I am trying to understand the definition of orientation as it is done in the book "Vector Calculus and Differential Forms: A Unified Approach" by Hubbard. What I really like about it is that he starts of with unit normals and unit tangents for surfaces and curves respectiely. Then orientation is defined using k-forms. It is called the "Unified approach". Then he says that these notions of orientation coincide.
Questions:
What does "coinicide" mean? What I know is that two k-forms that are positive multiples of each other dertermine the same orientation (An introduction to Smooth Manifolds, Tu,p.239). But this uses equivalence classes and is a bit more technical.
How does a normal vector of a surface determine an orientation in the sense that I get a basis for $T_{p}M$?
How can I find a normal vector N from knowing $\omega(v_{1},v_{2})$. In other words why does $\omega(v_{1},v_{2})$ help to find an orientation in the sense of definition 6.5.6?
Why is the determinant not used in definition 6.5.6? In definition 6.6.6 it is used.
I find the notion of orientation really confusing. Maybe in the 5th edition it is clearer defined. Unfortunately I don't have the latest version.
- Excerpt 1
- Excerpt 2
- Excerpt 3
How does a normal vector of a surface determine an orientation in the sense that I get a basis for $T_p M$?
Why do you mention a basis here? According to the definition you gave, an orientation is defined by a non-zero $2$-form on $T_p M$ (and two $2$-forms give the same orientation if they differ by a positive scalar).
The answer is: given a normal vector $\vec{n}$, you can define a non-zero $2$-form $\omega_\vec{n}$ by $$ \omega_\vec{n}(\vec{v}_1,\vec{v}_2) = \det(\vec{n},\vec{v}_1,\vec{v}_2). $$
What does "coincide" mean?
In the notation above, you could say (i.e. define) that the orientations defined by $\vec{n}$ and $\omega$ coincide if and only if the orientations defined by $\omega_\vec{n}$ and $\omega$ coincide. This in turn is equivalent to the fact that $\omega_\vec{n}$ and $\omega$ differ by a positive scalar.
With this convention, the first excerpt you give is a tautology.
How can I find a normal vector $\vec{n}$ from knowing $\omega(\vec{v}_1,\vec{v}_2)$? In other words, why does $\omega(\vec{v}_1,\vec{v}_2)$ help to find an orientation in the sense of definition 6.5.6?
It's easier to answer the opposite question directly - given $\vec{n}$, you can produce a $2$-form $\omega_\vec{n}$ as described above. Conversely, if you are given $\omega$, then you can find an appropriate $\vec{n}$. Since there are exactly two normal vectors $\vec{n}_1,\vec{n}_2$ (differing by a sign), then one of them has to agree with $\omega$ - if $\vec{n}_1$ doesn't agree with $\omega$, then $\vec{n}_2$ has to.
Added later: Relation with another notion of orientability. In the source you gave, choosing an orientation amounts to choosing a non-zero $2$-form (up to a positive scalar). However, in your comments you have referred to another notion, in which an orientation means that some bases are positively oriented, and some are negatively oriented (as always, there are two possible choices for an orientation).
You can easily connect these two notions. Given a nonzero $2$-form $\omega$, you can say that a basis $\vec{v}_1,\vec{v}_2$ is positively oriented if and only if $\omega(\vec{v}_1,\vec{v}_2) > 0$.