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I want to calculate the cellular homology of $\mathbb{C}P^4/\mathbb{C}P^2$, but I have trouble finding a suitable CW-decomopistion.

What is a CW-complex decomposition of $\mathbb{C}P^4/\mathbb{C}P^2$?

Because $\mathbb{C}P^4 = S^5/S^1,\,\,\mathbb{C}P^2 = S^3/S^1,$ so $$ \mathbb{C}P^4/\mathbb{C}P^2 = S^5/S^3 \approx\ ?$$

Michael Albanese
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lony
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    $\mathbb{CP}^4 \neq S^5/S^1$. There is a fibre bundle $S^1 \to S^9 \to \mathbb{CP}^4$; maybe you mean $S^9/S^1$. Likewise, there is a fibre bundle $S^1 \to S^5 \to \mathbb{CP}^2$. – Michael Albanese Dec 06 '16 at 15:36

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From Hatcher's Algebraic Topology, page 8:

Quotients. If $(X, A)$ is a CW pair consisting of a cell complex $X$ and a subcomplex $A$, then the quotient space $X/A$ inherits a natural cell complex structure from $X$. The cells of $X/A$ are the cells of $X − A$ plus one new $0$-cell, the image of $A$ in $X/A$. For a cell $e^n_{\alpha}$ of $X - A$ attached by $\varphi_{\alpha} : S^{n-1} \to X^{n-1}$, the attaching map for the corresponding cell in $X/A$ is the composition $S^{n-1} \to X^{n-1} \to X^{n-1}/A^{n-1}$.

Note that $\mathbb{CP}^n$ is a CW complex with a single cell in every even dimension betweeen $0$ and $2n$ (inclusive). The attaching map of the $(2k+2)$-cell is the projection map for the $S^1$-fibre bundle $S^{2k+1} \to \mathbb{CP}^k$. In particular, for $m < n$, $\mathbb{CP}^m$ is a subcomplex of $\mathbb{CP}^n$ consisting of the cells in dimensions between $0$ and $2m$ (inclusive).

So $\mathbb{CP}^n/\mathbb{CP}^m$ is a CW complex with a single $0$-cell, and a single cell in every even dimension between $2m+2$ and $2n$ (inclusive).

In particular, $\mathbb{CP}^4/\mathbb{CP}^2$ is a CW complex with a $0$-cell, a $6$-cell, and an $8$-cell. The $6$-cell is attached to the $0$-cell via a constant map, so the six-skeleton is $S^6$. The seven-skeleton is also $S^6$ and the $8$-cell is attached to it via the map $S^7 \to \mathbb{CP}^3 \to \mathbb{CP}^3/\mathbb{CP}^2 = S^6$. It follows from the answers to this question that this map is not homotopic to a constant map.

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Michael Albanese
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  • @ Michael Albanese. The 6-cell is attached to the 0-cell via a constant map, i.e, $S^5\to \mathbb{C}P^2\to \mathbb{C}P^2/ \mathbb{C}P^1 = e^0$ ?? – lony Dec 07 '16 at 03:04
  • @lony: Not quite: $S^5 \to \mathbb{CP}^2 \to \mathbb{CP}^2/\mathbb{CP}^2 = e^0$. – Michael Albanese Dec 07 '16 at 03:19
  • @ Michael Albanese. Hence CW-decomposition of $$\mathbb{C}P^4/\mathbb{C}P^2 = \bigcup\limits_{n=0}^8X^n,$$ where $$ X^0 = X^1 = \cdots X^{5} = e^0, X^{6} = e^0\sqcup e^6, X^{7} = X^8 = e^8?$$ – lony Dec 07 '16 at 03:31
  • For $0 \leq i \leq 5$, $X^i = e^0$, $X^6 = X^7 = S^6$, $X^8 = S^6\cup_{\varphi} e^8$ where $\varphi$ is the attaching map $S^7 \to S^6$. – Michael Albanese Dec 07 '16 at 03:52
  • @ Michael Albanese. Thank you very much! Now, if $X = S^n\cup_f e^{2n},$ where $f$ is the attaching map $S^{2n-1}\to S^n, n >1$ then CW-decomposition of $$X = \bigcup\limits_{k = 0}^{2n}X^k,$$ where $$X^0 = X^1 = \cdots X^{n-1} = e^0, X^n = S^n, X^{n+1} = \cdots = X^{2n} = e^0\sqcup e^n\sqcup e^{2n}?$$ – lony Dec 07 '16 at 04:05
  • @lony: Kind of. You shouldn't write $e^0\sqcup e^n \sqcup e^{2n}$ though because $\sqcup$ denotes disjoint union. I would instead write $S^n\cup_fe^{2n}$. – Michael Albanese Dec 07 '16 at 16:25
  • @ Michael Albanese. I think that $X = S^n \bigcup\limits_{f}e^{2n} = S^n \bigcup\limits_{f}D^{2n},$ where $D^{2n}$ is 2n-dimensional closed disk , so $X = e^0\sqcup e^n\sqcup e^{2n}$ – lony Dec 08 '16 at 01:01
  • @lony: The symbol you're using, namely $\sqcup$, usually means disjoint union, in which case $e^0\sqcup e^n\sqcup e^{2n}$ has three connected components ($e^0$, $e^n$, and $e^{2n}$). In this space, the cells aren't attached to each other. – Michael Albanese Dec 08 '16 at 01:23
  • @ Michael Albanese. $X = S^{n}\bigcup\limits_{f}e^{2n}\neq S^{n}\bigcup\limits_{f}D^{2n}$ ? – lony Dec 08 '16 at 01:46
  • @lony: No, $e^{2n} = D^{2n}$ so $S^n\cup_f e^{2n} = S^n\cup_f D^{2n}$. – Michael Albanese Dec 08 '16 at 15:00
  • @ Michael Albanese. I think that $e^{2n} \approx \mbox{int}(D^{2n}).$ I want to ask $S^6 \vee S^8 = e^0 \sqcup e^6\sqcup e^8$ ? – lony Dec 09 '16 at 00:47
  • @long: $e^{2n}$ and $D^{2n}$ usually both denote the closed $2n$-disk. I don't know if $S^6\vee S^8$ and $\mathbb{CP}^4/\mathbb{CP}^2$ are homotopy equivalent. They have the same (trivial) cohomology ring structure. – Michael Albanese Dec 10 '16 at 14:48
  • @ Michael Albanese. why they have the same (trivial) cohomology ring structure? Can you explain for me? Thank you very much! – lony Dec 11 '16 at 12:27
  • @lony: The space has cohomology only in degrees $0$, $6$, and $8$. The cup product of any two cohomology classes of positive degree is necessarily zero because the corresponding group is zero. Think about the cup product of two elements of $H^6(\mathbb{CP}^4, \mathbb{CP}^2; \mathbb{Z})$ for example. – Michael Albanese Dec 12 '16 at 18:56
  • @ Michael Albanese.Because $$H^6(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{Z}) = H^6((\mathbb{C}P^4, \mathbb{C}P^2); \mathbb{Z}) \cong \mathbb{Z},$$ hence $x\smile y\in H^{12}((\mathbb{C}P^4, \mathbb{C}P^2); \mathbb{Z}) = 0,\forall x, y\in H^6((\mathbb{C}P^4, \mathbb{C}P^2); \mathbb{Z}) $ ? – lony Dec 13 '16 at 02:27
  • @lony: Precisely. Alternatively, you can see that $H^6(\mathbb{CP}^4/\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}$ using cellular cohomology. – Michael Albanese Dec 13 '16 at 03:59
  • @ Michael Albanese Hence, $H^*(\mathbb{C}P^4/\mathbb{C}P^2, R) \cong H^n(\mathbb{C}P^4/\mathbb{C}P^2, R) \cong R$ where $n = 0, 6, 8$ and $R$ is a ring. – lony Dec 13 '16 at 04:14
  • @lony: Usually $H^$ is used to denote the full cohomology ring which is the direct sum of the groups $H^n$. That is $$H^(\mathbb{CP}^4/\mathbb{CP}^2; R) = \bigoplus_n H^n(\mathbb{CP}^4/\mathbb{CP}^2; R) \cong R \oplus R \oplus R.$$ But yes, $H^n(\mathbb{CP}^4/\mathbb{CP}^2; R) \cong R$ for $n = 0, 6, 8$. – Michael Albanese Dec 13 '16 at 04:24
  • @ Michael Albanese Thank you very much! I want to ask because the project map $p: S^6 \vee S^8 \to S^6$ a continuous, so $$p^:\mathbb{Z}\cong H^6(S^6; \mathbb{Z}) \to H^6( S^6 \vee S^8; \mathbb{Z})\cong H^6(S^6; \mathbb{Z}) \oplus H^6(S^8; \mathbb{Z})\cong \mathbb{Z}.$$ Is isomorphic. I know that $$ H^6(S^6; :\mathbb{Z}) \cong [S^6, K( \mathbb{Z}, 6)]; \quad H^6(S^6\vee S^8; :\mathbb{Z}) \cong [S^6\vee S^8, K( \mathbb{Z}, 6)]$$ but I don't understand why $p^$ is a isomorphic. Can you explain for me? Thanks! – lony Dec 13 '16 at 05:31
  • @lony: Do you understand Mariano's answer to your earlier question? If not, I suggest you do that first. Once you do, you should be able to answer this question too. – Michael Albanese Dec 13 '16 at 16:09