Cup Product and Ring Structure

The Cup Product

Cup Product

Let $R$ be a ring. The cup product is a map $⌣:C^k(X;R)\times C^l(X;R)\to C^{k+l}(X;R)$ given by the following formula. For $\phi \in C^k(X;R)$, $\psi \in C^l(X;R)$, and a singular $(k+l)$-simplex $\sigma :{\Delta }^{k+l}\to X$, we define: $(\phi ⌣\psi )(\sigma )=\phi (\sigma {|}_{[v_0,\dots ,v_k]})\cdot \psi (\sigma {|}_{[v_k,\dots ,v_{k+l}]})$ Here, $\sigma {|}_{[v_0,\dots ,v_k]}$ is the restriction of $\sigma$ to the “front face” (a $k$-simplex), $\sigma {|}_{[v_k,\dots ,v_{k+l}]}$ is the restriction to the “back face” (an $l$-simplex), and the product is taken in the ring $R$.

The cup product interacts with the coboundary operator $\delta$ via the following rule.

Coboundary Formula for Cup Product

For $\phi \in C^k(X;R)$ and $\psi \in C^l(X;R)$, the following Leibniz-type rule holds: $d^{k+l+1}(\phi ⌣\psi )=(d^{k+1}\phi )⌣\psi +(-1{)}^k\phi ⌣(d^{l+1}\psi ).$

An important consequence of this formula is that the cup product of two cocycles ($\delta \phi =0$, $\delta \psi =0$) is again a cocycle. Furthermore, the product of a cocycle and a coboundary is a coboundary. This implies that the cup product descends to a well-defined product on cohomology: $⌣:H^k(X;R)\times H^l(X;R)\to H^{k+l}(X;R)$

Lemma

The cohomology ring $H^{\ast }(X;R)={⨁}_k{H}^k(X;R)$ is a unital, associative, graded-commutative ring under the cup product. The unit $1\in H^0(X;R)$ is represented by the cocycle $\phi \in C^0(X;R)$ which maps every 0-simplex (a point in $X$) to the multiplicative identity $1_R\in R$.

Proof of Unital Property Let $1\in H^0(X;R)$ be the unit element, represented by the cocycle $\phi$ where $\phi (x)=1_R$ for any point $x\in X$. Let $[\psi ]\in H^n(X;R)$ be represented by the cocycle $\psi \in C^n(X;R)$. For any $n$-simplex $\sigma :{\Delta }^n\to X$ with vertices $[v_0,\dots ,v_n]$, we have: $(\phi ⌣\psi )(\sigma )=\phi (\sigma {|}_{[v_0]})\cdot \psi (\sigma {|}_{[v_0,\dots ,v_n]})$Since $\sigma {|}_{[v_0]}$ is a 0-simplex, $\phi (\sigma {|}_{[v_0]})=1_R$. Therefore, $(\phi ⌣\psi )(\sigma )=1_R\cdot \psi (\sigma {|}_{[v_0,\dots ,v_n]})=\psi (\sigma )$This shows that $\phi ⌣\psi =\psi$, so $[1]⌣[\psi ]=[\psi ]$. A similar calculation shows $[\psi ]⌣[1]=[\psi ]$, so it is a two-sided unit.

The Cohomology Ring

Properties of the Cup Product

Taking all cohomology groups together, $H^{\ast }(X;R)={⨁}_{k\ge 0}{H}^k(X;R)$, the cup product has the following properties:

• Unital: There is a multiplicative identity $1\in H^0(X;R)$.
• Distributive: For cohomology classes $x,y,z$, we have $(x+y)\cup z=x\cup z+y\cup z$ and $x\cup (y+z)=x\cup y+x\cup z$.
• Associative: For cohomology classes $x,y,z$, we have $(x\cup y)\cup z=x\cup (y\cup z)$.

These properties mean that $H^{\ast }(X;R)$ equipped with the cup product forms a ring.

Cohomology Ring

The cohomology ring of a space $X$ with coefficients in a ring $R$ is the graded ring $(H^{\ast }(X;R),+,⌣)$.

Induced Homomorphisms

Let $f:X\to Y$ be a continuous map. The induced map on cohomology $f^{\ast }:H^{\ast }(Y;R)\to H^{\ast }(X;R)$ is a homomorphism of graded rings. That is, it preserves the ring structure: $f^{\ast }(1)=1,f^{\ast }(x⌣y)=f^{\ast }(x)⌣f^{\ast }(y).$

e.g. The cohomology ring of the torus $T^2=S^1\times S^1$ with integer coefficients is isomorphic to the exterior algebra on two generators of degree 1. Recall our $\Delta$-complex structure for the torus: The chain complex is $0\to \mathbb{Z}\{U,L\}\stackrel{d_2}{\to }\mathbb{Z}\{a,b,c\}\stackrel{d_1}{\to }\mathbb{Z}\{v\}\to 0,$where $d_2(U)=a-c+b$, $d_2(L)=b-c+a$, and $d_1(a)=d_1(b)=d_1(c)=0$. So we have the homology groups: $H_0=\mathbb{Z}\{v\},H_1=\mathbb{Z}\{\bar{a},\bar{b}\},H_2=\mathbb{Z}\{U-L\}.$Dualize to get the cochain complex: $0\leftarrow \mathbb{Z}\{U^{\ast },L^{\ast }\}\stackrel{d^2}{\leftarrow }\mathbb{Z}\{a^{\ast },b^{\ast },c^{\ast }\}\stackrel{d^1}{\leftarrow }\mathbb{Z}\{v^{\ast }\}\leftarrow 0,$where $d^2$ maps $a^{\ast }\mapsto U^{\ast }+L^{\ast },b^{\ast }\mapsto U^{\ast }+L^{\ast },c^{\ast }\mapsto -U^{\ast }-L^{\ast },$and $d^1$ is the zero map (we can think of $d^1$ and $d^2$ as the transpose of $d_1$ and $d_2$). Thus, the cohomology groups are: $H^0=\mathbb{Z}\{v^{\ast }\},H^1=\mathbb{Z}\{a^{\ast }+c^{\ast },b^{\ast }+c^{\ast }\},H^2=\mathbb{Z}\{{\bar{U}}^{\ast }\}.$

If $\alpha ,\beta$ are generators for $H^1(T^2;\mathbb{Z})$, then: $H^{\ast }(T^2;\mathbb{Z})\cong {\Lambda }_{\mathbb{Z}}(\alpha ,\beta )$

Graded Commutativity

The cohomology ring is not strictly commutative in general, but it satisfies a property called super-commutativity or graded commutativity.

Graded Commutativity of the Cup Product

Let $R$ be a commutative ring. For any cohomology classes $x\in H^k(X;R)$ and $y\in H^l(X;R)$, the following relation holds: $x⌣y=(-1{)}^{kl}(y⌣x)$

Remark

An interesting consequence of graded commutativity is that for any element $x$ of odd degree $k=2m+1$ in a graded commutative ring $A$, we have $x\cup x=(-1{)}^{k\cdot k}(x\cup x)=(-1{)}^{(2m+1{)}^2}(x\cup x)=-(x\cup x)$ This implies $2(x\cup x)=0$. If the ring $A$ has no 2-torsion (e.g., if the coefficient ring is $\mathbb{Z}$ or $\mathbb{Q}$), this means $x^2=0$.

Examples of Cohomology Rings

Exterior Algebra

The exterior algebra on generators $x_1,\dots ,x_n$ over a ring $R$, denoted ${\Lambda }_R(x_1,\dots ,x_n)$, is an algebra where the generators anti-commute ($x_i{x}_j=-x_j{x}_i$) and each generator squares to zero ($x_i^2=0$). For example:

• ${\Lambda }_R(x)$ is additively $R\{1,x\}$ with $x^2=0$.
• ${\Lambda }_R(x,y)$ is additively $R\{1,x,y,xy\}$ with $x^2=0$, $y^2=0$, and $yx=-xy$.

e.g. Projective Spaces The cohomology rings of real and complex projective spaces are fundamental examples.

• The infinite complex projective space ${\mathbb{CP}}^{\infty }$ has a cohomology ring that is a polynomial ring on a single generator $x\in H^2({\mathbb{CP}}^{\infty };\mathbb{Z})$: $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})\cong \mathbb{Z}[x]$
• The infinite real projective space ${\mathbb{RP}}^{\infty }$ with $\mathbb{Z}/2\mathbb{Z}$ coefficients also has a polynomial ring structure on a generator $x\in H^1({\mathbb{RP}}^{\infty };\mathbb{Z}/2\mathbb{Z})$: $H^{\ast }({\mathbb{RP}}^{\infty };\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[x]$
• For the finite-dimensional real projective space ${\mathbb{RP}}^2$, the ring is a truncated polynomial ring: $H^{\ast }({\mathbb{RP}}^2;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[x]/(x^3)$