# 数学代写|拓扑学代写Topology代考|MTH3002

#### Doug I. Jones

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## 数学代写|拓扑学代写Topology代考|The Hopf fibration and polarization

One additional important example of a fiber bundle that comes up often in physics is the Hopf bundle. Recall that $S U(2)$ is the group of $2 \times 2$ special unitary matrices, the unitary matrices with determinant equal to $+1$. An element $s$ of $S U(2)$ can be viewed as a two-by-two complex matrix of the form:
$$s=\left(\begin{array}{cc} \alpha & -\beta^{} \ \beta & \alpha^{} \end{array}\right),$$
where $|\alpha|^{2}+|\beta|^{2}=1$. The reader can readily verify that
$$s^{\dagger} s=1 \quad \text { and } \operatorname{Det}(s)=1 .$$
Notice that the pair of complex numbers $\alpha$ and $\beta$ define a unit vector in $\mathbb{R}^{4}$; in other words, a point on the surface of the three-dimensional unit sphere $S^{3}$. We therefore have an isomorphism between the unitary group and the three-sphere, $S U(2) \sim S^{3}$.
Starting from $s \in S U(2)$, we now define a three-dimensional vector $x \in \mathbb{R}^{3}$ whose components are given by
$$x^{i}=\frac{1}{2} \operatorname{Tr}\left(\sigma_{i} S \sigma_{3} S^{-1}\right),$$
where $\sigma_{i}$ are the three Pauli matrices. This vector is a unit vector, $|x|^{2}=1$, which means that the mapping $\pi: s \rightarrow \boldsymbol{x}$ is a map between spheres,
$$\pi: S^{3} \rightarrow S^{2} \text {. }$$
This is a two-to-one mapping, since $s$ and $-s$ both project onto the same $\boldsymbol{x} \in S^{2}$. Explicitly, the mapping between $\boldsymbol{x}$ and $s$ is given by:
$$\begin{gathered} \alpha=\cos |x|+i x^{3} \sin |x| \ \beta=\left(i x^{1}-x^{2}\right) \sin |x| \end{gathered}$$
or equivalently,
\begin{aligned} &x^{1}=\operatorname{Re}\left(2 \alpha^{} \beta\right) \ &x^{2}=\operatorname{Im}\left(2 \alpha^{} \beta\right) \end{aligned} $x^{3}=|\alpha|^{2}-|\beta|^{2} .$

## 数学代写|拓扑学代写Topology代考|Euler characteristic

For a manifold $M$, the Euler-Poincaré characteristic (often simply called the Euler characteristic) can be defined via triangulations. Every sufficiently well-behaved space can be given a triangulation; we describe the idea for two dimensions, but a similar process can be carried out in higher dimensions.

Cover a two-dimensional surface $\mathcal{S}$ by a network of line segments or curved arcs called edges, which join up at a set of points, called vertices. Together, the lines and edges form a graph on the surface (figure 5.1). Arrange the vertices and edges so that the entire surface is covered by a set of triangles enclosed by the edges, with the vertices lying at the corners of the triangles. These triangles are called faces. Let $E, F$, and $V$ represent the number of edges, faces, and vertices in this triangulation. Then the Euler characteristic for the surface is defined by the formula
$$\chi(S)=F-F+V$$
Notice that the sign alternates with dimension: the even dimensional vertices and faces come with plus signs, the odd dimensional edges come with a minus sign. For a more general manifold $M$, the pattern will continue with higher dimensional objects added into the alternating sum: $\chi(M)=\sum_{n}(-1)^{n} b_{n}$, where $b_{n}$ is the $n$th Betti number (the rank of the $n$th homology group).

The Euler characteristic is in fact independent of the triangulation: any triangulation that can be drawn on the surface will give the same result. This is easy enough to see through examples. Starting from a surface with some triangulation already given on it, and consider what happens when the triangulation is changed by adding an additional vertex and corresponding edges, as in figure 5.2. When we add the new vertex, this requires adding three new edges, which in turn replaces the original face inside which the vertex was added by three smaller faces. So
$$V \rightarrow V+1, \quad E \rightarrow E+3, \quad F \rightarrow F+2,$$
which results in
$$\chi=F-E+V \rightarrow(F+2)-(E+3)+(V+1)=F-E+V=\chi .$$

# 拓扑学代考

## 数学代写|拓扑学代写Topology代考|The Hopf fibration and polarization

$$s=\left(\begin{array}{lll} \alpha & -\beta \beta & \alpha \end{array}\right),$$

$$s^{\dagger} s=1 \quad \text { and } \operatorname{Det}(s)=1 .$$

$$x^{i}=\frac{1}{2} \operatorname{Tr}\left(\sigma_{i} S \sigma_{3} S^{-1}\right),$$

$$\pi: S^{3} \rightarrow S^{2} .$$

$$\alpha=\cos |x|+i x^{3} \sin |x| \beta=\left(i x^{1}-x^{2}\right) \sin |x|$$

$$x^{1}=\operatorname{Re}(2 \alpha \beta) \quad x^{2}=\operatorname{Im}(2 \alpha \beta)$$
$$x^{3}=|\alpha|^{2}-|\beta|^{2} .$$

## 数学代写|拓扑学代写Topology代考|Euler characteristic

$$\chi(S)=F-F+V$$

$$V \rightarrow V+1, \quad E \rightarrow E+3, \quad F \rightarrow F+2,$$

$$\chi=F-E+V \rightarrow(F+2)-(E+3)+(V+1)=F-E+V=\chi$$

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