Gaussian curvature of Mobius Strip. I read about differential geometry of curves (by do Carmo) and I had seen that to find the Gaussian curvature, I will need a second fundamental form (which means I need an orientable surface, or else I cannot define the unitary normal vector field on it.) But I also know Gaussian curvature is an intrinsic property which means it can be expressed in terms of first fundamental form. As I know, a Mobius strip is not an orientable surface. So, how do I calculate the Gaussian curvature of a Mobius Strip?
 A: It depends on the metrics you put on the mobius strip. If you see it as embedded into $\mathbb{R}^3$ with some function, then it has an induced metric.
Locally it is orientable and you can compute the curvature as it is a local calculation.
If you see it as the quotient of the square with reverse identification of two sides, then it is a flat riemannian manifold, with zero curvature.
A: I hope Dldier_'s answers your doubts, so let me address one problem implicitly mentioned in your question: 


*

*As mentioned by Dldier_, curvature is a local thing, so one can just consider a smaller part of the Mobius strip, which is orientable. 

*If you choose the orientation, you have a unit normal field $\vec{n}$ (compatible with the orientation) and you probably consider the second fundamental form as the real-valued function $$ A_p(u,v) = -\langle \nabla_{u} \vec{n}, v \rangle_p. $$ A change in orientation results in changing the sign of $\vec{n}$ and hence also of $A$. But not of the curvature! This could suggest that the sign convention was artificial in the first place. 

*Instead, one can define the second fundamental as vector-valued, not real-valued: $$ A_p(u,v) = -\langle \nabla_{u} \vec{n}, v \rangle_p \vec{n}_p. $$ This has its advantages: 


*

*by definition, $A$ does not depend on the chosen orientation; 

*even better, the orientation is not involved at all! you can define $A$ also for non-orientable surfaces; 

*defining $A$ this way, it is easier to grasp the definition of $A$ for submanifolds of arbitrary codimension. 


