The maps go down in degree for cellular chain complexes! I.e., \begin{align*} C_*(X) = (0 \leftarrow C_0 \leftarrow C_1 \leftarrow\cdots ) .\end{align*}
If \(X, Y\) are CW complexes with \(p_X(t), p_Y(t)\) the generating functions for the number of cells (so \([t^n] p_X(t) \coloneqq a_n\) is the number of \(n{\hbox{-}}\)cells in \(X\)), then the generating function for the product is \begin{align*} p_{X\times Y}(t) = p_X(t)p_Y(t) .\end{align*} Categorified, this comes from a quasi-isomorphism \begin{align*} C_*^{{ \mathrm{cell}}}(X\times Y) \cong C_*^{{ \mathrm{cell}}}(X) \otimes_{\mathbf{Z}}C_*^{{ \mathrm{cell}}}(Y) .\end{align*} so \begin{align*} C_n^{{ \mathrm{cell}}}(X \times Y ) \cong \bigoplus_{i+j=n} C_i^{ \mathrm{cell}}(X) \times C_j^{{ \mathrm{cell}}}(Y) .\end{align*} The boundary maps are thus given by \begin{align*} {{\partial}}(a, b) &\coloneqq{{\partial}}_X a \otimes b + (-1)^{{\left\lvert {a} \right\rvert}} a \otimes{{\partial}}_Y b .\end{align*}
This is nontrivial, it’s the content of the Eilenberg-Zilber theorem.
Let \(X= S^a \times S^b\), so \(p_{S^a}(t) = 1 + t^a\) and \(p_{S^b}(t) = 1 + t^b\), then \(p_X(t) = 1 + t^a + t^b + t^{a+b}\), so \(X\) has
- One 1-cell
- One \(a{\hbox{-}}\)cell
- One \(b{\hbox{-}}\)cell
- One \((a+b){\hbox{-}}\)cell
Compute all of the possible cases for \(H_*(S^a\times S^b)\), where \(a, b\geq 0\) are not necessarily distinct.
CW Structures on Common Spaces
\(S^n = e^0 \cup e^n\): a point and an \(n{\hbox{-}}\)cell.
\({\mathbf{RP}}^n = e^1 \cup e^2 \cup \cdots \cup e^n\): one cell in each dimension.
\(\mathbb{CP}^n =e^2 \cup e^4 \cup \cdots e^{2n}\)
Fundamental domains
Examples of Simplicial Complexes
To write down a simplicial complex, label the vertices with increasing integers. Then each \(n\)-cell will correspond to a set of \(n+1\) of these integers - throw them in a list.
Torus
Klein Bottle and {\mathbf{RP}}^2
For counterexamples, note that this fails to be a triangulation of \(T\):
Not a Torus
This fails - for example, the specification of a simplex \([1,2,1]\) does not uniquely determine a triangle in the this picture.
Cellular Homology
- \(S^n\) has the CW complex structure of 2 \(k\)-cells for each \(0\leq k \leq n\).
How to compute:
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Write cellular complex \begin{align*}0 \to C^n \to C^{n-1} \to \cdots C^2 \to C^1 \to C^0 \to 0\end{align*}
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Compute differentials \({\partial}_{i}: C^i \to C^{i-1}\)
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Note: if \(C^0\) is a point, \({\partial}_{1}\) is the zero map.
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Note: \(H_{n} X = 0 \iff C^n = \emptyset\).
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Compute degrees: Use \({\partial}_{n}(e_{i}^n) = \sum_{i} d_{i} e_{i}^{n-1}\) where \begin{align*}d_{i} = \deg(\text{Attach }e_{i}^n \to \text{Collapse } X^{n-1}{\hbox{-}}\text{skeleton}),\end{align*} which is a map \(S^{n-1} \to S^{n-1}\).
Alternatively, choose orientations for both spheres. Then pick a point in the target, and look at points in the fiber. Sum them up with a weight of +1 if the orientations match and -1 otherwise.
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Note that \({\mathbf{Z}}^m \xrightarrow{f} {\mathbf{Z}}^n\) has an \(n\times m\) matrix
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Row reduce, image is span of rows with pivots. Kernel can be easily found by taking RREF, padding with zeros so matrix is square and has all diagonals, then reading down diagonal - if a zero is encountered on \(n\)th element, take that column vector as a basis element with \(-1\) substituted in for the \(n\)th entry.
For example: \begin{align*} \begin{matrix} \mathbf1&2&0&2\\0&0&\mathbf1&-1\\0&0&0&\mathbf0 \end{matrix} \to \begin{matrix} \mathbf1&2&0&2\\0&\mathbf0&0&0\\0&0&\mathbf1&-1\\0&0&0&\mathbf0 \end{matrix} \begin{matrix} \mathbf1&2&0&2\\0&\mathbf0&0&0\\0&0&\mathbf1&-1\\0&0&0&\mathbf0 \end{matrix} \\ \ker = \begin{matrix} 2\\-1\\0\\0 \end{matrix} \begin{matrix} 3\\0\\-1\\-1 \end{matrix}\\ \operatorname{im}= \left\langle{a+2b+2d,c-d}\right\rangle .\end{align*}
- Or look at elementary divisors, say \(n_{i}\), then the image is isomorphic to \(\bigoplus n_{i} {\mathbf{Z}}\)
Constructing a CW Complex with Prescribed Homology
Given \(G = \bigoplus G_{i}\), and want a space such that \(H_{i} X = G\)? Construct \(X = \bigvee X_{i}\) and then \(H_{i} (\bigvee X_{i}) = \bigoplus H_{i} X_{i}\). Reduces problem to: given a group \(H\), find a space \(Y\) such that \(H_{n}(Y) = G\). By the structure theorem of finitely generated abelian groups, it suffices to know how to do this for \({\mathbf{Z}}\) and \({\mathbf{Z}}/n{\mathbf{Z}}\), since their powers are just obtained by wedging (previous remark). Recipe:
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Attach an \(e^n\) to a point to get \(H_{n} = {\mathbf{Z}}\)
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Attach an \(e^{n+1}\) with attaching map of degree \(d\) to get \(H_{n} = {\mathbf{Z}}_{d}\)
Exercises
Let \(X\) be \(S^1\) with two 2-cells attached via \(z\mapsto z^5\) and \(z\mapsto z^3\).
- Compute \(\pi_1(X)\) and \(H^*(X)\) as a ring.
- Show that \(X\) is not homeomorphic to \(S^2\).
- \(X\) is path-connected, so \(H_0(X) = {\mathbf{Z}}\).
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\(X\) is a CW complex with only one 1-cell, so \(\pi_1(X^{(1)}) = {\mathbf{Z}}\) for the 1-skeleton.
- Use that \(3{\mathbf{Z}}+ 5{\mathbf{Z}}= 1{\mathbf{Z}}\) as a ring, so the attached cells kill off all 1-homotopy and \(\pi_1(X) = 1\).
- By Hurewicz, \(H_1(X) = 1\).
- By UCT, \(H^1(X) = H_1(X) {}^{ \vee }= \mathop{\mathrm{Hom}}_{\mathbf{Z}}(1, {\mathbf{Z}}) = 1\)
- By UCT, \(H^2(X) = H_2(X) {}^{ \vee }\)
- We can compute \(H_2\) directly: \begin{align*} C^*(X) = (0 \to C_2(X) \xrightarrow{d_2} C_1(X) \xrightarrow{d_1} C_0(X) ) = (0 \to {\mathbf{Z}} { \left[ \scriptstyle {e_1, e_2} \right] } \xrightarrow{\substack{e_1\mapsto 3e \\ e_2 \mapsto 5e} } {\mathbf{Z}} { \left[ \scriptstyle {e} \right] } \xrightarrow{d_1} {\mathbf{Z}} { \left[ \scriptstyle {{\operatorname{pt}}} \right] } ) .\end{align*}
- \(\ker d_2 = \left\langle{ 5e_1, -3e_2 }\right\rangle\), so \(H_2 = {\mathbf{Z}}\).
- Thus \(H^*(X) = { {\bigwedge}^{\scriptscriptstyle \bullet}} _{\mathbf{Z}}(x)\) where \({\left\lvert {x} \right\rvert} = 2\).
- \(X\not\cong S^2\): delete a point \(p\) in the interior of the 2-cell corresponding to \(z^3\), then use that \(S^2\setminus\left\{{{\operatorname{pt}}}\right\} \cong {\mathbf{R}}^2\) is contractible but \(\pi_1 X\setminus\left\{{{\operatorname{pt}}}\right\} = {\mathbf{Z}}/5\).