San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 Differential Geometry from a Differential Forms' Point of View

Differential geometry is like a fabulous city standing at the confluence of two rivers, drawing upon the riches of both river systems. Only differential geometry stands at the confluence of three streams of mathematics; analysis, geometry and algebra. Its beauty comes from the combining of elements of these three streams.

## Space

The stage for differential geometry is Euclidean three space. A point in three space is specified by its three componenets, (x, y, z) or alternatively (x1, x2, x3). Functions may be defined over regions of this three space. The range of a function could be real numbers. That is to say, at every point P=(x,y,z) of a region a real number is given; i.e., f(P). The range of a function could be a vector, an element with three components and a member of a vector space V; i.e., v(P). This type of function is called a vector field. An example of a vector field is the wind direction and magnitude. A field might have two or three vectors specified at every point of a region. An interesting case of this type is where three basic vectors of a vector space are defined at every point of the region. This is called a frame field.

## Curves

Although one might think that the way to define curves in three space is simply as a set of points this is not the approach that has proven fruitful. The preferred way to define curves is in terms of parametrization. A function is defined over some interval for a parameter, say t, that gives the coordinates of a point; i.e., a function f is specified such that I:t → P or (x(t), y(t), z(t)) is given for t belonging to some interval I. As t varies over its interval a path in three space is traced which is a curve.

If the parametrized curve is interpreted as the path of a particle as a function of time then some of the aspects of the curve have an interesting physical interpretation. The vector given by:

#### v = (dx(t)/dt, dy(t)/dt, dz(t)/dt)

is the velocity of the particle. The speed of the particle is the magnitude of the vector v. A unit speed vector u(t) can be defined by dividing the vector v by its magnitude; i.e., u(t) = v(t)/|v(t)|. This vector u(t) then has the property that u(t)·u(t) = 1. If this equation is differentiated with respect to t the result is

#### 2(du(t)/dt)·u(t)= 0,

which says that the derivative of u(t), du(t)/dt, is always perpendicular to u(t).

At this point it is desirable to switch to a special parametrization of the curve. The natural parameter for the curve is arc length. If s is arc length, distance along the curve, from the start of the curve then

#### ds/dt = |v(t)| = ((dx/dt)2+(dy/dt)2+(dz/dt)2)1/2

If this equation is solved to obtain arc length as a function of t, say s = F(t) and F has an inverse then t could be replaced in the parameterization by F-1(s) and the curve would be given by (x(s), y(s), z(s)). Now we find that the vector in the direction of movement as the parameter s changes is a unit vector; i.e.,

#### T(s) = (dx(s)/ds, dy(s)/ds, dz(s)/ds).

T(s) is the unit vector which is tangent to the curve. It then follows that dT(s)/ds·T(s)= 0; i.e., dT(s)/ds is perpendicular (or orthogonal or normal) to T(s). The magnitude of the vector dT(s)/ds is called the curvature of the curve and is denoted κ. That is to say, κ(s) = |dT(s)/ds|. The unit vector in the direction of dT(s)/ds is defined as the normal to the curve and denoted as N(s) where N(s) = (dT(s)/ds)/κ. Note that T(s) and N(s) are orthogonal; i.e., T(s)·N(s) = 0.

The unit vector that is perpendicular to both T(s) and N(s) is called the binormal B(s) of the curve and is given by the cross product of T(s) and N(s); i.e., B(s) = T(s)xN(s).

The set of vectors {T(s), N(s), B(s)} at any point P of the curve is an orthonormal basis and consequently any vector V can be represented in terms of this set; i.e.,
V = v1T(s)+v2N(s)+v3B(s). Furthermore the coefficients in this representation can be easily expressed in terms of the dot product of the vector V with the orthonormal basis vectors; i.e., v1 = V·T(s), v2 = V·N(s), and v3 = V·B(s).

There are six independent conditions which are satisfied by an orthonormal frame field on a curve. The differentiation of each of these conditions gives us information revelevant for finding the representation of vectors. Below is a table of those conditions and the implication resulting from differentiation.

Condition      Differentiation      Implication
T(s)·T(s) = 12T(s)·dT/ds = 0 T(s)·dT/ds = 0
T(s)·N(s) = 0T(s)·dN/ds + dT/ds·N(s) = 0 dT/ds·N(s) =
-T(s)·dN(s)/ds
T(s)·B(s) = 0T(s)·dB/ds + dT/ds·B(s) = 0 dT/ds·B(s) =
-T(s)·dB(s)/ds
N(s)·N(s) = 12N(s)·dN/ds = 0 N(s)·dN/ds = 0
N(s)·B(s) = 0N(s)·dB/ds + dN/ds·B(s) = 0 dN/ds·B(s) =
-N(s)·dB(s)/ds
B(s)·B(s) = 12B(s)·dB/ds = 0 B(s)·dB/ds = 0

Now let us determine the representations of dT/ds, dN/ds and dB/ds in terms of T(s), N(s) and B(s). We know be definition that dT/ds = κN(s) so this one is easy. But as a preliminary exercise we can verify its representation.

• First, we know that T(s)·dT(s)/ds = 0.
• Second, we have that
N(s)·dT(s)/ds = N(s)·(κN(s)) = κN(s)·N(s) = κ since N(s) is a unit vector.
• Third, we have B(s)·dT(s)/ds = - B(s)·κN(s) = κB(s)·N(s) = 0, from the orthogonality of B(s) and N(s).

Thus we verify that dT(s)/ds = 0(T(s))+κ(N(s))+0(B(s)) = κN(s).

For dN(s)/ds we need T(s)·dN(s)/ds, N(s)·dN(s)/ds and B(s)·dN(s)/ds. We know that T(s)·dN(s)/ds is equal to -N(s)·dT(s)/ds = -κN(s)·N(s) = -κ. We know that N(s)·dN(s)/ds = 0. Finally, B(s)·dN(s)/ds is a term we cannot further evaluate, but we can give it a more familiar designation. It will be called τ. Thus the representation of dN(s)/ds is -κT(s) + 0(N(s)) τB(s).

For dB(s)/ds we need T(s)·dB(s)/ds, N(s)·dB(s)/ds and B(s)·dB(s)/ds. We know that T(s)·dB(s)/ds is equal to -B(s)·dT(s)/ds = -κB(s)·N(s) = 0. N(s)·dB(s)/ds is equal to -B(s)·dN(s)/ds which has been given the designation τ. Thus the representation of dB(s)/ds is -κT(s)-τN(s)+0(B(s). When thesee representations are expressed in matrix form the result is:

 | dT(s)/ds | | 0 κ 0 | | T(s) | | dN(s)/ds | = | -κ 0 τ | | N(s) | | dB(s)/ds | | 0 -τ 0 | | B(s) |

These are the Frenet formulas, first discovered by Frenet (documented in his doctoral dissertation in 1847 but not published in a mathematics journal until 1852) and independently discovered again by Serret and published in a mathematics journal in 1851. The matrix which gives the rate of changes of the frame field in terms of the frame field itself is a skew-symmetric matrix; i.e., element (j,i) is the negative of element (i,j) and thus the elements on the diagonal must be zero.

Sources:

• Barrett O'Neil, Elementary Differential Geometry, Academic Press, 1966.
• Dirk Struik, Lectures on Classical Differential Geometry, Second Edition, Dover Publications, 1961.