2
$\begingroup$

Suppose that we want to find out cellular homology groups of $RP^n=:X$.

I know that $RP^n$ is obtained by attaching $k$ cell to $RP^{k-1}$ for every $k$ such that $0\le k\le n$.

So the chain complexes are $C_j(X)=H_j(X^j, X^{j-1})=Z$ for every $j$ such that $0\le j\le n$.

How do I find the boundary maps $d_j: C_j(X)\to C_{j-1}(X)$ using cellular homology? I have heard that it is very effective way to calculate homology groups but

Hatcher gives the following formula for $d_j$:

$d_j(e_\alpha^j)= \sum d_{\alpha\beta}e_\beta^{j-1},$ where $d_{\alpha\beta}$ is deg($S_\alpha^{j-1}\to X^{j-1}\to S_\beta^{j-1}),$ i.e., degree of the composition of the attaching map of $e_\alpha^j$ with the quotient map collapsing $X^{j-1}-e_\beta^{j-1}$ to a point.

I don't understand this definition basically the part "collapsing $X^{j-1}-e_\beta^{j-1}$ to a point." Why is this identification required?

Another thing that I don't understand is what makes cellular homology efficient for computation. How do we calculate the boundary maps in the example?

Can you please give detailed illustrations as to how to compute the boundary maps? Hatcher's book lacks details.

Personally I find singular homology much more practical computationally.

Thanks.

$\endgroup$
9
  • 5
    $\begingroup$ Can you explain why you find singular homology to be more practical computationally? In particular, how do you get around the fact that the singular chain groups (and cycle groups and boundary groups) generally have uncountable dimension? That makes it hard to write down a matrix for the boundary map...... $\endgroup$
    – Lee Mosher
    Commented May 7, 2023 at 18:20
  • $\begingroup$ @LeeMosher: In the examples, that I have seen: one just finds chain groups $C_n$'s as Z{generators} and then there is a straightforward formula to calculate the boundary maps $d_n$ and then simply $H_n()= Ker d_n/Im d_{n+1}$. I find this to be very simple and straightforward. On the other hand, in cellular homology I don't understand how to compute $d_n$'s using its definition. That's why I feel more comfortable with the singular homology and at the same time I wonder what makes cellular homology more computable. $\endgroup$
    – Koro
    Commented May 7, 2023 at 18:27
  • 1
    $\begingroup$ Perhaps, then, your question is less about "computability" than "definability"? $\endgroup$
    – Lee Mosher
    Commented May 7, 2023 at 18:30
  • $\begingroup$ @LeeMosher: The definition is fixed (I'm using Hatcher's definition as stated in my post), I want to know how to use that definition to understand 'why cellular homology is computationally more efficient than singular homology' with the help of some examples. $\endgroup$
    – Koro
    Commented May 7, 2023 at 18:32
  • 1
    $\begingroup$ Okay, so, your question is really about computability of the chain maps? Or, perhaps, the efficiency of the formulas for the chain maps? $\endgroup$
    – Lee Mosher
    Commented May 7, 2023 at 18:35

1 Answer 1

2
$\begingroup$

We might consider three ways of computing the homology of a CW complex:

  • singular
  • simplicial
  • cellular

For a typical CW complex (anything with cells with dimension greater than 0), the singular chain groups will be uncountable in each dimension. They are essentially uncomputable: $C_1(X)$ is the free abelian group with basis given by all continuous maps $[0,1] \to X$, for instance. You may understand the boundary maps in this chain complex, but it is very hard to do any nontrivial homology computation with the singular complex. So we are reduced to comparing simplicial and cellular homology.

To compute the cellular homology, first you need to choose a CW decomposition of your space. If you have a representation of your space as a simplicial complex (or a $\Delta$-complex), then you can use that representation to define the CW structure: each $n$-simplex becomes an $n$-cell, and the attaching maps will be determined by the boundary maps of the simplices. The cellular chain complex will be identical to the simplical chain complex, and therefore the two will be identical to compute. Therefore cellular homology is at least as efficient to compute as simplicial, because simplicial homology is a special case.

So the question becomes: are there some CW complexes for which cellular homology is actually easier? Sure: an $n$-sphere can be constructed with a single 0-cell and a single $n$-cell, so the cellular chain complex has a copy of $\mathbb{Z}$ in dimension $0$ and another in dimension $n$, zeroes otherwise. If $n \neq 1$, then there are no possible nonzero differentials, so the homology is the same as the chain complex. (If $n=1$, there is a possible nonzero differential, but it turns out to be zero in this case, too.)


Regarding the question about collapsing the complement of a particular cell to a point: you are trying to compute a map $$ H_n(X^n, X^{n-1}) \to H_{n-1}(X^{n-1}, X^{n-2}). $$ The former has the $n$-cells as a basis; the latter the $(n-1)$-cells. For each $n$-cell in $X$, you want to pick out how it is attached to each individual $(n-1)$-cell, so you collapse the complement of that $(n-1)$-cell to a point. Do this for each $(n-1)$-cell to determine the component of the cellular boundary map on the given $n$-cell.

A good example to think about is a torus constructed by identifying the sides of a rectangle together in the usual way. This has one $0$-cell, two $1$-cells, and one $2$-cell, and you should try to use Hatcher's description on p. 141 to compute the cellular boundary maps — they should all be zero.


For the particular case of $\mathbb{R}P^n$, Hatcher discusses this in Example 2.42. What questions do you have about that example?

$\endgroup$
4
  • $\begingroup$ Thanks a lot for the answer +1. I understand some of your third last para. It's written better than Hatcher's. About the example 2.42, I don't understand the calculation of the boundary maps (sometimes 2, sometimes 0 etc. and why are there cases n= odd, even etc.). $\endgroup$
    – Koro
    Commented May 7, 2023 at 21:47
  • $\begingroup$ There are differences between $n$ odd and $n$ even because the degree of the antipodal map depends on the parity of $n$. In the picture en.wikipedia.org/wiki/Real_projective_plane#/media/… (from wikipedia), use this CW decomposition: a single $0$-cell $v$ (the bottom right corner); a single $1$-cell $e$ (the concatenation of $A$ followed by $B$); and a single $2$-cell (the interior). The boundary of the $1$-cell is $v-v=0$. The boundary of the $2$-cell is $e + e$ — look at the orientation as you glue the interior to the 1-skeleton. $\endgroup$ Commented May 8, 2023 at 0:28
  • $\begingroup$ How is n cell added, n>2? How is orientation given in such a case? $\endgroup$
    – Koro
    Commented May 8, 2023 at 7:08
  • $\begingroup$ As Hatcher explains, each $n$-cell is glued to the $(n-1)$-skeleton by a double covering, and you can view the two coverings as the identity map on half, the antipodal map on the other half. This continues the pattern in dimensions 1 and 2. $\endgroup$ Commented May 8, 2023 at 15:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .