# Generic fibers of the Gauss map are linear spaces

Problem

I'm currently working on a problem in Cartan for Beginners by J.M. Landsberg and Thomas Ivey (Exercise 2.5.4.3). This question is stated:

Show that the generic fibers of the Gauss map are (open subsets of) linear spaces, and thus flat surfaces are ruled by lines.

I'm having a little trouble figuring out what the question is asking.

Questions

1. What does the question mean by "generic fibers of the Gauss map?" The Gauss map was defined in the text as the map taking $x$ to $e_3(x)$, where $x$ is an element of an oriented surface $M$, and $e_3$ the normal vector associated with an orthonormal moving frame $(e_1(x),e_2(x),e_3(x))$ for $M$.

From my understanding, the fiber of this map would simply be a collection of points in the manifold, (for example if $M$ is the sphere $S^2$, the the fiber of $(0,0,1)$ under the Gauss map would simply be $(0,0,1)$) but I don't see how this could possibly be a linear space, so I must be understanding incorrectly.

On the other hand, by "generic fiber of the Gauss map," he could be referring to the the fiber of the projection $(x,e_1,e_2,e_3)\mapsto e_3$, where $x\in M$ and $(e_1,e_2,e_3)$ is an orthonormal frame associated with the point $x$. This would make more sense, but this makes showing that it is a linear space trivial (if by linear space he means vector space).

2. What does a "linear space" exactly mean? A vector space? A plane? a line?

3. what does ''generic" mean exactly? I believe this to be an algebraic geometry term, however it has not been defined anywhere in the text. Does this mean just the fiber of an arbitrary point?

Sorry if I seem dense by needing clarification on half the words in the statement of the problem, but I would like to know the exact question being asked. For a flat surface this problem seems to make sense, because for a cylinder (for example) the collection of points having the same image under the gauss map makes a straight line. However for a sphere or something less trivial, it's just a singleton. Any help is greatly appreciated. Thank you.

• Can you also include the exercise? (Yes, in general the preimage of a Gauss map is not a linear space) – user99914 Sep 13 '15 at 7:22
• @JohnMa the exercise is the statement quoted. Is there any further information you would like? – Blake Sep 13 '15 at 7:23

Yes, it's a bit glib of a statement. The case with which you started is fine — points are $0$-dimensional vector spaces. There is a high-dimensional version of this statement, but I'm assuming (I don't actually know this book, although I know its authors) that you're dealing just with surfaces. Try this statement: Suppose $M\subset\Bbb R^3$ is a flat surface ($K=0$) with no planar points. — Translation: The Gauss map has rank $1$ everywhere. — Then prove that the fibers of the Gauss map are open subsets of lines (affine $1$-dimensional subspaces of $\Bbb R^3$).
Here's a hint to get you started: Choose a moving frame $e_1,e_2,e_3$ with $e_1(x)$ giving a basis for $\ker de_3(x)$. [Note that by the constant rank assumption this makes sense.]