Prove that $CP^3/CP^1$ is homotopy equivalent to $S^4\vee S^6$

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The following is an endeavor to establish the homotopy equivalence between $CP^3/CP^1$ and $S^4\vee S^6$. May I kindly request a review of my proof's accuracy? I intend to employ Steenrod operators in my argument.

To prove that $CP^3/CP^1$ is homotopy equivalent to $S^4\vee S^6$, we can use the fact that the cohomology ring of $CP^3/CP^1$ is isomorphic to the cohomology ring of $S^4\vee S^6$.

To do this, we will use the Steenrod squares. Let $X$ be a topological space and let $\mathbb{F}_2$ be the field with two elements. For any $n\geq 0$, the $n$th Steenrod square $Sq^n$ is a cohomology operation that takes a cohomology class in $H^k(X;\mathbb{F}_2)$ to a cohomology class in $H^{k+n}(X;\mathbb{F}_2)$.

Using the properties of Steenrod squares, we can show that $H^*(CP^3/CP^1;\mathbb{F}_2)$ is generated by the classes $x\in H^2(CP^3/CP^1;\mathbb{F}_2)$ and $y\in H^4(CP^3/CP^1;\mathbb{F}_2)$, subject to the relations $Sq^2(x)=y$ and $Sq^4(x)=0$.

On the other hand, $H^*(S^4\vee S^6;\mathbb{F}_2)$ is generated by the classes $u\in H^4(S^4\vee S^6;\mathbb{F}_2)$ and $v\in H^6(S^4\vee S^6;\mathbb{F}_2)$, subject to the relation $Sq^2(u)=v$.

We can define a map $f: CP^3/CP^1\rightarrow S^4\vee S^6$ by sending $x$ to $u$ and $y$ to $v$. This map is well-defined because of the relations between the Steenrod squares in the two cohomology rings.

To show that $f$ is a homotopy equivalence, we need to construct a map $g:S^4\vee S^6\rightarrow CP^3/CP^1$ such that $f\circ g$ and $g\circ f$ are homotopic to the identity maps on $S^4\vee S^6$ and $CP^3/CP^1$, respectively.

To define $g$, we can use the fact that $CP^3/CP^1$ is diffeomorphic to $S^2\times S^4$. We can define a map $g_1:S^4\rightarrow S^2\times S^4$ by sending $x\in S^4$ to $(p,q)$, where $p$ is the unique point on $S^2$ such that $x$ lies on the line connecting $p$ and the north pole of $S^4$, and $q$ is the point on $S^4$ that lies on this line at a distance of $\pi/2$ from $p$.

We can then define a map $g_2:S^6\rightarrow S^2\times S^4$ by sending $x\in S^6$ to $(p,q)$, where $p$ is the unique point on $S^2$ such that $x$ lies on the line connecting $p$ and the north pole of $S^6$, and $q$ is the point on $S^4$ that lies on this line at a distance of $\pi/2$ from $p$. Note that $S^6$ can be obtained from $S^4$ by attaching a cell of dimension 3 or by attaching a cell of dimension 7, so we can use the same construction for both $g_1$ and $g_2$.

We can then define $g:S^4\vee S^6\rightarrow CP^3/CP^1$ by sending $u$ to $(1,0)$ and $v$ to $(0,1)$. We can show that $f\circ g$ is homotopic to the identity map on $S^4\vee S^6$ by constructing a homotopy $H:S^4\vee S^6\times [0,1]\rightarrow S^4\vee S^6$ between $f\circ g$ and the identity map, and we can show that $g\circ f$ is homotopic to the identity map on $CP^3/CP^1$ using a similar argument.

Therefore, we have shown that $CP^3/CP^1$ is homotopy equivalent to $S^4\vee S^6$.

Best Answer

Your calculation of the steenrod operation is inaccurate, and there are several places where the statement your claimed is false or without justification. Here is (perhaps) a simpler argument.

We will analyze the attaching map of the 6-cell of $\Bbb CP^3/\Bbb CP^1$, which is the composition of the attaching map of the 6-cell in $\Bbb CP^3$ and the quotient map, i.e., $$\bar\varphi:S^5\to\Bbb CP^2\to\Bbb CP^2/\Bbb CP^1=S^4$$ So, $\bar\varphi$ represents something in $\pi_5(S^4)\cong\Bbb Z_2$, which means that exactly one of the following has to happen.

  1. $\bar\varphi$ is null-homotopic.
  2. $\bar\varphi\simeq\Sigma^2\eta$, where $\eta:S^3\to S^2$ is the Hopf map. If this is the case, then we would have $\Bbb CP^3/\Bbb CP^1\simeq C_{\Sigma^2\eta}$, where $C_{\Sigma^2\eta}$ (the mapping cone) is homeomorphic to $\Sigma^2C_\eta=\Sigma^2\Bbb CP^2$. An explicit computation shows that $Sq^2$ is non-trivial on $\Sigma^2\Bbb CP^2$ but is trivial on $\Bbb CP^3/\Bbb CP^1$, so this is a contradiction.

Therefore, it must be the case that $\bar\varphi$ is null-homotopic, which implies that the 6-cell is attached trivially, so $\Bbb CP^3/\Bbb CP^1\simeq S^4\vee S^6$ by comparing cellular structures.

Steenrod operations on those spaces:

For $S^4\vee S^6$, we consider the retraction $r:S^4\vee S^6\to S^4$ which induces an injection on $H^4$. With $\Bbb Z_2$ coefficient, $r^\ast$ has to be an isomorphism, so the generator $u\in H^4(S^4\vee S^6;\Bbb Z_2)$ is given by $u=r^\ast(u')$, where $u'$ generates $H^4(S^4;\Bbb Z_2)$. Hence, $Sq^2(u)=r^\ast Sq^2(u')=0$.

For $\Bbb CP^3/\Bbb CP^1$, we consider the quotient map $p:\Bbb CP^3\to\Bbb CP^3/\Bbb CP^1$. This induces isomorphisms on $H^4$ and $H^6$. The Steenrod operation on $\Bbb CP^n$ is given by $$Sq^{2k}(\alpha^m)=\binom{m}{k}\alpha^{m+k},\ |\alpha|=2$$ By substituting $k=1, m=2$ and pulling back via the quotient map, we see that $Sq^2(x)=0$ for $x\in H^4(\Bbb CP^3/\Bbb CP^1;\Bbb Z_2)$.

The action of Steenrod squares on $\Sigma^2\Bbb CP^2$ follows from its invariance under suspension.

It's also quite interesting how this generalizes to $\Bbb CP^{n}/\Bbb CP^{n-2}$ for $n\ge 3$.

Suppose $x$ generates $H^{2n-2}(\Bbb CP^n/\Bbb CP^{n-2};\Bbb Z_2)$ and $y$ generates $H^{2n}(\Bbb CP^n/\Bbb CP^{n-2};\Bbb Z_2)$, then $$Sq^2(x)=(n-1)y=\begin{cases}y & \text{ if }n\equiv 0\pmod{2}\\ 0 & \text{ if }n\equiv 1\pmod{2}\end{cases}$$ Since the $(k-3)$-fold suspended Hopf map generates $\pi_{k+1}(S^k)\cong\Bbb Z_2$ for $k>3$, we may apply the same method as before to classify the attaching map of the top cell, so the homotopy type of this quotient space actually depends on the parity of $n$. For $n\ge 3$,

  • If $n$ is odd, then $\Bbb CP^n/\Bbb CP^{n-2}\simeq S^{2n-2}\vee S^{2n}$
  • If $n$ is even, then $\Bbb CP^n/\Bbb CP^{n-2}\simeq\Sigma^{2n-4}\Bbb CP^2$
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