MTH3008 Lecture 18

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Last time we finished with Ricci’s theorem - the covariant derivative of the metric tensor $g_{ik}$ vanishes, giving us the useful identity $\frac{\partial g_{ik}}{\partial x^{\ell }}={\Gamma }_{ik\ell }+{\Gamma }_{ki\ell }$. Today we push further into Chapter 7 by asking a fundamental geometric question: how do we detect curvature? The answer leads us to the Riemann-Christoffel Tensor and, briefly, the Ricci Tensor.

Defining Spaces and Curvature

So far we have worked almost entirely in ordinary Riemann-Christoffel Tensor ${\mathbb{R}}^3$, whether in Cartesian or generalised coordinates. But other spaces exist - the surface of a sphere, for instance, is a two-dimensional non-Euclidean space.

The question is: how do we distinguish between different spaces? One answer is Riemann-Christoffel Tensor:

• Euclidean space has zero curvature,
• the surface of a sphere has positive curvature,
• Riemann-Christoffel Tensor has negative curvature.

The goal for today: find a tensorial condition that tells us when a space is Euclidean.

Partial Vs Covariant Derivatives

Let $\mathbf{A}$ be a vector field with components $A_i(x^1,x^2,x^3)$. Since each $A_i$ is just a function, its partial derivatives always commute:

$$\frac{\partial }{\partial x^k}\left(\frac{\partial A_i}{\partial x^j}\right)=\frac{\partial }{\partial x^j}\left(\frac{\partial A_i}{\partial x^k}\right).$$

But a vector field is not merely a collection of component functions - it is a geometric object whose components change under coordinate transformations. The partial derivative $\frac{\partial A_i}{\partial x^j}$ does not transform as a tensor under general coordinate changes.

This is exactly why we need the Covariant Differentiation:

$$A_{i,j}:={\nabla }_j{A}_i=\frac{\partial A_i}{\partial x^j}-{\Gamma }_{ij}^k{A}_k.$$

The partial derivative differentiates the components; the Christoffel symbols correct for the change of basis; the result transforms as a tensor.

Note

In Euclidean space with Cartesian coordinates, ${\Gamma }_{ij}^k=0$, so ${\nabla }_j{A}_i=\frac{\partial A_i}{\partial x^j}$. This is why partial and covariant derivatives appear identical in standard vector calculus.

Non-Commutativity of Second Covariant Derivatives

Second partial derivatives always commute. Second covariant derivatives, written

$$A_{i,kj}:={\nabla }_k{\nabla }_j{A}_i,$$

do not commute in general:

$${\nabla }_k{\nabla }_j{A}_i\ne {\nabla }_j{\nabla }_k{A}_i,\text{i.e.}{A}_{i,jk}\ne A_{i,kj}.$$

This failure of commutativity is precisely what curvature measures. We will show that $A_{i,jk}-A_{i,kj}$ is a tensor - the Riemann-Christoffel tensor - and that a space is Euclidean if and only if this tensor vanishes.

The Riemann-Christoffel Tensor

Definition

We define the Riemann-Christoffel tensor $R_{ijk}^r$ by

$$\boxed{{\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i=R_{ijk}^r{A}_r}$$

The plan: derive a coordinate formula for $R_{ijk}^r$ purely in terms of Christoffel symbols.

Derivation

Start from the general formula for covariant differentiation of a second-rank covariant tensor:

$${\nabla }_j{T}_{ik}=\frac{\partial T_{ik}}{\partial x^j}-{\Gamma }_{ij}^p{T}_{pk}-{\Gamma }_{kj}^p{T}_{ip}.$$

Apply this to $T_{ik}={\nabla }_k{A}_i$:

$${\nabla }_j{\nabla }_k{A}_i=\frac{\partial ({\nabla }_k{A}_i)}{\partial x^j}-{\Gamma }_{ij}^p{\nabla }_k{A}_p-{\Gamma }_{kj}^p{\nabla }_p{A}_i.$$

Now substitute ${\nabla }_k{A}_i=\frac{\partial A_i}{\partial x^k}-{\Gamma }_{ki}^r{A}_r$ into every occurrence:

$${\nabla }_j{\nabla }_k{A}_i=\frac{\partial }{\partial x^j}\left(\frac{\partial A_i}{\partial x^k}-{\Gamma }_{ki}^r{A}_r\right)-{\Gamma }_{ij}^p\left(\frac{\partial A_p}{\partial x^k}-{\Gamma }_{kp}^r{A}_r\right)-{\Gamma }_{kj}^p\left(\frac{\partial A_i}{\partial x^p}-{\Gamma }_{pi}^r{A}_r\right).$$

Expanding fully:

$${\nabla }_j{\nabla }_k{A}_i=\frac{{\partial }^2{A}_i}{\partial x^j\partial x^k}-{\Gamma }_{ki}^r\frac{\partial A_r}{\partial x^j}-A_r\frac{\partial {\Gamma }_{ki}^r}{\partial x^j}-{\Gamma }_{ij}^p\frac{\partial A_p}{\partial x^k}-{\Gamma }_{kj}^p\frac{\partial A_i}{\partial x^p}+\left({\Gamma }_{ij}^p{\Gamma }_{kp}^r+{\Gamma }_{kj}^p{\Gamma }_{pi}^r\right)A_r.$$

Interchanging $j$ and $k$ gives the analogous formula for ${\nabla }_k{\nabla }_j{A}_i$. Now form the difference ${\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$ and examine three types of terms separately.

Second derivatives of $A$: These give $\frac{{\partial }^2{A}_i}{\partial x^j\partial x^k}-\frac{{\partial }^2{A}_i}{\partial x^k\partial x^j}=0$ by commutativity of partial derivatives.

First derivatives of $A$: Collecting the terms involving $\frac{\partial A_r}{\partial x^j}$, $\frac{\partial A_p}{\partial x^k}$, $\frac{\partial A_i}{\partial x^p}$ and their $j\leftrightarrow k$ counterparts, each pair cancels by the symmetry ${\Gamma }_{ik}^{\ell }={\Gamma }_{ki}^{\ell }$ of the Christoffel symbols.

Remaining terms (Christoffel symbols only): After cancellation we are left with

$${\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i=\left(\frac{\partial {\Gamma }_{ki}^r}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^r}{\partial x^k}\right)A_r+\left({\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r\right)A_r.$$

Coordinate Formula

Comparing with the definition ${\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i=R_{ijk}^r{A}_r$, we read off:

$$\boxed{R_{ijk}^r=\frac{\partial {\Gamma }_{ki}^r}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^r}{\partial x^k}+{\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r}$$

Warning

The index placement on $R_{ijk}^r$ matters. The ordering of $\Gamma$ terms and partial derivatives changes depending on which convention you follow. Always check against the defining equation $R_{ijk}^r{A}_r={\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$ to verify signs.

Tensor Character

Important

$R_{ijk}^r$ is a tensor. The left-hand side ${\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$ is a difference of tensors, hence a tensor. Since $A_r$ is arbitrary, the quotient rule gives that $R_{ijk}^r$ is a tensor.

Condition for Euclidean Space

For Euclidean space we need zero curvature, i.e. $A_{i,jk}=A_{i,kj}$ for all $A_i$. Since $A_{i,jk}-A_{i,kj}=R_{ijk}^r{A}_r$, this holds if and only if:

$$\boxed{R_{ijk}^r=0⟺\text{the space is Euclidean}}$$

Example: The Unit Sphere

Example

We verify that the surface of the unit sphere ($r=1$) has non-vanishing Riemann-Christoffel tensor, confirming it is not Euclidean.

The position vector in spherical coordinates $(x^1,x^2)=(\varphi ,\theta )$ is

$$\mathbf{r}=\sin \varphi \cos \theta {\mathbf{i}}_1+\sin \varphi \sin \theta {\mathbf{i}}_2+\cos \varphi {\mathbf{i}}_3,$$

giving the local basis vectors

$${\mathbf{e}}_1=\frac{\partial \mathbf{r}}{\partial \varphi }=\cos \varphi \cos \theta {\mathbf{i}}_1+\cos \varphi \sin \theta {\mathbf{i}}_2-\sin \varphi {\mathbf{i}}_3,$$ $${\mathbf{e}}_2=\frac{\partial \mathbf{r}}{\partial \theta }=-\sin \varphi \sin \theta {\mathbf{i}}_1+\sin \varphi \cos \theta {\mathbf{i}}_2.$$

Step 1: Metric Tensor

From $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$:

$$g_{11}=1,g_{12}=0,g_{22}={\sin }^2\varphi .$$

The contravariant components are

$$g^{11}=1,g^{22}=\frac{1}{{\sin }^2\varphi },g^{12}=0.$$

Step 2: Christoffel Symbols of the First Kind

Using ${\Gamma }_{ijk}=\frac{1}{2}\left(\frac{\partial g_{ij}}{\partial x^k}+\frac{\partial g_{ik}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^i}\right)$, the only non-zero metric derivative is

$$\frac{\partial g_{22}}{\partial x^1}=\frac{\partial {\sin }^2\varphi }{\partial \varphi }=2\sin \varphi \cos \varphi .$$

So ${\Gamma }_{ijk}=0$ unless exactly two indices are $2$ and one is $1$. Working through:

$${\Gamma }_{111}={\Gamma }_{112}={\Gamma }_{121}={\Gamma }_{211}={\Gamma }_{222}=0,$$ $${\Gamma }_{122}=-\sin \varphi \cos \varphi .$$

Using Ricci’s theorem: $0=\frac{\partial g_{12}}{\partial x^2}={\Gamma }_{122}+{\Gamma }_{212}$, so

$${\Gamma }_{212}={\Gamma }_{221}=\sin \varphi \cos \varphi .$$

Step 3: Christoffel Symbols of the Second Kind

From ${\Gamma }_{jk}^i=g^{i\ell }{\Gamma }_{\ell jk}$:

$${\Gamma }_{11}^1={\Gamma }_{12}^1={\Gamma }_{11}^2={\Gamma }_{22}^2=0,$$ $${\Gamma }_{22}^1=g^{11}{\Gamma }_{122}+g^{12}{\Gamma }_{222}=-\sin \varphi \cos \varphi ,$$ $${\Gamma }_{12}^2={\Gamma }_{21}^2=g^{21}{\Gamma }_{112}+g^{22}{\Gamma }_{212}=\cot \varphi .$$

Step 4: Riemann-Christoffel Tensor

Most components vanish because of the zeros among the Christoffel symbols:

$$R_{111}^1=R_{112}^1=R_{121}^1=R_{211}^1=R_{122}^1=R_{221}^1=R_{222}^1=0,$$ $$R_{111}^2=R_{211}^2=R_{221}^2=R_{122}^2=R_{222}^2=0.$$

The key non-zero component is $R_{212}^1$:

$$R_{212}^1=\frac{\partial }{\partial x^1}{\Gamma }_{22}^1-\frac{\partial }{\partial x^2}{\Gamma }_{21}^1+{\Gamma }_{22}^p{\Gamma }_{p1}^1-{\Gamma }_{21}^p{\Gamma }_{p2}^1.$$

The only surviving terms are:

$$R_{212}^1=\frac{\partial }{\partial \varphi }{\Gamma }_{22}^1-{\Gamma }_{21}^2{\Gamma }_{22}^1$$ $$=\frac{\partial }{\partial \varphi }(-\sin \varphi \cos \varphi )-\frac{\cos \varphi }{\sin \varphi }(-\sin \varphi \cos \varphi )$$ $$=(-{\cos }^2\varphi +{\sin }^2\varphi )+{\cos }^2\varphi$$ $$\boxed{R_{212}^1={\sin }^2\varphi }$$

Since $R_{212}^1\ne 0$, the unit sphere is indeed a curved (non-Euclidean) space - exactly as expected.

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Pre-Lecture Notes from mth3008 lecture notes 18.pdf

• Curvature distinguishes spaces: Euclidean $\to$ zero curvature; sphere $\to$ positive; hyperbolic $\to$ negative
• Partial derivatives commute, but covariant derivatives do not in general - the failure is measured by curvature
• Riemann-Christoffel tensor defined by $R_{ijk}^r{A}_r={\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$
• Coordinate formula: $R_{ijk}^r=\frac{\partial {\Gamma }_{ki}^r}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^r}{\partial x^k}+{\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r$
• Derivation strategy: expand ${\nabla }_j{\nabla }_k{A}_i$ using the covariant derivative of a rank-2 tensor, substitute ${\nabla }_k{A}_i=\frac{\partial A_i}{\partial x^k}-{\Gamma }_{ki}^r{A}_r$, then subtract ${\nabla }_k{\nabla }_j{A}_i$; second-derivative terms and first-derivative terms cancel by symmetry of $\Gamma$
• Tensor character follows from the quotient rule since $A_r$ is arbitrary
• Euclidean condition: $R_{ijk}^r=0$ throughout the space
• Unit sphere example: $(x^1,x^2)=(\varphi ,\theta )$, metric $g_{11}=1$, $g_{22}={\sin }^2\varphi$, $g_{12}=0$; non-zero Christoffel symbols ${\Gamma }_{22}^1=-\sin \varphi \cos \varphi$, ${\Gamma }_{12}^2=\cot \varphi$; key result $R_{212}^1={\sin }^2\varphi \ne 0$, confirming positive curvature
• Next lecture: the Ricci tensor (contraction of the Riemann-Christoffel tensor)