second fundamental form

In classical differential geometrythe second fundamental form is a symmetric bilinear formdefined on a differentiable surface $M$ embedded in ${ℝ}^3$, which in some sensemeasures 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 ${ℝ}^3$, it may be consideredas a point on the sphere $S^2\subset {ℝ}^3$. Then wehave a map $n:M\to S^2$. It is called the normal mapor Gauss map.

The second fundamental form is the tensor field $ℐℐ$ on $M$ definedby

$$ℐ{ℐ}_p(\xi ,\eta )=-⟨\mathrm{D}{n}_p(\xi ),\eta ⟩,\xi ,\eta \in {\mathrm{T}}_{p}M,$$(1)

where $⟨,⟩$ is the dot product of ${ℝ}^3$,and we consider the tangent planes of surfaces in ${ℝ}^3$to be subspaces of ${ℝ}^3$.

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

$$⟨n,\frac{\partial \sigma }{\partial v}⟩=0,$$

differentiating with respect to $u$ using the product rule gives

(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:

$$⟨n,\frac{{\partial }^2\sigma }{\partial u\partial v}⟩=⟨n,\frac{{\partial }^2\sigma }{\partial v\partial u}⟩=ℐℐ(\frac{\partial \sigma }{\partial v},\frac{\partial \sigma }{\partial u}).$$(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 $ℐℐ$ is symmetric.∎

In view of Proposition 1,it is customary to regard the second fundamental formas a quadratic form,as it done with the first fundamental form.Thus, the second fundamental form is referred to with the following expression:

$$Ld{u}^2+2Mdudv+Nd{v}^2.$$

Compare with the tensor notation

$$ℐℐ=Ldu\otimes du+Mdu\otimes dv+Mdv\otimes du+Ndv\otimes dv.$$

Or in matrix form (with respect to the coordinates $u,v$),

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 naturalto consider the component of ${\gamma }^{\prime \prime }(s)$ in the normal $n(\gamma (s))$.Precisely, this quantity is

$$⟨{\gamma }^{\prime \prime }(s),n(\gamma (s))⟩,$$

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 accelerationrequired 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

$$⟨{\gamma }^{\prime \prime }(s),n(p)⟩=ℐℐ({\gamma }^{\prime }(s),{\gamma }^{\prime }(s)).$$

Proof.

From the equation

$$⟨n(\gamma (s)),{\gamma }^{\prime }(s)⟩=0,$$

differentiate with respect to $s$:

$⟨n(\gamma (s)),{\gamma }^{\prime \prime }(s)⟩$$=-⟨{\displaystyle \frac{d}{ds}}n(\gamma (s)),{\gamma }^{\prime }(s)⟩$

$=-⟨\mathrm{D}n({\gamma }^{\prime }(s)),{\gamma }^{\prime }(s)⟩$

$=ℐℐ({\gamma }^{\prime }(s),{\gamma }^{\prime }(s)).\mathit{∎}$

It is now time to mention an important consequence of Proposition 1:the fact that $ℐℐ$ is symmetric means that$-\mathrm{D}n$ is self-adjoint with respect to the inner product $ℐ$(the first fundamental form). So, if $-\mathrm{D}n$ is expressedas a matrix with orthonormal coordinates (with respect to $ℐ$),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 usthat, $-\mathrm{D}{n}_p$ being self-adjoint, it has an orthonormal basis ofeigenvectors $e_1,e_2$ with corresponding eigenvalues ${\kappa }_1\le {\kappa }_2$.These eigenvalues are called the principal curvaturesof $M$ 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

$$⟨{\gamma }^{\prime \prime }(s),n(p)⟩={\kappa }_1{\cos }^2\theta +{\kappa }_2{\sin }^2\theta ,p=\gamma (s).$$

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

$${\gamma }^{\prime }(s)=\cos \theta e_1+\sin \theta e_2$$

for some angle $\theta$.Then

$⟨{\gamma }^{\prime \prime }(s),n(p)⟩$$=ℐℐ({\gamma }^{\prime }(s),{\gamma }^{\prime }(s))$

$=⟨-\mathrm{D}{n}_p({\gamma }^{\prime }(s)),{\gamma }^{\prime }(s)⟩$

$=⟨{\kappa }_1\cos \theta e_1+{\kappa }_2\sin \theta e_2,\cos \theta e_1+\sin \theta e_2⟩$

$={\kappa }_1{\cos }^2\theta +{\kappa }_2{\sin }^2\theta .\mathit{∎}$

Matrix representations of second fundamental form and Weingarten map

At this point, we should find the explicit prescriptionsfor 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

$=ℐℐ(\xi ,\xi )=⟨-\mathrm{D}n(\xi ),\xi ⟩$

$={\left(Q\left(\begin{array}{c}\hfill {\xi }_u\hfill \\ \hfill {\xi }_v\hfill \end{array}\right)\right)}^{\mathrm{T}}QW\left(\begin{array}{c}\hfill {\xi }_u\hfill \\ \hfill {\xi }_v\hfill \end{array}\right)$

where $Q$ is the matrix that changes from $u,v$-coordinates to orthonormalcoordinates for ${\mathrm{T}}_{p}M$ — this is necessary to compute the inner product.But

because $Q$ is the matrixwith columns $\partial \sigma /\partial u$ and $\partial \sigma /\partial v$expressed in orthonormal coordinates.

(More to be written…)