On the book Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo the following definition can be found:
My question is very concrete: is there a typographical mistake in the formula boxed in red?
Possible typographical mistake in the definition of the normal curvature
differential-geometry
Related Solutions
For smooth manifolds the tangent space given by equivalence classes of smooth curves and the set of derivations on smooth functions is equivalent. You can give an isomorphism between these constructions. Each has its advantages.
For example, in the equivalence class of curves set-up the differential is really simple; $dF(\gamma) = F \circ \gamma$ the mapping $F: \mathcal{M} \rightarrow \mathcal{N}$ pushes the curve on $\gamma$ on $\mathcal{M}$ to the curve $F \circ \gamma$ on $\mathcal{N}$. To show $d(F \circ G)=dF \circ dG$ for manifolds we can argue $$ d(F \circ G)(\gamma) = F( G (\gamma)) = F( dG(\gamma)) = dF(dG(\gamma)) = (dF \circ dG)(\gamma) $$ I think the proof in the derivation view is less slick.
Derivations naturally arise from coordinates systems as the partial derivatives with respect to the manifold coordinates. The differential again pushes vectors from one manifolds tangent space to another. Or from the manifold to itself, but where the coordinate system has two overlapping charts. Believe it or not, you've already done this calculation in multivariate calculus. Changing coordinates from cartesian to polar for example: $$ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} +\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} = \cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$$ This can be understood as the vector $\frac{\partial}{\partial x}$ being pushed forward to the vector $\cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$ by the differential of the transition map. In other words, there are constructions for which the derivation formulation of tangent space nicely dovetails. Usually the derivation is sold as a sort of directional derivative like object. The formula $X(f) = \sum_{i=1}^n X^i \frac{\partial f}{\partial x^i}$ is like $D(f)(\vec{X}) = (\nabla f) \cdot \vec{X}$.
In either case, it is a bit unsettling to trade seemingly static objects like geometric vectors for classes of curves or differential operators. There is at least one other popular view, the contravariant vector formulation of tangent vectors.
- a contravariant vector is typically described in physics by saying something like $v^{\mu}$ is a contravariant vector because under a change to barred coordinates we find $\bar{v}^{\mu'} = \sum_{\nu=1}^n\Lambda^{\mu'}_{\nu} v^{ \nu}$ where the coordinates transform via $\bar{x}^{\mu'} =\sum_{\nu=1}^n\Lambda^{\mu'}_{\nu} x^{ \nu}$. This is the rule underwhich $X^i$ of the derivation changes. That fact follows from the chain rule for manifolds. This viewpoint has great advantages for those folks who do detailed calculations in one particular system of coordinates; a.k.a. physicists.
It should be commented that when we consider $C^k$ manifolds this correspondence breaks down.See page 49 in Conlon's 2nd edition of Differentiable Manifolds. In his Lemma 2.2.20 he builds functions which are used to frame the isomorphism between curves and derivations. The construction fails for $C^k$ manifolds. Moreover, the reason for this is that the space of derivations on $C^k$-germs of functions is infinite dimensional.
I'd love to say you can pick one to understand. However, the nature of the subject of differential geometry requires you to be conversant in all these views.
The question is about curvature and if the surface can isometrically transformed to the plane. The cone has zero curvature so it is only locally isometric with another surface of zero curvature. The cone and also be unfolded ? or opened up to lie on the plane. Eg, you can make a cone from a piece of paper, but not a torus or a sphere. I think the exercise is in reference to Minding's theorem.
Best Answer
No, this is absolutely correct. You're looking at the portion of the curvature vector $k\mathbf n$ that is normal to the surface (i.e., in the direction of $\mathbf N$). That is, you take $(k\mathbf n)\cdot \mathbf N = k(\mathbf n\cdot \mathbf N) = k\cos\theta$.