second fundamental form
In classical differential geometry the second fundamental form^{} is a symmetric bilinear form^{} defined on a differentiable^{} surface $M$ embedded in ${\mathbb{R}}^{3}$, which in some sense measures the curvature^{} of $M$ in space.
To construct the second fundamental form requires a small digression. After the digression we will discuss how it relates to the curvature of $M$.
Construction of the second fundamental form
Consider the tangent planes^{} ${\mathrm{T}}_{p}M$ of the surface $M$ for each point $p\in M$. There are two unit normals^{} to ${\mathrm{T}}_{p}M$. Assuming $M$ is orientable, we can choose one of these unit normals (http://planetmath.org/MutualPositionsOfVectors), $n(p)$, so that $n(p)$ varies smoothly with $p$.
Since $n(p)$ is a unit vector^{} in ${\mathbb{R}}^{3}$, it may be considered as a point on the sphere ${S}^{2}\subset {\mathbb{R}}^{3}$. Then we have a map $n:M\to {S}^{2}$. It is called the normal map or Gauss map.
The second fundamental form is the tensor field $\mathcal{I}\mathcal{I}$ on $M$ defined by
$$\mathcal{I}{\mathcal{I}}_{p}(\xi ,\eta )=-\u27e8\mathrm{D}{n}_{p}(\xi ),\eta \u27e9,\xi ,\eta \in {\mathrm{T}}_{p}M,$$ | (1) |
where $\u27e8,\u27e9$ is the dot product^{} of ${\mathbb{R}}^{3}$, and we consider the tangent planes of surfaces in ${\mathbb{R}}^{3}$ to be subspaces^{} of ${\mathbb{R}}^{3}$.
The linear transformation $\mathrm{D}{n}_{p}$ is in reality the tangent^{} mapping $\mathrm{D}{n}_{p}:{\mathrm{T}}_{p}M\to {\mathrm{T}}_{n(p)}{S}^{2}$, but since ${\mathrm{T}}_{n(p)}{S}^{2}={\mathrm{T}}_{p}M$ by the definition of $n$, we prefer to think of $\mathrm{D}{n}_{p}$ as $\mathrm{D}{n}_{p}:{\mathrm{T}}_{p}M\to {\mathrm{T}}_{p}M$.
The tangent map $\mathrm{D}n$, is often called the Weingarten map.
Proposition 1.
The second fundamental form is a symmetric form.
Proof.
This is a computation using a coordinate chart $\sigma $ for $M$. Let $u,v$ be the corresponding names for the coordinates^{}. From the equation
$$\u27e8n,\frac{\partial \sigma}{\partial v}\u27e9=0,$$ |
differentiating with respect to $u$ using the product rule^{} gives
$$\begin{array}{cc}\hfill \u27e8n,\frac{{\partial}^{2}\sigma}{\partial u\partial v}\u27e9& =-\u27e8\frac{\partial n}{\partial u},\frac{\partial \sigma}{\partial v}\u27e9\hfill \\ & =-\u27e8\mathrm{D}n\left(\frac{\partial \sigma}{\partial u}\right),\frac{\partial \sigma}{\partial v}\u27e9\hfill \\ & =\mathcal{I}\mathcal{I}(\frac{\partial \sigma}{\partial u},\frac{\partial \sigma}{\partial v}).\hfill \end{array}$$ | (2) |
(The second equality follows from the definition of the tangent map $\mathrm{D}n$.) Reversing the roles of $u,v$ and repeating the last derivation, we obtain also:
$$\u27e8n,\frac{{\partial}^{2}\sigma}{\partial u\partial v}\u27e9=\u27e8n,\frac{{\partial}^{2}\sigma}{\partial v\partial u}\u27e9=\mathcal{I}\mathcal{I}(\frac{\partial \sigma}{\partial v},\frac{\partial \sigma}{\partial u}).$$ | (3) |
Since $\partial \sigma /\partial u$ and $\partial \sigma /\partial v$ form a basis for ${\mathrm{T}}_{p}M$, combining (2) and (3) proves that $\mathcal{I}\mathcal{I}$ is symmetric^{}. ∎
In view of Proposition^{} 1, it is customary to regard the second fundamental form as a quadratic form^{}, as it done with the first fundamental form^{}. Thus, the second fundamental form is referred to with the following expression^{1}^{1} Unfortunately the coefficient $M$ here clashes with our use of the letter $M$ for the surface (manifold), but whenever we write $M$, the context should make clear which meaning is intended. The use of the symbols $L,M,N$ for the coefficients of the second fundamental form is standard, but probably was established long before anyone thought about manifolds.:
$$Ld{u}^{2}+2Mdudv+Nd{v}^{2}.$$ |
Compare with the tensor notation
$$\mathcal{I}\mathcal{I}=Ldu\otimes du+Mdu\otimes dv+Mdv\otimes du+Ndv\otimes dv.$$ |
Or in matrix form (with respect to the coordinates $u,v$),
$$\mathcal{I}\mathcal{I}=\left(\begin{array}{cc}\hfill L\hfill & \hfill M\hfill \\ \hfill M\hfill & \hfill N\hfill \end{array}\right).$$ |
Curvature of curves on a surface
Let $\gamma $ be a curve lying on the surface $M$, parameterized by arc-length. Recall that the curvature $\kappa (s)$ of $\gamma $ at $s$ is ${\gamma}^{\prime \prime}(s)$. If we want to measure the curvature of the surface, it is natural to consider the component of ${\gamma}^{\prime \prime}(s)$ in the normal $n(\gamma (s))$. Precisely, this quantity is
$$\u27e8{\gamma}^{\prime \prime}(s),n(\gamma (s))\u27e9,$$ |
and is called the normal curvature^{} of $\gamma $ on $M$.
So to study the curvature of $M$, we ignore the component of the curvature of $\gamma $ in the tangent plane of $M$. Also, physically speaking, the normal curvature is proportional to the acceleration required to keep a moving particle on the surface $M$.
We now come to the motivation for defining the second fundamental form:
Proposition 2.
Let $\gamma $ be a curve on $M$, parameterized by arc-length, and $\gamma \mathit{}\mathrm{(}s\mathrm{)}\mathrm{=}p$. Then
$$\u27e8{\gamma}^{\prime \prime}(s),n(p)\u27e9=\mathcal{I}\mathcal{I}({\gamma}^{\prime}(s),{\gamma}^{\prime}(s)).$$ |
Proof.
From the equation
$$\u27e8n(\gamma (s)),{\gamma}^{\prime}(s)\u27e9=0,$$ |
differentiate with respect to $s$:
$\u27e8n(\gamma (s)),{\gamma}^{\prime \prime}(s)\u27e9$ | $=-\u27e8{\displaystyle \frac{d}{ds}}n(\gamma (s)),{\gamma}^{\prime}(s)\u27e9$ | ||
$=-\u27e8\mathrm{D}n({\gamma}^{\prime}(s)),{\gamma}^{\prime}(s)\u27e9$ | |||
$=\mathcal{I}\mathcal{I}({\gamma}^{\prime}(s),{\gamma}^{\prime}(s)).\mathit{\u220e}$ |
It is now time to mention an important consequence of Proposition 1: the fact that $\mathcal{I}\mathcal{I}$ is symmetric means that $-\mathrm{D}n$ is self-adjoint^{} with respect to the inner product $\mathcal{I}$ (the first fundamental form). So, if $-\mathrm{D}n$ is expressed as a matrix with orthonormal coordinates (with respect to $\mathcal{I}$), then the matrix is symmetric. (The minus sign in front of $\mathrm{D}n$ is to make the formulas^{} work out nicely.)
Certain theorems in linear algebra tell us that, $-\mathrm{D}{n}_{p}$ being self-adjoint, it has an orthonormal basis of eigenvectors^{} ${e}_{1},{e}_{2}$ with corresponding eigenvalues^{} ${\kappa}_{1}\le {\kappa}_{2}$. These eigenvalues are called the principal curvatures of $M$ at $p$. The eigenvectors ${e}_{1},{e}_{2}$ are the principal directions. The terminology is justified by the following theorem:
Theorem 1 (Euler’s Theorem).
The normal curvature of a curve $\gamma $ has the form
$$\u27e8{\gamma}^{\prime \prime}(s),n(p)\u27e9={\kappa}_{1}{\mathrm{cos}}^{2}\theta +{\kappa}_{2}{\mathrm{sin}}^{2}\theta ,p=\gamma (s).$$ |
It follows that the minimum possible normal curvature is ${\kappa}_{\mathrm{1}}$, and the maximum possible is ${\kappa}_{\mathrm{2}}$.
Proof.
Since ${e}_{1},{e}_{2}$ form an orthonormal basis for ${\mathrm{T}}_{p}M$, we may write
$${\gamma}^{\prime}(s)=\mathrm{cos}\theta {e}_{1}+\mathrm{sin}\theta {e}_{2}$$ |
for some angle $\theta $. Then
$\u27e8{\gamma}^{\prime \prime}(s),n(p)\u27e9$ | $=\mathcal{I}\mathcal{I}({\gamma}^{\prime}(s),{\gamma}^{\prime}(s))$ | ||
$=\u27e8-\mathrm{D}{n}_{p}({\gamma}^{\prime}(s)),{\gamma}^{\prime}(s)\u27e9$ | |||
$=\u27e8{\kappa}_{1}\mathrm{cos}\theta {e}_{1}+{\kappa}_{2}\mathrm{sin}\theta {e}_{2},\mathrm{cos}\theta {e}_{1}+\mathrm{sin}\theta {e}_{2}\u27e9$ | |||
$={\kappa}_{1}{\mathrm{cos}}^{2}\theta +{\kappa}_{2}{\mathrm{sin}}^{2}\theta .\mathit{\u220e}$ |
Matrix representations of second fundamental form and Weingarten map
At this point, we should find the explicit prescriptions for calculating the second fundamental form and the Weingarten map.
Let $\sigma $ be a coordinate chart for $M$, and $u,v$ be the names of the coordinates. For a test vector $\xi \in {\mathrm{T}}_{p}M$, we write ${\xi}_{u}$ and ${\xi}_{v}$ for the $u,v$ coordinates of $\xi $.
We compute the matrix $W$ for $-\mathrm{D}n$ in $u,v$-coordinates. We have
$\left(\begin{array}{cc}\hfill {\xi}_{u}\hfill & \hfill {\xi}_{v}\hfill \end{array}\right)\left(\begin{array}{cc}\hfill L\hfill & \hfill M\hfill \\ \hfill M\hfill & \hfill N\hfill \end{array}\right)\left(\begin{array}{c}\hfill {\xi}_{u}\hfill \\ \hfill {\xi}_{v}\hfill \end{array}\right)$ | $=\mathcal{I}\mathcal{I}(\xi ,\xi )=\u27e8-\mathrm{D}n(\xi ),\xi \u27e9$ | ||
$={\left(Q\left(\begin{array}{c}\hfill {\xi}_{u}\hfill \\ \hfill {\xi}_{v}\hfill \end{array}\right)\right)}^{\mathrm{T}}QW\left(\begin{array}{c}\hfill {\xi}_{u}\hfill \\ \hfill {\xi}_{v}\hfill \end{array}\right)$ | |||
$=\left(\begin{array}{cc}\hfill {\xi}_{u}\hfill & \hfill {\xi}_{v}\hfill \end{array}\right)({Q}^{\mathrm{T}}Q)W\left(\begin{array}{c}\hfill {\xi}_{u}\hfill \\ \hfill {\xi}_{v}\hfill \end{array}\right),$ |
where $Q$ is the matrix that changes from $u,v$-coordinates to orthonormal coordinates for ${\mathrm{T}}_{p}M$ — this is necessary to compute the inner product. But
$${Q}^{\mathrm{T}}Q=\left(\begin{array}{cc}\hfill E\hfill & \hfill F\hfill \\ \hfill F\hfill & \hfill G\hfill \end{array}\right)=\mathcal{I}\mathit{\hspace{1em}}\text{(the first fundamental form),}$$ |
because $Q$ is the matrix with columns $\partial \sigma /\partial u$ and $\partial \sigma /\partial v$ expressed in orthonormal coordinates.
(More to be written…)
References
- 1 Michael Spivak. A Comprehensive Introduction to Differential Geometry^{}, volumes I and II. Publish or Perish, 1979.
- 2 Andrew Pressley. Elementary Differential Geometry. Springer-Verlag, 2003.
Title | second fundamental form |
Canonical name | SecondFundamentalForm |
Date of creation | 2013-03-22 15:29:02 |
Last modified on | 2013-03-22 15:29:02 |
Owner | stevecheng (10074) |
Last modified by | stevecheng (10074) |
Numerical id | 5 |
Author | stevecheng (10074) |
Entry type | Definition |
Classification | msc 53A05 |
Related topic | FirstFundamentalForm |
Related topic | ShapeOperator |
Related topic | NormalSection |
Related topic | NormalCurvatures |
Defines | normal curvature |
Defines | principal direction |
Defines | principal curvature |
Defines | Weingarten matrix |
Defines | Weingarten map |
Defines | Gauss map |