Surface

### Surface area and its variation

A function that gives surface area is one of the basic funcationals and it is expressed as:

(1)
\begin{align} a=\int_a{da}, \end{align}

where $a$ is the area, $\int_a$ represents an integration over the surface domain and $da$ is an area element.
While this expression is easy to understand, we cannot perform anything with it unless we know the detailed expression of $da$.
Hence, we want to rewrite it as

(2)
\begin{align} a=\int_a{f(\theta^1,\theta^2)d\theta^1d\theta^2} \end{align}

and identify the concrete expression of $f(\theta^1,\theta^2)$.
Consequently, we obtain

(3)
\begin{align} f=\sqrt{g_{11}g_{22}-g_{12}^2}, \end{align}

where $g_{ij}$ is a metric.
This has been historically expressed as

(4)
\begin{align} f=\sqrt{EF-G^2}. \end{align}

If we suppose that $g_{ij}$ represents a matrix as well as representing its components, i.e.,

(5)
\begin{align} g_{ij}=\left[\begin{array}{cc}g_{11}g_{12}\\g_{21}g_{22}\end{array}\right], \end{align}

we can write

(6)
\begin{align} f=\sqrt{\mathrm{det}g_{ij}}. \end{align}

The variation of the surface area can be written as

(7)
\begin{align} \delta a=\int_{a}{\delta f d\theta^{1}d\theta^{2}}. \end{align}

Thus, we calculate the variation of $f$.

(8)
\begin{align} \delta f =\delta\sqrt{\mathrm{det}g_{ij}}=\frac{1}{2}\frac{\delta\mathrm{det}g_{ij}}{\sqrt{\mathrm{det}g_{ij}}} \end{align}
(9)
\begin{align} \because \delta y^{p}=py^{p-1}\delta y. \end{align}

Thus, we calculate the variation of $\mathrm{det}g_{ij}$.
This calculation is extremely complicated but it is like as follows:

(10)
\begin{align} \delta\mathrm{det}g_{ij}=\mathrm{det}g^{ij}\frac{\delta\mathrm{det}g_{ij}}{\mathrm{det}g^{ij}}=\mathrm{det}g^{ij}\mathrm{det}\left(\hat{g}^{ik}\delta g_{kj}\right). \end{align}

Here, $\hat{\cdot}$ on $\hat{g}^{jk}$ means that $g^{jk}$ does not subject variation.
The calculation continues as follows:

(11)
\begin{align} =\mathrm{det}g_{ij}\left(\mathrm{det}\left(\delta^i_{\cdot j}+\hat{g}^{ik}\delta g_{kj}\right)-1\right) \end{align}

and

(12)
\begin{align} =\mathrm{det}g_{ij}g^{ij}\delta g_{ji}. \end{align}

As the result, we obtain

(13)
\begin{align} \delta a=\frac{1}{2}\int{g^{ij}\delta g_{ji}\sqrt{\mathrm{det}g_{ij}}d\theta^1d\theta^2} \end{align}
(14)
\begin{align} \therefore\delta a=\frac{1}{2}\int{g^{ij}\delta g_{ji}\mathrm{da}}. \end{align}

This is dimension-free expression so it can be easily generalized to an arbitrary dimension.

(15)
\begin{align} \delta\sqrt{\det{g_{ij}}}=\frac{1}{2}g^{\alpha\beta}\delta g_{\alpha\beta}\sqrt{\det{g_{ij}}} \end{align}

is a relation that is frequently used in relativity.
Here, because we have

(16)
\begin{align} \frac{1}{2}\delta g_{ij}=\frac{1}{2}\left(\delta \boldsymbol{g}_i \cdot \boldsymbol{g}_j+\boldsymbol{g}_i\cdot\delta \boldsymbol{g}_j\right), \end{align}

we obtain

(17)
\begin{align} \delta a=\frac{1}{2}\left(\int{ g^{ij}\delta\boldsymbol{g}_i\cdot \boldsymbol{g}_j\mathrm{da}}+\int{ g^{ij}\delta\boldsymbol{g}_j\cdot \boldsymbol{g}_i\mathrm{da}}\right)=\frac{1}{2}\left(\int{\delta \boldsymbol{g}_i\cdot \boldsymbol{g}^i\mathrm{da}}+\int{\delta \boldsymbol{g}_j\cdot \boldsymbol{g}^j\mathrm{da}}\right) \end{align}
(18)
\begin{align} =\int{\delta \boldsymbol{g}_i\cdot \boldsymbol{g}^i\mathrm{da}} \end{align}
(19)
\begin{align} =\int{\frac{\partial\delta\boldsymbol{u}}{\partial \theta^i}\cdot \boldsymbol{g}^i\mathrm{da}} \end{align}
(20)
\begin{align} =\int{\mathrm{div}\delta\boldsymbol{u}\mathrm{da}}. \end{align}

### Taylor expansion of a surface

Using dyadic products, we can write the Taylor expansion of a surface as

(21)
\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}

Here we defined $d\boldsymbol{x},\boldsymbol{I},\boldsymbol{H}$ by

(22)
\begin{align} d\boldsymbol{x}\equiv d\theta^i\boldsymbol{g}_i, \end{align}
(23)
\begin{align} \boldsymbol{I}= \boldsymbol{x}\otimes\nabla=\frac{\partial \boldsymbol{x}}{\partial \theta^i}\otimes\boldsymbol{g}^i=\boldsymbol{g}_i\otimes\boldsymbol{g}^i=\boldsymbol{g}^i\otimes\boldsymbol{g}_i \end{align}

and

(24)
\begin{align} \boldsymbol{H}= \nabla\otimes\left(\boldsymbol{x}\otimes\nabla\right)=\boldsymbol{g}^\alpha\otimes\frac{\partial}{\partial \theta^\alpha}\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right). \end{align}

Actually, if one uses the second fundamental form and normalized normal vector $\boldsymbol{n}$, this Tayler expansion can also be written in a simpler form as

(25)
\begin{align} =\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}

Indeed this is more readable representation but we derive this in the second fundamental form and only derive the representation expressed with dyadic product in this page.
Original Taylor expansion
We first try to recall the normal Taylor expansion.
First, we suppose that

(26)
\begin{align} f(x_0+\epsilon)=f(x_0)+a\epsilon+b\epsilon^2+c\epsilon^3+\cdots. \end{align}

Second, we take differentiation of both sides with respect to $\epsilon$ recursively.
(Before that) let us make a convention, such that we write

(27)
\begin{align} f'(x_0+\epsilon)=\frac{df}{d\epsilon} \end{align}

and

(28)
\begin{align} f''(x_0+\epsilon)=\frac{d^2f}{d\epsilon^2}. \end{align}

Then, we can write

(29)
\begin{align} f'=a+2b\epsilon+3c\epsilon^2+\cdots, \end{align}
(30)
\begin{align} f''=2b+6c\epsilon+\cdots, \end{align}

and

(31)
\begin{align} f'''=6c+\cdots. \end{align}

By substituting $\epsilon=0$ to them, we obtain

(32)
$$a=f',$$
(33)
\begin{align} b=\frac{1}{2}f'', \end{align}

and

(34)
\begin{align} c=\frac{1}{6}f'''. \end{align}

When those are substituted to the first equation, we obtain

(35)
\begin{align} f(x_0+\epsilon)=f(x_0)+f'\epsilon+\frac{1}{2}f''\epsilon^2+\frac{1}{6}f'''\epsilon^3+\cdots. \end{align}

This procedure is the original Taylor expansion.
Taylor expansion of a surface
In the following, $\hat{\cdot}$ means that $\cdot$ is fixed.
Basically, a differentiation is defined with respect to one-parameter only. Hence we assume that a "fixed" coordinates

(36)
\begin{align} \left\{\hat{\theta}^1,\hat{\theta}^2\right\} \end{align}

and direction of differentiation

(37)
\begin{align} d\hat{\boldsymbol{x}}(\epsilon)=\epsilon\left(d\hat{\theta^i}\hat{\boldsymbol{g}}_i\right) \end{align}

is given. Note that there is only one parameter, $\epsilon$, in those expressions.

(38)
\begin{align} \hat{\boldsymbol{x}}(\epsilon)\equiv\boldsymbol{x}(\hat{\theta}^1+\epsilon d\hat{\theta}^1,\hat{\theta}^2+\epsilon d\hat{\theta}^2), \end{align}

where the right-hand side is the position vector that represents the shape of the surface and the left-side hand is a one-parameter vector-valued function that was determined by fixing coordinates and direction of small movement on the surface.
Note that $\hat{\cdot}$ indicates that all the parameters are fixed except $\epsilon$.
Thus, functions became one-parameter functions so we can perform the normal procedure of the Taylor expansion.
By the way, let us suppose

(39)
\begin{align} \hat{\boldsymbol{x}}(\epsilon)=\hat{\boldsymbol{x}}(0)+a\epsilon+b\epsilon^2+c\epsilon^3+\cdots. \end{align}

Here note that

(40)
\begin{align} \hat{\boldsymbol{x}}(0)=\boldsymbol{x}(\hat{\theta}^1,\hat{\theta}^2). \end{align}

By way of trial, we perform the differentiation of the left-hand side with respect to $\epsilon$. Because of the chain-rule, we obtain

(41)
\begin{align} \frac{d\hat{\boldsymbol{x}}}{d\epsilon}=\frac{\partial\boldsymbol{x}}{\partial\theta^1}\frac{d\theta^1}{d\epsilon}+\frac{\partial\boldsymbol{x}}{\partial\theta^2}\frac{d\theta^2}{d\epsilon}=\frac{\partial \boldsymbol{x}}{\partial \theta^1}d\hat{\theta}^1+\frac{\partial \boldsymbol{x}}{\partial \theta^2}d\hat{\theta}^2. \end{align}

Here, we used

(42)
\begin{align} \theta^1(\epsilon)=\hat{\theta}^1+\epsilon d\hat{\theta}^1 \end{align}

and

(43)
\begin{align} \theta^2(\epsilon)=\hat{\theta}^2+\epsilon d\hat{\theta}^2. \end{align}

In addition, because we can write

(44)
\begin{align} \otimes\nabla d\hat{\boldsymbol{x}}=\epsilon\left(\frac{\partial}{\partial \theta^1} d\hat{\theta^1}+\frac{\partial}{\partial \theta^2}d\hat{\theta^2}\right) \end{align}

in genral, in the following calculations we can extensively exploit the following relation:

(45)
\begin{align} \frac{d\hat{\ \ }}{d \epsilon}\epsilon=\otimes\nabla\cdot d\hat{\boldsymbol{x}}. \end{align}

It is not mistaken that $\epsilon$ is not written as $d\epsilon$. This is OK because of the conventions we prepared in the above.
Consequently, we can conclude that differentiation with respect to $\epsilon$ and directional differentiation in $d\hat{\boldsymbol{x}}$ are essentialy identical.
By using those, we can write the first order differentiation as

(46)
\begin{align} \frac{d\hat{\boldsymbol{x}}}{d\epsilon}\epsilon=\boldsymbol{x}\otimes\nabla\cdot d\hat{\boldsymbol{x}}=\boldsymbol{I}\cdot d\hat{\boldsymbol{x}}. \end{align}

The second order differentiation is written as

(47)
\begin{align} \frac{d\hat{\boldsymbol{x}}}{d\epsilon^2}\epsilon^2=d\hat{\boldsymbol{x}}\cdot\nabla\otimes\left(\boldsymbol{I}\cdot d\hat{\boldsymbol{x}}\right). \end{align}

Although it's complicated appearance, because $d\hat{\boldsymbol{x}}$ is only a function of $\epsilon$, we can rewrite

(48)
\begin{align} \frac{d\hat{\boldsymbol{x}}}{d\epsilon^2}\epsilon^2=d\hat{\boldsymbol{x}}\cdot\boldsymbol{H}\cdot d\hat{\boldsymbol{x}}. \end{align}

We are almost in the final stage.
We recursively differentiate Eq. (39).

(49)
\begin{align} \frac{d\hat{\boldsymbol{x}}}{d\epsilon}=a+2b\epsilon+3c\epsilon^2\cdots \end{align}
(50)
\begin{align} \frac{d^2\hat{\boldsymbol{x}}}{d\epsilon^2}=2b+6c\epsilon\cdots \end{align}

Hence, by substituting $\epsilon=0$, we obtain

(51)
\begin{align} a\epsilon=\boldsymbol{I}\cdot d\boldsymbol{x} \end{align}

and

(52)
\begin{align} b\epsilon^2=\frac{1}{2}d\hat{\boldsymbol{x}}\cdot\boldsymbol{H}\cdot d\hat{\boldsymbol{x}}. \end{align}

Finally, we obtain

(53)
\begin{align} \hat{\boldsymbol{x}}(\epsilon)=\hat{\boldsymbol{x}}(0)+\boldsymbol{I}\cdot d\hat{\boldsymbol{x}}(\epsilon)+\frac{1}{2}d\hat{\boldsymbol{x}}(\epsilon)\cdot\boldsymbol{H}\cdot d\hat{\boldsymbol{x}}(\epsilon)+\cdots \end{align}

Without loss of generality, we can replace $d\hat{\boldsymbol{x}}$ with $d\boldsymbol{x}$. This means that, we can generalize the above derivation to arbitrary directions. Hence, we can dettatch $\hat{}$ and obtain

(54)
\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}

Here, as mentioned previously, we define

(55)
\begin{align} d\boldsymbol{x}= d\theta^i\boldsymbol{g}_i \end{align}
(56)
\begin{align} \boldsymbol{I}= \boldsymbol{x}\otimes\nabla=\frac{\partial \boldsymbol{x}}{\partial \theta^i}\otimes\boldsymbol{g}^i=\boldsymbol{g}_i\otimes\boldsymbol{g}^i=\boldsymbol{g}^i\otimes\boldsymbol{g}_i \end{align}
(57)
\begin{align} \boldsymbol{H}= \nabla\otimes\left(\boldsymbol{x}\otimes\nabla\right)=\boldsymbol{g}^\alpha\otimes\frac{\partial}{\partial \theta^\alpha}\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right). \end{align}

Thus we obtained the Taylor expansion of a surface expressed with dyadic products.

### The second fundamental form

We examine the second term of the Taylor expansion of a surface represented by

(58)
\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}

i.e.,

(59)
\begin{align} \frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{H}\cdot d\boldsymbol{x}. \end{align}

We make a convention such that

(60)
\begin{align} \boldsymbol{I}=\boldsymbol{x}\otimes\nabla=\boldsymbol{g}_i\otimes\boldsymbol{g}^i, \end{align}
(61)
\begin{align} \boldsymbol{H}=\nabla\otimes\boldsymbol{I}=\boldsymbol{g}^\alpha\otimes\frac{\partial \boldsymbol{g}_i\otimes\boldsymbol{g}^i}{\partial \theta^\alpha} \end{align}

and

(62)
\begin{align} d\boldsymbol{x}=d\theta^i\boldsymbol{g}_i. \end{align}

First, it is clear that

(63)
\begin{align} \frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{H}\cdot d\boldsymbol{x}=\frac{1}{2}d\theta^\alpha\frac{\partial \boldsymbol{g}_i\otimes\boldsymbol{g}^i}{\partial \theta^\alpha}\cdot d\boldsymbol{x}. \end{align}

This follows as

(64)
\begin{align} =\frac{1}{2}d\theta^\alpha\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\otimes\boldsymbol{g}^i\cdot d\boldsymbol{x}+\frac{1}{2}d\theta^\alpha\boldsymbol{g}_i\otimes\frac{\partial \boldsymbol{g}^i}{\partial \theta^\alpha}\cdot d\boldsymbol{x} \end{align}
(65)
\begin{align} =\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}d\theta^\alpha　d\theta^i+\frac{1}{2}\left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^\alpha}\cdot \boldsymbol{g}_i\right)\boldsymbol{g}_id\theta^\alpha　d\theta^i. \end{align}

One might think that this equation would be mush simpler if the second term is absent. However, it is mistaken and this equation becomes simpler because it contains the second term.
Let us recall the relation between covariant and contravariant basis, i.e., $\boldsymbol{g}_i\cdot\boldsymbol{g}^j=\delta_{i}^{\cdot j}$. This quantity takes either 0 or 1, hence

(66)
\begin{align} \frac{\partial\left( \boldsymbol{g}_i\cdot\boldsymbol{g}^j\right)}{\partial \theta^\alpha}=0 \end{align}

is true for any combination. This is expanded as

(67)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j+\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha}=0, \end{align}

and follows

(68)
\begin{align} -\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j=\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha}. \end{align}

Hence, even in a special case such that $j=i$,

(69)
\begin{align} -\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^i=\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^i}{\partial \theta^\alpha} \end{align}

is true.
Substituting this to Eq. (65), we obtain

(70)
\begin{align} =\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}d\theta^\alpha d\theta^i-\frac{1}{2}\left(\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^i\right)\boldsymbol{g}_id\theta^\alpha d\theta^i. \end{align}

Now, the second term have a meaning that it computes the contravariant component of the first term and takes a linear combination of covariant basis.
Subtraction of the second term from first term means that it represents the orthogonal component of the first term, i.e, the component that does not belong to the tangent plane.
We denote a unit normal vector $\boldsymbol{n}$, i.e., it is the third base vector.
The norm of this vector is 1, and it is perpendicular to the tangent plane; hence we don't need to distinguish whether it is a covariant or a contravariant base vector.
By using this unit normal vector, we can describe a three-dimensional unit vector as

(71)
\begin{align} \boldsymbol{D}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2+\boldsymbol{n}\otimes\boldsymbol{n}. \end{align}

Note that the unit tensor defined on the surface is represented by

(72)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2. \end{align}

In other words,

(73)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n}. \end{align}

By using unit normal vector and three-dimensional unit tensor,

(74)
\begin{align} \frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{H}\cdot d\boldsymbol{x}=\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}d\theta^\alpha d\theta^i-\frac{1}{2}\left(\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^i\right)\boldsymbol{g}_id\theta^\alpha d\theta^i \end{align}

can be rewritten as

(75)
\begin{align} =\left(\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\boldsymbol{n}d\theta^\alpha d\theta^i\right)\boldsymbol{n}. \end{align}

We have certain variation of this formula such as

(76)
\begin{align} =\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\left(\boldsymbol{n}\otimes\boldsymbol{n}\right)d\theta^\alpha d\theta^i \end{align}

and

(77)
\begin{align} =\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\left(\boldsymbol{D}-\boldsymbol{I}\right)d\theta^\alpha　d\theta^i. \end{align}

Those three expressions denote exactly the same thing.
Here, when we define

(78)
\begin{align} h_{ij}=\frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n}, \end{align}

The term in the parenthesis in Eq. (75) can be simply written as

(79)
\begin{align} \mathrm{II}\equiv\frac{1}{2}h_{ij}d\theta^i d\theta^j. \end{align}

This is historically called the second fundamental form and it is a quadratic form.
Conventionally, we sometimes write

(80)
\begin{align} L=\frac{\partial \boldsymbol{g}_1}{\partial \theta^1}\cdot\boldsymbol{n} \end{align}
(81)
\begin{align} M=\frac{\partial \boldsymbol{g}_1}{\partial \theta^2}\cdot\boldsymbol{n} \end{align}
(82)
\begin{align} M=\frac{\partial \boldsymbol{g}_2}{\partial \theta^1}\cdot\boldsymbol{n} \end{align}
(83)
\begin{align} N=\frac{\partial \boldsymbol{g}_2}{\partial \theta^2}\cdot\boldsymbol{n} \end{align}

and the second fundamental form is expressed as

(84)
\begin{align} II=Ld\theta^1d\theta^1+2Md\theta^1d\theta^2+Nd\theta^2 d\theta^2. \end{align}

Here, the identity of two $M$s becomes clear because of the following two calculations.

(85)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}=\frac{\partial^2 \boldsymbol{x}}{\partial \theta^i\partial \theta^j} \end{align}
(86)
\begin{align} \frac{\partial \boldsymbol{g}_j}{\partial \theta^i}=\frac{\partial^2 \boldsymbol{x}}{\partial \theta^i\partial \theta^j} \end{align}

There are another way of defining second fundamental form.
Because the normal vector and covariant basis of the surface are always perpendicular, we have

(87)
\begin{align} \boldsymbol{g}_i\cdot\boldsymbol{n}=0. \end{align}

Differentiating this with coordinate parameters gives

(88)
\begin{align} \frac{\partial \boldsymbol{g}_i\cdot\boldsymbol{n}}{\partial \theta^j}=0. \end{align}

Hence,

(89)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot \boldsymbol{n}+\boldsymbol{g}_i\frac{\partial \boldsymbol{n}}{\partial \theta^j}=0. \end{align}

By means of transposition, we obtain

(90)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot \boldsymbol{n}=-\boldsymbol{g}_i\frac{\partial \boldsymbol{n}}{\partial \theta^j}. \end{align}

Therefore, we have

(91)
\begin{align} L=-\boldsymbol{g}_1\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^1} \end{align}
(92)
\begin{align} M=-\boldsymbol{g}_1\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^2} \end{align}
(93)
\begin{align} M=-\boldsymbol{g}_2\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^1} \end{align}
(94)
\begin{align} N=-\boldsymbol{g}_2\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^2}. \end{align}

Finally we obtain

(95)
\begin{align} II=Ld\theta^1d\theta^1+2Md\theta^1d\theta^2+Nd\theta^2 d\theta^2. \end{align}

Incidentally, using $h_{ij}$, it's likely that we can define a rank-3 tensor by

(96)
\begin{align} \boldsymbol{h}=h_{ij}\boldsymbol{g}^i\otimes\boldsymbol{n}\otimes\boldsymbol{g}^j. \end{align}

Let us examine whether this is identical with $\boldsymbol{H}$.
From the definition, we have

(97)
\begin{align} \boldsymbol{H}=\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}\otimes\boldsymbol{g}^{i}}{\partial\theta^{\alpha}}. \end{align}

By the property of unit tensor, we have

(98)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}^{i}\otimes\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}. \end{align}

Performing differentiation, it follows

(99)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\otimes\boldsymbol{g}^{i}+\boldsymbol{g}^{\alpha}\otimes\boldsymbol{g}_{i}\otimes\frac{\partial\boldsymbol{g}^{i}}{\partial\theta^{\alpha}}. \end{align}

Denoting a unit normal vector $\boldsymbol{n}$ we obtain

(100)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\otimes\boldsymbol{g}^{i}+\boldsymbol{g}^{\alpha}\otimes\boldsymbol{g}_{i}\otimes\left(\frac{\partial\boldsymbol{g}^{i}}{\partial\theta^{\alpha}}\cdot\boldsymbol{g}_{j}\right)\boldsymbol{g}^{j}+\boldsymbol{g}^{\alpha}\otimes\boldsymbol{g}_{i}\otimes\left(\frac{\partial\boldsymbol{g}^{i}}{\partial\theta^{\alpha}}\cdot\boldsymbol{n}\right)\boldsymbol{n}. \end{align}

Here it is possible to rewrite the equation to

(101)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\otimes\boldsymbol{g}^{i}-\boldsymbol{g}^{\alpha}\otimes\boldsymbol{g}_{i}\otimes\left(\frac{\partial\boldsymbol{g}_{j}}{\partial\theta^{\alpha}}\cdot\boldsymbol{g}^{i}\right)\boldsymbol{g}^{j}-\boldsymbol{g}^{\alpha}\otimes\boldsymbol{g}_{i}\otimes\left(\frac{\partial\boldsymbol{n}}{\partial\theta^{\alpha}}\cdot\boldsymbol{g}^{i}\right)\boldsymbol{n}. \end{align}

Because the second term is commutative in $i,j$, we obtain

(102)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\otimes\boldsymbol{g}^{i}-\boldsymbol{g}^{\alpha}\otimes\left(\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\cdot\boldsymbol{g}^{j}\right)\boldsymbol{g}_{j}\otimes\boldsymbol{g}^{i}-\boldsymbol{g}^{\alpha}\otimes\left(\frac{\partial\boldsymbol{n}}{\partial\theta^{\alpha}}\cdot\boldsymbol{g}^{i}\right)\boldsymbol{g}_{i}\otimes\boldsymbol{n}. \end{align}

Hence, the first and second terms can be unified.
By using

(103)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_{1}\otimes\boldsymbol{g}^{1}+\boldsymbol{g}_{2}\otimes\boldsymbol{g}^{2}, \end{align}
(104)
\begin{align} \boldsymbol{D}=\boldsymbol{g}_{1}\otimes\boldsymbol{g}^{1}+\boldsymbol{g}_{2}\otimes\boldsymbol{g}^{2}+\boldsymbol{n}\otimes\boldsymbol{n} \end{align}

and

(105)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n}, \end{align}

we obtain

(106)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}\cdot\left(\boldsymbol{D}-\boldsymbol{I}\right)\otimes\boldsymbol{g}^{i}+\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{n}}{\partial\theta^{\alpha}}\cdot\boldsymbol{I}\otimes\boldsymbol{n} \end{align}
(107)
\begin{align} =h_{i\alpha}\boldsymbol{g}^{\alpha}\otimes\boldsymbol{n}\otimes\boldsymbol{g}^{i}+\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{n}}{\partial\theta^{\alpha}}\cdot\boldsymbol{I}\otimes\boldsymbol{n} \end{align}
(108)
\begin{align} \therefore \boldsymbol{H}=\boldsymbol{h}+\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{n}}{\partial\theta^{\alpha}}\cdot\boldsymbol{I}\otimes\boldsymbol{n}. \end{align}

Based on the above discussion, we notice that $h_{ij}$ is actually the component of the first term of the above equation and $h$ is different from $H$.
However, $d\boldsymbol{x}\cdot \boldsymbol{h}\cdot d\boldsymbol{x}$ and $d\boldsymbol{x}\cdot \boldsymbol{H}\cdot d\boldsymbol{x}$ are identical.

Using the second fundamental form and unit normal vector, the Taylor expansion of a surface can be represented by

(109)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)=\boldsymbol{x}(\theta^1,\theta^2)+d\boldsymbol{x}+\frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{h}\cdot d\boldsymbol{x}+\cdots \end{align}
(110)
\begin{align} =\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}

This might be much simpler and useful representation than that in Taylor expansion of a surface.

### 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

(111)
\begin{align} \mathrm{I}\equiv dx^idx^jg_{ij} \end{align}

or

(112)
\begin{align} \mathrm{I}=Ed\theta^1d\theta^1+2Fd\theta^1d\theta^2+Gd\theta^2 d\theta^2. \end{align}

Here,

(113)
\begin{align} g_{ij}=\boldsymbol{g}_i\cdot\boldsymbol{g}_j \end{align}

and

(114)
$$E=g_{11},F=g_{12}=g_{21},G=g_{22}.$$

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

(115)
\begin{align} \mathrm{II}\equiv h_{ij}d\theta^\alpha d\theta^i \end{align}

or

(116)
\begin{align} \mathrm{II}=Ld\theta^1d\theta^1+2Md\theta^1d\theta^2+Nd\theta^2 d\theta^2. \end{align}

Here,

(117)
\begin{align} h_{ij}=\frac{1}{2}\frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n} \end{align}

and

(118)
$$L=h_{11},M=h_{12}=h_{21},N=h_{22}.$$

The Taylor expansion of a surface is represented by

(119)
\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

(120)
\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

(121)
\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

(122)
\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

(123)
\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

(124)
\begin{align} \boldsymbol{r}(\theta^1,\theta^2), \end{align}

we obtain

(125)
\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

(126)
\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 $f(\theta^1,\theta^2)$,

(127)
\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

(128)
\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

(129)
\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

(130)
\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

(131)
\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

(132)
\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):

(133)
\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

(134)
\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

(135)
\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

(136)
\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

(137)
\begin{align} \boldsymbol{b}=b^i\boldsymbol{g}_i\mid g_{ij}b^ib^j=1. \end{align}

(138)
\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

(139)
$$g_{ij}b^i b^j=1$$

as a constraint condition and take the variation of

(140)
$$h_{ij}b^ib^j$$

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

(141)
\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

(142)
\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:

(143)
\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

(144)
\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

(145)
\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.(143), it follows:

(146)
\begin{align} \Leftrightarrow h_{ij}b^i\delta b^j+\lambda(g_{ij} b^i\delta b^j)=0 \end{align}
(147)
\begin{align} \therefore h_{ij}b^i+\lambda g_{ij} b^i=0 \end{align}

Inverting the sign of $\lambda$, we obtain

(148)
\begin{align} h_{ij}b^i-\lambda g_{ij} b^i=0. \end{align}

Note that this resembles well with characteristic polynomial.

(149)
\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}
(150)
\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,

(151)
\begin{align} (L-\lambda E)(N-\lambda G)-(M-\lambda F)^2=0 \end{align}
(152)
\begin{align} \Leftrightarrow LN+\lambda^2EG-\lambda(LG+EN)-M^2-\lambda^2F^2+2\lambda M F=0 \end{align}
(153)
\begin{align} \Leftrightarrow \lambda^2(EG-F^2)+\lambda(2MF-LG-EN) +LN-M^2=0. \end{align}

The solutions are

(154)
\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

(155)
\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

(156)
\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

(157)
\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

(158)
\begin{align} H=\frac{1}{2}(\lambda_1+\lambda_2)=\frac{LG-2MF+EN}{2(EG-F^2)} \end{align}

and Gaussian curvature

(159)
\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

(160)
\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

(161)
\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

(162)
\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

(163)
\begin{align} h_{ij}b^i-\lambda b^j=0. \end{align}

It follows

(164)
\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

(165)
\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

(166)
\begin{align} (w^1_{\cdot 1}-\lambda)(w^2_{\cdot 2}-\lambda)-w^1_{\cdot 2}w^2_{\cdot 1}=0 \end{align}
(167)
\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

(168)
\begin{align} \lambda_1=\frac{(w^1_{\cdot 1}+w^2_{\cdot 2})+\cdots}{2} \end{align}

and

(169)
\begin{align} \lambda_2=\frac{(w^1_{\cdot 1}+w^2_{\cdot 2})-\cdots}{2}. \end{align}

Therefore

(170)
\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

(171)
\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

(172)
\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

(173)
\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

(174)
\begin{align} \boldsymbol{h}=h_{ij}\boldsymbol{g}^j\otimes \boldsymbol{g}^i. \end{align}

Using a tangent vector

(175)
\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:

(176)
\begin{align} \Leftrightarrow\boldsymbol{h}\cdot\boldsymbol{b}=\lambda\boldsymbol{b}. \end{align}

This follows

(177)
\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.,

(178)
\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

(179)
\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.

(180)
\begin{align} T_{ij}=\boldsymbol{g}_i\cdot\boldsymbol{T}\cdot\boldsymbol{g}_j, \end{align}

we obtain

(181)
\begin{align} \Leftrightarrow \mathrm{det}\left(h_{ij}-\lambda g_{ij}\right)=0. \end{align}

On the other hand, if we adopt

(182)
\begin{align} T^i_{\cdot j}=\boldsymbol{g}^i\cdot\boldsymbol{T}\cdot\boldsymbol{g}_j, \end{align}

we obtain

(183)
\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

(184)
\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

(185)
\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

(186)
\begin{align} \mathrm{det}\left(h_{ij}-\lambda g_{ij}\right)=0. \end{align}

### Mean curvature

In this section, we prove that the divergence of a unit tensor of a surface is the double of the mean curvature vector.
When this is proved, consequently, we can confirm that a minimal surface is a surface which its mean curvature is zero everywhere.

(187)
\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}

Here, $d\boldsymbol{x},\boldsymbol{I},\boldsymbol{H}$ are respectively defined by

(188)
\begin{align} d\boldsymbol{x}\equiv d\theta^i\boldsymbol{g}_i \end{align}
(189)
\begin{align} \boldsymbol{I}\equiv \boldsymbol{x}\otimes\nabla=\frac{\partial \boldsymbol{x}}{\partial \theta^i}\otimes\boldsymbol{g}^i=\boldsymbol{g}_i\otimes\boldsymbol{g}^i=\boldsymbol{g}^i\otimes\boldsymbol{g}_i \end{align}
(190)
\begin{align} \boldsymbol{H}\equiv \nabla\otimes\left(\boldsymbol{x}\otimes\nabla\right)=\boldsymbol{g}^\alpha\otimes\frac{\partial}{\partial \theta^\alpha}\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right). \end{align}

The derivation is written in Taylor expansion of a surface.

Now, an arbitrary tangent vector on a surface can be represented by $\boldsymbol{I}\cdot d\boldsymbol{x}=d\boldsymbol{x}$; hence when we omit the higher terms, we obtain

(191)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)\simeq\boldsymbol{x}(\theta^1,\theta^2)+d\boldsymbol{x}. \end{align}

This represents a tangent plane.
We define

(192)
\begin{align} \boldsymbol{N}\equiv\boldsymbol{g}_1\cdot\boldsymbol{H}\cdot\boldsymbol{g}^1+\boldsymbol{g}_2\cdot\boldsymbol{H}\cdot\boldsymbol{g}^2=\boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i. \end{align}

Let us rewrite the right-hand side as

(193)
\begin{align} \mathrm{tr}\boldsymbol{H}\equiv\boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i. \end{align}

Such an operation is called contraction. Particularly, the above operation is called a trace.
Interestingly, the following relation holds:

(194)
\begin{align} \mathrm{div}\boldsymbol{I}=\boldsymbol{N}. \end{align}

This can be verified by comparing

(195)
\begin{align} \mathrm{div}\boldsymbol{I}\equiv\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i \end{align}

and

(196)
\begin{align} \boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i=\boldsymbol{g}_i\cdot\boldsymbol{g}^\alpha\otimes\frac{\partial\boldsymbol{I}}{\partial \theta^\alpha}\cdot\boldsymbol{g}^i=\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i. \end{align}

In summary, the following relations holds:

(197)
\begin{align} \boldsymbol{N}=\mathrm{div}\boldsymbol{I}=\mathrm{tr}\boldsymbol{H}. \end{align}

It is very interesting that $\boldsymbol{N}$ represents a normal (orthogonal to the tangent plane) vector and its norm is the double of a quantity called mean curvature.
We first prove that it represents a normal vector.
The calculation is like as follows:

(198)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i=\frac{\partial \left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right)}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha=\frac{\partial\left( \boldsymbol{g}^i\otimes\boldsymbol{g}_i\right)}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha=\frac{\partial\boldsymbol{g}^i}{\partial \theta^\alpha}\otimes\boldsymbol{g}_i\cdot\boldsymbol{g}^\alpha+\boldsymbol{g}^i\otimes\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha \end{align}

Hence,

(199)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}+\boldsymbol{g}^i\otimes\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha \end{align}

ここで、少し複雑な法則を用います。共変基底、反変基底の関係$\boldsymbol{g}_i\cdot\boldsymbol{g}^j=\delta_{i}^{\cdot j}$を思い出しましょう。いずれの場合も0か1ですから、任意の共変基底と反変基底、任意の座標パラメータ$\theta^\alpha$について

(200)
\begin{align} \frac{\partial\left( \boldsymbol{g}_i\cdot\boldsymbol{g}^j\right)}{\partial \theta^\alpha}=0 \end{align}

が成り立ちます。従って、

(201)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j+\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha}=0 \end{align}

さらに、

(202)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha} \end{align}

(203)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot \boldsymbol{g}^j=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^j} \end{align}

が成り立ちます。代入して

(204)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}-\boldsymbol{g}^j\otimes\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j \end{align}

さて、第二項ですが、$\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j$がスカラーですから、

(205)
\begin{align} \left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j\right)\boldsymbol{g}^j \end{align}

と書けます。従って、

(206)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}-\left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j\right)\boldsymbol{g}^j \end{align}

となります。

(207)
\begin{align} \boldsymbol{n} \end{align}

を定義します。大きさが1で、接平面と直交していますから、共変基底、反変基底の区別を付ける必要がありません。
この単位法線ベクトルを用いると空間の単位テンソル

(208)
\begin{align} \boldsymbol{D}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2+\boldsymbol{n}\otimes\boldsymbol{n} \end{align}

を記述できます。

(209)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2 \end{align}

と比べてみると、両者の関係がよくわかります。つまり、

(210)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n} \end{align}

です。すると

(211)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}(\boldsymbol{D}-\boldsymbol{I}) \end{align}

と書けます。他に

(212)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}(\boldsymbol{n}\otimes\boldsymbol{n}) \end{align}

(213)
\begin{align} \boldsymbol{N}=\left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{n}\right)\boldsymbol{n} \end{align}

とかけますが、いずれも同じことを表しています。つまり

(214)
\begin{align} \frac{\partial \boldsymbol{g}^1}{\partial \theta^1}+\frac{\partial \boldsymbol{g}^2}{\partial \theta^2} \end{align}

から接平面成分が削除され、法線成分のみ取り出したものが$\boldsymbol{N}$です。

さてこの法線成分について

(215)
\begin{align} \frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot \boldsymbol{n}=-\boldsymbol{g}^i\frac{\partial\boldsymbol{n}}{\partial \theta^i}=-g^{ij}\boldsymbol{g}_j\frac{\partial\boldsymbol{n}}{\partial \theta^i} \end{align}

について計算を続けます。

(216)
\begin{align} \mathrm{II}\equiv\frac{1}{2}h_{ij}d\theta^i d\theta^j \end{align}

において

(217)
\begin{align} h_{ij}=\frac{\partial \boldsymbol{g}_j}{\partial \theta^i}\cdot\boldsymbol{n}=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^j} \end{align}

だったことを思い出すと、

(218)
$$=g^{ij}h_{ij}$$

と書けます。

(219)
\begin{align} \frac{\partial g^{ij}\boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n}=\left(\frac{\partial g^{ij}}{\partial \theta^j}\boldsymbol{g}_i\right)\cdot\boldsymbol{n}+h_{ij}g^{ij} \end{align}

として、第一項が零であることを用いても同じ結果を得ます。
さて、曲面上の曲線と主曲率で平均曲率は

(220)
\begin{align} H=\frac{1}{2}(\lambda_1+\lambda_2)=\frac{LG-2MF+EN}{2(EG-F^2)} \end{align}

として与えられましたが、同時に

(221)
\begin{align} H=\frac{g^{ij}h_{ij}}{2} \end{align}

とも書けました。

page revision: 65, last edited: 13 May 2015 05:54