Real projective space $\mathbb{R}P^n$ embeds in a natural way in complex projective space $\mathbb{C}P^n$. (Using standard projective coordinates on $\mathbb{C}P^n$, $\mathbb{R}P^n$ is the subspace consisting of points that have a representative in which all coordinates are real.)

I know (very standard) cell decompositions for $\mathbb{C}P^n$ and $\mathbb{R}P^n$ that realize them as CW-complexes: $\mathbb{C}P^n$ has one cell in every even dimension while $\mathbb{R}P^n$ has one cell in every dimension. So in this standard CW-complex structure on $\mathbb{C}P^n$, $\mathbb{R}P^n$ does not occur as a subcomplex.

What I would like to know is this:

Is there a cell decomposition of $\mathbb{C}P^n$ that reveals $\mathbb{R}P^n$ as a subcomplex? What is it?

[An addition to the bounty notice. In case it makes a difference I will consider a non-trivial but non-general case as well, say $n=3$ or some such, for bounty. A general answer (affirmative or negative) is obviously preferred. Equally obviously, I'm not speaking for Ben. Have fun, JL]

Jyrki Lahtonen
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Ben Blum-Smith
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    I think the answer is yes. The idea is that you presume that you have done it for dimension $n$, then you attach a dimension n+1 disk along the copy of $\mathbb{R}P^n$. Then you make an appropriate decomposition of the disk $D^{2n+2}$ that contains $D^{n+1}$ as a cell complex, and you use this this to guide how you attach the various cells in your decomposition of $D^{2n+2}$. Of course, the base case is easy since it just is $S^1$ included into $S^2$ as a meridian. – Connor Malin Jan 02 '20 at 04:59
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    You can see the inductive step if you look at $n=0$. – Connor Malin Jan 02 '20 at 05:05
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    @ConnorMalin Why not an official answer? – Paul Frost Jan 02 '20 at 10:27
  • Because I certainly don’t think what I’ve written is rigorous, and I feel like making it rigorous will be very painful. – Connor Malin Jan 02 '20 at 20:45
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    @ConnorMalin - to make sure I understand this idea: say we have it done in dimension $n$. You're suggesting to realize the needed additional cell for $\mathbb{R}P^{n+1}$ as a subcomplex of the needed additional cell for $\mathbb{C}P^{n+1}$. Right? (cont'd) – Ben Blum-Smith Jan 05 '20 at 08:54
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    If I've got you correctly, then why do you feel confident there's a way to do this so the attaching map for $D^{2n+2}$ restricts appropriately to $D^{n+1}$? – Ben Blum-Smith Jan 05 '20 at 08:54
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    This question received a bounty from [the Pearl Dive](https://chat.stackexchange.com/rooms/102837/pearl-dive). – Jyrki Lahtonen Jan 26 '20 at 12:43

1 Answers1


I believe the following works. Consider a cell decomposition made of subsets of $\mathbb{CP}^n$ of the form

$$\{ (x_1+s_1 y_1 i : x_2+s_2 y_2 i : \ldots : x_k+s_k y_k i : 1 : 0 : \ldots : 0) \mid x_i \in \mathbb{R}, y_i \in \mathbb{R}_{>0} \},$$

where $s_i \in \{-1,0,+1\}$. We label such a cell simply by $s_1 s_2 s_3 \ldots s_k$ (if $k=0$ we denote it by $\varnothing$).

The dimension of each cell is $k + \sum_{j=1}^k |s_j|$ (because each $s_j$ is associated with one degree of freedom $x_j$ if it is zero and with two $(x_j, y_j)$ if it is nonzero). It is easy to see that the boundary of the cell $s_1 s_2 s_3 \ldots s_k$ consists of cells where either one or more of the $s_j$ are removed, or some nonzero $s_j$ are replaced by $0$.

The real projective space $\mathbb{RP}^n$ is then explicitly realized as a subcomplex with cells $\varnothing, 0, 00, 000,$ etc.


  • The cell decomposition for $\mathbb{CP}^1$ (which is the Riemann sphere) consists of two hemispheres $+$ and $-$ separated by an "equator" $0$ with a marked point $\varnothing$.

  • The cell decomposition for $\mathbb{CP}^2$ consists of the following set of 13 cells:

$$\{ \varnothing, 0, +, -, 00, +0, -0, 0+, 0-, ++, +-, -+, -- \}.$$

The number of $D$-dimensional cells with $D = 0, 1, 2, 3, 4$ is 1, 1, 3, 4 and 4 respectively. To check that this makes sense, we can for example compute the Euler characteristic: $\chi = 1-1+3-4+4=3$, as expected (recall that $\chi(\mathbb{CP}^n)=n+1$).

  • The cell decomposition for $\mathbb{CP}^3$ has 40 cells in total:

$$\{ \varnothing, 0, +, -, 00, +0, -0, 0+, 0-, ++, +-, -+, --, 000, +00, 0+0, 00+, -00, 0-0, 00-, ++0, +0+, 0++, +-0, +0-, 0+-, -0+, -0+, 0-+, --0, -0-, 0--, +++, -++, +-+, ++-, --+, -+-, +--, --- \}.$$

The Euler characteristic is $\chi = 1-1+3-5+10-12+8=4$, which again checks out.

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  • This looks very promising! (+1) I buy that these are cells, and that they partition $\mathbb{C}P^n$, and that $\mathbb{R}P^n$ is the subcomplex mentioned. I'm basically convinced the whole construction is correct, but something is bothering me about your description of the boundaries. Given a cell $s_1\dots s_k$, I see that any cell in which nonzero $s_j$'s are replaced by 0's lies on its boundary. I also see that removing a *suffix* of $s_j$'s gives a cell on the boundary. But I'm not sure I see how removal of $s_j$'s yields a boundary cell in general. Does $-$ lie in the boundary of $+-$? – Ben Blum-Smith Jan 28 '20 at 00:43
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    @BenBlum-Smith Thank you! My reasoning was as follows: $+-$ is the set of representatives $(x_1+y_1 i : x_2-y_2 i : 1)$, or equivalently $(\frac{x_1+y_1 i}{x_2-y_2 i} : 1 : \frac{1}{x_2-y_2 i})$. Taking the limit as $x_i,y_i \to \infty$ in such a way that $\frac{x_1+y_1 i}{x_2-y_2 i} = \frac{x_1x_2 - y_1y_2 + (x_1y_2 + y_1x_2) i}{x_2^2+y_2^2}$ is bounded, we obtain a certain point $(a+b i : 1 : 0)$ in a boundary. – pregunton Jan 28 '20 at 09:23
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    @BenBlum-Smith Depending on how we take the limit, we can make $b = \lim \frac{x_1y_2 + y_1x_2}{x_2^2+y2^2}$ have a positive, negative or zero sign, since $x_1$ and $x_2$ can be chosen arbitrarily from $\mathbb{R}$. Thus the boundary can be $+$, $-$ or $0$ (the last one corresponds to the operation of removing a $s_i$ and making the other zero). – pregunton Jan 28 '20 at 09:24
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    @BenBlum-Smith Though now I see that the same reasoning seems to apply to $-$ as a boundary of $++$, so my description of the boundaries is incomplete. I'll have to think about it. – pregunton Jan 28 '20 at 09:38
  • Thank you, this clarifies. I'll think more about it too. I would love a clear picture of the incidence relations. – Ben Blum-Smith Jan 29 '20 at 05:17
  • Out of curiosity, does this also work for complex projective space $\mathbb CP^{n}$ and quaternion projective space $\mathbb HP^{n}$? – Andrews Feb 08 '20 at 06:40
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    @Andrews Good question! Replacing $i$ by $j$ in the definition and making the coordinates complex doesn't quite work because $\mathbb{C}$ has no notion of "positive"; one needs to be more careful to make well-defined cells (with interior homeomorphic to an open $n$-ball). If I'm not mistaken, instead of $s_i y_i$ being positive/negative/zero, one possibility is to consider separate cases for $y_i$ in the upper/lower half-plane, $y_i$ in the positive/negative real half-line, and $y_i=0$ (so a cell would have to be labeled with a sequence of 5 possible "signs" $\uparrow, \downarrow, +, -, 0$). – pregunton Feb 08 '20 at 18:20
  • @pregunton And one last question about generalization, does this work for quaternion projective plane $\mathbb HP^2$ and octonion projective plane $\mathbb OP^{2}$? Although there's no octonion projective space for $(n>2)$ since associativity of multiplication is needed for relation $(z_{0}, \ldots, z_{n}) \sim \lambda (z_{0}, \ldots, z_{n})$ to be an equivalence relation, we can still define octonion projective plane $\mathbb OP^{2}$ via Hopf map $S^{15} \to S^{8}$. The details are in Example 4.47, page 378 of Hatcher's book *Algebraic topology*. – Andrews Feb 09 '20 at 02:03
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    @Andrews If I remember correctly the octonion projective plane can also be defined in terms of associative triples of octonions. Since any two octonions generate an associative subalgebra we can safely take $(a:b:1)$, $(c:1:0)$ and $(1:0:0)$ as our representatives with $a,b,c \in \mathbb{O}$, and the generalization should work fine. We replace $j$ by $\ell$ (=$(0,1)$ in the Cayley-Dickson construction), make the coordinates $x_i, y_i$ quaternionic, and consider separate cases for $y_i$ in the upper/lower quaternionic half-$4$-space, upper/lower half-$3$-space, etc. (9 "signs" in total). – pregunton Feb 09 '20 at 08:32
  • @Andrews For completeness we could consider also $\mathbb{HP}^1$ inside $\mathbb{OP}^1$, but that case is easy because they are both spheres. – pregunton Feb 09 '20 at 08:33
  • Marking as answered, since it seems pretty clear the construction works even without an understanding of the boundary maps. – Ben Blum-Smith Feb 24 '20 at 22:24