MTH3008 Lecture 11

In the last lecture we finished vectors in generalised coordinates, focusing on how covariant components and contravariant components transform and how the metric tensor mediates between them. Today we upscale all of that to second-rank tensor and higher-order objects in non-Cartesian coordinates, keeping very close track of where each index lives. The main new subtlety is the appearance of mixed tensor components and how symmetry behaves under index raising and lowering.

Recap: Dual Bases and Vector Components

Dual bases and components

Two bases $e_1,e_2,e_3$ and $e^1,e^2,e^3$ are dual if $e^i\cdot e_k=0$ for $i\ne k$ and $e^i\cdot e_i=1$. Any vector $\mathbf{A}$ can be expanded as $\mathbf{A}=A^i{e}_i=A_i{e}^i$, where $A^i$ are the contravariant components and $A_i$ are the covariant components.

Given a non-orthonormal basis $\{e_i\}$, the dual basis $\{e^i\}$ is defined by $e^i\cdot e_k={\delta }_k^i$, and one way to construct $e^i$ in ${\mathbb{R}}^3$ is

$$e^i=\frac{e_j\times e_k}{e_i\cdot (e_j\times e_k)},$$

with $(i,j,k)$ a cyclic permutation of $(1,2,3)$. This ensures that the matrix of dot products between $\{e^i\}$ and $\{e_k\}$ is exactly the identity, so extracting components via contraction works as in the orthonormal case.

The metric tensor on this basis is defined by

$$g_{ik}=e_i\cdot e_k,g^{ik}=e^i\cdot e^k,$$

And we use it to convert between covariant and contravariant components:

$$\boxed{A_i=g_{ik}{A}^k,A^i=g^{ik}{A}_k.}$$

Vector vs suffix notation

In these notes $\mathbf{A}$ and its components $A^i,A_i$ are the same geometric object. The whole point of tensor notation is to separate the geometric object from how it looks in a particular coordinate system, so mixing up $A_i$ and $A^i$ in a calculation will usually break tensorial transformation rules.

Second-rank Tensors: Covariant, Contravariant and Mixed Components

A second-rank tensor $\mathbf{A}$ on a $3$‑dimensional space is determined by $3\times 3=9$ components once we pick a basis. In a generalised coordinate system, we can represent those components in four natural ways:

• Covariant components $A_{ik}$,
• Contravariant components $A^{ik}$,
• Mixed components $A_i^k$,
• Mixed components $A_k^i$.

Dot-notation for mixed components

Slides use a dot to emphasise index order, e.g. $A_{\cdot k}^i$ or $A_i^{\cdot k}$, which we will write informally as $A_i^k$ and $A_k^i$. The dot reminds you which index is covariant and which is contravariant in each position.

Transformations of Second-rank Components

Given a change of basis with matrices $L_k^{i^{\prime }}$ for contravariant components and $L_{i^{\prime }}^k$ for covariant components, the transformation laws are:

$$\begin{array}{rl}{A}_{ik}^{\prime }& =L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{A}_{\ell m},\\ A^{\prime ik}& =L_{\ell }^{i^{\prime }}{L}_m^{k^{\prime }}{A}^{\ell m},\\ A_i^{\prime k}& =L_{i^{\prime }}^{\ell }{L}_m^{k^{\prime }}{A}_{\ell }^m,\\ A_k^{\prime i}& =L_{\ell }^{i^{\prime }}{L}_{k^{\prime }}^m{A}_m^{\ell },\end{array}$$

Where summation over repeated indices is understood.

Raising and lowering indices for rank 2

The different flavours of components of a second-rank tensor are related by the metric: $A_{ik}=g_{i\ell }{g}_{km}{A}^{\ell m}=g_{k\ell }{A}_i^{\ell }=g_{i\ell }{A}_k^{\ell },$ $A^{ik}=g^{i\ell }{g}^{km}{A}_{\ell m}=g^{i\ell }{A}_{\ell }^k=g^{k\ell }{A}_{\ell }^i,$ $A_i^k=g^{k\ell }{A}_{i\ell }=g_{i\ell }{A}^{\ell k},$ $A_k^i=g^{i\ell }{A}_{\ell k}=g_{k\ell }{A}^{i\ell }.$ In particular, once you know any one of $A_{ik},A^{ik},A_i^k,A_k^i$ and the metric, you know them all.

Symmetry lives at fixed index positions

Symmetry or antisymmetry is defined with respect to indices in the same slot (both lower, or both upper). For example, $A_{ik}^{\cdot \cdot \ell }$ is symmetric in $i,k$ if $A_{ik}^{\cdot \cdot \ell }=A_{ki}^{\cdot \cdot \ell }$, while $B_{ik}^{\cdot \cdot \ell }$ is antisymmetric if $B_{ik}^{\cdot \cdot \ell }=-B_{ki}^{\cdot \cdot \ell }$. Raising or lowering one of these indices can destroy symmetry because it changes which metric components appear.

Tensor Character via Contraction: Examples

Example 1: Contraction with an Arbitrary Vector

Covariant rank 3 from a contraction

Let $B^k$ be an arbitrary contravariant vector. Suppose $A_{ijk}{B}^k$ transforms as a second‑rank covariant tensor. Then $A_{ijk}$ must itself transform as a third‑rank covariant tensor.

The assumption ”$A_{ijk}{B}^k$ is a covariant tensor of rank $2$” means that under a change of coordinates we have

$$A_{ijk}^{\prime }{B}^{\prime k}=L_{i^{\prime }}^{\ell }{L}_{j^{\prime }}^m{A}_{\ell mn}{B}^n,$$

for all choices of $B^n$. Since $B^k$ is contravariant, its components satisfy $B^{\prime k}=L_n^{k^{\prime }}{B}^n$, or equivalently $B^n=L_{k^{\prime }}^n{B}^{\prime k}$.

Substituting this into the transformation of $A_{ijk}{B}^k$ gives

$$A_{ijk}^{\prime }{B}^{\prime k}=L_{i^{\prime }}^{\ell }{L}_{j^{\prime }}^m{A}_{\ell mn}{L}_{k^{\prime }}^n{B}^{\prime k}.$$

Now $B^{\prime k}$ is arbitrary, so we can cancel it from both sides to get the transformation law

$$\boxed{A_{ijk}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{j^{\prime }}^m{L}_{k^{\prime }}^n{A}_{\ell mn},}$$

Which is exactly the covariant rank‑3 tensor rule.

Example 2: Adding Covariant and Contravariant Vectors

Why $A_i+B^i$ is not a tensor

Let $A_i$ be a covariant tensor (a covector) and $B^i$ a contravariant tensor (a vector). Show that $A_i+B^i$ does not transform as a tensor.

We know

$$A_i^{\prime }=L_{i^{\prime }}^k{A}_k,B^{\prime i}=L_k^{i^{\prime }}{B}^k.$$

Therefore, the sum transforms as

$$A_i^{\prime }+B^{\prime i}=L_{i^{\prime }}^k{A}_k+L_k^{i^{\prime }}{B}^k.$$

There is no way to factor out a single transformation matrix $L$ that acts on $A_k+B^k$, because one term uses the covariant rule and the other uses the contravariant rule. So $A_i+B^i$ does not obey any tensor transformation law and hence is not a tensor.

Pitfall: mixing index types

You can contract indices of different type (e.g. $A_i{B}^i$ is a scalar), but you cannot add objects with different index types and expect a tensor. Always check that each free index appears with the same covariant/contravariant status on every term.

Practical Question: Mixed Third-order Tensor

Contraction producing a mixed third-order tensor

Let $A_{ik\ell }$ be a covariant tensor of order $3$ and $B^{pqmn}$ a contravariant tensor of order $4$. Show that $C_i^{mn}:=A_{ik\ell }{B}^{k\ell mn}$ is a mixed tensor of order $3$ (one covariant and two contravariant indices).

Sketch: under a change of basis, $A_{ik\ell }$ transforms with three factors of the covariant transformation matrix and $B^{pqmn}$ with four factors of the contravariant one. When we form $A_{ik\ell }{B}^{k\ell mn}$ we contract over $k,\ell$, which pairs two covariant indices from $A$ with two contravariant indices from $B$. This gives three free indices $(i,m,n)$, with $i$ covariant and $m,n$ contravariant, and the transformation rule is exactly that of a $(1,2)$‑type tensor $C_i^{mn}$.

Worked Coordinate-change Example for a Second-rank Tensor

We now look at a fully explicit change of basis to see how to compute covariant, contravariant and mixed components in practice.

Second-rank tensor under a non-orthonormal change of basis

Start with a Cartesian coordinate system $K$ with orthonormal basis $\{i_1,i_2,i_3\}$ and a second-order tensor whose components in this basis are $[A_{ik}]=[A^{ik}]=[A_i^k]=[A_k^i]=\left(\begin{array}{ccc}2& 1& 3\\ 2& 3& 4\\ 1& 2& 1\end{array}\right).$ Let $K^{\prime }$ be a new coordinate system with basis vectors $e_1=i_1,e_2=i_1+i_2,e_3=i_1+i_2+i_3$. Compute all four types of components in $K^{\prime }$.

Step 1: Build the Transformation Matrix

The basis change can be written in matrix form as

$$\left(\begin{array}{c}{e}_1\\ e_2\\ e_3\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)\left(\begin{array}{c}{i}_1\\ i_2\\ i_3\end{array}\right).$$

Comparing this with $e_{i^{\prime }}=L_{i^{\prime }}^j{i}_j$, we read off the coefficients

$$L_{1^{\prime }}^1=1,L_{1^{\prime }}^2=0,L_{1^{\prime }}^3=0;L_{2^{\prime }}^1=1,L_{2^{\prime }}^2=1,L_{2^{\prime }}^3=0;L_{3^{\prime }}^1=1,L_{3^{\prime }}^2=1,L_{3^{\prime }}^3=1.$$

We can collect these into the matrix

$$L=[L_{i^{\prime }}^j]=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right).$$

Step 2: Covariant Components via Matrix Multiplication

The covariant components transform as

$$A_{ik}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{A}_{\ell m}.$$

If we view $A=[A_{ij}]$ and $L=[L_{i^{\prime }}^j]$ as matrices, this is exactly the matrix product

$$A^{\prime }=LA{L}^T.$$

In this example we compute

$$A^{\prime }=LA{L}^T=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)\left(\begin{array}{ccc}2& 1& 3\\ 2& 3& 4\\ 1& 2& 1\end{array}\right)\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}2& 3& 6\\ 4& 8& 15\\ 5& 11& 19\end{array}\right).$$

So in $K^{\prime }$ the covariant components of $A$ are $A_{ik}^{\prime }$ given by this new matrix.

Step 3: Metric Tensor and Raised Indices

The new basis vectors $\{e_i\}$ are not orthonormal, so the metric tensor in $K^{\prime }$ is nontrivial:

$$G=[g_{ik}]=[e_i\cdot e_k]=\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right).$$

This matrix is symmetric and invertible, and its inverse is $[g^{ik}]$.

To get the contravariant components from the covariant ones, we use $A^{\prime ik}=g^{i\ell }{g}^{km}{A}_{\ell m}^{\prime }$. At the matrix level this becomes

$$[A^{\prime ik}]=G^{-1}{A}^{\prime }{G}^{-T},$$

And in this particular example the slides show that it suffices to compute

$$[A^{\prime ik}]=G{A}^{\prime }{G}^T$$

Because here $G$ is built according to that convention. Similarly,

$$[A_i^{\prime k}]=A^{\prime }{G}^T=A^{\prime }G,[A_k^{\prime i}]=G{A}^{\prime }.$$

The takeaway is that once we have $A^{\prime }$ and the metric $G$, we can obtain all other index placements by matrix multiplications with $G$ and $G^{-1}$.

Pre-Lecture Notes from Mth3008 Lecture Notes 11.pdf|University Notes

• Recap of local coordinate transformations for vectors: dual basis construction, covariant vs contravariant components, transformation rules, and the role of the metric tensor $g_{ik}$ and $g^{ik}$.
• In a generalised (non-Cartesian) coordinate system, a first-rank tensor is determined by either its three covariant components $A_i$ or its three contravariant components $A^i$, with $A_i=g_{ik}{A}^k$ and $A^i=g^{ik}{A}_k$.
• For second-rank tensors, we track four types of components: covariant $A_{ik}$, contravariant $A^{ik}$, and mixed $A_i^k$, $A_k^i$, each with its own transformation law under basis change.
• The metric tensor $g_{ik}$ and its inverse $g^{ik}$ give explicit formulae relating these components; contractions such as $A_{ijk}{B}^k$ can raise or lower the order while preserving tensor character, but sums like $A_i+B^i$ usually break tensorial transformation rules.
• Practical recipes: compute new covariant components via $A^{\prime }=LA{L}^T$, then use the metric $G=[g_{ik}]$ to raise and lower indices and build contravariant or mixed components. Next time we focus on associated tensors, deeper properties of the metric, and higher-order tensors in generalised coordinates.