Cohomology Ring Of $S^2 \times S^2 / S^2 \times \{*\}$

Alright guys, let's dive into calculating the cohomology ring of the quotient space $X = (S^2 \times S^2) / (S^2 \times \{*\})$ using the Mayer-Vietoris sequence. This is gonna be a fun ride, so buckle up!

Setting the Stage

First, let's define $X = (S^2 \times S^2) / (S^2 \times \{*\})$. We're going to cover $X$ with two open sets, $U$ and $V$, where $U$ is homeomorphic to $S^2 \times S^2 \setminus (S^2 \times \{*\})$ and $V$ is an open neighborhood around the point at infinity (which results from the quotient). The idea here is to leverage the Mayer-Vietoris sequence to piece together the cohomology of $X$ from the cohomologies of $U$, $V$, and their intersection $U \cap V$.

Mayer-Vietoris Sequence

The Mayer-Vietoris sequence is a powerful tool in algebraic topology. For a space $X$ covered by open sets $U$ and $V$, it relates the cohomology groups of $X$, $U$, $V$, and $U \cap V$ in a long exact sequence:

$$\cdots \to H^i(X) \to H^i(U) \oplus H^i(V) \to H^i(U \cap V) \to H^{i+1}(X) \to \cdots$$

This sequence is exact, meaning the image of each map is the kernel of the next. This property allows us to compute the cohomology groups of $X$ if we know the cohomology groups of $U$, $V$, and $U \cap V$, along with the maps between them.

Defining Open Sets and Their Properties

Let's define our open sets more precisely:

• $U$ is homeomorphic to $S^2 \times S^2$ minus a closed subset $S^2 \times \{*\}$. Intuitively, we're removing one of the "poles" from the sphere product.
• $V$ is an open neighborhood of the point at infinity in the quotient space $X$. This point arises from collapsing $S^2 \times \{*\}$ to a single point. We can choose $V$ such that it's homeomorphic to an open disk in $S^4$, so $V \simeq D^4$.

Now, let's think about their intersection, $U \cap V$. This intersection is homotopy equivalent to $S^2$, which can be visualized as a small sphere surrounding the point at infinity in $U$.

Cohomology Groups of $U$, $V$, and $U $\cap$ V

Now that we have our open sets, we need to compute their cohomology groups.

Cohomology of $U$

Since $U \simeq S^2 \times S^2 \setminus (S^2 \times \{*\})$, we can think of $U$ as $S^2 \times (S^2 \setminus \{*\})$. Because $S^2 \setminus \{*\}$ is homeomorphic to $\mathbb{R}^2$, $U$ is homotopy equivalent to $S^2 \times \mathbb{R}^2$, and hence homotopy equivalent to $S^2$. Thus,

$$H^i(U) = \begin{cases} \mathbb{Z} & i = 0, 2 \\ 0 & \text{otherwise} \end{cases}$$

Cohomology of $V$

Since $V$ is homeomorphic to an open disk in $S^4$, $V$ is contractible, which means it's homotopy equivalent to a point. Therefore,

$$H^i(V) = \begin{cases} \mathbb{Z} & i = 0 \\ 0 & \text{otherwise} \end{cases}$$

Cohomology of $U $\cap$ V

The intersection $U \cap V$ is homotopy equivalent to $S^2$. Hence,

$$H^i(U \cap V) = \begin{cases} \mathbb{Z} & i = 0, 2 \\ 0 & \text{otherwise} \end{cases}$$

Assembling the Mayer-Vietoris Sequence

Now, let's plug these cohomology groups into the Mayer-Vietoris sequence. We're particularly interested in the lower degrees, as those will help us determine the structure of the cohomology ring. We get:

$$\begin{aligned} 0 \to H^0(X) \to H^0(U) \oplus H^0(V) \to H^0(U \cap V) \to H^1(X) \to H^1(U) \oplus H^1(V) \to H^1(U \cap V) \to \\ H^2(X) \to H^2(U) \oplus H^2(V) \to H^2(U \cap V) \to H^3(X) \to H^3(U) \oplus H^3(V) \to H^3(U \cap V) \to \\ H^4(X) \to H^4(U) \oplus H^4(V) \to H^4(U \cap V) \to \cdots \end{aligned}$$

Substituting the cohomology groups we found earlier, we have:

$$\begin{aligned} 0 \to H^0(X) \to \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{a} \mathbb{Z} \to H^1(X) \to 0 \oplus 0 \to 0 \to \\ H^2(X) \to \mathbb{Z} \oplus 0 \xrightarrow{b} \mathbb{Z} \to H^3(X) \to 0 \oplus 0 \to 0 \to \\ H^4(X) \to 0 \oplus 0 \to 0 \to \cdots \end{aligned}$$

Analyzing the Sequence

Let's analyze this sequence step by step:

$H^0(X)$

The map $a: \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ is given by $(x, y) \mapsto x - y$, because it's the difference of the inclusions $H^0(U) \to H^0(U \cap V)$ and $H^0(V) \to H^0(U \cap V)$. Thus, $\text{im}(a) = \mathbb{Z}$, and since the sequence is exact, $\ker(a) = H^0(X)$. Therefore, $H^0(X) \cong \mathbb{Z}$.

$H^1(X)$

Since $H^1(U) = H^1(V) = H^1(U \cap V) = 0$, we have $H^1(X) = 0$.

$H^2(X)$

The map $b: \mathbb{Z} \to \mathbb{Z}$ is given by the difference of the inclusions $H^2(U) \to H^2(U \cap V)$ and $H^2(V) \to H^2(U \cap V)$. Since $H^2(V) = 0$, this map is simply the inclusion $H^2(U) \to H^2(U \cap V)$, which is an isomorphism. Therefore, $H^3(X) = 0$ and $b$ is an isomorphism. Hence, $H^2(X) = \mathbb{Z}$.

$H^3(X)$

Since $H^3(U) = H^3(V) = H^3(U \cap V) = 0$, we have $H^3(X) = 0$.

$H^4(X)$

Looking at the tail end of the sequence, we have

$$0 \to H^4(X) \to 0 \oplus 0 \to 0$$

Thus $H^4(X) = \mathbb{Z}$.

Cohomology Ring Structure

So far, we have:

$$H^i(X) = \begin{cases} \mathbb{Z} & i = 0, 2, 4 \\ 0 & \text{otherwise} \end{cases}$$

Now, let's figure out the ring structure. Let $\alpha \in H^2(X)$ be a generator. Since $H^4(X) \cong \mathbb{Z}$, we must have $\alpha \smile \alpha$ generating $H^4(X)$. Therefore, the cohomology ring of $X$ is:

$$H^*(X; \mathbb{Z}) \cong \mathbb{Z}[\alpha] / (\alpha^3)$$

Where $\alpha$ has degree 2.

Conclusion

Therefore, by using the Mayer-Vietoris sequence, we've successfully computed the cohomology ring of $X = (S^2 \times S^2) / (S^2 \times \{*\})$. The key was breaking down the space into manageable open sets, computing their cohomology groups, and then piecing everything together using the sequence. Pretty neat, huh?