Curves on a surface and principal curvatures

Now that first fundamental form, second fundamental form and Taylor expansion of a surface all become present.
We remind that the first fundamental form can be represented as

(1)
\begin{align} \mathrm{I}\equiv dx^idx^jg_{ij} \end{align}

or

(2)
\begin{align} \mathrm{I}=Ed\theta^1d\theta^1+2Fd\theta^1d\theta^2+Gd\theta^2 d\theta^2. \end{align}

Here,

(3)
\begin{align} g_{ij}=\boldsymbol{g}_i\cdot\boldsymbol{g}_j \end{align}

and

(4)
$$E=g_{11},F=g_{12}=g_{21},G=g_{22}.$$

Also, we remind that the second fundamental form can be represented as

(5)
\begin{align} \mathrm{II}\equiv h_{ij}d\theta^\alpha d\theta^i \end{align}

or

(6)
\begin{align} \mathrm{II}=Ld\theta^1d\theta^1+2Md\theta^1d\theta^2+Nd\theta^2 d\theta^2. \end{align}

Here,

(7)
\begin{align} h_{ij}=\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n} \end{align}

and

(8)
$$L=h_{11},M=h_{12}=h_{21},N=h_{22}.$$

The Taylor expansion of a surface is represented by

(9)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)=\boldsymbol{x}(\theta^1,\theta^2)+\boldsymbol{I}\cdot d\boldsymbol{x}+\frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{H}\cdot d\boldsymbol{x}+\cdots \end{align}

or

(10)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)=\boldsymbol{x}(\theta^1,\theta^2)+d\theta^i\boldsymbol{g}_i+\frac{1}{2}h_{ij}d\theta^i d\theta^j\boldsymbol{n}+\cdots. \end{align}

Now, we consider a curve on a surface, which is defined by

(11)
\begin{align} c(t)=\left\{\theta^1(t),\theta^2(t)\right\}. \end{align}

First, by differentiating it with respect to $t$, we obtain velocity vector as

(12)
\begin{align} \dot{c}(t)\equiv\frac{d c}{dt}=\left\{\frac{d\theta^1(t)}{dt},\frac{d\theta^2(t)}{dt}\right\}. \end{align}

Let's write this as

(13)
\begin{align} =\left\{\dot{\theta^1},\dot{\theta^2}\right\}. \end{align}

Before continuing the discussion, let us check if we can perform differentiation in such a simple way (because it is actually not obvious).
The substance of the curve can be represented by means of position vector.
Substituting $c(t)$ to

(14)
\begin{align} \boldsymbol{r}(\theta^1,\theta^2), \end{align}

we obtain

(15)
\begin{align} \boldsymbol{r}(t)=\boldsymbol{r}(\theta^1(t),\theta^2(t)). \end{align}

This is the substance of the curve. By differentiating this with $t$, we obtain

(16)
\begin{align} \frac{d\boldsymbol{r}(\theta^1(t),\theta^2(t))}{dt}=\frac{\partial\boldsymbol{r}}{\partial\theta^1}\frac{d\theta^1}{dt}+\frac{\partial\boldsymbol{r}}{\partial\theta^2}\frac{d\theta^2}{dt}=\frac{d\theta^i}{dt}\boldsymbol{g}_i. \end{align}

Hence, the definition of velocity is representing contravariant component of a vector, which means that it is a geometrical entity.
Because the substance is a vector, it is assumed that further differentiations contain differentiations of base vectors.
Actually, with respect to an arbitrary function $f(\theta^1,\theta^2)$,

(17)
\begin{align} \frac{d f}{dt}=\frac{\partial f}{\partial \theta^1}\frac{d\theta^1}{dt}+\frac{d\theta^2}{dt}\frac{\partial f}{\partial \theta^2} \end{align}

holds. Hence, (without specifying any concrete function,) between two differential operators, there holds the following relation

(18)
\begin{align} \frac{d}{dt}=\frac{d\theta^1}{dt}\frac{\partial}{\partial \theta^1}+\frac{d\theta^2}{dt}\frac{\partial}{\partial \theta^2}. \end{align}

Because the velocity vector is not a unit vector, let us normalize it. The norm of the velocity is represented by

(19)
\begin{align} \left|\dot{c}(t)\right|=\sqrt{g_{ij}\frac{d \theta^i}{dt}\frac{d \theta^j}{dt}}. \end{align}

Hence, the normalized velocity vector is represented by

(20)
\begin{align} \bar{\dot{c}}=\frac{\dot{c}(t)}{\left|\dot{c}(t)\right|}. \end{align}

Here, note that the substance of this vector is

(21)
\begin{align} \bar{\dot{c}}=\frac{\dot{\theta}^i}{\sqrt{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}}\boldsymbol{g}_i. \end{align}

Now, let us derive the curvature vector by differentiating the velocity vector.
A curvature is a change rate of normalized velocity vector per unit length.
Hence, first we differentiate it by $t$ and obtain

(22)
\begin{align} \frac{d\bar{\dot{c}}}{dt}=\frac{d}{dt}\frac{\dot{\theta}^i}{\sqrt{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}}\boldsymbol{g}_i+\frac{\dot{\theta}^i}{\sqrt{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}}\frac{d\boldsymbol{g}_i}{dt}. \end{align}

Next, by dividing this by $\sqrt{g_{ij}\dot{\theta^i}\dot{\theta^j}}$, we obtain the following vector field (note that this is not a tangent vector field):

(23)
\begin{align} \bar{\ddot{c}}(t)\equiv\frac{1}{\left|\dot{c}(t)\right|}\frac{d}{dt}\frac{\dot{c}(t)}{\left|\dot{c}(t)\right|}=\left(\frac{1}{\left|\dot{c}(t)\right|}\frac{d}{dt}\frac{\dot{\theta}^i}{\left|\dot{c}(t)\right|}\right)\boldsymbol{g}_i+\frac{\dot{\theta}^i}{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}\frac{d\boldsymbol{g}_i}{dt}, \end{align}

which represents the curvature vector of the curve. Although this equation is very complicated, this is a "very" correct expression.
When the curvature vector of the curve is pointing the orthogonal direction of the tangent plane everywhere, the curve is referred to as a geodesic. Additionally, we call the entity that remains after removing the orthogonal component from the curvature vector geodesic curvature and geodesics are curvature of which geodesic curvature is zero everywhere.
The main topic of this section is not to discuss about geodesic curvature but the orthogonal component.
Obviously the first term vanishes and we obtain

(24)
\begin{align} \bar{\ddot{c}}(t)\cdot\boldsymbol{n}=\frac{\dot{\theta}^i}{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}\frac{d\boldsymbol{g}_i}{dt}\cdot\boldsymbol{n}. \end{align}

This follows

(25)
\begin{align} \bar{\ddot{c}}(t)\cdot\boldsymbol{n}=\frac{\dot{\theta}^i}{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}\left(\frac{\partial\boldsymbol{g}_i}{\partial\theta^j}\frac{d\theta^j}{dt}\right)\cdot\boldsymbol{n}. \end{align}

Consequently, using the second fundamental form $h_{ij}$, we write

(26)
\begin{align} \bar{\ddot{c}}(t)\cdot\boldsymbol{n}=\frac{h_{ij}\dot{\theta}^i\dot{\theta}^j}{g_{\alpha\beta}\dot{\theta}^\alpha \dot{\theta}^\beta}. \end{align}

What is extremely important here is that this expression is independent from the shape of the curve but only dependent on the velocity vector at each point.
This means that, on a given point of a surface, we can obtain the above defined curvature by specifying a direction (which is specified by a tangent vector), without specifying any curve.
Hence, we notice that this curvature is independent from the chosen curve and it represents a geometric property of the surface itself. In other word, this curvature is a function of direction.
As a means of giving the direction, if we use a normalized tangent vector defined by

(27)
\begin{align} \boldsymbol{b}=b^i\boldsymbol{g}_i\mid g_{ij}b^ib^j=1. \end{align}

(28)
\begin{align} h(\boldsymbol{b})\equiv h_{ij}b^ib^j \end{align}

can be defined at each point of the surface.
Now, we discuss the maximum and minimum values of this quadratic form when various normalized vectors are substituted to it.
We refer those values to as principal curvatures. To derive the principal curvatures we use Lagrange multiplier method in the following.
This method takes

(29)
$$g_{ij}b^i b^j=1$$

as a constraint condition and take the variation of

(30)
$$h_{ij}b^ib^j$$

with respect to $b^1,b^2$. Because there are two parameters and one condition, the number of the independent parameter is 1. However, reducing the number of parameters to 1 is essentially very difficult. As an alternative way, in Lagrange multiplier method, we solve

(31)
\begin{align} h_{ij}b^ib^j+\lambda(g_{ij}b^i b^j-1)\rightarrow \mathrm{stationary.} \end{align}

By taking the variation with respect to $b^i$, we obtain

(32)
\begin{align} \Leftrightarrow h_{ij}b^j\delta b^i+h_{ij}b^i\delta b^j+\lambda(g_{ij} b^j\delta b^i+g_{ij}b^i \delta b^j)=0. \end{align}

Here, because $h_{ij}=h_{ji}$,$g_{ij}=g_{ji}$, we obtain the stationary condition as follows:

(33)
\begin{align} \Leftrightarrow 2h_{ij}b^i\delta b^j+2\lambda(g_{ij} b^i\delta b^j)=0. \end{align}

If $\delta b^1,\delta b^2$ is a direction that satisfies the constraint condition, (i.e, if we rotate the vector by keeping its length as a unit length), we obtain

(34)
\begin{align} \delta\left( g_{ij}b^ib^j\right)=0\Leftrightarrow 2g_{ij}b^i\delta b^j=0. \end{align}

Thus, the second term in the stationary condition vanishes and the stationary condition becomes

(35)
\begin{align} 2h_{ij}b^i\delta b^j=0. \end{align}

Because this means that $h_{ij}b^ib^j$ does not change if we change the (unitized) direction a little bit, i.e., it is taking the maximum or minimum value, we proved that the Lagrange multiplier method is an appropriate approach to obtain the directions that give the maximum and minimum values of $h_{ij}b^ib^j$.
Let us go back to Eq.(33), it follows:

(36)
\begin{align} \Leftrightarrow h_{ij}b^i\delta b^j+\lambda(g_{ij} b^i\delta b^j)=0 \end{align}
(37)
\begin{align} \therefore h_{ij}b^i+\lambda g_{ij} b^i=0 \end{align}

Inverting the sign of $\lambda$, we obtain

(38)
\begin{align} h_{ij}b^i-\lambda g_{ij} b^i=0. \end{align}

Note that this resembles well with characteristic polynomial.

(39)
\begin{align} \left(\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]-\lambda\left[\begin{array}{ccc}E&F\\F&G\end{array}\right]\right)\left[\begin{array}{c}b^1\\ \\ b^2\end{array}\right]=\left[\begin{array}{c}0 \\0\end{array}\right] \end{align}
(40)
\begin{align} \Leftrightarrow\mathrm{det}\left(\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]-\lambda\left[\begin{array}{ccc}E&F\\F&G\end{array}\right]\right)=0 \end{align}

Therefore,

(41)
\begin{align} (L-\lambda E)(N-\lambda G)-(M-\lambda F)^2=0 \end{align}
(42)
\begin{align} \Leftrightarrow LN+\lambda^2EG-\lambda(LG+EN)-M^2-\lambda^2F^2+2\lambda M F=0 \end{align}
(43)
\begin{align} \Leftrightarrow \lambda^2(EG-F^2)+\lambda(2MF-LG-EN) +LN-M^2=0. \end{align}

The solutions are

(44)
\begin{align} \lambda_1=\frac{-2MF+LG+EN+\sqrt{(2MF-LG-EN)^2-4(EG-F^2)(LN-M^2)}}{2(EG-F^2)} \end{align}

and

(45)
\begin{align} \lambda_2=\frac{-2MF+LG+EN-\sqrt{(2MF-LG-EN)^2-4(EG-F^2)(LN-M^2)}}{2(EG-F^2)}. \end{align}

Note that

(46)
\begin{align} h_{ij}b^i=\lambda b^ig_{ij} \end{align}

holds for each solution and corresponding normalized tangent vector $\boldsymbol{b}=b^i\boldsymbol{g}_i$.

Taking inner product of both sides with $\boldsymbol{b}$ gives

(47)
\begin{align} h_{ij}b^ib^j=\lambda b^ig_{ij}b^j=\lambda. \end{align}

Hence, $\lambda_1,\lambda_2$ are presenting principal curvatures.
From these, mean curvature

(48)
\begin{align} H=\frac{1}{2}(\lambda_1+\lambda_2)=\frac{LG-2MF+EN}{2(EG-F^2)} \end{align}

and Gaussian curvature

(49)
\begin{align} K=\lambda_1\lambda_2=\frac{(LN-M^2)}{(EG-F^2)} \end{align}

can be defined.
When they are generalized to an arbitrary dimensional space, $K$ can be represented by

(50)
\begin{align} K=\frac{\mathrm{det}(h_{ij})}{\mathrm{det}(g_{ij})}. \end{align}

Although $H$ does not have such a simpler form, actually there is. Let us multiply $g^{ij}$ with

(51)
\begin{align} \left(\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]-\lambda\left[\begin{array}{ccc}E&F\\F&G\end{array}\right]\right)\left[\begin{array}{c}b^1\\ \\ b^2\end{array}\right]=\left[\begin{array}{c}0 \\0\end{array}\right] \end{align}

from the left side. Then, because $g^{ij}g_{jk}=\delta^i_k$, we obtain

(52)
\begin{align} \left(\left[\begin{array}{ccc}g^{11}&g^{12}\\g^{12}&g^{22}\end{array}\right]\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\right)\left[\begin{array}{c}b^1\\ \\ b^2\end{array}\right]=\left[\begin{array}{c}0 \\0\end{array}\right]. \end{align}

This means that, the principal curvatures are, in a normal sense, the solutions of a characteristic polynomial

(53)
\begin{align} h_{ij}b^i-\lambda b^j=0. \end{align}

It follows

(54)
\begin{align} \Leftrightarrow\mathrm{det}\left(\left[\begin{array}{ccc}g^{11}&g^{12}\\g^{12}&g^{22}\end{array}\right]\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\right)=0. \end{align}

Thus, we discuss a general formula that takes the form

(55)
\begin{align} \Leftrightarrow\mathrm{det}\left(\left[\begin{array}{ccc}w^1_{\cdot 1}&w^1_{\cdot 2}\\w^2_{\cdot 1}&w^2_{\cdot 2}\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\right)=0. \end{align}

Because

(56)
\begin{align} (w^1_{\cdot 1}-\lambda)(w^2_{\cdot 2}-\lambda)-w^1_{\cdot 2}w^2_{\cdot 1}=0 \end{align}
(57)
\begin{align} \Leftrightarrow \lambda^2-\lambda(w^1_{\cdot 1}+w^2_{\cdot 2})+w^1_{\cdot 1}w^2_{\cdot 2}-w^1_{\cdot 2}w^2_{\cdot 1}=0, \end{align}

we obtain

(58)
\begin{align} \lambda_1=\frac{(w^1_{\cdot 1}+w^2_{\cdot 2})+\cdots}{2} \end{align}

and

(59)
\begin{align} \lambda_2=\frac{(w^1_{\cdot 1}+w^2_{\cdot 2})-\cdots}{2}. \end{align}

Therefore

(60)
\begin{align} \frac{\lambda_1+\lambda_2}{2}=\frac{w^1_{\cdot 1}+w^2_{\cdot 2}}{2}. \end{align}

Based on this, the mean curvature can be expressed as

(61)
\begin{align} H=\frac{g^{ij}h_{ij}}{2}, \end{align}

which can be used for any dimensional space.
Here, we note that $g^{ij}h_{ij}$ is representing the sum of diagonal components of a matrix

(62)
\begin{align} \left[\begin{array}{ccc}g^{11}&g^{12}\\g^{12}&g^{22}\end{array}\right]\left[\begin{array}{ccc}L&M\\M&N\end{array}\right]=g^{ij}h_{jk}, \end{align}

and called the trace of this matrix. A trace of a matrix is defined by

(63)
\begin{align} \delta_i^{\cdot k}g^{ij}h_{jk}=g^{ij}h_{ij}. \end{align}

Such an operation, i.e., taking the sum by coupling a pair of indices like $i=k$, is called a contraction.

By the way, let us check what those forms presented above take when they are represented with dyadic products.
We discuss eigenvalue and eigen-vectors of a rank-2 tensor represented by

(64)
\begin{align} \boldsymbol{h}=h_{ij}\boldsymbol{g}^j\otimes \boldsymbol{g}^i. \end{align}

Using a tangent vector

(65)
\begin{align} \boldsymbol{b}=b^i\boldsymbol{g}_i, \end{align}

the conditions that a pair of eigenvalue $\lambda$ and eigenvector $\boldsymbol{b}$ must satisfy is represented as follows:

(66)
\begin{align} \Leftrightarrow\boldsymbol{h}\cdot\boldsymbol{b}=\lambda\boldsymbol{b}. \end{align}

This follows

(67)
\begin{align} \Leftrightarrow\left(\boldsymbol{h}-\lambda\boldsymbol{I}\right)\cdot\boldsymbol{b}=0. \end{align}

Here, $\boldsymbol{I}$ is a unit matrix, i.e.,

(68)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_i\otimes\boldsymbol{g}^i=g_{ij}\boldsymbol{g}^i\otimes\boldsymbol{g}^j=g^{ij}\boldsymbol{g}_i\otimes\boldsymbol{g}_j=\boldsymbol{g}^i\otimes\boldsymbol{g}^j. \end{align}

Hence, when we have

(69)
\begin{align} \Leftrightarrow\mathrm{det}\left(\boldsymbol{h}-\lambda\boldsymbol{I}\right)=0, \end{align}

there are four ways to explicitly represent it using components of tensors.

(70)
\begin{align} T_{ij}=\boldsymbol{g}_i\cdot\boldsymbol{T}\cdot\boldsymbol{g}_j, \end{align}

we obtain

(71)
\begin{align} \Leftrightarrow \mathrm{det}\left(h_{ij}-\lambda g_{ij}\right)=0. \end{align}

On the other hand, if we adopt

(72)
\begin{align} T^i_{\cdot j}=\boldsymbol{g}^i\cdot\boldsymbol{T}\cdot\boldsymbol{g}_j, \end{align}

we obtain

(73)
\begin{align} \Leftrightarrow \mathrm{det}\left(g^{ij}h_{jk}-\lambda\delta^i_{\cdot k}\right)=0. \end{align}

Consquently, we notice that both truly derive the eigenvalues of $\boldsymbol{h}$, and the cause of the different representations resulted from the fact that a rank-2 tensor have four types of components.
Additionally, in general, the solution of

(74)
\begin{align} \mathrm{det}\left(w^i_{\cdot j}-\lambda \delta^i_{\cdot j}\right)=0, \end{align}

i.e., $\lambda_1,\cdots,\lambda_n$, satisfies

(75)
\begin{align} \mathrm{tr}\left(w^i_{\cdot j}\right)\equiv \sum{w^i_{\cdot i}}=\sum{\lambda_i}. \end{align}

Note that this does not hold with

(76)
\begin{align} \mathrm{det}\left(h_{ij}-\lambda g_{ij}\right)=0. \end{align}
page revision: 68, last edited: 15 May 2015 11:22