Curvature forms as exterior covariant derivative?

I have read on several forums like this one, that given a connection form $ \omega$ on a principal bundle and its curvature form $ \Omega$ , I can state that $ \Omega=d_\omega\omega$ alike I do in the case of torsion, namely that torsion is the covariant differential of a differential form. To me that does not make sense, because $ \Omega=d\omega+\frac{1}{2}[\omega,\omega]$ and there is no representation in which that $ \frac{1}{2}$ can arise.

Any explanations?

Calculating points of a curve given by its curvature [on hold]

I try to draw a curve using it’s curvature. The curvature is given in a function: f(x), f(0) meaning the curvature at the start of the curve. This (linear) function describes the curve from start to finish and is given as: curvStart to curvEnd. Negative curvature = turn to the right, positive curvature = turn to the left.

I also have the starting coordinates for a curve (x and y). What I want to do now is getting points that I can plot on a 2D plane from these informations.

For example: Curvature from -0.5 to 0, starting coordinates (0,0) and I want 3 points would result in a point at the start (0,0), a point at the end of the curve and one exactly in the middle of the curve.

I hope you can understand what I want to do and I am very sorry for my bad explanation, I am lacking the terms and the understanding of the topic to describe it properly. Sadly most “tutorials” online only care about the curve to curvature part.

Directional Curvature

What is Directional Curvature and how can I achieve it for any function? A common approach with an example would be much appreciated.

(Reference: I am reading “The Non-convex Geometry of Low-rank Matrix Optimization” paper and in section 1.2, Weighted PCA part, I got stuck. Link to the paper: https://academic.oup.com/imaiai/article/8/1/51/4951409)

Thanks in advance!

Explicit computation of connection & curvature matrix

I have recently learned the generalized Gauss-Bonnet theorem, which states that:

\begin{equation} \int_M \text{Pf}(\Omega) = (2\pi)^n\chi(M), \end{equation} where $ n$ is half the dimension of an even dimensional, compact, Riemannian manifold.

Here, $ \Omega$ is the curvature matrix of 2-forms determined by the Riemannian metric $ g$ and some metric compatible connection $ \nabla$ , and $ \text{Pf}(\Omega)$ is the Pfaffian.

By Chern-Weil, we know that our choice of $ \nabla$ does not make any difference.

Question: The above integral, if the dimension of the manifold in question is 2, better reduce to the Gauss-Bonnet theorem that we know and love: \begin{equation} \int_M KdA = 2\pi\chi(M), \end{equation} where $ K$ is the Gaussian curvature. But I am not sure how I can carry out the computation necessary to get there…

More specifically, I know that if the dimension is 2, then $ \text{Pf}(\Omega)$ is gonna be a 2 form, more precisely, some number times $ \Omega_1^2$ , the upper-right entry of $ 2\times 2$ curvature matrix. If all were to work, this 2-form better be the form $ KdA$ .

By Chern-Weil, we may as well assume that the connection in question is Levi-Civita. Then the Theorema Egregium allows us to write $ K$ in terms of $ g$ and the associated Christoffel symbols.

My problem is that I don’t know how to carry out this explicit computation… Could you help me with this?

When are principal lines of curvature geodesics?

Let $ S$ be a smooth surface embedded in $ \mathbb{R}^3$ . When are (some of) the principal lines of curvature geodesics on $ S$ ? Perhaps on the ellipsoid below, the (blue) central cycle, a max principal line, is a geodesic? And perhaps the (red) min principal line connecting the two umbilical points is a geodesic?


         
          Image from Jorge Sotomayor.1


Is there any $ S$ all of whose principal lines of curvature are geodesics?


1Sotomayor, Jorge. “Historical Comments on Monge’s Ellipsoid and the Configuration of Lines of Curvature on Surfaces Immersed in $ {\mathbb R}^ 3$ .” arXiv Abstract (2004). São Paulo Journal of Mathematical Sciences 2, 1 (2008), 99–143.

Lemma $3.2$ – Mean curvature flow singularities for mean convex surfaces

This is a lemma of the paper “Mean curvature flow singularities for mean convex surfaces” by Gerhard Huisken and Carlo Sinestrari (the paper is available here):

$ \textbf{Lemma 3.2.}$ Suppose $ (1 + \eta) H^2 \leq |A|^2 \leq c_0 H^2$ for some $ \eta, c_0 > 0$ at some point of $ \mathscr{M}_t$ , then we also have

(i) $ -2Z \geq \eta H^2|A|^2$ ;

(ii) $ |H \nabla_i h_{kl} – \nabla_i H h_{kl}|^2 \geq \frac{\eta^2}{4n(n-1)^2c_0} H^2 |\nabla H^2|$

My doubt is concerning to item $ (ii)$ and below is the argument given by the authors

We have (see [10, Lemma $ 2.3$ (ii)], reference [10] is available here)

$ |H \ \nabla_i h_{kl} – \nabla_i H \ h_{kl}|^2 \geq \frac{1}{4} |\nabla_i H \ h_{kl} – \nabla_k H \ h_{il}|^2 = \frac{1}{2} (|A|^2 |\nabla H|^2 – |\nabla^i H h_{il}|^2).$

Let us denote with $ \lambda_1, \cdots, \lambda_n$ the eigenvalues of $ A$ in such a way that $ \lambda_n$ is an eigenvalue with the largest modulus. Then we have $ |\nabla^i H \ h_{il}|^2 \leq \lambda_n^2 |\nabla H|^2$ and

\begin{align*} |H \ \nabla_i h_{kl} – \nabla_i H \ h_{kl}|^2 &\geq \frac{1}{2} \sum\limits_{i=1}^{n-1} \lambda_i^2 |\nabla H|^2 = \sum\limits_{i=1}^{n-1} \lambda_i^2 \lambda_n^2 \frac{|\nabla H|^2}{2\lambda_n^2}\ &\geq \sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \frac{|\nabla H|^2}{2(n-1)|A|^2}\ &\geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2 \frac{|\nabla H|^2}{n(n-1)|A|^2}\ &= \frac{(|A|^2 – H^2)^2}{4n(n-1)|A|^2} |\nabla H|^2 \geq \frac{\eta^2 H^2}{4n(n-1)c_0} |\nabla H|^2. \square \end{align*}

I would like to understand the following equality and inequalities:

a) $ \frac{1}{4} |\nabla_i H \ h_{kl} – \nabla_k H \ h_{il}|^2 = \frac{1}{2} (|A|^2 |\nabla H|^2 – |\nabla^i H h_{il}|^2)$ ;

b) $ |H \ \nabla_i h_{kl} – \nabla_i H \ h_{kl}|^2 \geq \frac{1}{2} \sum\limits_{i=1}^{n-1} \lambda_i^2 |\nabla H|^2$ ;

c) $ \sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \frac{|\nabla H|^2}{2(n-1)|A|^2} \geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2 \frac{|\nabla H|^2}{n(n-1)|A|^2}$ .

My thoughts:

$ a$ and $ b$ ) I just consider use normal coordinates, but this doesn’t helps me because the right hand side in $ a$ and $ b$ would be zero.

$ c$ ) I just try to prove that $ \sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2$ , but I can’t prove this because I don’t know if all eigenvalues are non-negative. Indeed, I even don’t know if the $ H > 0$ because I didn’t see the hypothesis that the hypersurfaces is mean convex on the paper until this lemma.

Thanks in advance!

Signed curvature of catenary involving turning/tangential angle

Suppose we want to find the signed curvature of the catenary $ $ \gamma(t)=(t,\cosh t)$ $ where $ \mathcal{k}_n=\frac{d\phi}{ds}$ and $ \phi(s)$ is the turning angle of $ \gamma$ such that$ $ \dot\gamma(s)=(\cos\phi(s), \sin\phi(s))$ $

We proceed: $ $ s=\int_{s_0}^{s}|\dot\gamma(t)|dt=\int_{s_0}^s\sqrt{1+\sinh^2t}dt=\sinh t$ $ so if $ \phi$ is the angle between $ \dot\gamma$ and the $ x$ -axis, then $ $ \tan\phi=\sinh t=s\implies\sec^2\phi\frac{d\phi}{ds}=1\implies k_s=\frac{1}{1+s^2}$ $

Can someone explain how we proceed in that last line of calculation? Why does $ \tan\phi=\sinh t=s? $

Curvature of plane curves

My text defines the curvature of a plane curve as $ <\ddot{x},N>$ where $ N$ is the normal to the normalized tangent of $ x$ and $ x$ is the curve. I thought the $ \ddot{x}$ also was perpendicular to $ x$ , making this projection kind of odd. Can someone see where I go wrong? Isnt projectuon of parallel lines a wierd thing?

Example of a Manifold which has One Non-zero Component of Ric corresponding to Scalar Curvature

I am wondering if there is a simple example of a manifold such that, given a value for the scalar curvature $ R$ , I can find a manifold such that the Ricci tensor has all zero components except for one component which takes the value $ R$ .

I feel like this can be achieved using a warped product of two metrics to separate out one coordinate and then just solve the differential equations so that the first coefficient vanishes, but obviously the coefficient of the non-zero component needs to be $ R$ .

What curve minimizes distance from the origin, given length and total curvature?

Let $ \textit{F}$ be the family of $ C^1$ curves in $ \mathbb{R}^2$ of fixed length $ \bar{l}$ and fixed tangent’s turning angle $ \bar{k}$ .

What are the curves in $ \textit{F}$ minimizing the Euclidean distance between the starting and the ending point? Is the arc of circle of proper radius one of those?