MTH3008 Lecture 15

Quote

…

Lecture 14 wrapped up Chapter 6: Tensor Algebra with the outer product, non-commutativity of the tensor product, and contraction. We now open Chapter 7: Tensor Fields - where the bases themselves start to move.

Chapter 7 Overview

Chapter 7 covers:

1. Preliminary (vector fields, local bases)
2. Covariant differentiation
3. Christoffel symbols
4. Covariant differentiation of tensors
5. Ricci’s theorem
6. Riemann-Christoffel tensor
7. Ricci tensor

This lecture handles items 1-3.

Preliminary: Vector Fields and Local Bases

Up to now, most bases we have used were independent of position. The canonical basis of ${\mathbb{R}}^3$ is the standard example:

$${\mathbf{e}}_1=(1,0,0),{\mathbf{e}}_2=(0,1,0),{\mathbf{e}}_3=(0,0,1).$$

This is a fixed basis - it does not depend on any parameter.

A Local Basis assigns a vector to every point in a subset of space. Physical examples include velocity fields (fluid flow) and gravitational fields.

The key new idea: we can define a local basis ${\mathbf{e}}_1(\mathbf{x}),{\mathbf{e}}_2(\mathbf{x}),{\mathbf{e}}_3(\mathbf{x})$ that depends on position.

Derivatives of Tensor Fields - Recap

We already know how to differentiate tensor fields of rank zero:

• Gradient: $\nabla f={\nabla }_{i}f$, or ${\nabla }_i{u}_j=\frac{\partial u_j}{\partial x^i}$
• Divergence: $\nabla \cdot \mathbf{u}={\nabla }_i{u}^i$
• Curl: $\nabla \times \mathbf{u}={ϵ}_{ijk}{\nabla }_j{u}_k$

We can also differentiate higher-rank tensors. For instance, $\frac{\partial {\sigma }_{jk}}{\partial x^i}$ is the derivative of the conductivity tensor - itself a rank 3 tensor.

Covariant Differentiation

Covariant differentiation is a way to differentiate vectors and tensors in curved space while ensuring the result is still a tensor. The intuition: when you transport a vector along a curved surface, you need to adjust for the curvature. Covariant differentiation performs exactly that adjustment.

Cartesian Coordinates

Consider a Cartesian coordinate system with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$ and a vector field $\mathbf{A}=\mathbf{A}(\mathbf{r})$ with components $A_1(\mathbf{r}),A_2(\mathbf{r}),A_3(\mathbf{r})$:

$$\mathbf{A}=A_1(\mathbf{r}){\mathbf{i}}_1+A_2(\mathbf{r}){\mathbf{i}}_2+A_3(\mathbf{r}){\mathbf{i}}_3=A_j(\mathbf{r}){\mathbf{i}}_j.$$

The differential of $\mathbf{A}$, by the product rule, is:

$$d\mathbf{A}=d(A_j(\mathbf{r}){\mathbf{i}}_j)={\mathbf{i}}_{j}d{A}_j(\mathbf{r})+A_j(\mathbf{r})d{\mathbf{i}}_j.$$

Since the Cartesian basis does not vary from point to point, $d{\mathbf{i}}_j=0$ for all $j$. So:

$$\boxed{d\mathbf{A}={\mathbf{i}}_{j}d{A}_j(\mathbf{r})=(d{A}_1(\mathbf{r}),d{A}_2(\mathbf{r}),d{A}_3(\mathbf{r}))}$$

The differential of $\mathbf{A}$ depends only on the differentials of its components. Life is simple in Cartesian coordinates.

General Coordinates with a Fixed Basis

Now consider a generalised coordinate system with a fixed basis ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ (constants). A vector $\mathbf{A}$ expands as:

$$\mathbf{A}={\mathbf{e}}^1{A}_1(\mathbf{r})+{\mathbf{e}}^2{A}_2(\mathbf{r})+{\mathbf{e}}^3{A}_3(\mathbf{r})={\mathbf{e}}_1{A}^1(\mathbf{r})+{\mathbf{e}}_2{A}^2(\mathbf{r})+{\mathbf{e}}_3{A}^3(\mathbf{r}),$$

where $A_1,A_2,A_3$ are the Covariant and Contravariant Components and $A^1,A^2,A^3$ are the Covariant and Contravariant Components.

Because the basis is fixed ($d{\mathbf{e}}_j=d{\mathbf{e}}^j=0$), the differentials reduce to:

$$d\mathbf{A}=d(A_j(\mathbf{r}){\mathbf{e}}^j)={\mathbf{e}}^{j}d{A}_j(\mathbf{r}),$$ $$d\mathbf{A}=d(A^j(\mathbf{r}){\mathbf{e}}_j)={\mathbf{e}}_{j}d{A}^j(\mathbf{r}).$$

Same story as Cartesian: fixed basis means the differential only sees the components.

Generalised Coordinates with a Local Basis

Now the real challenge. Suppose we have a generalised coordinate system with a local basis ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ that varies from point to point:

$${\mathbf{e}}_j={\mathbf{e}}_j(x^1,x^2,x^3),{\mathbf{e}}^j={\mathbf{e}}^j(x^1,x^2,x^3).$$

Any vector $\mathbf{A}=\mathbf{A}(\mathbf{r})$ still expands as $\mathbf{A}=A_j{\mathbf{e}}^j=A^j{\mathbf{e}}_j$, but now it is not true that $d{\mathbf{e}}_j=0$. The product rule gives two terms:

$$\boxed{d\mathbf{A}={\mathbf{e}}^{j}d{A}_j+A_{j}d{\mathbf{e}}^j={\mathbf{e}}_{j}d{A}^j+A^{j}d{\mathbf{e}}_j}$$

Warning

In a local basis, $d{\mathbf{e}}_j\ne 0$. Forgetting the second term - the one from the varying basis - is the single most common mistake in covariant differentiation. Every identity that follows exists to handle this extra piece.

Partial Differentiation in a Local Basis

Since $d\mathbf{A}=\frac{\partial \mathbf{A}}{\partial x^k}d{x}^k$, we can read off the partial derivative:

$$\frac{\partial \mathbf{A}}{\partial x^k}=\frac{\partial (A_j{\mathbf{e}}^j)}{\partial x^k}=\frac{\partial A_j}{\partial x^k}{\mathbf{e}}^j+A_j\frac{\partial {\mathbf{e}}^j}{\partial x^k}$$ $$\frac{\partial \mathbf{A}}{\partial x^k}=\frac{\partial (A^j{\mathbf{e}}_j)}{\partial x^k}=\frac{\partial A^j}{\partial x^k}{\mathbf{e}}_j+A^j\frac{\partial {\mathbf{e}}_j}{\partial x^k}$$

Both expressions have the same structure: the ordinary partial derivative of the components, plus a correction term from the changing basis.

Worked Example

Example

Consider a local basis ${\mathbf{e}}^1=2x^1\mathbf{i}+x^1\mathbf{j}+x^3\mathbf{k}$, ${\mathbf{e}}^2=x^2\mathbf{j}+x^2\mathbf{k}$, ${\mathbf{e}}^3=x^3\mathbf{k}$. Find $\frac{\partial \mathbf{A}}{\partial x^1}$ for $\mathbf{A}=x^2\mathbf{i}+2x^1\mathbf{j}+x^1{x}^3\mathbf{k}$.

We use $\frac{\partial \mathbf{A}}{\partial x^1}=\frac{\partial A_j}{\partial x^1}{\mathbf{e}}^j+A_j\frac{\partial {\mathbf{e}}^j}{\partial x^1}$.

First term - differentiate the components:

$$\frac{\partial A_j}{\partial x^1}{\mathbf{e}}^j=\frac{\partial A_1}{\partial x^1}{\mathbf{e}}^1+\frac{\partial A_2}{\partial x^1}{\mathbf{e}}^2+\frac{\partial A_3}{\partial x^1}{\mathbf{e}}^3=0+2{\mathbf{e}}^2+x^3{\mathbf{e}}^3.$$

Substituting the basis vectors: $2(x^2\mathbf{j}+x^2\mathbf{k})+x^3(x^3\mathbf{k})=2x^2\mathbf{j}+(2x^2+(x^3{)}^2)\mathbf{k}$.

Second term - differentiate the basis:

$$A_j\frac{\partial {\mathbf{e}}^j}{\partial x^1}=A_1\frac{\partial {\mathbf{e}}^1}{\partial x^1}+A_2\frac{\partial {\mathbf{e}}^2}{\partial x^1}+A_3\frac{\partial {\mathbf{e}}^3}{\partial x^1}=A_1(2\mathbf{i}+\mathbf{j})+0+0=2A_1\mathbf{i}+A_1\mathbf{j}.$$

Combining:

$$\frac{\partial \mathbf{A}}{\partial x^1}=2A_1\mathbf{i}+(2x^2+A_1)\mathbf{j}+(2x^2+(x^3{)}^2)\mathbf{k}.$$

The Covariant Derivative as a Tensor

Each partial derivative $\frac{\partial \mathbf{A}}{\partial x^k}$ is itself a vector. Denote it ${\mathbf{v}}_k=\frac{\partial \mathbf{A}}{\partial x^k}$. Like any vector, ${\mathbf{v}}_k$ has contravariant and covariant components:

$${\mathbf{v}}_k=(A_k{)}^i{\mathbf{e}}_i=(A_k{)}_i{\mathbf{e}}^i.$$

We adopt the shorthand:

$$A_{\cdot k}^i=(A_k{)}^i,A_{i,k}=(A_k{)}_i.$$

These components $A_{\cdot k}^i$ and $A_{i,k}$ form a second-rank tensor called the Covariant Differentiation.

Extracting the Components

To find $A_{i,k}$, expand ${\mathbf{v}}_k$ in the dual basis and project:

$$\frac{\partial \mathbf{A}}{\partial x^k}=A_{1,k}{\mathbf{e}}^1+A_{2,k}{\mathbf{e}}^2+A_{3,k}{\mathbf{e}}^3.$$

Taking the dot product with ${\mathbf{e}}_i$ and using ${\mathbf{e}}^j\cdot {\mathbf{e}}_i={\delta }_i^j$:

$$\boxed{A_{i,k}=\frac{\partial \mathbf{A}}{\partial x^k}\cdot {\mathbf{e}}_i}$$

Similarly:

$$\boxed{A_{\cdot k}^i=\frac{\partial \mathbf{A}}{\partial x^k}\cdot {\mathbf{e}}^i}$$

Summary Formulas for Covariant Differentiation

Substituting the local-basis expression for $\frac{\partial \mathbf{A}}{\partial x^k}$ into the projection formulas:

$$\boxed{A_{i,k}=\frac{\partial A_i}{\partial x^k}+A_j\frac{\partial {\mathbf{e}}^j}{\partial x^k}\cdot {\mathbf{e}}_i}$$ $$\boxed{A_{\cdot k}^i=\frac{\partial A^i}{\partial x^k}+A^j\frac{\partial {\mathbf{e}}_j}{\partial x^k}\cdot {\mathbf{e}}^i}$$

The first term is the ordinary partial derivative of the component. The second term corrects for the fact that the basis itself is changing.

Christoffel Symbols

The correction terms above contain dot products of basis derivatives with basis vectors. These appear so frequently that they earn their own notation.

Christoffel Symbols of the Second Kind

Important

The Christoffel Symbols are defined by: ${\Gamma }_{jk}^i={\mathbf{e}}^i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}$ and the related identity ${\Gamma }_{ik}^j=-{\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}^j}{\partial x^k}$. Each symbol has 27 components (3 choices for each of the three indices).

With this notation, the covariant derivatives become:

$$\boxed{A_{i,k}=\frac{\partial A_i}{\partial x^k}-{\Gamma }_{ik}^j{A}_j}$$ $$\boxed{A_{\cdot k}^i=\frac{\partial A^i}{\partial x^k}+{\Gamma }_{jk}^i{A}^j}$$

Warning

Watch the sign: the covariant-component formula has a minus sign in front of $\Gamma$, while the contravariant-component formula has a plus sign. Mixing these up is an easy route to wrong answers.

Fixed Basis as a Special case

If the basis is fixed, then $\frac{\partial {\mathbf{e}}_j}{\partial x^k}=0$, so all Christoffel symbols vanish: ${\Gamma }_{jk}^i=0$. The covariant derivatives collapse to ordinary partial derivatives:

$$A_{i,k}=\frac{\partial A_i}{\partial x^k},A_{\cdot k}^i=\frac{\partial A^i}{\partial x^k}.$$

This confirms that covariant differentiation generalises ordinary differentiation - the two agree when the basis is constant.

Christoffel Symbols as Expansion Coefficients

The Christoffel symbols ${\Gamma }_{jk}^i$ are the expansion coefficients of the vector $\frac{\partial {\mathbf{e}}_j}{\partial x^k}$ with respect to the basis ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$:

$$\boxed{\frac{\partial {\mathbf{e}}_j}{\partial x^k}={\Gamma }_{jk}^i{\mathbf{e}}_i}$$

To prove this, write $\frac{\partial {\mathbf{e}}_j}{\partial x^k}=A_{jk}^i{\mathbf{e}}_i$ and dot both sides with ${\mathbf{e}}^m$:

$${\mathbf{e}}^m\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}=A_{jk}^i({\mathbf{e}}^m\cdot {\mathbf{e}}_i)=A_{jk}^i{\delta }_i^m=A_{jk}^m.$$

But ${\mathbf{e}}^m\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}={\Gamma }_{jk}^m$ by definition, so $A_{jk}^m={\Gamma }_{jk}^m$.

Christoffel Symbols of the First Kind

We can also expand $\frac{\partial {\mathbf{e}}_j}{\partial x^k}$ in the dual basis. The resulting coefficients are the Christoffel Symbols:

$$\frac{\partial {\mathbf{e}}_j}{\partial x^k}={\Gamma }_{ijk}{\mathbf{e}}^i⟹\boxed{{\Gamma }_{ijk}={\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}}$$

Note

First kind: all indices downstairs, dot with ${\mathbf{e}}_i$. Second kind: one index upstairs, dot with ${\mathbf{e}}^i$. The position of the index on $\Gamma$ mirrors which basis vector you project onto.

Relation to the Metric Tensor

The two kinds of Christoffel symbol are related via the metric tensor. Starting from the first-kind definition and substituting the second-kind expansion:

$${\Gamma }_{ijk}={\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}={\mathbf{e}}_i\cdot ({\Gamma }_{jk}^{\ell }{\mathbf{e}}_{\ell })=({\mathbf{e}}_i\cdot {\mathbf{e}}_{\ell }){\Gamma }_{jk}^{\ell }=g_{i\ell }{\Gamma }_{jk}^{\ell }.$$

Similarly:

$${\Gamma }_{jk}^i={\mathbf{e}}^i\cdot ({\Gamma }_{\ell jk}{\mathbf{e}}^{\ell })=g^{i\ell }{\Gamma }_{\ell kj}.$$

So the metric tensor raises and lowers the first index of the Christoffel symbol, just as it does for tensor components:

$$\boxed{{\Gamma }_{ijk}=g_{i\ell }{\Gamma }_{jk}^{\ell },{\Gamma }_{jk}^i=g^{i\ell }{\Gamma }_{\ell kj}}$$

Symmetry of Christoffel Symbols

Both kinds of Christoffel symbol are symmetric in the lower two indices $j$ and $k$:

$$\boxed{{\Gamma }_{ijk}={\Gamma }_{ikj},{\Gamma }_{jk}^i={\Gamma }_{kj}^i}$$

The proof for the first kind uses ${\mathbf{e}}_i=\frac{\partial \mathbf{r}}{\partial x^i}$:

$${\Gamma }_{ijk}={\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}=\frac{\partial \mathbf{r}}{\partial x^i}\cdot \frac{\partial }{\partial x^k}\frac{\partial \mathbf{r}}{\partial x^j}=\frac{\partial \mathbf{r}}{\partial x^i}\cdot \frac{\partial }{\partial x^j}\frac{\partial \mathbf{r}}{\partial x^k}={\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}_k}{\partial x^j}={\Gamma }_{ikj},$$

where we swapped the order of the mixed partial (equality of mixed partials). The second-kind symmetry follows identically.

This symmetry means ${\Gamma }_{jk}^i$ has at most $3\times 6=18$ independent components (not 27), since the 6 comes from the symmetric pair $(j,k)$.

Christoffel Symbols Are not Tensors

Despite having indices, Christoffel symbols do not transform as tensors. Under a coordinate transformation $x^i\to x^{\prime i}$ with transformation matrix $L$, the first-kind symbol transforms as:

$${\Gamma }_{ijk}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{j^{\prime }}^m{L}_{k^{\prime }}^n{\Gamma }_{\ell mn}+L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^n\frac{\partial L_{j^{\prime }}^m}{\partial x^n}{g}_{\ell m}.$$

The second-kind symbol transforms as:

$${{\Gamma }_{jk}^i}^{\prime }=L_{\ell }^{i^{\prime }}{L}_{j^{\prime }}^m{L}_{k^{\prime }}^n{\Gamma }_{mn}^{\ell }+L_m^{i^{\prime }}{L}_{k^{\prime }}^n\frac{\partial L_{j^{\prime }}^m}{\partial x^n}.$$

In both cases, the second term (involving a derivative of $L$) spoils the tensor transformation law. The Christoffel symbol picks up an extra piece that a genuine tensor never would.

Warning

Christoffel symbols have indices and obey certain algebraic rules, but they are not tensors. They transform with an additional inhomogeneous term. This is precisely because they encode how the basis changes - information that is coordinate-system-dependent, not intrinsic.

---

Pre-Lecture Notes from mth3008 lecture notes 15.pdf

• Chapter 7: Tensor Fields opens with seven topics; this lecture covers Preliminary, Covariant Differentiation, and Christoffel Symbols
• A vector field assigns a vector to each point in a region; a local basis ${\mathbf{e}}_j(x^1,x^2,x^3)$ varies with position, unlike a fixed basis
• In Cartesian or any fixed basis, $d{\mathbf{e}}_j=0$, so $d\mathbf{A}={\mathbf{i}}_{j}d{A}_j$ - the differential depends only on the components
• In a local basis, the product rule gives $d\mathbf{A}={\mathbf{e}}^{j}d{A}_j+A_{j}d{\mathbf{e}}^j$, producing a correction term from the changing basis
• The covariant derivative components are $A_{i,k}=\frac{\partial \mathbf{A}}{\partial x^k}\cdot {\mathbf{e}}_i$ and $A_{\cdot k}^i=\frac{\partial \mathbf{A}}{\partial x^k}\cdot {\mathbf{e}}^i$
• Christoffel symbols of the second kind: ${\Gamma }_{jk}^i={\mathbf{e}}^i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}$; these are the expansion coefficients of $\frac{\partial {\mathbf{e}}_j}{\partial x^k}$ in the basis ${\mathbf{e}}_i$
• Christoffel symbols of the first kind: ${\Gamma }_{ijk}={\mathbf{e}}_i\cdot \frac{\partial {\mathbf{e}}_j}{\partial x^k}$; related to the second kind via ${\Gamma }_{ijk}=g_{i\ell }{\Gamma }_{jk}^{\ell }$
• Both kinds are symmetric in $j,k$: ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ and ${\Gamma }_{ijk}={\Gamma }_{ikj}$ (from equality of mixed partials)
• Christoffel symbols are not tensors - their transformation law contains an extra derivative term beyond the standard tensor law
• Covariant derivative formulas: $A_{i,k}=\frac{\partial A_i}{\partial x^k}-{\Gamma }_{ik}^j{A}_j$ and $A_{\cdot k}^i=\frac{\partial A^i}{\partial x^k}+{\Gamma }_{jk}^i{A}^j$
• Next lecture: more on Christoffel symbols and covariant differentiation of tensors