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# 物理代写|拓扑物理代写Physical topology代考|Computation of Homology Groups Of Surfaces

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## 物理代写|拓扑物理代写Physical topology代考|The Cylinder

We consider the following triangulation given in Fig. $4.3$ for the compact cylin$\operatorname{der} S^{1} \times[0,1]$. Let us denote it by K.

Now we observe that the 0 -th chain group $C_{0}(K)$ is the abelian group generated by the vertices $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$. Thus, we formally write
$$C_{0}(K)=\left[v_{0}\right] \mathbb{Z} \oplus\left[v_{1}\right] \mathbb{Z} \oplus\left[v_{2}\right] \mathbb{Z} \oplus\left[v_{3}\right] \mathbb{Z} \oplus\left[v_{4}\right] \mathbb{Z} \oplus\left[v_{5}\right] \mathbb{Z}$$
In other words $C_{0}(K) \cong \mathbb{Z}^{5}$. Similarly, we note that the chain group $C_{1}(K)$ is generated by all the edges shown in the figure. Thus, we have
$$\begin{gathered} C_{1}(K)=\left[v_{0}, v_{1}\right] \mathbb{Z} \oplus\left[v_{0}, v_{2}\right] \mathbb{Z} \oplus\left[v_{0}, v_{3}\right] \mathbb{Z} \oplus\left[v_{0}, v_{4}\right] \mathbb{Z} \ \oplus\left[v_{1}, v_{3}\right] \mathbb{Z} \oplus\left[v_{1}, v_{5}\right] \mathbb{Z} \oplus \ldots \oplus\left[v_{4}, v_{5}\right] \mathbb{Z} \end{gathered}$$

## 物理代写|拓扑物理代写Physical topology代考|Torus

The torus $T$ can be obtained as the quotient space of a rectangle with opposite sides identified in a manner that preserves orientations on them. Thus, we can triangulate the torus as shown in the Fig. 4.5.

Instead of going through the tedious computations, we argue geometrically to compute the homology groups of torus T.

We first observe that all the homology groups for $H_{r}(T)=0$ for $r \geq 3$ as there are no non-trivial $r$-chains for $r \geq 3$. Next, recall that $H_{0}(T)=C_{0} / B_{0}$. Again any two vertices can be joined by an oriented path. Therefore, there is only one 0 -cycle upto 0 -boundaries. Hence, $H_{0}(T) \cong \mathbb{Z}$.

There are eighteen triangles in the above diagram. To get a 2-cycle, we observe that each edge is shared by precisely two adjacent triangles. Thus, the coefficients of these triangles must be the same for cancellations to take place. Any two triangles can be joined by a sequence of adjacent triangles. Therefore, the coefficients of all triangles must be the same. This shows that the subgroup of 2-cycles is generated by the cycle that is obtained by taking the sum of all (oriented) triangles as shown in the diagram. Thus, $H_{2}(T) \cong \mathbb{Z}$.

## 物理代写|拓扑物理代写PHYSICAL TOPOLOGY代考|The Projective Plane

The projective plane is obtained by taking a closed unit disk in the plane and identifying diametrically opposite points on the boundary circle. The resulting topological space can not be given an orientation. We consider the triangulation shown in Fig. $4.6$ for the projective plane.

By now, an alert reader might have acquired a geometric feel for guessing the homology groups and supporting the guesses with rigorous arguments.
First, observe that $H_{r}\left(\mathbb{R} P^{2}\right)=0$ for $r \geq 3$ as there are no non-trivial chains in dimension greater than or equal to three. Now $H_{0}\left(\mathbb{R} P^{2}\right) \cong \mathbb{Z}$ as all the vertices can be connected by a path of edges. Next, we note that $H_{2}\left(\mathbb{R} P^{2}\right)=0$ as any edge is shared by precisely two adjacent triangles. Further, any two triangles can be joined by a sequence of adjacent triangles. Therefore, the coefficients must be the same for cancellations to happen along non-peripheral edges. The coefficients along the peripheral edges $\left[v_{0}, v_{1}\right],\left[v_{1}, v_{2}\right]$ and $\left[v_{2}, v_{0}\right]$ get added instead of cancelling. This is due to the identification along the boundary. Therefore, one may argue that the only 2-cycle is the trivial 2-cycle (the element 0$)$. Hence $H_{2}\left(\mathbb{R} P^{2}\right)=0$.

## 物理代写|拓扑物理代写PHYSICAL TOPOLOGY代考|THE CYLINDER

$$C_{0}(K)=\left[v_{0}\right] \mathbb{Z} \oplus\left[v_{1}\right] \mathbb{Z} \oplus\left[v_{2}\right] \mathbb{Z} \oplus\left[v_{3}\right] \mathbb{Z} \oplus\left[v_{4}\right] \mathbb{Z} \oplus\left[v_{5}\right] \mathbb{Z}$$

$$\begin{gathered} C_{1}(K)=\left[v_{0}, v_{1}\right] \mathbb{Z} \oplus\left[v_{0}, v_{2}\right] \mathbb{Z} \oplus\left[v_{0}, v_{3}\right] \mathbb{Z} \oplus\left[v_{0}, v_{4}\right] \mathbb{Z} \ \oplus\left[v_{1}, v_{3}\right] \mathbb{Z} \oplus\left[v_{1}, v_{5}\right] \mathbb{Z} \oplus \ldots \oplus\left[v_{4}, v_{5}\right] \mathbb{Z} \end{gathered}$$

## Matlab代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。