate each $ϵ$ to recognise $ϵ_{ijk} a_{j} b_{k} = (a \times b)_{i}$ .
Bundle dummy pairs as dot products: $a_{j} b_{j} = a \cdot b$ .
Reassemble using the free index as the result’s slot.

Example. $δ_{ij} c_{j} + ϵ_{kji} a_{k} b_{j} = d_{ℓ} e_{m} c_{i} b_{ℓ} c_{m}$ .

LHS: $δ_{ij} c_{j} = c_{i}$ ; $ϵ_{kji} = ϵ_{ikj}$ (cyclic) $\Rightarrow ϵ_{ikj} a_{k} b_{j} = (a \times b)_{i}$ .
RHS: $(d_{ℓ} b_{ℓ}) (e_{m} c_{m}) c_{i} = (d \cdot b) (e \cdot c) c_{i}$ .
$\Rightarrow c + a \times b = (d \cdot b) (e \cdot c) c$ .

Differentiating in suffix - product rule + $\partial_{k} x_{i} = δ_{ki}$ .

Example. $\partial_{k} (a_{ij} x_{i} x_{j})$ with $a_{ij} = a_{ji}$ constant:

Pull out constant: $= a_{ij} \partial_{k} (x_{i} x_{j})$ .
Product rule on $x_{i} x_{j}$ : $= a_{ij} [(\partial_{k} x_{i}) x_{j} + x_{i} (\partial_{k} x_{j})] = a_{ij} [δ_{ki} x_{j} + x_{i} δ_{kj}]$ .
Collapse $δ$ s: $= a_{kj} x_{j} + a_{ik} x_{i}$ .
Use symmetry $a_{ik} = a_{ki}$ and relabel $i \to j$ : $= a_{kj} x_{j} + a_{kj} x_{j} = 2 a_{kj} x_{j}$ .

§3. Vector Calculus Identities (suffix proofs)

Identity	Proof sketch
$\nabla \times (\nabla f) = 0$	$ϵ_{ijk} \partial_{j} \partial_{k} f = 0$ - kill ( $ϵ$ antisym, $\partial\partial$ sym in $j, k$ )
$\nabla \cdot (\nabla \times u) = 0$	$ϵ_{ijk} \partial_{i} \partial_{j} u_{k} = 0$ - kill in $i, j$
$\nabla \cdot (f u) = f \nabla \cdot u + u \cdot \nabla f$	$\partial_{i} (f u_{i}) = f \partial_{i} u_{i} + u_{i} \partial_{i} f$
$\nabla \times (f u) = f \nabla \times u + \nabla f \times u$	$ϵ_{ijk} \partial_{j} (f u_{k}) = f ϵ_{ijk} \partial_{j} u_{k} + ϵ_{ijk} (\partial_{j} f) u_{k}$
$\nabla \times (β \nabla β) = 0$	Product rule: $ϵ_{ijk} \partial_{j} (β \partial_{k} β) = ϵ_{ijk} (\partial_{j} β) (\partial_{k} β) + β ϵ_{ijk} \partial_{j} \partial_{k} β$ . Both kill ( $(\partial_{j} β) (\partial_{k} β)$ sym in $(j, k)$ ; $\partial_{j} \partial_{k} β$ sym).
$u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$	$ϵ_{ijk} u_{j} ϵ_{k ℓ m} v_{ℓ} w_{m} = ϵ_{kij} ϵ_{k ℓ m} \dots$ apply $ϵϵ$
$(u \times v) \cdot (w \times z) = (u \cdot w) (v \cdot z) - (v \cdot w) (u \cdot z)$	$ϵ_{ijk} ϵ_{i ℓ m} u_{j} v_{k} w_{ℓ} z_{m}$ apply $ϵϵ$
$(u \times (v \times w)) \cdot z = (u \cdot w) (v \cdot z) - (u \cdot v) (w \cdot z)$	BAC-CAB then dot $z$ (pull scalar prefactors first)
$\nabla f (r) = f^{'} (r) r / r$	$\partial_{i} f (r) = f^{'} (r) \partial_{i} r = f^{'} (r) x_{i} / r$
$det A = 0$ if two rows equal	$det A = ϵ_{ijk} A_{1 i} A_{2 j} A_{3 k}$ ; sym $A_{1 i} A_{1 j} \times$ antisym $ϵ_{ijk} \Rightarrow 0$

§4. Cartesian Transformations

Basis axioms

Basis of $R^{3}$ = three linearly independent vectors spanning the space.
Orthogonal: $e_{i} \cdot e_{j} = 0$ for $i \neq = j$ .
Orthonormal: orthogonal + $∣ e_{i} ∣ = 1$ .
Global basis: same vectors at every point (Cartesian, rigid).
Local basis: vectors vary with position (curvilinear - §7).

Rotation matrix $L_{ij}$ (between two orthonormal bases)

Definition: $L_{ij} = e_{i}^{'} \cdot e_{j}$ .
Basis: $e_{i}^{'} = L_{ij} e_{j}$ .
Coordinates: $x_{i}^{'} = L_{ij} x_{j}$ ; inverse $x_{i} = L_{ji} x_{j}^{'}$ .
Jacobian: $\partial x_{i}^{'} / \partial x_{j} = L_{ij}$ , $\partial x_{i} / \partial x_{j}^{'} = L_{ji}$ .
Chain rule: $\partial / \partial x_{i}^{'} = L_{ij} \partial / \partial x_{j}$ .

Properties

Orthogonality (two forms, both used to collapse contracted $L$ s):
- $L_{ij} L_{kj} = δ_{ik}$ - sum on the second index ( $L L^{T} = I$ ).
- $L_{ji} L_{jk} = δ_{ik}$ - sum on the first index ( $L^{T} L = I$ ). Equivalently $L_{ja} L_{jb} = δ_{ab}$ with dummy first.
$L^{- 1} = L^{T}$ .
$∣ det L ∣ = 1$ ; proper rotation $\Rightarrow det L = + 1$ .

Tensor transformation rule (Cartesian rank- $r$ - one $L$ per free index)

$T_{i_{1} \dots i_{r}}^{'} = L_{i_{1} j_{1}} \dots L_{i_{r} j_{r}} T_{j_{1} \dots j_{r}}$
Rank 0 (scalar): $T^{'} = T$ .
Rank 1 (vector): $a_{i}^{'} = L_{ij} a_{j}$ .
Rank 2: $T^{'} = L T L^{T}$ .
Rank = number of free indices; $3^{r}$ components in 3D.

Recipe - prove $X$ is a tensor

Replace each factor in $X^{'}$ by its primed form, fresh dummy per factor.
Group all $L$ s on the left, original tensors on the right.
Collapse paired $L$ s via orthogonality (above): $L_{ka} L_{k c} = δ_{a c}$ (dummy = first index of both).
Use $δ$ s to contract dummies on the originals.
Count surviving $L$ s: one per free index $\Rightarrow$ tensor of that rank.

Example - $T_{ijij}$ scalar:

$T_{ijij}^{'} = L_{ia} L_{jb} L_{i c} L_{j d} T_{ab c d} = δ_{a c} δ_{b d} T_{ab c d} = T_{abab}$ .

Example - $(u \cdot \nabla) u$ vector (suffix $u_{j} \partial_{j} u_{i}$ ):

$u_{j}^{'} \partial_{j}^{'} u_{i}^{'} = (L_{ja} u_{a}) (L_{jb} \partial_{b}) (L_{i c} u_{c}) = L_{i c} δ_{ab} (orthogonality) L_{ja} L_{jb} u_{a} \partial_{b} u_{c} = L_{i c} (u_{a} \partial_{a} u_{c})$ .
One $L_{i c}$ on free $i$ $\Rightarrow$ vector ✓.

Show $X$ is NOT a tensor - derive transformation, identify junk term no rule produces.

E.g. $\partial V_{i} / \partial x^{j}$ : $\partial_{j^{'}} V_{i^{'}}^{'} = L^{k}_{i^{'}} L^{ℓ}_{j^{'}} \partial_{ℓ} V_{k} + V_{k} \partial_{j^{'}} L^{k}_{i^{'}}$ . First term obeys cov rank-2 law; second is junk from non-constant $L$ .

§5. Generalised Coords: Dual Basis, Components, Rank-2 Transforms

Two $L$ flavours (index position matters now)

$L^{j}_{i^{'}} = \partial x^{j} / \partial x^{' i}$ - matches lower indices.
$L^{i^{'}}_{j} = \partial x^{' i} / \partial x^{j}$ - matches upper indices.
Inverse identity: $L^{i}_{k^{'}} L^{k^{'}}_{j} = δ_{j}^{i}$ (chain rule on identity $x^{i} (x^{'} (x))$ ).

General transformation rule - one $L$ per free index, kind matched to its level.

Each upper free index gets $L^{i^{'}}_{a}$ . Each lower gets $L^{a}_{i^{'}}$ .
$V_{i^{'}}^{'} = L^{k}_{i^{'}} V_{k}$ ; $V^{' i^{'}} = L^{i^{'}}_{k} V^{k}$ ; $T^{' ij}_{k} = L^{i^{'}}_{a} L^{j^{'}}_{b} L^{c}_{k^{'}} T^{ab}_{c}$ .

Dual basis - unique $e^{i}$ with $e^{i} \cdot e_{k} = δ_{k}^{i}$ .

Cross-product formula: $e^{i} = (e_{j} \times e_{k}) / V$ , where $V = e_{1} \cdot (e_{2} \times e_{3})$ , $(i, j, k)$ cyclic.
Original from dual: $e_{i} = g_{ij} e^{j}$ (or same recipe with $V^{'} = 1/ V$ ).
Orthonormal $\Rightarrow e^{i} = e_{i}$ .
Always verify: $e^{i} \cdot e_{k} = δ_{k}^{i}$ on the diagonal.
Cross product formula: see §2 (cover-row / cofactor expansion).

Vector expansion - $A = A^{i} e_{i} = A_{i} e^{i}$ .

Contravariant (in original basis): $A^{i} = A \cdot e^{i}$ .
Covariant (in dual basis): $A_{i} = A \cdot e_{i}$ .
Orthogonal shortcut: $A_{i} = g_{ii} A^{i}$ (no sum).

Metric tensor

Covariant: $g_{ij} = e_{i} \cdot e_{j}$ .
Contravariant: $g^{ij} = e^{i} \cdot e^{j} = [g_{ij}]^{- 1}$ (matrix inverse).
Mixed: $g^{i}_{j} = δ_{j}^{i}$ - just relabels, never raises/lowers.
Symmetric, positive on diagonal.
Orthogonal coord $\Rightarrow g_{ij}$ diagonal, $g^{ii} = 1/ g_{ii}$ (no sum).

Raise / lower - only fully cov/contra metrics do real work.

$A_{i} = g_{ik} A^{k}$ ; $A^{i} = g^{ik} A_{k}$ .
Inner product: $A \cdot B = A^{i} B_{i} = A_{i} B^{i} = g_{ik} A^{i} B^{k} = g^{ik} A_{i} B_{k}$ .

Four flavours of rank-2 - all numerically equal in Cartesian $K$ (since $g = δ$ ); differ in $K^{'}$ .

$A_{ik} = g_{i ℓ} g_{km} A^{ℓ m}$
$A^{ik} = g^{i ℓ} g^{km} A_{ℓ m}$
$A^{i}_{k} = g^{i ℓ} A_{ℓ k}$
$A_{i}^{k} = g^{k ℓ} A_{i ℓ}$

Transformation of rank-2 - suffix + matrix form.

Rows of $L = [L^{j}_{i^{'}}]$ = new basis $e_{i}^{'}$ in Cartesian (read off directly).
Rows of $M^{*} = [L^{i^{'}}_{j}]$ = new dual $e^{' i}$ in Cartesian.

Want	Suffix	Matrix
$P_{i^{'} k^{'}}^{'}$ (cov)	$L^{ℓ}_{i^{'}} L^{m}_{k^{'}} P_{ℓ m}$	$L P L^{T}$
$P^{' i^{'} k^{'}}$ (contra)	$L^{i^{'}}_{ℓ} L^{k^{'}}_{m} P^{ℓ m}$	$M^{} P (M^{})^{T}$
$P^{' \cdot k^{'}}_{i^{'}}$ (mixed)	$L^{ℓ}_{i^{'}} L^{k^{'}}_{m} P^{\cdot m}_{ℓ}$	$L P (M^{*})^{T}$
$P^{' i^{'}}_{\cdot k^{'}}$ (mixed)	$L^{i^{'}}_{ℓ} L^{m}_{k^{'}} P^{ℓ}_{\cdot m}$	$M^{*} P L^{T}$

Prefer the matrix form; never invert $g_{i^{'} k^{'}}$ unless asked.

§6. Tensor Algebra

Op	Definition	Rank
Addition	$(A + B)_{ik} = A_{ik} + B_{ik}$ , same rank & structure	preserves
Outer product	$(A \otimes B)_{ij \dots k ℓ \dots} = A_{ij \dots} B_{k ℓ \dots}$	sums
Contraction	set two indices equal (curvilinear: 1 upper + 1 lower)	drops by 2
Inner product	outer + contract	net

Mismatched structure (e.g. $A_{i} + B^{i}$ ) is NOT a tensor.

Outer product of matrices - $M$ ( $m_{1} \times m_{2}$ ) $\otimes$ $N$ ( $n_{1} \times n_{2}$ ) = $(m_{1} n_{1}) \times (m_{2} n_{2})$ block; entry $M_{ij} \to$ block $M_{ij} \cdot N$ .

Example. $M = (- 1 0 11), N = (11 - 1 2) \Rightarrow M \otimes N = - 1 - 1 00 1 - 2 00 1111 - 1 2 - 1 2$ .

Contraction - set two indices of one tensor equal; rank drops by 2 (curvilinear: pair 1 upper + 1 lower).

Trace (rank 2 $\to$ scalar): $T_{ii} = T_{11} + T_{22} + T_{33}$ . E.g. $T = 147258369 \Rightarrow T_{ii} = 1 + 5 + 9 = 15$ .
Rank 3 $\to$ rank 1: $A_{iji}$ sums over $i$ (free index $j$ ). Curvilinear analogue: $A^{i}_{ji}$ or $A^{i}_{ij}$ .
Sensitivity to which indices: $B_{i} A_{ij}$ vs $B_{j} A_{ij}$ differ. $B_{i} A_{ij}$ (dummy = first index $i$ → sums down columns of $A$ ) $= B^{T} A$ ; $B_{j} A_{ij}$ (dummy = second index $j$ → sums across rows of $A$ ) $= A B$ . E.g. $A = 131000214, B = (1, 1, 0)$ : $B_{i} A_{ij} = (1 + 3 + 0, 0, 2 + 1 + 0) = (4, 0, 3)$ ; $B_{j} A_{ij} = (1 + 0 + 0, 3 + 0 + 0, 1 + 0 + 0) = (1, 3, 1)$ .

Inner product - outer product then contract; net rank = (sum of ranks) $- 2$ .

$A_{i} B_{i} = A \cdot B$ (rank $1 + 1 \to 0$ ). E.g. $A = (1, 2, 3), B = (4, - 1, 2) \Rightarrow A_{i} B_{i} = 4 - 2 + 6 = 8$ .
$T_{ik} A_{k} = T A$ (rank $2 + 1 \to 1$ , free $i$ ). E.g. $T = 101 02 - 1 214, A = (1, - 1, 1) \Rightarrow T_{ik} A_{k} = (3, - 1, 6)$ .
$T_{ik} A_{k} B_{i} = B^{T} T A$ (rank $2 + 1 + 1 \to 0$ , scalar). With $B = (1, - 2, 0)$ : $B_{i} T_{ik} A_{k} = 3 - (- 2) + 0 = 5$ .
$A_{ij} B^{ij}$ (rank $2 + 2 \to 0$ ): sum of entrywise products.

Contraction of two tensors is a tensor

Example. $A_{ij} B_{ik ℓ}$ rank-3: $A_{ij}^{'} B_{ik ℓ}^{'} = L_{jb} L_{k d} L_{ℓ e} δ_{a c} A_{ab} B_{c d e} = L_{jb} L_{k d} L_{ℓ e} (A_{c b} B_{c d e})$ - one $L$ per surviving free index ✓.

Symmetry from $ϵ$

$A_{ijk} = ϵ_{ij ℓ} B_{k ℓ}$ : swap $i \leftrightarrow j$ gives $A_{jik} = - A_{ijk}$ - antisym in $(i, j)$ .
Other pairs need direct check.

Levi-Civita inversion - $B_{ijk} = ϵ_{ij ℓ} A_{ℓ k}$ .

Multiply by $ϵ_{ijm}$ , apply $ϵ_{ijm} ϵ_{ij ℓ} = 2 δ_{m ℓ}$ .
$\Rightarrow A_{mk} = \frac{1}{2} ϵ_{ijm} B_{ijk}$ .

Quotient rule - if $T_{ij} b_{j}$ is a tensor for every vector $b$ , then $T_{ij}$ is a tensor.

§7. Curvilinear / Local-basis Geometry

Coordinate basis

$e_{i} = \partial r / \partial x^{i}$ - tangent to $x^{i}$ -coordinate curve; varies with position.

Metric

$g_{ij} = e_{i} \cdot e_{j}$ , symmetric.
Orthogonal coord $⟺$ off-diagonals all zero.
Arc length: $(d s)^{2} = g_{ij} d x^{i} d x^{j}$ ; orthogonal $\Rightarrow (d s)^{2} = g_{ii} (d x^{i})^{2}$ .
Lamé / scale factors: $h_{i} = g_{ii} = ∣ e_{i} ∣$ .

Christoffel symbols - encode how the basis changes with position.

Second kind: $Γ^{i}_{jk} = e^{i} \cdot \partial_{k} e_{j}$ , so $\partial_{k} e_{j} = Γ^{i}_{jk} e_{i}$ .
First kind: $Γ_{ijk} = e_{i} \cdot \partial_{k} e_{j} = g_{i ℓ} Γ^{ℓ}_{jk}$ .

Properties

Symmetric in last two: $Γ^{i}_{jk} = Γ^{i}_{kj}$ , $Γ_{ijk} = Γ_{ikj}$ (from $\partial_{j} \partial_{k} r = \partial_{k} \partial_{j} r$ ).
NOT a tensor - extra inhomogeneous term in transformation.
Constant basis (rigid global) $\Rightarrow \partial_{k} e_{j} = 0$ $\Rightarrow$ all $Γ = 0$ . Any $K^{'}$ with basis as fixed linear combos of $i_{n}$ gives all $Γ = 0$ .

Formula from metric

$Γ_{ijk} = \frac{1}{2} (\partial_{k} g_{ij} + \partial_{j} g_{ik} - \partial_{i} g_{jk})$
$Γ^{i}_{jk} = g^{i ℓ} Γ_{ℓ jk}$
Orthogonal $g$ : $Γ^{i}_{jk} = Γ_{ijk} / g_{ii}$ (no sum).

Pattern table - orthogonal $g$ (only these nonzero)

Pattern of $(i, j, k)$	$Γ_{ijk}$
$i = j = k$	$+ \frac{1}{2} \partial_{i} g_{ii}$
Doubled- $i$ ( $i = j \neq = k$ or $i = k \neq = j$ )	$+ \frac{1}{2} \partial_{lone} g_{ii}$
Lone- $i$ ( $i \neq = j = k$ )	$- \frac{1}{2} \partial_{i} g_{jj}$
All distinct	$0$

Mnemonic: doubled → +, deriv of $g_{ii}$ wrt the lone index. Lone → −, deriv of $g_{jj}$ wrt $i$ .

Worked example - parabolic (mock Q4). $r = σ τ i_{1} + \frac{1}{2} (σ^{2} - τ^{2}) i_{2} + ϕ i_{3}$ , $(x^{1}, x^{2}, x^{3}) = (σ, τ, ϕ)$ .

Basis ( $e_{i} = \partial r / \partial x^{i}$ ): $e_{1} = (τ, σ, 0)$ , $e_{2} = (σ, - τ, 0)$ , $e_{3} = (0, 0, 1)$ .
Metric ( $g_{ij} = e_{i} \cdot e_{j}$ ): $g_{11} = τ^{2} + σ^{2} = σ^{2} + τ^{2}$ ; $g_{22} = σ^{2} + τ^{2}$ ; $g_{33} = 1$ ; $g_{12} = τ σ + σ (- τ) = 0$ ; $g_{13} = g_{23} = 0$ . Orthogonal.
Arc length (orthogonal $\Rightarrow (d s)^{2} = g_{ii} (d x^{i})^{2}$ ): $(d s)^{2} = (σ^{2} + τ^{2}) (d σ^{2} + d τ^{2}) + d ϕ^{2}$ . Lamé ( $h_{i} = g_{ii}$ ): $h_{1} = h_{2} = σ^{2} + τ^{2}$ , $h_{3} = 1$ .
Metric derivs: $\partial_{σ} g_{11} = \partial_{σ} g_{22} = 2 σ$ ; $\partial_{τ} g_{11} = \partial_{τ} g_{22} = 2 τ$ ; all others 0 (including everything with $ϕ$ or $g_{33}$ ).
All $Γ_{ijk}$ by pattern (table in §7):
- $i = j = k$ ( $+ \frac{1}{2} \partial_{i} g_{ii}$ ): $Γ_{111} = σ$ ; $Γ_{222} = τ$ ; $Γ_{333} = 0$ .
- Doubled- $i$ ( $+ \frac{1}{2} \partial_{lone} g_{ii}$ ): $Γ_{112} = Γ_{121} = τ$ ; $Γ_{212} = Γ_{221} = σ$ ; the rest (involving $\partial_{ϕ}$ or $g_{33}$ ) all 0.
- Lone- $i$ ( $- \frac{1}{2} \partial_{i} g_{jj}$ ): $Γ_{122} = - σ$ ; $Γ_{211} = - τ$ ; rest 0.
- All distinct: all 0.
Final nonzero list: $Γ_{111} = σ$ , $Γ_{222} = τ$ , $Γ_{112} = Γ_{121} = τ$ , $Γ_{212} = Γ_{221} = σ$ , $Γ_{122} = - σ$ , $Γ_{211} = - τ$ ; all others 0.

Covariant derivative - fixes non-tensoriality of $\partial$ .

$A_{i, k} = \partial_{k} A_{i} - Γ^{j}_{ik} A_{j}$ (lower index: $- Γ$ ).
$A^{i}_{, k} = \partial_{k} A^{i} + Γ^{i}_{jk} A^{j}$ (upper index: $+ Γ$ ).
Higher rank: one $Γ$ per index.

Ricci’s theorem - the metric is covariantly constant: $g_{ik, ℓ} = 0$ .

Proof: $g_{ik, ℓ} = \partial_{ℓ} g_{ik} - Γ^{m}_{i ℓ} g_{mk} - Γ^{m}_{k ℓ} g_{im} = \partial_{ℓ} g_{ik} - Γ_{ki ℓ} - Γ_{ik ℓ}$ . By the metric formula, $Γ_{ik ℓ} + Γ_{ki ℓ} = \partial_{ℓ} g_{ik}$ , so $g_{ik, ℓ} = 0$ ✓.
Rearrangement: $\partial_{ℓ} g_{ik} = Γ_{ik ℓ} + Γ_{ki ℓ}$ .
Orthogonal coords ( $g_{ik} = 0$ for $i \neq = k$ ): RHS must vanish $\Rightarrow Γ_{ik ℓ} = - Γ_{ki ℓ}$ for $i \neq = k$ (first-kind antisym in first two indices). NOT true for second kind.

Riemann-Christoffel tensor - true tensor, built from $Γ$ .

$R^{r}_{ijk} = \partial_{j} Γ^{r}_{ki} - \partial_{k} Γ^{r}_{ij} + Γ^{p}_{ik} Γ^{r}_{p j} - Γ^{p}_{ij} Γ^{r}_{p k}$
Antisym in last two: $R^{r}_{ijk} = - R^{r}_{ikj}$ ; so $R^{r}_{ijj} = 0$ .
Flatness: $R^{r}_{ijk} \equiv 0$ everywhere $⟺$ space is Euclidean (flat).
Sphere ( $g_{11} = 1, g_{22} = sin^{2} ϕ$ ): $R^{1}_{212} = sin^{2} ϕ \neq = 0$ (curved).
Cylindrical (disguised flat $R^{3}$ ): all components vanish.
Ricci tensor $R_{ij} = R^{k}_{ikj}$ , symmetric rank-2; $R_{ij} = 0$ necessary but not sufficient for flat.

§8. Playbook & Sanity Checks

Question phrasing → section

Vector ↔ suffix conversion / $\partial$ -in-suffix simplify → §2
$\nabla \times (\nabla f)$ , $\nabla \times (β \nabla β)$ , $\nabla f (r)$ , triple-cross etc. → §3
Prove $X$ is / isn’t a tensor → §4 (Cartesian) or §5 (general)
Outer product $M \otimes N$ , contractions $B_{i} A_{ij}$ vs $B_{j} A_{ij}$ → §6
Dual basis, $V_{i}$ , $V^{i}$ , $L^{j}_{i^{'}}$ , transform $P_{ik}$ → $K^{'}$ → §5
Symmetry of constructed $A_{ijk} = ϵ_{ij ℓ} B_{k ℓ}$ , Levi-Civita inversion → §6
Christoffel for constant basis (all 0) → §7
Curvilinear: basis / metric / arc / $Γ$ → §7
Riemann / flatness → §7

Pitfalls

Triple-repeated index (e.g. $a_{i} b_{i} c_{i}$ ) is illegal.
Free indices must match across all terms and both sides.
Distinct dummies for independent sums.
Cyclically rotate $ϵ$ so the shared index is in position 1 before applying the $ϵϵ$ identity.
$Γ$ is NOT a tensor.
Ricci $= 0$ does not imply flat - only Riemann $= 0$ does.
All four forms of $[P]$ are numerically equal in Cartesian $K$ ; they differ in $K^{'}$ .
Christoffel pattern: lone- $i$ → negative; doubled- $i$ → positive.
Constant basis (rigid linear combos of $i_{n}$ ) $\Rightarrow$ all $Γ = 0$ regardless of orthogonality.

Sanity checks before final answer

Dual basis: verify $e^{i} \cdot e_{k} = δ_{k}^{i}$ .
Components: reconstruct $V = V^{i} e_{i} = V_{i} e^{i}$ and compare.
Tensor proof: count surviving $L$ s - must equal claimed rank.
Christoffel: $Γ^{i}_{jk} = Γ^{i}_{kj}$ ; orthogonal $\Rightarrow$ first-kind antisym in first two for $i \neq = k$ .
Metric: symmetric, positive on diagonal; orthogonal $\Rightarrow$ off-diagonals all 0.
Riemann: antisym in last two $\Rightarrow R^{r}_{ijj} = 0$ .

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mth3008 final exam cheat sheet

MTH3008 Tensor Analysis - Final Exam Cheat Sheet

§1. Suffix-notation Foundations

§2. Suffix ↔ Vector

§3. Vector Calculus Identities (suffix proofs)

§4. Cartesian Transformations

§5. Generalised Coords: Dual Basis, Components, Rank-2 Transforms

§6. Tensor Algebra

§7. Curvilinear / Local-basis Geometry

§8. Playbook & Sanity Checks

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Kintsugi Zuihitsu

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Explorer

mth3008 final exam cheat sheet

MTH3008 Tensor Analysis - Final Exam Cheat Sheet

§1. Suffix-notation Foundations

§2. Suffix ↔ Vector

§3. Vector Calculus Identities (suffix proofs)

§4. Cartesian Transformations

§5. Generalised Coords: Dual Basis, Components, Rank-2 Transforms

§6. Tensor Algebra

§7. Curvilinear / Local-basis Geometry

§8. Playbook & Sanity Checks

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