Problem 12
Question
Show that the spaces \(\left(S^{1} \times \mathrm{CP}^{\infty}\right) /\left(S^{1} \times\left\\{x_{0}\right\\}\right)\) and \(S^{3} \times \mathrm{CP}^{\infty}\) have isomorphic cohomology rings with \(\mathbb{Z}\) or any other coefficients. IAn exercise for \(\$ 4 . \mathrm{L}\) is to show these two spaces are not homotopy equivalent.]
Step-by-Step Solution
Verified Answer
The spaces share isomorphic cohomology rings with generators and relations from \(\mathrm{CP}^{\infty}\).
1Step 1: Identify the Spaces and Their Characteristics
We have two spaces: \(\frac{S^{1} \times \mathrm{CP}^{\infty}}{S^{1} \times \{x_0\}}\) and \(S^{3} \times \mathrm{CP}^{\infty}\). Recognize that \(\mathrm{CP}^{\infty}\) is the infinite complex projective space, which has a well-known cohomology ring structure with generators and relations.
2Step 2: Understand Cohomology of the Spaces
Cohomology with coefficients in \(\mathbb{Z}\) for \(S^1\): \(H^n(S^1; \mathbb{Z}) = \mathbb{Z}\) if \(n = 0\), otherwise 0. For \(\mathrm{CP}^{\infty}\), \(H^*(\mathrm{CP}^{\infty}; \mathbb{Z}) \sim \mathbb{Z}[x]\) where \(\deg(x) = 2\). Consider potential effect of the quotient involved in the first space.
3Step 3: Cohomology of Quotient Space
The first space is a quotient which collapses \(S^1\). This impacts the cohomology: only classes in \(\mathrm{CP}^{\infty}\) survive. Thus, cohomology for the space is dominated by that of \(\mathrm{CP}^{\infty}\), i.e., \(H^*( \mathrm{CP}^{\infty}; \mathbb{Z})\).
4Step 4: Consider Product Space's Cohomology
For \(S^3 \times \mathrm{CP}^{\infty}\), use Künneth formula: \(H^*(S^3;\mathbb{Z})\) is \(\mathbb{Z}\) for 0 and 3, and 0 otherwise. Thus, \(H^*(S^3 \times \mathrm{CP}^{\infty}; \mathbb{Z}) \sim H^*(S^3) \otimes H^*(\mathrm{CP}^{\infty}) \). Product gives the same rings generated by respective cycles.
5Step 5: Verify Isomorphism of Cohomology Rings
Observe isomorphic structure: Both spaces' cohomology have \(x\) as generator with degree 2 (from \(\mathrm{CP}^{\infty}\)), implying the cohomology rings' isomorphism. Thus, they share polynomial algebra structure over \(\mathbb{Z}\), even with different space representations.
Key Concepts
Infinite Complex Projective SpaceKünneth FormulaCohomology Rings
Infinite Complex Projective Space
The Infinite Complex Projective Space, denoted as \( \mathrm{CP}^{\infty} \), is an important construct in algebraic topology. It can be thought of as the limit of finite-dimensional complex projective spaces. In simpler terms, it represents all possible lines through the origin in an infinite-dimensional complex space. This space is very powerful because it has a straightforward and well-known cohomology structure, which makes it incredibly useful in theoretical applications.
The cohomology ring of \( \mathrm{CP}^{\infty} \) is interesting because it forms a polynomial ring. Specifically, it is denoted as \( H^*(\mathrm{CP}^{\infty}; \mathbb{Z}) \cong \mathbb{Z}[x] \) where \( \deg(x) = 2 \).
This indicates that the cohomology groups are generated by a single element \( x \) of degree 2, thus creating a ring structure which extends indefinitely. The polynomial aspect means each power of \( x \) corresponds to a different class in the cohomology, creating a simple yet profound structure that is mirrored across various applications in topology. This is a key aspect when comparing cohomology rings, such as those in the exercise.
The cohomology ring of \( \mathrm{CP}^{\infty} \) is interesting because it forms a polynomial ring. Specifically, it is denoted as \( H^*(\mathrm{CP}^{\infty}; \mathbb{Z}) \cong \mathbb{Z}[x] \) where \( \deg(x) = 2 \).
This indicates that the cohomology groups are generated by a single element \( x \) of degree 2, thus creating a ring structure which extends indefinitely. The polynomial aspect means each power of \( x \) corresponds to a different class in the cohomology, creating a simple yet profound structure that is mirrored across various applications in topology. This is a key aspect when comparing cohomology rings, such as those in the exercise.
Künneth Formula
The Künneth Formula plays a vital role when dealing with the cohomology of product spaces. It provides a method to compute the cohomology of a product space based on the cohomologies of the individual spaces.
When you have two topological spaces \( X \) and \( Y \), the Künneth Formula helps determine the structure of the cohomology of their product \( X \times Y \). It states that if \( H^*(X; A) \) and \( H^*(Y; A) \) are free modules over a ring \( A \), then the module \( H^*(X \times Y; A) \) can be expressed as:
\[H^*(X \times Y; A) \cong H^*(X; A) \otimes_A H^*(Y; A)\]
This tells us how the cohomology ring of \( X \times Y \) develops a product ring structure derived from the individual cohomology rings of \( X \) and \( Y \). For the spaces in the exercise, like \( S^3 \times \mathrm{CP}^{\infty} \), it helps verify that the product space's cohomology forms a similar cohomology ring structure as the quotient space in the exercise, despite being a different space altogether.
When you have two topological spaces \( X \) and \( Y \), the Künneth Formula helps determine the structure of the cohomology of their product \( X \times Y \). It states that if \( H^*(X; A) \) and \( H^*(Y; A) \) are free modules over a ring \( A \), then the module \( H^*(X \times Y; A) \) can be expressed as:
\[H^*(X \times Y; A) \cong H^*(X; A) \otimes_A H^*(Y; A)\]
This tells us how the cohomology ring of \( X \times Y \) develops a product ring structure derived from the individual cohomology rings of \( X \) and \( Y \). For the spaces in the exercise, like \( S^3 \times \mathrm{CP}^{\infty} \), it helps verify that the product space's cohomology forms a similar cohomology ring structure as the quotient space in the exercise, despite being a different space altogether.
Cohomology Rings
Cohomology rings provide a deep insight into the topology of a space by algebraically encoding information about its structure and features. These rings are formed by the cohomology groups of a space with a graded ring structure, where the multiplication operation can reveal interactions between different cohomology classes.
In the context of our exercise, showing that two spaces have isomorphic cohomology rings means they share similar algebraic structures, derived from their topological properties, even if the spaces themselves are not the same. If two spaces have isomorphic cohomology rings, it implies that their topological behaviors, at least from a cohomological perspective, are identical.
For the spaces \( \left(S^{1} \times \mathrm{CP}^{\infty}\right)/\left(S^{1} \times\{x_{0}\}\right) \) and \( S^{3} \times \mathrm{CP}^{\infty} \), although distinct in their geometric construction, their cohomology rings with integer coefficients align due to structures and actions in their respective rings of cohomology classes. This correspondence showcases the beautiful intricacies of topology, where different geometric shapes can share similar algebraic signatures.
In the context of our exercise, showing that two spaces have isomorphic cohomology rings means they share similar algebraic structures, derived from their topological properties, even if the spaces themselves are not the same. If two spaces have isomorphic cohomology rings, it implies that their topological behaviors, at least from a cohomological perspective, are identical.
For the spaces \( \left(S^{1} \times \mathrm{CP}^{\infty}\right)/\left(S^{1} \times\{x_{0}\}\right) \) and \( S^{3} \times \mathrm{CP}^{\infty} \), although distinct in their geometric construction, their cohomology rings with integer coefficients align due to structures and actions in their respective rings of cohomology classes. This correspondence showcases the beautiful intricacies of topology, where different geometric shapes can share similar algebraic signatures.
Other exercises in this chapter
Problem 11
If \(T^{n}\) is the \(n\) -dimensional torus, the product of \(n\) circles, show that the Pontryagin ring \(H_{*}\left(T^{n} ; \mathbb{Z}\right)\) is the exteri
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Using cup products, show that every map \(s^{k+\ell} \rightarrow S^{k} \times S^{\ell}\) induces the trivial homomorphism \(H_{k+\ell}\left(S^{k+\ell}\right) \r
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Describe \(H^{*}\left(\mathrm{CP}^{\infty} / \mathrm{CP}^{1} ; \mathbb{Z}\right)\) as a ring with finitely many multiplicative generators. How does this ring co
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Show that the coproduct in the Hopf algebra \(H_{*}(X ; R)\) dual to \(H^{*}(X ; R)\) is induced by the diagonal map \(X \rightarrow X \times X, x \mapsto(x, x)
View solution