# Possible typographical mistake in the definition of the normal curvature

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?

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 $$(k\mathbf n)\cdot \mathbf N = k(\mathbf n\cdot \mathbf N) = k\cos\theta$$.

This complicated setup starts with a surface $$S$$ in $$\mathbb R^3$$ and a point $$P$$ with a normal unit vector to the surface $$\vec N_S.$$

Each point on the surface has an associated orthogonal vector (shortened in the diagram above to let the vector at $$P$$ stand out). The surface $$S$$ has domain boundaries between $$-1 and $$-1 and is governed by the equation:

$$f(x,y)=-x^2+\cos(x)+\cos(y)$$

The normal vector to the surface at any given point $$\vec N_S(P)$$ was calculated as:

\begin{align} \vec N(t)&=\left (-\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} ,-\frac{\partial f}{\partial y}\frac{\partial y}{\partial t},1\right)\\[3ex] &=\left(2x+\sin(x),\sin(y),1\right) \end{align}

On $$S$$ a space curve $$C\in \mathbb R^3$$ was parameterized by $$t$$ with $$-1 as:

$$C(t)=(t,t^2,f(x,y))$$

On this space curve a tangent vector can be defined at each point as the curve derivative:

$$\vec T(t)=(1,2t,-2t-\sin(t)-2t\sin(t^2))$$

In the Frenet Serret or TNB frame $$\vec T$$ would be of unit $$1,$$ and defined as $$\vec T=\frac{\vec r'(t)}{\vert \vec r'(t)\vert}=\vec r'(s),$$ the latter part indicating that there is no need to normalize if the curve is parameterized by arc length $$(s).$$

A second orthogonal vector called the normal vector to the curve $$C$$ at $$P$$ can be calculated as the derivative to the tangent vector, provided that is parameterized by arc length as

$$\vec n(s)=\frac{\vec T'(s)}{\vert T'(s)\vert}$$

with $$k(s)=\vert T'(s)\vert=\frac{\vert T'(t)\vert}{\vert r'(t)\vert}$$ corresponding to the curvature of $$C$$ at $$P.$$ However, this is not straightforward to compute given the square roots to normalize the derivatives, and as reflected here. A way to circumvent this problem is to generate a vector via the wedge products

$$\vec n= \left( C'(t)\times C''(t)\right)\times C'(t)$$

and then proceeding to normalize it.

This vector can be used to generate the osculating circle, knowing that the curvature can also be calculated as $$k(t)=\frac{C'(t)\wedge C''(t)}{\vert C'(t)\vert^3},$$ and that the radius of the osculating circle is $$r=1/k:$$

In the animation $$\vec B(t)$$ completes the Frenet-Serret triad. $$\vec B(t)$$ is the binormal unit vector, the cross product of $$\vec T$$ and $$\vec n.$$ It is worth noting that since the derivative of the tangent vector $$C'(t)$$ is normalized its derivative $$C''(t)$$ is orthogonal. Together, $$\{\vec T, \vec n, \vec B\}$$ form an orthonormal basis for $$\mathbb R^3.$$

The derivatives of $$\vec T$$ and $$\vec B$$ are in the span of $$\vec n:$$ The derivative of the tangent vector can be expressed as $$T'(s)=k(s) \vec n.$$ As a scalar product, $$k(s) =\langle \vec n(s), T'(s) \rangle.$$ Similary, the derivative of the binormal vector $$\vec B$$ can be expressed as $$B'(s)=\tau (s) \vec n,$$ where $$\tau (s)$$ is the torsion of the curve $$C$$ at $$s,$$ which can also be expressed as a scalar product as $$\tau (s) =\langle \vec n(s), B'(s) \rangle.$$ That it makes sense for $$\tau$$ to denote the torsion of the curve can be seen by noticing how it stays constant if the curve is planar:

Here is an animation to illustrate the Frenet triad moving along the curve in relation to vector field $$\vec N:$$

The normal vector to the curve is in the span of the normal vector to the surface at any points along a geodesic curve:

Delving in the topic of the OP, $$k_n=k\cos\theta$$ is a scalar value with $$\theta$$ corresponding to the angle between $$\vec N_S(t)$$ and $$\vec n(t):$$

Since both $$\vec N$$ and $$\vec n$$ are unitary, the $$\cos(\theta)=\langle \vec N, \vec n\rangle$$ is given by the scalar product of the vectors.

The unit vector $$\vec n$$ multiplied by the scalar value of the curvature $$k$$ yields a vector $$k\vec n,$$ whose projection on $$\vec N$$ is $$k_n\vec N:$$

$$K_n=k\cos\theta$$ is called the normal curvature of $$C$$ at $$P.$$

The geodesic curvature, $$k_g,$$ is the curvature of the curve projected onto the surface tangent plane. The geodesic curvature measures how far the curve is from being a geodesic:

And the punch line of the story is that if a vector $$\vec v \in T_P S$$ has norm $$\vert v \vert=1,$$ the second fundamental form applied to the vector, i.e. $$\vec v^\top \mathbf{{II}_P} \vec v$$ equals the normal curvature of any curve through $$P$$ at velocity $$\vec v.$$

The second fundamental form corresponds to the Hessian of a surface with a chart $$f\left(u,v, h(u,v)\right),$$ while the trace of the Hessian is the Laplacian. This makes intuitive sense, since the normal vector at a point $$\vec N$$ of a graph of a function is the gradient (vector of first derivatives), $$\nabla F(x,t,f(x,y))=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1\right)(*),$$ and the second fundamental form involved the derivative of the normal, $$dN.$$ The second fundamental form can be expressed as a symmetrical matrix, which when applied to the derivative of a curve $$\alpha(0)$$ coursing through $$p\in S$$ at $$t=0$$ will result in (see here):

\begin{align} \mathrm{II}_p(\alpha'(0),\alpha'(0))&=-\langle dN_p(\alpha'(0),\alpha'(0) \rangle\\ &=-\langle N'(0),\alpha'(0) \rangle\\ &\underset{*}{=}\bbox[5px,border:2px solid black]{\langle N(p),\alpha''(0) \rangle}\\ &=\langle N(p),k(p) \vec n(p) \rangle\\ &=k\langle \vec N_p,\vec n(p)\rangle\\ &=k_n(p) &=k\cos\theta \end{align}

The boxed result being Euler's theorem: the acceleration of curve $$\alpha$$ at point $$p$$ dotted with the normal to the surface at the same point is the second fundamental form.

$$(*)$$ in an arc length parameterized curve: \begin{align} &\langle N(s),\alpha'(s) \rangle =0\\ &\implies \langle N'(s),\alpha'(s) \rangle+\langle N(s),\alpha''(s) \rangle=0\\ \end{align}

The associated vectors to the minimum $$k_1(p)$$ and maximum $$k_2(p)$$ eigenvalues of the $$\mathrm{II}_p$$ restricted to vectors of norm $$1$$ in $$T_pS$$ will form an orthonormal basis of $$T_pS,$$ because $$\mathrm{II}$$ is a symmetric matrix. $$\{k_1,k_2\}$$ are the principal curvatures of the surface at $$p.$$

A unit vector $$\vec v\in T_pS$$ in the tangent space can be thus represented in relation to the angle with these orthonormal basis vectors with $$\vec v:$$

$$\vec v=\cos \varphi \vec e_1 + \sin \varphi e_2$$

and applying the quadratic of the second fundamental form to $$\vec v:$$

\begin{align} k_n=\mathrm{II}_p(\vec v)&=-\langle dN_p(\vec v), \vec v \rangle\\ &=-\langle dN_p(\cos \varphi \vec e_1 + \sin \varphi e_2), \cos \varphi \vec e_1 + \sin \varphi e_2 \rangle\\ &=\bbox[5px,border:2px solid black]{-\cos^2 \varphi k_1 - \sin^2 \varphi k_2} \end{align}

which is Euler's curvature formula.

The surface $$z=f(x,y)$$ is identical to $$F(x,y,z)=0,$$ where $$F(x,y,z)=f(x,y)-z.$$ Hence $$\left(\frac{\partial}{\partial x} F, \frac{\partial}{\partial y} F, \frac{\partial}{\partial z} F \right)=\left(\frac{\partial}{\partial x} f(x,y), \frac{\partial}{\partial y} f(x,y),-1 \right).$$