We will begin with a sphere of radius $R$ referred to coordinates illustrated in the following figure. This example will offer a great deal of insight as we will analyze it with the help of two different ambient coordinate systems: Cartesian and spherical. Since a spherical coordinate system is perfectly suited to the geometry of a sphere, its use will simplify our calculations despite the greater complexity of the ambient metric tensor and Christoffel symbol.

In this Chapter, we will denote matrices corresponding to a system by putting brackets around the associated indicial symbol. For example, $\left[ S_{\alpha\beta}\right]$ will denote the matrix associated with the covariant metric tensor and $\left[ Z_{\alpha}^{i}\right]$ will denote the matrix corresponding to the shift tensor.

Let us operate with spherical coordinates $\theta,\varphi$ on the surface and Cartesian coordinates $x,y,z$ in the ambient space. Recall the equations of the surface Recall that all versions of the ambient metric tensor, i.e. $Z_{ij}$, $Z^{ij}$, and $\delta_{j}^{i}$, correspond to the $3\times3$ identity matrix, i.e. Thus, the placement of the ambient indices has no effect on the values of any of the variants.

The shift tensors $Z_{\alpha}^{i}$ and $Z_{i\alpha}$ are obtained by differentiating the equations of the surface with respect to each of the surface variables, $\theta$ and $\varphi$. Therefore, we have The covariant metric tensor $S_{\alpha\beta}$ is given by the equation and is therefore represented by the matrix which yields Since $\left[ S_{\alpha\beta}\right]$ is diagonal, the covariant basis is orthogonal. Also, you probably recognize $\left[ S_{\alpha\beta}\right]$ as the bottom right $2\times2$ submatrix of the metric tensor associated with spherical coordinates in a three-dimensional Euclidean space. It is left as an exercise to explain why this is so by observing the relationship between $S_{\alpha}$ and the ambient covariant basis corresponding to spherical coordinates.

The contravariant metric tensor $S^{\alpha\beta}$ corresponds to $\left[ S_{\alpha\beta}\right] ^{-1}$, i.e. The area element $\sqrt{S}$ is the square root of the determinant of $\left[ S_{\alpha\beta}\right]$, i.e. If you recall, $\sqrt{S}$ is used in converting geometric integrals over the surface into arithmetic integrals. For example, the total area of the sphere of radius $R$ is given by

Next, let us calculate the shift tensors $Z^{i\alpha}$ and $Z_{i}^{\alpha}$ with raised surface indices. Since, we have where justifying the order of the terms in the matrix products is left as an exercise. Note that, technically, the above products should involve $\left[ S^{\alpha\beta}\right] ^{T}$, however the transpose is not necessary since $\left[ S^{\alpha\beta}\right]$ is symmetric. Carrying out the matrix products, we find

Let us now turn our attention to calculating the components $N^{i}$ of the unit normal. This can be accomplished in two ways. First, recall the normalization condition which, in matrix form, reads It tells us that $\left[ N^{i}\right]$ is contained in the null space of the matrix $\left[ Z_{i\alpha}\right] ^{T}$. Thus, $\left[ N^{i}\right]$ can be calculated by determining the null space of $\left[ Z_{i\alpha }\right] ^{T}$ and then normalizing it to unit length. It is left as an exercise to show that the transposes of the matrices corresponding to all four forms of the shift tensor, i.e. $Z_{\alpha}^{i}$, $Z_{i\alpha}$, $Z_{i}^{\alpha}$, and $Z^{i\alpha}$, have the same null space. Furthermore, calculating $N^{i}$ by this approach is also left as an exercise.

Second, the components $N^{i}$ can be calculated by evaluating the cross product of the vectors $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$, once again followed by normalization to unity. As we discussed in Chapter 3, the components of $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$ are contained in the columns of the matrix In anticipation of the normalization step, we can drop the factors of $R$ from $\left[ Z_{i\alpha}\right]$ in the cross product calculation. Therefore, we have The length of the resulting vector is $\sin\theta$. Thus, dividing by $\sin\theta$, we find or Finally, note that for a sphere, the components $N^{i}$ could have also been easily guessed from geometric considerations as we know that $\mathbf{N}$ is a unit vector pointing in the radial direction and is therefore collinear with the position vector $\mathbf{R}$ if the latter emanates from the center of the sphere.

The last remaining object to be calculated is the surface Christoffel symbol $\Gamma_{\alpha\beta}^{\gamma}$. Recall that the relationship between the surface and the ambient Christoffel symbols reads which reduces to when the ambient space is referred to affine coordinates, which is the case here. The above identity has three free indices $\alpha$, $\beta$, and $\gamma$ and, therefore, cannot be captured by a matrix identity. In order to continue using matrix algebra, we need to fix the value of one of the free indices. We will choose $\beta$ for this role and consider $\beta=1$ and $\beta=2$ separately. Since $S^{1}=\theta$ and $S^{2}=\varphi$, we have The elements of $\partial Z_{\alpha}^{i}/\partial\theta$ are obtained by differentiating $\left[ Z_{\alpha}^{i}\right]$ with respect to $\theta$, i.e. Since we find Meanwhile, the elements of $\partial Z_{\alpha}^{i}/\partial\varphi$ are obtained by differentiating $\left[ Z_{\alpha}^{i}\right]$ with respect to $\varphi$, i.e. As before, since we find In summary, the three nonzero elements of $\Gamma_{\alpha\beta}^{\gamma}$ are

This completes our analysis of a sphere of radius $R$ in Cartesian ambient coordinates. We will now repeat the same analysis in spherical ambient coordinates and it will be particularly interesting to observe what changes and what stays the same.

Let us now perform the parallel analysis using the same coordinates $\theta,\varphi$ on the surface and spherical coordinates $r,\theta _{1},\varphi_{1}$ in the ambient space. Recall that for this combination of coordinate systems, a sphere of radius $R$ is described by the exceedingly simple equations The ambient covariant metric tensor $Z_{ij}$, on the other hand, is more complicated than in the case of Cartesian coordinates. It was calculated in Section TBD of Introduction to Tensor Calculus where we showed that it corresponds to the matrix Thus, the contravariant metric tensor $Z^{ij}$ corresponds to The shift tensor $Z_{\alpha}^{i}$ is obtained by differentiating the equations of the surface with respect to $\theta$ and $\varphi$ and has the exceedingly simple form Since the metric tensor $\left[ Z_{ij}\right]$ does not correspond to the identity matrix, the shift tensor $Z_{i\alpha}$ corresponds to a different matrix compared to $Z_{\alpha}^{i}$. Since we have which yields

The covariant metric tensor $S_{\alpha\beta}$ is related to the shift tensors $Z_{i\alpha}$ and $Z_{\beta}^{i}$ by the equation Therefore, we have Crucially, we have arrived at the same values as before. This is entirely to be expected since $S_{\alpha\beta}$ does not depend on the choice of the ambient coordinates. Thus, the contravariant metric tensor $S^{\alpha\beta}$ and the area element $\sqrt{S}$ also have the same value as before and need not be documented here.

Now we can calculate the two remaining forms of the shift tensor. The contravariant form $Z^{i\alpha}$ is given by and therefore corresponds to the matrix product $\left[ Z_{\alpha} ^{i}\right] \left[ S^{\alpha\beta}\right]$ which yields Finally, $Z_{i}^{\alpha}$ is given by and therefore corresponds to the product $\left[ Z_{ij}\right] \left[ Z^{i\alpha}\right]$ which yields It is interesting to note that $Z_{\alpha}^{i}$ and $Z_{i}^{\alpha}$ have identical values, which is not true in general. It is left as an exercise to explain why it makes sense in this case by considering the relationships between the contravariant surface and ambient space bases.

The components $N^{i}$ of the normal are, once again, easy to guess, since the (exterior) unit normal $\mathbf{N}$ coincides with the ambient basis vector $\mathbf{Z}_{1}$ Therefore, the components $N^{i}$ are given by Note that this result could have also been obtained either from the null space of the matrix or by calculating the cross product of $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$.

Finally, although we already know the elements of the Christoffel symbol $\Gamma_{\alpha\beta}^{\gamma}$, it will be interesting to calculate them again since this time, it is the first term on the right in the identity that survives. Indeed, the second term vanishes since the elements of the shift tensor $Z_{\alpha}^{i}$ are constants. Therefore, we have

Once again, matrix algebra cannot capture all aspects of the above identity. We therefore again have to fix the values of some of the indices. Let us choose $\gamma$ and consider $\gamma=1$ and $\gamma=2$ separately. For $\gamma=1$, we have while for $\gamma=2$, we have Recall that Thus, the only nonzero element in $Z_{k}^{1}$ is $Z_{2}^{1}=1$ and the only nonzero element of $Z_{k}^{2}$ is $Z_{3}^{2}=1$. Thus expressions for $\Gamma_{\alpha\beta}^{1}$ and $\Gamma_{\alpha\beta}^{2}$ are given by the simplified equations which can be handled by matrix algebra. Recall that the nonzero elements of the ambient Christoffel symbol $\Gamma_{ij}^{k}$ are Organize the values of the systems $\Gamma_{ij}^{2}$ and $\Gamma_{ij}^{3}$ on the surface of the sphere into matrices, i.e. With the help of these matrices, we have and Collecting the nonzero elements of $\Gamma_{\alpha\beta}^{\gamma}$, we observe that we have arrived at the same values as before, i.e. This completes our analysis of a sphere of radius $R$.

Consider a cylinder of radius $R$ referred to coordinates $z,\theta$ illustrated in the following figure. As was the case with a sphere, a cylinder can also be effectively analyzed by referring the ambient space either to Cartesian or cylindrical coordinates. We will choose Cartesian coordinates $x,y,z_{1}$ and save cylindrical coordinates for the exercises. Recall that the equations of the surface read Once again, the ambient metric tensors $Z_{ij}$ and $Z^{ij}$ corresponds to the $3\times3$ identity matrix and thus, once again, the placement of the ambient indices has no effect on the values of any of the variants.

The shift tensors $Z_{\alpha}^{i}$ and $Z_{i}^{\alpha}$ are obtained by differentiating the equations of the surface with respect to $\theta$ and $z$, which yields The surface metric tensor $S_{\alpha\beta}$ is given by and therefore while The area element $\sqrt{S}$ is the square root of the determinant of $\left[ S_{\alpha\beta}\right]$, i.e.

Notice that the elements of the metric tensors have constant values, which is a signature feature of affine coordinates. The fact that a cylinder admits such a coordinate system distinguishes it from all the other two-dimensional surfaces discussed in this Chapter. This fact means that, in a certain sense, a cylinder represents a Euclidean space. On the one hand, this makes sense, since a cylinder can be cut along a line of constant $\theta$ and uncurled, without any distortion, into a flat strip. On the other hand, a cylinder is a curved surface that cannot accommodate straight lines, except in one special direction. Understanding this dichotomy will be one of the goals of our study of curvature.

Let us now return to the calculation of the remaining differential objects. The shift tensors $Z^{i\alpha}=S^{\alpha\beta}Z_{\beta}^{i}$ and $Z_{i}^{\alpha}=S^{\alpha\beta}Z_{i\beta}$ correspond to the matrix product $\left[ Z_{\alpha}^{i}\right] \left[ S^{\alpha\beta}\right]$ which yields

As was the case with the sphere, the components $N^{i}$ of the unit normal can be determined from the null space of $\left[ Z_{i\alpha}\right]$, by calculating the cross product of $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$, or simply by intuiting from general geometric considerations. In any case, the components $N^{i}$ correspond to the matrix Since the elements of the metric tensors are constant, the Christoffel symbol $\Gamma_{\alpha\beta}^{\gamma}$ vanishes at all points, i.e. which can also be described as a signature characteristic of an affine coordinate system.

This completes our analysis of a cylinder and we will now turn our attention to the more general surfaces of revolution.

A sphere and a cylinder are both examples of surfaces of revolution. Thus, the analysis presented in this Section represents a generalization of the preceding two Sections.

We will refer the ambient space to Cartesian coordinates and will invite the reader to repeat the analysis in cylindrical coordinates as an exercise. In Cartesian coordinates $x,y,z$, the equations of a surface of revolution read Since we have already performed several similar analyses, we will begin to omit some of the details in the upcoming calculations. Filling in those details will be left as exercises.

The shift tensors $Z_{\alpha}^{i}$ and $Z_{i\alpha}$ are obtained by differentiating the equations of the surface with respect to $\theta$ and $\gamma$, which yields where $G_{\gamma}$ and $H_{\gamma}$ denote the derivatives of the functions $G$ and $H$.

For the covariant and the contravariant metric tensors, we have

The area element $\sqrt{S}$ is given by

For the remaining forms $Z^{i\alpha}$ and $Z_{i}^{\alpha}$ of the shift tensor, we have

For the components $N^{i}$ of the normal, which can be determined by any of the approaches described above, we have

The nonzero elements of the Christoffel symbol $\Gamma_{\beta\gamma}^{\alpha}$ are The findings of this Section will pay instant dividends as we turn our attention to another shape of revolution: a torus.

A torus with radii $R$ and $r$ is a shape of revolution obtained by rotating a circle of radius $r$ around an axis that is a distance $R$ from the center of the circle. Note that, unlike other Sections where the symbol $r$ denotes one of the ambient coordinates, in this Section, $r$ is a constant parameter. Within its plane, the circle is described by the equations The following figure illustrates the resulting surface coordinates $\theta$ and $\varphi$. Note that both coordinates vary from $0$ to $2\pi$. When the ambient space is referred to Cartesian coordinates $x,y,z$, the equations of the torus read All of the following identities can be obtained as special cases of the equations from the previous Section on surfaces of revolution. Therefore, we will catalogue the results without further clarifications.

The shift tensors $Z_{\alpha}^{i}$ and $Z_{i\alpha}$:

The metric tensors $S_{\alpha\beta}$ and $S^{\alpha\beta}$, and the area element $\sqrt{S}$: With the help of the area element $\sqrt{S}$, we can calculate the total area of the torus, i.e.

The shift tensors $Z^{i\alpha}$ and $Z_{i}^{\alpha}$:

The components $N^{i}$ of the normal:

And, finally, the nonzero elements of the Christoffel symbol $\Gamma _{\alpha\beta}^{\gamma}$:

This completes our analysis of two-dimensional surfaces embedded in a three-dimensional Euclidean space and we now turn our attention to curves in a two-dimensional Euclidean plane.

The theory that we have developed for surfaces applies, without any changes, to one-dimensional curves in a two-dimensional Euclidean plane. In fact, with the exception of the normal which requires the surface to be a hypersurface, all of the fundamental differential objects considered in this Chapter can be calculated for a curve embedded in a space of any dimension. We will, however, limit ourselves to curves in the plane here, and save the study of curves in a three-dimensional space for Chapter 8.

A one-dimensional curve is parameterized by a single coordinate $S^{1}$ which we will denote by the letter $\gamma$. First, refer a two-dimensional plane to Cartesian coordinates $x,y$ and consider a general curve given by the equations In the equations that follow, the symbols $x$ and $y$ will refer to the functions $x\left( \gamma\right)$ and $y\left( \gamma\right)$ while $x_{\gamma}$, $x_{\gamma\gamma}$, $y_{\gamma}$, and $y_{\gamma\gamma}$ will refer to the derivatives $x^{\prime}\left( \gamma\right)$, $x^{\prime \prime}\left( \gamma\right)$, $y^{\prime}\left( \gamma\right)$, and $y^{\prime\prime}\left( \gamma\right)$.

The ambient metric tensors $Z_{ij}$ and $Z^{ij}$ correspond to the $2\times2$ identity matrix, i.e.

The shift tensors $Z_{\alpha}^{i}$ and $Z_{i\alpha}$ are obtained by differentiating the equations of the curve with respect to $\gamma$, i.e.

The surface metric tensor $S_{\alpha\beta}=Z_{\alpha}^{i}Z_{i\beta}$ corresponds to the matrix product $\left[ Z_{\alpha}^{i}\right] ^{T}\left[ Z_{i\beta}\right]$. The result is a $1\times1$ matrix with a single entry. Nevertheless, we will continue to use brackets to denote the matrices. We have The contravariant surface metric tensor $S^{\alpha\beta}$ is the matrix inverse of $S_{\alpha\beta}$. Since $\left[ S_{\alpha\beta}\right]$ is a $1\times1$ matrix, its inverse consists of a single entry that is the reciprocal of the sole entry in $\left[ S_{\alpha\beta}\right]$, i.e. The determinant of the matrix $\left[ S_{\alpha\beta}\right]$ equals its only entry. Therefore, the length element $\sqrt{S}$ is given by the equation This is a familiar expression that we have encountered on numerous occasions when studying curves.

The remaining forms of the shift tensor are

The single entry of the Christoffel symbol $\Gamma_{\beta\gamma}^{\alpha}$ is given by