# 微分几何代写 Differential Geometry代考2023

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## 微分几何代写Differential Geometry

### 伪黎曼尼几何学Pseudo-Riemannian geometry代写

• Finsler manifold芬斯勒流形
• Symplectic geometry辛几何

## 微分几何的历史

The history and development of differential geometry as a discipline can be traced back at least to the ancient classics. It is closely related to the development of geometry, the concepts of space and form, and the study of topology, particularly manifolds. In this section we focus on the history of the application of infinitesimal methods to geometry, and then on the idea of tangent spaces, and finally on the development of the modern formalism of the discipline in terms of tensors and tensor fields.

## 微分几何相关课后作业代写

Let $c$ be a regular curve such that $|c(s)| \leq 1$ for all $s$. Suppose that there is a point $t$ where $|c(t)|=1$. Prove that the curvature at that point satisfies $|\kappa(t)| \geq 1$.

Let $c$ be a regular curve with arc length parameterization $s$. By definition, the curvature $\kappa(s)$ of the curve $c$ at the point $c(s)$ is given by $\kappa(s) = |\boldsymbol{c}”(s)|$, where $\boldsymbol{c}”(s)$ denotes the second derivative of $\boldsymbol{c}(s)$ with respect to $s$.

Since $|\boldsymbol{c}(s)| \leq 1$ for all $s$, we have $|\boldsymbol{c}'(s)| = 1$ for all $s$. Moreover, since $|\boldsymbol{c}(t)| = 1$, we have $|\boldsymbol{c}'(t)| = 0$, which implies that $\boldsymbol{c}”(t)$ is perpendicular to $\boldsymbol{c}'(t)$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|=\left|c^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|$.

Now, we observe that $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t)$ is the tangential component of $\boldsymbol{c}”(t)$ at the point $c(t)$, which is given by $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = \frac{d}{ds}\left(|\boldsymbol{c}'(s)|^2\right)\bigg|{s=t} = 2|\boldsymbol{c}'(t)|\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = 0$. Thus, we can write $\boldsymbol{c}”(t) = \boldsymbol{c}”(t){\perp} + \boldsymbol{c}”(t){\parallel}$, where $\boldsymbol{c}”(t){\perp}$ is the perpendicular component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$ and $\boldsymbol{c}”(t){\parallel}$ is the tangential component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$. Since $|\boldsymbol{c}'(t)| = 0$, we have $\boldsymbol{c}”(t){\parallel} = 0$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t){\perp}\right| \leq\left|\boldsymbol{c}^{\prime \prime}(t)\right|{\max }$

where $|\boldsymbol{c}”(t)|_{\max}$ denotes the maximum value of $|\boldsymbol{c}”(t)|$ over all vectors $\boldsymbol{v}$ perpendicular to $\boldsymbol{c}'(t)$ with $|\boldsymbol{v}| = 1$.

On the other hand, we know that the maximum value of $|\boldsymbol{c}”(t)|$ is attained when \$\boldsymbol{c}