Possible typographical mistake in the definition of the normal curvature

Question

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?

Answer 1

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

Answer 2

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 $-1<x<1$ and $-1<y<1$ 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 $-1<t<1$ 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:

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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.

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