User:SamHB/MVCalc2



This is the second lecture in multivariable calculus.

=Submanifolds=

In this lecture we will discuss lower-dimensional regions embedded in a higher-dimensional space, such as lines within 2- or 3-dimensional space, or surfaces within 3-dimensional space. The proper name for the "spaces" we have been discussing is manifold, which means "a set that can have coordinate systems".

The points in an embedded submanifold are described by sets of numbers, commonly called "parameters", though we will just call them coordinates. The assignment of numbers to the points is commonly called "parameterization", though they are really just coordinates. For example, a surface in 3-dimensional space might be described by two parameters or coordinates $$u\,$$ and $$v\,$$, so that the surface is described by functions:
 * $$x(u, v) \ \ \ y(u, v)\ \ \ z(u, v)\,$$

Similarly, a particle's path through space might be "parameterized by time", so that its position at any instant is given by 3 functions:
 * $$x(t) \ \ \ y(t)\ \ \ z(t)\,$$

We will treat the subject of describing submanifolds simply by choosing a coordinate system for the larger space, and then removing some coordinates from the system by setting them to constants. Once we have developed the formulas for working with submanifolds, we will be able to dispense with the extra coordinates.

For example, a circle of radius $$R\,$$, in a 2-dimensional plane, can be described by choosing polar coordinates $$(r, \theta)\,$$ on the plane, centered at the circle's center. Set the $$r\,$$ coordinate to a constant: $$r = R\,$$. It drops out of the coordinate system, leaving $$\theta\,$$ as the single coordinate.

The surface of a sphere, within 3-dimensional space, can be described by choosing spherical coordinates $$(r, \theta, \phi)\,$$, and setting $$r = R\,$$. The coordinates of the surface are $$(\theta, \phi)\,$$. An ellipse can be defined by choosing confocal coordinates and setting $$\rho\,$$ to some constant.

Out With the Jacobian, In With the Metric
There is some unfinished business from lecture 1 that needs to be attended to. We defined the Jacobian matrix in terms of the transformation between two coordinate systems, and we came close to considering the Jacobian between Cartesian coordinates and some coordinate system $$(u, v, w)\,$$ as "the Jacobian of the system $$(u, v, w)\,$$". We stopped short of doing that because it isn't true. The usual Jacobian $$[J\langle xyz/r\theta\phi\rangle]\,$$ for spherical coordinates is:
 * $$J_1\langle xyz/r\theta\phi\rangle = \begin{bmatrix} \sin\theta \cos\phi & r \cos\theta \cos\phi & - r \sin\theta \sin\phi \\ \sin\theta \sin\phi & r \cos\theta \sin\phi & r \sin\theta \cos\phi \\ \cos\theta & - r \sin\theta & 0 \end{bmatrix}$$

But a different Cartesian system, created, for example, by permuting $$x\,$$, $$y\,$$, and $$z\,$$, would have:
 * $$x = r \cos\theta\,$$
 * $$y = r \sin\theta \cos\phi\,$$
 * $$z = r \sin\theta \sin\phi\,$$

with a Jacobian matrix of:
 * $$J_2\langle xyz/r\theta\phi\rangle = \begin{bmatrix} \cos\theta & - r \sin\theta & 0 \\ \sin\theta \cos\phi & r \cos\theta \cos\phi & - r \sin\theta \sin\phi \\ \sin\theta \sin\phi & r \cos\theta \sin\phi & r \sin\theta \cos\phi \end{bmatrix}$$

There is no "Jacobian matrix of spherical coordinates $$(r, \theta, \phi)\,$$". But, fortunately for integration, there is a "Jacobian determinant of $$(r, \theta, \phi)\,$$". The determinants $$\mathbf{J_1}\,$$ and $$\mathbf{J_2}\,$$ are both $$r^2 \sin\theta\,$$.

This explains why we were able to pretend that a given coordinate system had a "Jacobian" when we were developing the integration formulas&mdash;we only used the determinant. If a coordinate system $$(u, v, w)\,$$ has two Jacobian matrices $$[J_1]\,$$ and $$[J_2]\,$$ for its transformations to two Cartesian coordinate systems, we have
 * $$[J_1] = [C][J_2]\,$$

where $$[C]\,$$ is the orthogonal matrix connecting the two Cartesian systems.

Now the determinant of an orthogonal matrix is $$\pm 1\,$$.
 * Why?   $$\ \ \ \ C^t = C^{-1}\,$$, so $$det(C) = det(C^t) = det(C^{-1}) = \frac{1}{det(C)}\,$$

Since we are being careful to maintain orientation, the determinant is always $$+1\,$$.
 * Some textbooks don't bother with that, and have to take the absolute value of the Jacobian determinant in integration formulas.

This means that $$det(J_1) = det(J_2)\,$$, so the integrations formulas work.

But taking the determinant of $$\mathbf{J}\,$$ throws away too much information. Is there some manipulation that we can perform that preserves more information but is still impervious to the choice of the "reference" Cartesian system?

This is sometimes called the "metric tensor" (for reasons that we definitely won't go into), or just the "metric".

The metric matrix, like the Jacobian, is referenced to two coordinate systems: $$[g\langle xyz/uvw\rangle]\,$$, though we will soon see that we don't need to name the Cartesian system.

Like the Jacobian, its determinant is indicated by a boldface lowercase $$\mathbf{g}\,$$ or $$\mathbf{g}\langle xyz/uvw\rangle\,$$.

Theorem: The metric matrix between a given coordinate system and some "reference" Cartesian system is independent of that Cartesian system.

Proof: The Jacobians relating the given "test" system to the two "reference" Cartesian systems are $$[J_1]\,$$ and $$[J_2]\,$$. We showed above (in a "digression") that $$[J_1] = [C][J_2]\,$$ where $$[C]\,$$ is an orthogonal matrix.

So
 * $$[g_1] = [J_1]^t [J_1] = ([C][J_2])^t ([C][J_2]) = [J_2]^t [C]^t [C] [J_2] = [J_2]^t [C]^{-1} [C] [J_2] = [J_2]^t [J_2] = [g_2]\,$$

This means that we really are justified in calling some matrix $$[g]\,$$ the "metric of the coordinate system", as soon as we can calculate it relative to any Cartesian coordinate system. And we no longer need to state what the "reference" system is. $$[g\langle uvw\rangle]\,$$ means the same as $$[g\langle xyz/uvw\rangle]\,$$.

Theorem: $$[g]\,$$ is a symmetric matrix.

Proof:
 * $$[g]^t = ([J]^t [J])^t = [J]^t {[J]^t}^t = [J]^t [J] = [g]\,$$

Theorem: $$\mathbf{J} = \sqrt{\mathbf{g}}\,$$

Proof: $$\det[g] = \det([J]^t [J]) = \det[J]^t \det[J] = (\det[J])^2\,$$

The square root of the metric determinant is much more commonly used than the determinant itself.

We can now reformulate the integration formula from the previous lecture:

Or, letting the $$q_i\,$$ be some Cartesian system, we get the formula for integrating in arbitrary coordinates relative to the metric:

And similarly in other dimensions.

Here are some metrics:

2-dimensional polar coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 \\[2.5ex] 0 & r^2 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2\ \ \ \ \sqrt{\mathbf{g}} = r\,$$

3-dimensional spherical coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & r^2 & 0 \\[2.5ex] 0 & 0 & r^2 \sin^2\theta \end{bmatrix}\ \ \ \ \mathbf{g} = r^4 \sin^2\theta\ \ \ \ \sqrt{\mathbf{g}} = r^2 \sin\theta\,$$

Toroidal coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & (S + r \cos\phi)^2 & 0 \\[2.5ex] 0 & 0 & r^2 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2 (S + r \cos\phi)^2\ \ \ \ \sqrt{\mathbf{g}} = r (S + r \cos\phi)\,$$

Confocal coordinates:
 * $$[g] = \begin{bmatrix} A^2 (\sinh^2\rho + \sin^2\theta) & 0 \\[2.5ex] 0 & A^2 (\sinh^2\rho + \sin^2\theta) \end{bmatrix}\ \ \ \ \mathbf{g} = A^4 (\sinh^2\rho + \sin^2\theta)^2\ \ \ \ \sqrt{\mathbf{g}} = A^2 (\sinh^2\rho + \sin^2\theta)\,$$

Submanifolds
Now that the preliminaries are out of the way, we can get to the subject of this lecture.

A submanifold is the result of setting some coordinates, in some coordinate system of the larger manifold, equal to some constants. The remaining coordinates are the coordinates of the submanifold.

A surface embedded in 3-dimensional space is the result of holding one coordinate constant. If we hold one coordinate constant in 3-dimensional Cartesian coordinates, we get the plane determined by the other 2 coordinates. But we can get more interesting surfaces by using more interesting coordinate systems. And we can get curves in 2- or 3-dimensional spaces by holding constant all cordinates except one. Here are some examples:


 * If the original coordinate system was spherical coordinates, the result of holding $$r = R\,$$ is the surface of a sphere of radius $$R\,$$.


 * A toroidal surface can be the result of holding $$r = R\,$$ in toroidal coordinates $$(r, \theta, \phi)\,$$. Its coordinates are $$\theta\,$$ and $$\phi\,$$.


 * A circle in the plane can be the result of holding $$r = R\,$$ in polar coordinates.


 * A helix in 3 dimensions can be made by introducing "helical coordinates" $$(r, \theta, s)\,$$:
 * $$x = r \cos\theta\,$$
 * $$y = r \sin\theta\,$$
 * $$z = P \theta + s\,$$
 * One then sets $$r = R\,$$ and $$s = 0\,$$. The result is a 1-dimensional manifold, that is, a curve, with coordinate $$\theta\,$$.  $$P\,$$ is the "pitch" of the helix, per radian.


 * Since we will need these soon, here are the Jacobian and metric for helical coordinates:
 * $$[J\langle xyz/r\theta s\rangle] = \begin{bmatrix} \cos\theta & - r \sin\theta & 0 \\[2.5ex] \sin\theta & r \cos\theta & 0 \\[2.5ex] 0 & P & 1 \end{bmatrix}$$


 * $$[g\langle r\theta s\rangle] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & r^2 + P^2 & P \\[2.5ex] 0 & P & 1 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2\ \ \ \ \sqrt{\mathbf{g}} = r\,$$

Does This Really Give Us Freedom to Choose the Coordinates on the Submanifold?
Yes, as long as we have freedom to choose any coordinates on the main manifold. And we do have that freedom&mdash;that's why we did not restrict ourselves to orthogonal coordinate systems.

Here is an example. Suppose we are interested in a surface which is a hemisphere of unit radius, centered at the origin, and above the $$x-y\,$$ plane. But instead of the obvious $$(\theta, \phi)\,$$ coordinates, we want to express a point's position in terms of its $$x\,$$ and $$y\,$$ coordinates, that is, the coordinates of the point on the $$x-y\,$$ plane directly below it. We create an artificial coordinate system $$(u, v, w)\,$$:
 * $$x = u\qquad\qquad\qquad\ \ \ \ u = x\,$$
 * $$y = v\qquad\qquad\qquad\ \ \ \ v = y\,$$
 * $$z = \sqrt{w^2 - u^2 - v^2}\ \ \ \ w = \sqrt{x^2 + y^2 + z^2}\,$$

This system is obviously nonstandard. We calculate:
 * $$[J\langle xyz/uvw\rangle] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & 1 & 0 \\[2.5ex] \displaystyle\frac{- u}{\sqrt{w^2 - u^2 - v^2}} & \displaystyle\frac{- v}{\sqrt{w^2 - u^2 - v^2}} & \displaystyle\frac{w}{\sqrt{w^2 - u^2 - v^2}} \end{bmatrix}$$


 * $$[g\langle uvw\rangle] = \begin{bmatrix} \displaystyle\frac{w^2 - v^2}{w^2 - u^2 - v^2} & \displaystyle\frac{uv}{w^2 - u^2 - v^2} & \displaystyle\frac{- uw}{w^2 - u^2 - v^2} \\[2.5ex] \displaystyle\frac{uv}{w^2 - u^2 - v^2} & \displaystyle\frac{w^2 - u^2}{w^2 - u^2 - v^2} & \displaystyle\frac{- vw}{w^2 - u^2 - v^2} \\[2.5ex] \displaystyle\frac{- uw}{w^2 - u^2 - v^2} & \displaystyle\frac{- vw}{w^2 - u^2 - v^2} & \displaystyle\frac{w^2}{w^2 - u^2 - v^2} \end{bmatrix}$$


 * $$\mathbf{g} = \displaystyle\frac{w^2}{w^2 - u^2 - v^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{w}{\sqrt{w^2 - u^2 - v^2}}\,$$

This is a very peculiar coordinate system. But it might be just what we need if we had to find the electric charge on a hemispherical object in which the charge density was given as a function of $$x\,$$ and $$y\,$$.

Now we set $$w = 1\,$$. We have:
 * $$x = u\,$$
 * $$y = v\,$$
 * $$z = \sqrt{1 - u^2 - v^2}\,$$

We get a 2-dimensional manifold, with a coordinates $$(u, v)\,$$. For all points of the manifold, we have $$\sqrt{x^2 + y^2 + z^2} = 1\,$$, that is, the points lie on the hemisphere.

Now we create the 2-dimensional metric for the hemisphere. To do this, we simply remove the rows and columns of the metric that referred to the coordinates that were removed from the submanifold. We get this metric:
 * $$[g\langle uv\rangle] = \begin{bmatrix} \displaystyle\frac{1 - v^2}{1 - u^2 - v^2} & \displaystyle\frac{uv}{1 - u^2 - v^2} & \\[2.5ex] \displaystyle\frac{uv}{1 - u^2 - v^2} & \displaystyle\frac{1 - u^2}{1 - u^2 - v^2} \end{bmatrix}\ \ \ \ \mathbf{g} = \displaystyle\frac{1}{1 - u^2 - v^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\,$$

This is the (2-dimensional) metric on the hemisphere, effectively in $$(x, y)\,$$ coordinates. We might say that the hemisphere has been parameterized by $$x\,$$ and $$y\,$$.

We might want to express these in the more familiar Cartesian Coordinates. In fact, the only reason we used completely different names $$(u, v, w)\,$$ for the new coordinate system, even though the two coordinates that we were actually interested in are $$x\,$$ and $$y\,$$, is that one must be extremely careful in calculating partial derivatives to obtain the Jacobian.


 * $$[g\langle xy\rangle] = \begin{bmatrix} \displaystyle\frac{1 - y^2}{1 - x^2 - y^2} & \displaystyle\frac{xy}{1 - x^2 - y^2} & \\[2.5ex] \displaystyle\frac{xy}{1 - x^2 - y^2} & \displaystyle\frac{1 - x^2}{1 - x^2 - y^2} \end{bmatrix}\ \ \ \ \mathbf{g} = \displaystyle\frac{1}{1 - x^2 - y^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{1}{\sqrt{1 - x^2 - y^2}}\,$$

We normally think of $$x\,$$ and $$y\,$$ as Cartesian coordinates, but, in this case, on the surface of a sphere, the coordinates are most definitely not Cartesian.

Of course, we can see the 3-dimensional space, including the $$z\,$$ coordinate, and we know that
 * $$\sqrt{\mathbf{g}} = \displaystyle\frac{1}{z}\,$$

but tiny creatures living on the hemispherical surface don't know that. They only see the $$x\,$$ and $$y\,$$ coordinates.

What is accomplished by getting the metric on the 2-dimensional manifold? We can integrate! The square root of the metric determinant is the correction factor that we have to put into an integral to integrate over the hemisphere. The integration is in terms of $$u\,$$ and $$v\,$$, or $$x\,$$ and $$y\,$$ if we clearly understand that these coordinates are not the actual Cartesian ones.

The area of a tiny patch of the hemisphere in these coordinates is
 * $$\sqrt{\mathbf{g}} \Delta u\ \Delta v = \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}} \Delta u\ \Delta v\,$$

So we can use this to obtain the surface area of the hemisphere by integration. Use the usual rule&mdash;plug in the square root of the metric matrix:
 * $$A = \int_{-1}^1 \int_{-\sqrt{1 - u^2}}^{\sqrt{1 - u^2}} \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv\ du\,$$

but
 * $$\int \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv = \tan^{-1}\frac{v}{\sqrt{1 - u^2 - v^2}}\,$$

so
 * $$\int_{-\sqrt{1 - u^2}}^{\sqrt{1 - u^2}} \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv = \tan^{-1}(\infty) - \tan^{-1}(-\infty) = \pi\,$$

so
 * $$A = \int_{-1}^1 \pi\ du = 2 \pi\,$$

Before proving this, we will look at a few more examples.

For the torus, set $$r = R\,$$ in toroidal coordinates, so the 2-dimensional patch has coordinates $$\theta\,$$ and $$\phi\,$$. After removing $$r\,$$, and setting it to the constant $$R\,$$, we have:
 * $$[g] = \begin{bmatrix} (S + R \cos\phi)^2 & 0 \\[2.5ex] 0 & R^2 \end{bmatrix}\ \ \ \ \mathbf{g} = R^2 (S + R \cos\phi)^2\ \ \ \ \sqrt{\mathbf{g}} = R (S + R \cos\phi)\,$$

The reader should convince himself/herself that, for a small "rectangular" patch on the surface of a torus, the "height" around the cross-sectional circle is $$R \Delta \phi\,$$, the "length" around the main circle is $$(S + R \cos\phi) \Delta \theta\,$$, and the area is $$R (S + R \cos\phi) \Delta\theta \Delta\phi\,$$.

For the helix, set $$r = R\,$$ and $$s = 0\,$$ in helical coordinates. It can be seen that the horizontal distance traversed for a small change in $$\theta\,$$ is $$R \Delta\theta\,$$, and the vertical distance is $$P \Delta\theta\,$$. (Remember that $$P\,$$ is the pitch, or vertical rise, per radian of $$\theta\,$$.) The actual distance along the helix itself is $$\sqrt{R^2 + P^2} \Delta\theta\,$$.

Here is the metric, once we have removed the 1st and 3rd rows and columns. The metric matrix is a 1x1 matrix:
 * $$[g] = \begin{bmatrix} R^2 + P^2 \end{bmatrix}\ \ \ \ \mathbf{g} = R^2 + P^2\ \ \ \ \sqrt{\mathbf{g}} = \sqrt{R^2 + P^2}\,$$

Geometrical Meaning of the Metric
To prove what we have been saying about the metric (and to prepare for the next lecture), we need to work out the geometric significance of the metric. It should come as no surprise that something with the wonderful properties we saw in the previous sections should have a simple and elegant geometrical interpretation.

Vector Components Defined Relative to a Coordinate System
Consider the 3 infinitesimal vectors, starting out tangent to the coordinate lines, going from the point $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$, $$(u, v+\Delta v, w)\,$$, and $$(u, v, w+\Delta w)\,$$.

Well, they can't go precisely to those points, because the coordinate lines are curved, and vectors have to be straight. But the notion of such a vector works in the infinitesimal limit. We will define "natural components" for these vectors, relative to our coordinate system, that will make them, in effect, "natural basis vectors".

We will say that:
 * The vector $$[\Delta u, 0, 0]\,$$ goes from $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$


 * The vector $$[0, \Delta v, 0]\,$$ goes from $$(u, v, w)\,$$ to $$(u, v+\Delta v, w)\,$$


 * The vector $$[0, 0, \Delta w]\,$$ goes from $$(u, v, w)\,$$ to $$(u, v, w+\Delta w)\,$$

That is, the 3 components of tiny vectors tell how much the 3 coordinates change when we move from a given point to another point. In the language of vector algebra, we have chosen a new basis for infinitesimal vectors, that is, a new description of vectors in terms of sets of 3 numbers. We have shown just the basis vectors, but of course a vector can point anywhere, and have any or all of its components nonzero.

Now these are actual geometric vectors. As such, they have actual $$x\,$$, $$y\,$$, and $$z\,$$ Cartesian components. The "natural curvilinear vector components" are different from the Cartesian components.


 * We need to emphasize this. Vector lengths, angles, dot products, and cross products are physical geometric concepts that are independent of the coordinate system.  But vectors have different components when measured according to different coordinate systems.  We will explain how, given the coordinate system, these geometric concepts may be calculated from the vector components, such that one always gets the right answer; in fact, this is the main result of this course.  One thing that we can say right now is that you must not assume that the length of a vector is the square root of the sum of the squares of its components, unless the coordinate system is Cartesian.  The length (and everything else!) comes from the knowledge of the coordinate system, which can be found in the metric.

Given a vector's natural curvilinear components, we can get its Cartesian components by using the Jacobian. When we moved from $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$, the $$x\,$$, $$y\,$$, and $$z\,$$ coordinates changed thusly:
 * $$\Delta x = \displaystyle\frac{\partial x}{\partial u} \Delta u = J_{11} \Delta u\,$$


 * $$\Delta y = \displaystyle\frac{\partial y}{\partial u} \Delta u = J_{21} \Delta u\,$$


 * $$\Delta z = \displaystyle\frac{\partial z}{\partial u} \Delta u = J_{31} \Delta u\,$$

So the Cartesian components of the vector with natural curvilinear components $$[\Delta u, 0, 0]\,$$ are $$[J_{11}\Delta u, J_{21}\Delta u, J_{31}\Delta u]\,$$.

Similarly, $$[0, \Delta v, 0]\,$$ has Cartesian components $$[J_{12}\Delta v, J_{22}\Delta v, J_{32}\Delta v]\,$$, and $$[0, 0, \Delta w]\,$$ has Cartesian components $$[J_{13}\Delta w, J_{23}\Delta w, J_{33}\Delta w]\,$$.

Now let's calculate the dot product of the natural vectors $$[\Delta u, 0, 0]\,$$ and $$[0, \Delta v, 0]\,$$.


 * The reader should look at the dot product page if necessary. That particular properties we need are linearity, and the "sum of pairwise products" rule&mdash;in Cartesian coordinates:
 * $$\vec{P} \cdot \vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z\,$$.

Once the components of a vector are expressed in Cartesian terms, the answer is obtained by the sum of pairwise products rule:
 * $$[\Delta u, 0, 0] \cdot[0, \Delta v, 0] = J_{11}\Delta u\ J_{12}\Delta v\ +\ J_{21}\Delta u\ J_{22}\Delta v\ +\ J_{31}\Delta u\ J_{32}\Delta v\,$$
 * $$= \left(J_{11} J_{12} + J_{21} J_{22} + J_{31} J_{32}\right)\Delta u\ \Delta v = [J^t J]_{12}\ \Delta u\ \Delta v = g_{12}\ \Delta u\ \Delta v\,$$

Similarly:
 * $$[\Delta u, 0, 0] \cdot[0, 0, \Delta w] = g_{13}\Delta u \Delta w\,$$
 * $$[\Delta u, 0, 0] \cdot[\Delta u, 0, 0] = g_{11}(\Delta u)^2\,$$

And so on. We can get the dot products of any pairs of natural coordinate basis vectors directly from the metric!

We can only do this for vectors that are of infinitesimal length, because, as we have seen, the vectors only follow the coordinate lines for infinitesimal distances. The vectors start out tangent to the coordinate lines, but the coordinate lines curve while the vectors stay straight. The rule here is: when vectors are considered to "point to" a point on the manifold, they must be infinitesimal. This doesn't mean that we can never deal with large vectors&mdash;when doing physical problems involving electric and magnetic fields, for example, the vectors can be any size we like. We just can't interpret such vectors as "pointing to" some location on the manifold.

A few things that we noticed before now become obvious.
 * The symmetry of the metric matrix follows from the symmetry of the dot product.
 * If the coordinate system is orthogonal, so that the coordinate lines are orthogonal at any point, then the tiny basis vectors are orthogonal, so their dot products are zero. This means that the off-diagonal entries in the metric matrix are zero in that case.
 * The length of a vector is the square root of its dot product with itself, so the length of the vector $$[\Delta u, 0, 0]\,$$ is $$\sqrt{g_{11}}\Delta u\,$$. If the coordinate system is orthogonal, so that we have Lamé coefficients, this is just $$h_1\Delta u\,$$.  The Lamé coefficients are the scale factors for the lengths of infinitesimal basis vectors, if the coordinate system is orthogonal.

But we can do even better. The metric matrix gives us directly (albeit laboriously) the dot products of any vectors, in terms of their natural components. This is because the dot product is linear in each argument:
 * $$\vec{P} \cdot (\lambda \vec{Q}) = \lambda (\vec{P} \cdot \vec{Q})\,$$
 * $$\vec{P} \cdot (\vec{Q} + \vec{R}) = \vec{P} \cdot \vec{Q} + \vec{P} \cdot \vec{R}\,$$

By breaking a vector into the sum of the basis vectors times its various components, and using linearity, we have:
 * $$\vec{P} \cdot \vec{Q} = \sum_{i, j = 1}^3 g_{ij} P_i Q_j\,$$

where $$P_i\,$$ and $$Q_j\,$$ are the natural components of the two vectors.

This is more easily calculated using matrix manipulations. Express one vector as a "row matrix", followed by the metric matrix, followed by the other vector as a "column matrix". When the three matrices are multiplied, the result is a 1x1 matrix. The value in that matrix is the dot product.
 * $$\vec{P} \cdot \vec{Q} = \begin{bmatrix} P & \text{as} & \text{row} \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} Q \\ \text{as} \\ \text{column} \end{bmatrix}\,$$.

The Metric on a Submanifold
It is now easy to see why we can use the method of striking rows and columns out of the metric matrix to get the metric for a submanifold. The elements of the metric matrix are just the dot products of basis vectors. A basis vector in a submanifold is also a basis vector in the larger manifold, with all of the dropped components equal to zero. The dot product of two vectors is the same whether we use the large metric matrix or the small one, because the extra matrix elements are multiplied by zero.

In fact, any vector in the submanifold is a vector in the larger manifold, with the extra components in the larger manifold equal to zero. So the smaller metric matrix gives the dot products of vectors in the submanifold, as required.

Finally, let's give a direct proof that the square root of the metric determinant gives the correct "area/length/volume element correction factor" in integrals.

We have tiny basis vectors
 * $$\vec{U} = [\Delta u, 0, 0]\,$$
 * $$\vec{V} = [0, \Delta v, 0]\,$$
 * $$\vec{W} = [0, 0, \Delta w]\,$$

We need to find the actual geometrical volume of the parallelepiped that they span. If they were orthogonal (which they would be if the coordinate system were orthogonal), they would span a rectangular region, and the volume would be simply the product of their geometrical lengths. But in the general case, we need to use the "parallelogram trick" to make them all perpendicular to each other. This is done by adding appropriate multiples of one vector to another. This does not change the volume, but the correct choice of multiplier can make the vectors orthogonal.

First, we subtract a multiple of $$\vec{U}\,$$ from $$\vec{V}\,$$, obtaining $$\vec{P}\,$$. Let that multiple be:
 * $$A \frac{\Delta v}{\Delta u}\,$$

We have:
 * $$\vec{P} = \vec{V} - A \frac{\Delta v}{\Delta u} \vec{U} = [0, \Delta v, 0] - A [\Delta v, 0, 0] = [-A, 1, 0] \Delta v\,$$

Then
 * $$\vec{P} \cdot \vec{U} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \Delta u \Delta v = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} g_{11} \\ g_{12} \\ g_{13} \end{bmatrix} \Delta u \Delta v = (g_{12} - A g_{11}) \Delta u \Delta v\,$$.

So, to get $$\vec{P} \perp \vec{U}\,$$, we need this equal to zero, so:
 * $$A = \displaystyle\frac{g_{12}}{g_{11}}\,$$

While we are "straightening" the vectors by subtracting multiples of one vector from another without changing the volume that they span, we perform a similar operation on the metric matrix, subtracting multiples of one row from another. A theorem of determinants tells us that this does not change the value of the determinant. We subtract $$A\,$$ times the first row from the second row.


 * $$\mathbf{g} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ g_{12} & g_{22} & g_{23} \\ g_{13} & g_{23} & g_{33} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ g_{12} - A g_{11} & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} & g_{23} & g_{33} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} & g_{23} & g_{33} \end{vmatrix}\,$$

A very important cancelation occurs.

Similarly, we create a vector $$\vec{Q}\,$$ by subtracting a multiple of $$\vec{U}\,$$ from $$\vec{W}\,$$.
 * $$\vec{Q} = \vec{W} - B \frac{\Delta w}{\Delta u} \vec{U} = [-B, 0, 1] \Delta w\,$$

The value to make $$\vec{Q}\,$$ perpendicular to $$\vec{U}\,$$ is:
 * $$B = \displaystyle\frac{g_{13}}{g_{11}}\,$$

Now $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{Q}\,$$ span the necessary volume, and $$\vec{P}\,$$ and $$\vec{Q}\,$$ are both perpendicular to $$\vec{U}\,$$, but not to each other.

We also subtract $$B\,$$ times the first row of the matrix that we are working on from the third row. As before, the determinant is still $$\mathbf{g}\,$$:
 * $$\mathbf{g} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} - B g_{11} & g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ 0 & g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix}\,$$

But the rules for calculating determinants say that we can now remove the first column:
 * $$\mathbf{g} = (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix}\,$$

Finally, we "straighten" $$\vec{Q}\,$$, to make it perpendicular to $$\vec{P}\,$$. We define:
 * $$\vec{R} = \vec{Q} - C \frac{\Delta w}{\Delta v} \vec{P} = [AC-B, -C, 1] \Delta w\,$$

Now:
 * $$\vec{R} \cdot \vec{P} = \begin{bmatrix} AC-B, -C, 1 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} -A \\ 1 \\ 0 \end{bmatrix} \Delta v \Delta w\,$$


 * $$= \begin{bmatrix} AC-B,1 -C, 1 \end{bmatrix} \begin{bmatrix} g_{12} - A g_{11} \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} \Delta v \Delta w = \begin{bmatrix} AC-B, -C, 1 \end{bmatrix} \begin{bmatrix} 0 \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} \Delta v \Delta w\,$$


 * $$= (-C (g_{22} - A g{12}) + (g_{23} - A g{13}) ) \Delta v \Delta w\,$$

To get $$\vec{R} \perp \vec{P}\,$$, this must be zero, so:
 * $$C = \displaystyle\frac{g_{23} - A g_{13}}{g_{22} - A g_{12}}\,$$

Also, $$\vec{U} \cdot \vec{R} = \vec{U} \cdot (\vec{Q} - C \frac{\Delta w}{\Delta v} \vec{P}) = \vec{U} \cdot \vec{Q} - C \frac{\Delta w}{\Delta v} \vec{U} \cdot \vec{P} = 0\,$$

So $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{R}\,$$ are all orthogonal, so the volume is the product of their lengths.

We will need these facts presently:
 * $$A g_{13} = B g_{12}\,$$
 * $$C(g_{22} - A g_{12}) = g_{23} - A g_{13}\,$$

We subtract $$C\,$$ times the first row of the (now 2x2) matrix that we are working on from the second row.
 * $$\mathbf{g} = (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{23} - B g_{12} - C(g_{22} - A g_{12}) & g_{33} - B g_{13} - C(g_{23} - A g_{13}) \end{vmatrix}\,$$


 * $$= (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ 0 & g_{33} - B g_{13} - C(g_{23} - A g_{13}) \end{vmatrix} = (g_{11})\ (g_{22} - A g_{12})\ (g_{33} - B g_{13} - C(g_{23} - A g_{13}))\,$$


 * $$= (g_{11})\ (g_{22} - A g_{12})\ ((AC-B) g_{13} - C g_{23} + g_{33})\,$$

Now calculate the lengths of $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{R}\,$$.


 * $$\frac{\vec{U} \cdot \vec{U}}{(\Delta u)^2} = \begin{bmatrix} 1, 0, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1, 0, 0 \end{bmatrix} \begin{bmatrix} g_{11} \\ g_{12} \\ g_{13} \end{bmatrix} = g_{11}\,$$


 * $$\frac{\vec{P} \cdot \vec{P}}{(\Delta v)^2} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} -A \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} g_{12} - A g_{11} \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} 0 \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} = g_{22} - A g_{12}\,$$


 * $$\frac{\vec{R} \cdot \vec{R}}{(\Delta w)^2} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} AC - B \\ -C \\ 1 \end{bmatrix} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} (AC - B) g_{11} - C g_{12} + g_{13} \\ (AC - B) g_{12} - C g_{22} + g_{23} \\ (AC - B) g_{13} - C g_{23} + g_{33} \end{bmatrix}\,$$

Making use of the previous facts:
 * $$\frac{\vec{R} \cdot \vec{R}}{(\Delta w)^2} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ (AC - B) g_{13} - C g_{23} + g_{33} \end{bmatrix} = (AC - B) g_{13} - C g_{23} + g_{33}\,$$

But these three dot products are just the factors of $$\mathbf{g}\,$$ that we worked out previously. So
 * $$\|\vec{U}\| \|\vec{P}\| \|\vec{R}\| = \sqrt{\mathbf{g}}\ \Delta u\ \Delta v\ \Delta w\,$$

Example: What is the surface area of a torus?
 * $$\int_0^{2\pi} \int_0^{2\pi} R (S + R \cos \phi)\ d\theta\ d\phi = 2\pi\int_0^{2\pi} (RS + R^2 \cos \phi)\ d\phi = 4 \pi^2 RS$$

Exercise: Here is one in which the integrand isn't just 1. The Gaussian curvature, per unit area, of a torus is $$K = \displaystyle\frac{\cos\phi}{R (S + R \cos \phi)}\,$$. (The calculation of this is rather advanced. It involves either Riemann's tensor or the "second fundamental form".)  We have expressed this in toroidal coordinates, of course. Note that it is positive on the outer (bulging) part, and negative on the inner (saddle-like) part. What is the total surface integral of the Gaussian curvature, $$\iint_T K dA\,$$?

=Footnotes and References=