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# 数学代写|拓扑学代写TOPOLOGY代考|Homology and Orientability

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## 数学代写|拓扑学代写TOPOLOGY代考|Homology and Orientability

Let $S$ be a compact, connected surface (without boundary). The homology detects the orientability of $S$, in the following way. Note that $H_2(\mathbb{T}) \cong \mathbb{Z}$, whereas $H_2(\mathbb{K})=0$. In general, the 2-dimensional homology of $S$ is $\mathbb{Z}$ if $S$ is orientable, and it’s 0 if $S$ is nonorientable. More generally, if $X$ is a compact, connected $n$-dimensional manifold (without boundary), then $H_n(X) \cong \mathbb{Z}$ if $X$ is orientable, and $H_n(X)=0$ if $X$ is nonorientable. We’ll only prove this for surfaces since we’ll work in terms of ID spaces.

Let’s first suppose that $S$ is orientable, and that we have an ID space for $S$, which is a polygon with edges identified in pairs. As we recall from Chapter 4, because $S$ is orientable, the edges are Type I edges, i.e. as we traverse the boundary of the ID space polygon, the two instances of that edge appear with opposite orientations.
Now, split the ID space for $S$ up into triangles so that we have a triangulation of $S$, into triangles $T_1, T_2, \ldots, T_r$. We orient each triangle $T_i$ in the counterclockwise orientation. Then a 2-chain $c$ is a sum $\sum_{i=1}^r a_i T_i$, where each $a_i \in \mathbb{Z}$. What does it mean for $c$ to be a 2-cycle? Take two triangles that share an edge in the interior of the polygon, say $T_i$ and $T_j$, which share edge $e$. These are the only two triangles containing that edge, so they are the only contributors to $e$ in $\partial_2(c)$. Thus $\partial_2\left(a_i T_i+\right.$ $a_j T_j$ ) must have a coefficient of 0 for $e$. The contribution from $a_i T_i$ is $a_i$, whereas the contribution from $a_j T_j$ is $-a_j$ (or the signs may both be swapped). Thus we find that a necessary condition for $c$ to be a 2-cycle is that $a_i=a_j$. Because we can apply this argument to an arbitrary interior edge, we find that all the $a_i$ ‘s must be equal, i.e. it must be the case that $c=\sum_{i=1}^r a T_i$ for some $a \in \mathbb{Z}$.

But is such a $c$ actually a cycle? The only thing that can go wrong is that the boundary edges of the polygon might not cancel. However, since $S$ is assumed to be orientable, each boundary edge appears once with a positive orientation and once with a negative orientation. Thus chains of the form $\sum_{i=1}^r a T_i$ are indeed cycles, and they are the only cycles in $S$. If $a \neq 0$, then they are not boundaries, because $C_3(S)=0$. Thus we find that $H_2(S)=\left\langle\sum_{i=1}^r T_i\right\rangle \cong \mathbb{Z}$.

## 数学代写|拓扑学代写TOPOLOGY代考|Smith Normal Form

It seems as though computing homology is easy and completely mechanical-so that the process is something that one could program a computer to do. But there is one step that is still difficult. Once we have computed $Z_i(X)$ and $B_i(X)$, we obtain some presentation for $H_i(X)$, but we would like to be able to identify it in a more convenient form. If $X$ has a finite triangulation, then $H_i(X)$ is a finitely generated abelian group, and we know what all the finitely generated abelian groups look like. But when we see a group like
$$\left\langle a_1, a_2, a_3, a_4 \mid 5 a_1-2 a_2+3 a_4, 3 a_1+2 a_2+2 a_3, 4 a_3-2 a_4, 9 a_2+6 a_3\right\rangle,$$
how do we write that nicely, in the form $\mathbb{Z}^k \times$ (finite group)?
Fortunately, there is a fairly simple algorithm for doing this. It will be convenient to write out the relations as a matrix. Each relation gets a row, and each generator gets a column, and the coefficients go in the matrix. Hence the matrix we get from the presentation (13.1) is
$$\left(\begin{array}{cccc} 5 & -2 & 0 & 3 \ 3 & 2 & 2 & 0 \ 0 & 0 & 4 & -2 \ 0 & 9 & 6 & 0 \end{array}\right) .$$

## 数学代写|拓扑学代写TOPOLOGY代考|Smith Normal Form

$$\left\langle a_1, a_2, a_3, a_4 \mid 5 a_1-2 a_2+3 a_4, 3 a_1+2 a_2+2 a_3, 4 a_3-2 a_4, 9 a_2+6 a_3\right\rangle,$$

$$\left(\begin{array}{llllllllllllllll} 5 & -2 & 0 & 3 & 3 & 2 & 2 & 0 & 0 & 0 & 4 & -2 & 0 & 9 & 6 & 0 \end{array}\right) .$$

## 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。