19
$\begingroup$

So, I have shown that the natural projection $\pi\colon \mathbb{CP^n}\rightarrow \mathbb{CP^n/CP^k}$ induces a monomorphism $\pi^*\colon H^*(\mathbb{CP^n/CP^k},\mathbb Z)\rightarrow H^*(\mathbb{CP^n},\mathbb Z) $. I would like to use this and the cohomology ring structure to show that we can't have a retract, but I am not exactly sure what the ring structure of $\mathbb{CP^2/CP^1}$ and $\mathbb{CP^4/CP^1}$ are.

I know that $H^*(\mathbb{CP^n},\mathbb Z) \cong \mathbb Z[\gamma]/(\gamma^{n+1})$, where $|\gamma|=2$, so is $H^*(\mathbb{CP^n/CP^k},\mathbb Z) \cong \mathbb Z[\gamma^{k+1},\ldots ,\gamma^{n}]/(\gamma^{n+1})$?

This would give me $H^*(\mathbb{CP^2/CP^1},\mathbb Z) \cong \mathbb Z[\gamma^{2}]/(\gamma^{4})$ and $\pi^*(\gamma^2)\neq 0 \in H^*(\mathbb{CP^4/CP^1},\mathbb Z)$.

If this is true, then I think I can use that:

$0=\pi^*(0)=\pi^*(\gamma^2 \cup \gamma^2)=\pi^*(\gamma^2) \cup \pi^*(\gamma^2)\neq0$, which gives a contradiction.

$\endgroup$
  • 2
    $\begingroup$ treat $H^*(\mathbb{CP}^n/\mathbb{CP}^k)$ as a subring of $H^*(\mathbb{CP}^n)$ (included via $\pi^*$) $\endgroup$ – user8268 Jun 23 '14 at 20:47
  • 1
    $\begingroup$ $\Bbb CP^2/\Bbb CP^1$ is homeomorphic to the $4$ sphere, and thus its cohomology ring is $\Bbb Z[t]/(t^2)$ with $t$ of degree $4$. If you use the long exact cohomology sequence $$\cdots\to \tilde{H}^*(\Bbb CP^4/\Bbb CP^1)\to H^*(\Bbb CP^4)\to H^*(\Bbb CP^1)\to \cdots$$ you can see that the cohomology ring of $\Bbb CP^4/\Bbb CP^1$ is $\Bbb Z[u,v]/(u^3,v^2,uv=vu=0)$ with $u$ in degree $4$ and $v$ in degree $6$. I'm not certain you can solve the problem only using the ring structure of cohomology. $\endgroup$ – Olivier Bégassat Jun 23 '14 at 21:28
  • $\begingroup$ @OlivierBégassat As you wrote $\mathbb{C}P^{2}/\mathbb{C}P^{1}$ is homeomorphic to $S^{4}$ which does not satisfies the fixed point property. As a possible alternative proof for the main question can one prove that $\mathbb{C}P^{4}/\mathbb{C}P^{1}$ has the fixed point property?(Note that a retract of a fixed point space is a fixed point space) $\endgroup$ – Ali Taghavi Jun 29 '14 at 5:48
  • 1
    $\begingroup$ For what it's worth, I believe that, care of the first answer here: math.stackexchange.com/questions/308318/…, that $\mathbb{C}P^3/\mathbb{C}P^1$ does retract onto $\mathbb{C}P^2/\mathbb{C}P^1$. Sketch of proof: According to the link, $\mathbb{C}P^3/\mathbb{C}P^1$ has the homotopy type of $S^4\vee S^6$, which clearly does retract on $S^4$. (I don't know how to make all the details work). The point is, this question could be quite subtle! $\endgroup$ – Jason DeVito Jan 30 '15 at 19:51
  • 3
    $\begingroup$ I'm confused- why doesn't Olivier's answer just answer the question. There is no nonzero element in degree 4 that squares to zero, so there's not even a nonzero map $H^*(\mathbb{C}P^2/\mathbb{C}P^1) \rightarrow H^*(\mathbb{C}P^4/\mathbb{C}P^1)$ let alone a retract. $\endgroup$ – Dylan Wilson Apr 26 '15 at 12:59
2
$\begingroup$

(I'm just answering this to get it off the unanswered list. It's CW because I'm not doing anything but expanding on information in the comments).

As Olivier notes, $H^\ast(\mathbb{C}P^4/\mathbb{C}P^1)\cong \mathbb{Z}[u,v]/u^3 = v^2 = uv = 0$ where $|u| = 4$ and $|v| = 6$. This follows, as Olivier says, via the long exact sequence of the pair $(\mathbb{C}P^1,\mathbb{C}P^4)$, together with the fact that the inclusion map $\mathbb{C}P^1\rightarrow \mathbb{C}P^4$ induces an isomorphism on $H^2$ and $H^0$, but is other wise the zero map.

Further, by the same technique, one easily sees that $H^\ast(\mathbb{C}^2/\mathbb{C}P^1)\cong \mathbb{Z}[t]/t^2$ where $|t| = 4$ (or one can also use the usual cell structure on $\mathbb{C}P^2$ to see that $\mathbb{C}P^2/\mathbb{C}P^1$ is homeomorphic to $S^4$).

Now, let $i:\mathbb{C}P^2/\mathbb{C}P^1\rightarrow \mathbb{C}P^4/\mathbb{C}P^1$ be the inclusion. Assume for a contradiction that $r:\mathbb{C}P^4/\mathbb{C}P^1\rightarrow \mathbb{C}P^2/\mathbb{C}P^1$ is a retraction.

Then we have the formula $r\circ i = Id_{\mathbb{C}P^2/\mathbb{C}P^1}$. In particular, $i^\ast r^\ast:H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)\rightarrow H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)$ is an isomorphism, which implies that $r^\ast:H^\ast(\mathbb{C}P^2/\mathbb{C}P^1)\rightarrow H^\ast(\mathbb{C}P^4/\mathbb{C}P^1)$ is an injection. In particular, $r^\ast$ is not the zero map on $H^4$.

Now consider $r^\ast(t)$. It must be some multiple of $u$, $r^\ast(t) = ku$, because $u$ generates $H^4(\mathbb{C}P^4/\mathbb{C}P^1)\cong \mathbb{Z}$. Because $r^\ast$ is not the zero map, $k\neq 0$.

But now we have a contradiction: $t^2 = 0$ so $0 = r^\ast(t^2) =r^\ast(t)^2 = (ku)^2 = k^2 u^2$ which forces $k = 0$. Thus, there can not be a retraction $r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.