The second fundamental form

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

(1)
\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.,

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

We make a convention such that

(3)
\begin{align} \boldsymbol{I}=\boldsymbol{x}\otimes\nabla=\boldsymbol{g}_i\otimes\boldsymbol{g}^i, \end{align}
(4)
\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

(5)
\begin{align} d\boldsymbol{x}=d\theta^i\boldsymbol{g}_i. \end{align}

First, it is clear that

(6)
\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

(7)
\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}
(8)
\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

(9)
\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

(10)
\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

(11)
\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 $j=i$,

(12)
\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. (8), we obtain

(13)
\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

(14)
\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

(15)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2. \end{align}

In other words,

(16)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n}. \end{align}

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

(17)
\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

(18)
\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

(19)
\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

(20)
\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

(21)
\begin{align} h_{ij}=\frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n}, \end{align}

The term in the parenthesis in Eq. (18) can be simply written as

(22)
\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

(23)
\begin{align} L=\frac{\partial \boldsymbol{g}_1}{\partial \theta^1}\cdot\boldsymbol{n} \end{align}
(24)
\begin{align} M=\frac{\partial \boldsymbol{g}_1}{\partial \theta^2}\cdot\boldsymbol{n} \end{align}
(25)
\begin{align} M=\frac{\partial \boldsymbol{g}_2}{\partial \theta^1}\cdot\boldsymbol{n} \end{align}
(26)
\begin{align} N=\frac{\partial \boldsymbol{g}_2}{\partial \theta^2}\cdot\boldsymbol{n} \end{align}

and the second fundamental form is expressed as

(27)
\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.

(28)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}=\frac{\partial^2 \boldsymbol{x}}{\partial \theta^i\partial \theta^j} \end{align}
(29)
\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

(30)
\begin{align} \boldsymbol{g}_i\cdot\boldsymbol{n}=0. \end{align}

Differentiating this with coordinate parameters gives

(31)
\begin{align} \frac{\partial \boldsymbol{g}_i\cdot\boldsymbol{n}}{\partial \theta^j}=0. \end{align}

Hence,

(32)
\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

(33)
\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

(34)
\begin{align} L=-\boldsymbol{g}_1\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^1} \end{align}
(35)
\begin{align} M=-\boldsymbol{g}_1\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^2} \end{align}
(36)
\begin{align} M=-\boldsymbol{g}_2\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^1} \end{align}
(37)
\begin{align} N=-\boldsymbol{g}_2\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^2}. \end{align}

Finally we obtain

(38)
\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

(39)
\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

(40)
\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

(41)
\begin{align} =\boldsymbol{g}^{\alpha}\otimes\frac{\partial\boldsymbol{g}^{i}\otimes\boldsymbol{g}_{i}}{\partial\theta^{\alpha}}. \end{align}

Performing differentiation, it follows

(42)
\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

(43)
\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

(44)
\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

(45)
\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

(46)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_{1}\otimes\boldsymbol{g}^{1}+\boldsymbol{g}_{2}\otimes\boldsymbol{g}^{2}, \end{align}
(47)
\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

(48)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n}, \end{align}

we obtain

(49)
\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}
(50)
\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}
(51)
\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

(52)
\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}
(53)
\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.

page revision: 38, last edited: 14 May 2015 07:53