For my Differential Geometry class, I've encountered an issue with a problem from do Carmo. Although I'm aware that similar questions have been previously addressed, my issue differs. I understand how to solve the problem but am encountering a specific challenge.
Firstly, let me outline the problem:
Let $\mathbf{x} = \mathbf{x}(u,v)$ be a regular parametrized surface. A parallel surface to $\mathbf{x}$ is a parametrized surface $$\overline{\mathbf{y}}(u,v) = \mathbf{x}(u,v) + a \mathbf{N}(u,v),$$ where $a$ is a real constant and $\mathbf{N}$ denotes the unit normal to $\mathbf{x}$.
(b) Show that the Gaussian and mean curvatures $\overline{K}$ and $\overline{H}$ of $\mathbf{y}$ are respectively given by $$\overline{K} = \frac{K}{1 - 2Ha + Ka^2},$$ and $$\overline{H} = \frac{H - Ka}{1 - 2Ha + Ka^2}.$$
From question (a), we derived: $$\overline{\mathbf{y}}_u \times \overline{\mathbf{y}}_v = (1 - 2 H a + K a^2) (\mathbf{x}_u \times \mathbf{x}_v).$$
This implies that the unit normal $\overline{\mathbf{N}}$ to $\overline{\mathbf{y}}$ is given by $$\overline{\mathbf{N}} = \frac{\overline{\mathbf{y}}_u \times \overline{\mathbf{y}}_v}{||\overline{\mathbf{y}}_u \times \overline{\mathbf{y}}_v||} = \frac{1 - 2 H a + K a^2}{|1 - 2 H a + K a^2|} \frac{\mathbf{x}_u \times \mathbf{x}_v}{||\mathbf{x}_u \times \mathbf{x}_v||} = \text{sgn}(1 - 2 H a + K a^2) \mathbf{N}.$$ Thus, $\overline{\mathbf{N}}$ and $\mathbf{N}$ are parallel as expected. However, the term $\text{sgn}(1 - 2 H a + K a^2)$ suggests they may not always point in the same direction. My confusion arises from this sign, as other solutions I've reviewed do not include the sign function, which doesn't make sense to me.
When I account for this term throughout my proof, I arrive at the given formula for $\overline{K}$, which makes sense since Gaussian curvature is independent of orientation. However, for the mean curvature $\overline{H}$, I derive: $$\overline{H} = \text{sgn}(1 - 2 H a + K a^2) \frac{H - Ka}{1 - 2Ha + Ka^2} = \frac{H - Ka}{|1 - 2Ha + Ka^2|}.$$
I am wondering if I've made an error in my reasoning.
Note that in class we are using the convention where the ordering of the pair $(u,v)$ matters in defining the normal unit vector. If I don't use this convention, I can get the right result by choosing the orientation that suits me, but it will depend on $u$ and $v$, which would be a bit like cheating.
I would say it is largely a convention that $1-2Ha+Ka^2$ is positive. But it makes sense from an evolution point of view as $a$ changes.
When $a=0$, the parallel surface is the original surface. When $a$ is small, then $1-2Ha+Ka^2>0$.
As you said, $$ \bar {\bf y}_u \times \bar {\bf y}_v = (1-2Ha+Ka^2) \, (\bar {\bf x}_u \times \bar {\bf x}_v). $$ For the surface $\bar {\bf y}_a$ to be regular, we need the above to be nonzero. So from a evolution point of view, the coefficient $1-2Ha+Ka^2$ starts as 1 and would stay positive, before the parallel surface becomes singular.
(See also the comment of Shifrin above.)