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.
Best Answer
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 $\overline K$.