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# 数学代写|拓扑学代写TOPOLOGY代考|MATH271 The Geometry of the Projective Plane and the Klein Bottle

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## 数学代写|拓扑学代写TOPOLOGY代考|The Geometry of the Projective Plane and the Klein Bottle

We now take a small diversion to discuss some interesting properties of the projective plane and the Klein bottle that we introduced in the previous chapter. Recall that these are abstract surfaces that exist in their own right, without reference to an embedding space like $\mathbb{R}^3$, but which nonetheless are locally homeomorphic to open sets in the plane.
The Projective Plane. We start by presenting another way of describing the projective plane. Consider the set of all lines through the origin in $\mathbb{R}^3$. It turns out that we can make a sort of space out of this set (which will turn out to be $\mathbb{R}^2$ ) as follows. Since the origin is on all these lines, and every other point is on exactly one of them, it makes sense to throw out the origin and look at $\mathbb{R}^3 \backslash{(0,0,0)}$. Now we consider two nonzero points $(a, b, c)$ and $\left(a^{\prime}, b^{\prime}, c^{\prime}\right)$ to be the same if they lie on the same line. Alternatively, we can rephrase this condition by saying that points $(a, b, c)$ and ( $\left.a^{\prime}, b^{\prime}, c^{\prime}\right)$ are considered the same if there is some (nonzero) number $\lambda \in \mathbb{R}$ so that $a^{\prime}=\lambda a, b^{\prime}=\lambda b$, and $c^{\prime}=\lambda c$. In this way, each point of $\mathbb{R}^3 \backslash{(0,0,0)}$ is glued to many other points in $\mathbb{R}^3 \backslash{(0,0,0)}$. Now the “space of all lines” that we’re studying becomes $\mathbb{R}^3 \backslash{(0,0,0)}$ but with points glued like this.

## 数学代写|拓扑学代写TOPOLOGY代考|Orientable and Nonorientable Surfaces

The Möbius strip, the Klein bottle, and the projective plane are examples of nonorientable surfaces (or nonorientable surfaces with boundary in the former case). In this section, we will define this notion more carefully.

The orientability of a compact surface or surface with boundary will be a boolean topological invariant-either a surface $S$ is orientable or it is nonorientable; and this remains true for the image of $S$ under any homeomorphism. One intuitive way to define this notion is by means of the “number of sides” that a surface embedded in $\mathbb{R}^3$ has. Take any such surface $S$ and pick a point $p \in S$. Let’s assume that $S$ is in fact a very thin three-dimensional shell rather than an idealized, infinitesimally thin, twodimensional membrane. Now pick a side of this shell at $p$ and start painting it blue near $p$. Keep painting such that every new place you paint is physically reachable from a place that you’ve already painted. At some point, you will not be able to reach any unpainted parts of the shell. At this point you ask: have you painted the whole shell? If your shell is spherical and has two distinct sides, the answer is no. If your shell is the Möbius strip and has only one distinct side (boundary edges don’t count) then the answer is yes. No means orientable, and yes means nonorientable!

The previous intuitive definition can actually be made rigorous in the case of surfaces embedded in $\mathbb{R}^3$. It requires a notion of “side,” i.e. for every $p \in S$ there is a neighborhood in $\mathbb{R}^3$ containing $p$ that can be subdivided into two subsets of $\mathbb{R}^3$ which we designate the “interior” and the “exterior” sides. We can encode the sidedness of $S$ by defining a unit normal vector for each $p \in S$. This is a vector of length one that is orthogonal to the tangent plane of $S$ – and note that $S$ must be a smooth surface! The region into which the normal vector points is the “exterior.” Now, $S$ will be called orientable if it is possible to define this normal vector in a consistent and continuous way. More precisely:

Definition 4.6 A smooth, i.e. differentiable, compact surface $S$ with or without boundary embedded in $\mathbb{R}^3$ is said to be orientable if there exists a continuous way to assign a unit normal vector at each $p \in S$. If no such assignment exists, then $S$ is nonorientable.

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