# The Gaussian and Mean Curvatures of a Parallel Surface

This is a homework problem from do Carmo. Given a regular parametrized surface $X(u,v)$ we define the parallel surface $Y(u,v)$ by $$Y(u,v)=X(u,v) + aN(u,v)$$ where $N(u,v)$ is the unit normal on $X$ and $a$ is a constant. I have been asked to compute the Gaussian and mean curvatures $\overline{K}$ and $\overline{H}$ of $Y(u,v)$ in terms of those of X, $K$ and $H$. Now, I know how to do this by brute force: calculate the coefficients of the first and second fundamental forms of $Y$ in terms of those of $X$. However, this is a lengthy and messy calculation. do Carmo says that $$\overline{K}=\frac{K}{1-2Ha+Ka^2}$$ and $$\overline{H}=\frac{H-Ka}{1-2Ha+Ka^2}.$$ The denominator of these fractions is actually something that arose earlier in the problem; I calculated $$Y_u\times Y_v=(1-2Ha+Ka^2)(X_u\times X_v).$$ So, it seems like I should be able to calcuate $\overline{K}$ and $\overline{H}$ from this initial step. Is there something I'm missing? Or, is it actually just a brute force calculation?

Thanks.

Yes, you can compute all the coefficients $e,f,g,E,F,G$ and get the gaussian and mean curvature and yes, it's tedious.

Here's another way:

From the first step we get : $Y_u\times Y_v=(1-2Ha+Ka^2)(X_u\times X_v)$, ie if $N$ and $\overline N$ are the normal vectors of $X$ and $Y$ respectively, then $\overline N\circ Y$ and $N\circ X$ coincide, since they're parallel. If these functions coincide then we have the following relations :

$$d\overline N(Y_u)=(\overline N\circ Y)_u=(N\circ X)_u=dN(X_u) \tag1$$ $$d\overline N(Y_v)=(\overline N\circ Y)_v=(N\circ X)_v=dN(X_v) \tag2$$

Let $\overline B$ be the matrix of $d\overline N$ with respect to $\{Y_u,Y_v\}$ and $B$ the matrix of $dN$ with respect to $\{X_u,X_v\}$.

Now, to compute $\overline K$ and $\overline H$ we need to find the expression of $\overline B$.

Put $$B=\begin{bmatrix}b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{bmatrix}$$

From the definition of $Y$ we have: $$Y_u=X_u+a\cdot N_u=(a\cdot b_{11}+1)\cdot X_u+a\cdot b_{21}\cdot X_v$$

$$Y_v=X_v+a\cdot N_v=a\cdot b_{12}\cdot X_u+(a\cdot b_{22}+1)\cdot X_v$$

From these equations we can get the "change of basis" matrix : $Q=\begin{bmatrix}a\cdot b_{11}+1 & a\cdot b_{12}\\ a\cdot b_{21} & a\cdot b_{22}+1\\ \end{bmatrix}$ from $\{X_u,X_v\}$ to $\{Y_u,Y_v\}$. Then from the initial relations $(1)$ and $(2)$, we have the following equation: $$B=Q\cdot \overline B$$

Since $Q$ is invertible: $$\overline B=Q^{-1}\cdot B$$ From this point you can compute the entries of $\overline B$ and calculate $\overline H$ and $\overline K$.

You can also notice that, since $Q^{-1}=(I+a\cdot B)^{-1}$, you have $\overline B=(I+a\cdot B)^{-1}\cdot B$. So, if $B$ has eigenvalues $-\lambda_1$ and $-\lambda_2$, then the eigenvalues of $\overline B$ are $\frac{-\lambda_1}{1-a\cdot \lambda_1}$ and $\frac{-\lambda_2}{1-a\cdot \lambda_2}$ and you can easily compute $\overline H$ and $\overline K$.

• Thank you, that is very clear – Anri Rembeci Apr 1 '13 at 1:25
• Sorry, how did you got the last eigenvalues? Was it just straight calculus or did you use some property? – Kernel Oct 20 '15 at 11:47
• @Kernel To obtain those eigenvalues, one uses the property that when B is a matrix over the reals and f is a rational function on the complex numbers defined at all eigenvalues of B, if b is an eigenvalue of B, then f(b) is an eigenvalue of the matrix f(B). When applied to B, the denominator of the rational function f indicates the inverse of the corresponding matrix (that f is defined at all the eigenvalues of B ensures this matrix is invertible). – jawheele Feb 15 '18 at 6:30

Here's a different approach, using the Weingarten map.

As above, since $Y$ and $X$ are parallel surfaces, their unit normals have to have the same direction at every point.

Therefore their Weingarten maps have to coincide. (The Weingarten map is the negative differential of the Gauss map.)

Let us call the Weingarten maps $L, \, \bar L$ respectively, and the principle curvatures $\kappa_i, \, \bar \kappa_i$, $i \in \{1,2\}$.

For orthonormal bases $\{\frac{\partial X}{\partial u} , \frac{\partial X}{\partial v} \}$, $\{\frac{\partial X}{\partial u} , \frac{\partial X}{\partial v} \}$ of $T_pX$, $T_{\bar p}Y$, respectively, $p \in X$, $\bar p \in Y$, the Weingarten maps satisfy the eigenvalue equations:

$$L\left(\frac{\partial X}{\partial x^i}\right) = \kappa_i \frac{\partial X}{\partial x^i}, \quad i \in \{1,2\}, \, x^i \in \{u,v\}$$ $$\bar L\left(\frac{\partial Y}{\partial x^i}\right) = \bar \kappa_i \frac{\partial Y}{\partial x^i}, \quad i \in \{1,2\}, \, x^i \in \{u,v\}$$

Now we use the fact, that they are equal,

$$L \left(\frac{\partial X}{\partial x^i}\right) = \bar L \left(\frac{\partial Y}{\partial x^i}\right) = \bar \kappa_i \frac{\partial Y}{\partial x^i} = \bar \kappa_i \left( \frac{\partial X}{\partial x^i} + a \frac{\partial N}{\partial x^i} \right), \quad i \in \{1,2\}, \, x^i \in \{u,v\}.$$

As mentioned above, $$L_{ij} = - \sum_{k=1}^2 \frac{\partial N_i}{\partial x^k} \left( \frac{\partial X_k}{\partial x^j} \right)^{-1} , \, x^j, x^k \in \{u, v\}, \, i \in \{1,2\}.$$

Therefore $$\frac{\partial N}{\partial x^i} = -\sum_{j=1}^2 L_{ij} \frac{\partial X}{\partial x^j} = - L\left(\frac{\partial X}{\partial x^i}\right) = - \kappa_i \frac{\partial X}{\partial x^i}, \, x^i, x^j \in \{u, v\} \, i,j \in \{1,2\}$$

and so we can put things together:

$$\bar \kappa_i \Big( \frac{\partial X}{\partial x^i} + a \frac{\partial N}{\partial x^i} \Big) = \bar \kappa_i \left( 1 - a \kappa_i \frac{\partial X}{\partial x^i} \right) , \quad i \in \{1,2\}, \, x^i \in \{u,v\}$$

$$\bar \kappa_i = \frac{\kappa_i}{1- a \kappa_i}$$

The mean curvature is just half the sum of the principal curvatures, the Gauss curvature their product.