Problem 14 - Algebraic Topology

Question

A map \(f: S^{n} \rightarrow S^{n}\) satisfying \(f(x)=f(-x)\) for all \(x\) is called an even map. Show that an even map \(S^{n} \rightarrow S^{n}\) must have even degree, and that the degree must in fact be zero when \(n\) is even. When \(n\) is odd, show there exist even maps of any given even degree. IHints: If \(f\) is even, it factors as a composition \(S^{n} \rightarrow \mathbb{R} P^{n} \rightarrow S^{n} .\) Using the calculation of \(H_{n}\left(\mathbb{R} P^{n}\right)\) in the text, show that the induced map \(H_{n}\left(S^{n}\right) \rightarrow H_{n}\left(\mathbb{R P}^{n}\right)\) sends a generator to twice a generator when \(n\) is odd. It may be helpful to show that the quotient map \(\mathbb{R P}^{n} \rightarrow \mathbb{R} \mathrm{P}^{n} / \mathbb{R} \mathrm{P}^{n-1}\) induces an isomorphism on \(H_{n}\) when \(n\) is odd.]

Step-by-Step Solution

Answer

For an even map, the degree is zero when \(n\) is even; for odd \(n\), it can be any even degree.

Step 1: Understanding Even Maps

An even map from the n-sphere, denoted as \(S^n\), is one where \(f(x)=f(-x)\) for all \(x\). This symmetrical property implies that \(f\) can be factored through the projective space \(\mathbb{R}P^n\). Thus, an even map \(f: S^n \to S^n\) can be written as the composition of two maps: \(S^n \to \mathbb{R}P^n\) followed by \(\mathbb{R}P^n \to S^n\).

Step 2: Homology Groups and Generators

Considering homology, the nth homology group of \(S^n\), \(H_n(S^n)\), is \(\mathbb{Z}\). For \(\mathbb{R}P^n\), the nth homology group, \(H_n(\mathbb{R}P^n)\), differs depending on \(n\). When \(n\) is odd, \(H_n(\mathbb{R}P^n)\) is also \(\mathbb{Z}\), but when \(n\) is even, it is 0.

Step 3: Calculating the Induced Maps in Homology

The map \(S^n \to \mathbb{R}P^n\) induces a map \(H_n(S^n) \to H_n(\mathbb{R}P^n)\). In homology, this map sends a generator of \(H_n(S^n)\) to twice a generator of \(H_n(\mathbb{R}P^n)\) when \(n\) is odd due to the \(f(x) = f(-x)\) condition. This results in the degree of the final map being a multiple of 2.

Step 4: Even Degree for Odd n

For odd \(n\), because \(H_n(\mathbb{R}P^n) = \mathbb{Z}\), it's possible to have maps of any even degree. Since the degree relates directly to how homology generators map, and given odd \(n\), even maps will have degrees such as 0, 2, 4, etc.

Step 5: Trivial Degree for Even n

For even \(n\), \(H_n(\mathbb{R}P^n)\) is 0 due to the standard cellular structure of \(\mathbb{R}P^n\). This effectively means that any map composed from an even map \(S^n \to \mathbb{R}P^n \to S^n\) for even \(n\) must have degree 0 as there are no non-zero maps possible in homology.

Key Concepts

n-sphere

The n-sphere, denoted as \(S^n\), is a generalization of the surface of a common sphere into n dimensions. Imagine the ordinary sphere in three dimensions, which is the 2-sphere \(S^2\). For n=1, the n-sphere \(S^1\) is simply a circle. It is crucial to remember that although these n-spheres are surfaces lying in Euclidean spaces, we only consider them in terms of points equidistant from a fixed central point, without necessarily visualizing them in higher-dimensional spaces. The n-sphere can be defined mathematically as the set of points in \(\mathbb{R}^{n+1}\) at a unit distance from the origin. Understanding n-spheres helps clarify the concept of even maps because it shows how symmetry can manifest in multiple dimensions. Knowing how to handle and visualize these higher-dimensional objects is foundational for grasping the wider implications in topology and geometry.

• \(S^0\) is two distinct points, often used to illustrate maps without dimensions.
• \(S^1\) is the circumference of a circle, frequently aiding in visualizing continuity in mapping.
• \(S^2\) is the surface of a 3D sphere, a good analogy for more intuitive understanding.

projective space

Projective space, particularly real projective space denoted by \(\mathbb{R}P^n\), is constructed by identifying points in a vector space that are scalar multiples of each other. It's like "squashing" all lines through the origin in one direction into a single point. This way of handling space means directions in space boil down to points in projective space. Projective spaces help in solving geometric problems where symmetrical properties are involved. This is essential when dealing with even maps on spheres, where \(f(x) = f(-x)\). The factored map \(S^n \to \mathbb{R}P^n\) allows the study of such symmetrical characteristics in homology.

• In \(\mathbb{R}P^2\), think of projecting a 3D sphere onto a plane but wrapping edges together.
• In general, \(\mathbb{R}P^n\) converts complexity into manageable, repeatable properties in topology.
• These properties are helpful when considering maps that rely on symmetry or continuous transformations.

homology groups

Homology groups are tools used in algebraic topology to analyze and classify topological spaces based on their shapes and features. They provide a way to "count" holes in different dimensions. For example, the nth homology group of a space, denoted \(H_n(X)\), helps to quantify n-dimensional holes, like voids in complex shapes. For spheres, \(H_n(S^n)\) is \(\mathbb{Z}\) because there is precisely one cycle that is not the boundary of any n+1-dimensional feature. Homology groups are crucial for understanding map degree calculations since they describe how maps transform these cycles. This understanding is particularly pertinent for even maps, where the induced homology maps must preserve the structure defined by the symmetry \(f(x) = f(-x)\).

• When \(n\) is even, \(H_n(\mathbb{R}P^n)\) is 0, indicating no n-dimensional "holes."
• For odd \(n\), both \(H_n(S^n)\) and \(H_n(\mathbb{R}P^n)\) equal \(\mathbb{Z}\), preserving cycle symmetry.
• Homology gives a systematic way to address the degree of maps by assessing transformations of generators.

map degree

The degree of a continuous map is a topological invariant that tells how many times the mapped domain wraps around its range. Formally, the degree is an integer reflecting the net number of times the space is covered. Calculating the degree involves algebraically adding preimages of any point with consideration to orientation. In the context of even maps between spheres, an even map \(S^n \to S^n\) must inherently stabilize the map degree as a multiple of two due to each transformation being symmetrical in nature. For even \(n\), this symmetry compels the degree to be zero because the covering and wrapping cancel completely. Conversely, for odd \(n\), the map's degree can be any even number, thanks to the stable values of the homology groups involved.

• Degree is 0 when an even map's structure nullifies topological wrappings for even \(n\).
• Any even degree is possible for odd \(n\), ensuring flexible transformations.
• Understanding degree assists in recognizing transformations that maintain or alter topological features.