7.1 Cup product in cohomology

The cup product is a powerful tool in cohomology that combines two classes to create a new one of higher degree. It's like a mathematical recipe that mixes ingredients to create something new and exciting. This operation helps us understand the structure of spaces and how they relate to each other.

By using the cup product, we can compute cohomology rings and find important invariants of spaces. It's like a secret code that reveals hidden properties of shapes and surfaces. This knowledge helps us solve puzzles about maps between spaces and discover geometric structures we couldn't see before.

Cup Product Operation

Definition and Construction

• The cup product is a binary operation that takes two cohomology classes and produces a new cohomology class of higher degree
• For cohomology classes $\alpha \in H^n(X;R)$ and $\beta \in H^m(X;R)$, their cup product is a class $\alpha \smile \beta \in H^{n+m}(X;R)$
• The cup product is induced by the diagonal map $\Delta: X \to X \times X$ and the cross product in cohomology
• The cup product can be defined using singular cohomology, simplicial cohomology, or cellular cohomology (singular, simplicial, cellular cohomology)

Naturality and Maps

• The cup product is natural with respect to continuous maps between topological spaces
  • If $f: X \to Y$ is a continuous map and $\alpha \in H^n(Y;R)$ and $\beta \in H^m(Y;R)$ are cohomology classes, then $f^*(\alpha \smile \beta) = f^*(\alpha) \smile f^*(\beta)$, where $f^*$ is the induced homomorphism on cohomology
  • This naturality property allows the cup product to be used in functorial constructions and to study maps between spaces
  • Example: For the inclusion map $i: S^n \to \mathbb{R}^{n+1}$, the induced homomorphism $i^*: H^*(\mathbb{R}^{n+1};R) \to H^*(S^n;R)$ preserves the cup product structure

Computing Cup Products

Cellular Cohomology

• In cellular cohomology, the cup product is computed using the cellular cochain complex
• For cellular cohomology classes represented by cocycles $\alpha \in C^n(X;R)$ and $\beta \in C^m(X;R)$, their cup product $\alpha \smile \beta$ is represented by the cocycle $(\alpha \smile \beta)(e^{n+m}) = \alpha(e^n)\beta(e^m)$ for each $(n+m)$-cell $e^{n+m}$
• The cup product of cellular cohomology classes is computed by multiplying the values of the cocycles on the cells of the CW complex
• The cellular boundary formula $\partial(e^{n+m}) = \sum_i [e^{n+m} : e^n_i]e^n_i + \sum_j [e^{n+m} : e^m_j]e^m_j$ is used to determine the cup product on the cellular cochain level

Examples and Calculations

• Example: Let $X = S^2 \vee S^4$ be the wedge sum of spheres. The cellular cohomology groups are $H^0(X;R) = R$, $H^2(X;R) = R$, $H^4(X;R) = R$, and $H^i(X;R) = 0$ for $i \neq 0,2,4$. Let $\alpha \in H^2(X;R)$ and $\beta \in H^4(X;R)$ be the generators. Then $\alpha \smile \alpha = 0$ and $\alpha \smile \beta = 0$ because there are no cells of dimension 4 or 6 in $X$
• Example: For the torus $T = S^1 \times S^1$, let $a,b \in H^1(T;R)$ be the generators corresponding to the two circles. Then $a \smile b = -b \smile a$ generates $H^2(T;R)$, and $a \smile a = b \smile b = 0$. The cohomology ring $H^*(T;R)$ is isomorphic to the exterior algebra $\Lambda_R[a,b]$

Properties of Cup Products

Algebraic Properties

• The cup product is associative: $(\alpha \smile \beta) \smile \gamma = \alpha \smile (\beta \smile \gamma)$ for cohomology classes $\alpha$, $\beta$, and $\gamma$
• The cup product is graded commutative: $\alpha \smile \beta = (-1)^{nm}(\beta \smile \alpha)$ for $\alpha \in H^n(X;R)$ and $\beta \in H^m(X;R)$
  • This graded commutativity reflects the sign convention in the definition of the cup product and the orientation of cells
  • Example: For the real projective plane $\mathbb{RP}^2$, let $\alpha \in H^1(\mathbb{RP}^2;\mathbb{Z}/2\mathbb{Z})$ be the generator. Then $\alpha \smile \alpha \neq 0$, but $\alpha \smile \alpha = -(\alpha \smile \alpha)$, so $2(\alpha \smile \alpha) = 0$, showing that $H^*(\mathbb{RP}^2;\mathbb{Z}/2\mathbb{Z})$ is not commutative
• The proofs of associativity and graded commutativity rely on the properties of the diagonal map and the cross product in cohomology

Interaction with Other Operations

• The cup product is distributive over addition: $(\alpha + \beta) \smile \gamma = \alpha \smile \gamma + \beta \smile \gamma$ and $\alpha \smile (\beta + \gamma) = \alpha \smile \beta + \alpha \smile \gamma$
• The identity element for the cup product is the cohomology class $1 \in H^0(X;R)$, satisfying $1 \smile \alpha = \alpha \smile 1 = \alpha$ for any cohomology class $\alpha$
  • The identity element corresponds to the constant map $X \to R$ with value 1
  • Example: For any space $X$, the cup product with the identity element induces isomorphisms $H^n(X;R) \cong H^0(X;R) \otimes H^n(X;R)$ and $H^n(X;R) \cong H^n(X;R) \otimes H^0(X;R)$

Applications of Cup Products

Cohomology Rings and Invariants

• The cup product can be used to compute the cohomology rings of various spaces, such as projective spaces, lens spaces, and Eilenberg-MacLane spaces
• The cohomology ring $H^*(\mathbb{CP}^n;\mathbb{Z})$ of the complex projective space $\mathbb{CP}^n$ is isomorphic to $\mathbb{Z}[x]/(x^{n+1})$, where $x$ is the generator of $H^2(\mathbb{CP}^n;\mathbb{Z})$ and the cup product corresponds to polynomial multiplication
  • Example: In $\mathbb{CP}^2$, the generator $x \in H^2(\mathbb{CP}^2;\mathbb{Z})$ satisfies $x \smile x = x^2 \neq 0$ but $x \smile x \smile x = x^3 = 0$
• The cup product provides a way to define and compute cohomological invariants of spaces, such as the cohomology ring, Betti numbers, and Poincaré polynomials
• Example: The Poincaré polynomial $P_X(t) = \sum_i \dim H^i(X;R) t^i$ encodes the dimensions of the cohomology groups and can be used to distinguish spaces up to homotopy equivalence

Obstructions and Non-Existence of Maps

• The cup product can detect the non-existence of certain continuous maps between spaces by comparing their cohomology rings
  • If $f: X \to Y$ is a continuous map and the induced homomorphism $f^*: H^*(Y;R) \to H^*(X;R)$ does not preserve the cup product structure, then $f$ cannot be a homotopy equivalence
  • Example: There is no continuous map $f: \mathbb{CP}^2 \to S^4$ that is a homotopy equivalence because $H^*(\mathbb{CP}^2;\mathbb{Z})$ is not isomorphic to $H^*(S^4;\mathbb{Z})$ as rings
• The cup product can be used to define cohomological obstructions to the existence of certain geometric structures, such as complex structures, almost complex structures, and symplectic structures
• Example: The Chern classes $c_i(E) \in H^{2i}(X;\mathbb{Z})$ of a complex vector bundle $E \to X$ are defined using the cup product and provide obstructions to the existence of nowhere-zero sections and triviality of the bundle

Higher Operations and Duality

• The cup product can be used to define higher cohomology operations, such as Steenrod squares and Massey products, which provide additional tools for studying the topology of spaces
  • Steenrod squares $Sq^i: H^n(X;\mathbb{Z}/2\mathbb{Z}) \to H^{n+i}(X;\mathbb{Z}/2\mathbb{Z})$ are cohomology operations that generalize the cup product and capture additional information about the cohomology of a space with $\mathbb{Z}/2\mathbb{Z}$ coefficients
  • Massey products are higher-order cohomology operations that generalize the cup product and provide obstructions to the formality of spaces and the realization of cohomology classes by geometric constructions
• The cup product is a key ingredient in the definition of the cup product pairing between cohomology and homology, which is used in Poincaré duality and the study of manifolds
  • For a closed orientable $n$-dimensional manifold $M$, the cup product pairing $H^k(M;R) \otimes H^{n-k}(M;R) \to H^n(M;R) \cong R$ is non-degenerate and induces isomorphisms $H^k(M;R) \cong H_{n-k}(M;R)$
  • Example: For the torus $T = S^1 \times S^1$, the cup product pairing $H^1(T;\mathbb{Z}) \otimes H^1(T;\mathbb{Z}) \to H^2(T;\mathbb{Z}) \cong \mathbb{Z}$ is given by $(a,b) \mapsto a \smile b$ and induces the isomorphisms $H^1(T;\mathbb{Z}) \cong H_1(T;\mathbb{Z})$ and $H^0(T;\mathbb{Z}) \cong H_2(T;\mathbb{Z})$