More (co)homological constructions

 Preliminary exam prep

Recall a previous post (2016-09-16, "Complexes and their homology") that focused on constructing topological spaces in different ways and recovering the homology. Here we complete that task, introducing cellular homology. Recall a cell complex (or CW complex) $X$ was a sequence of skeleta $X_k$ for $k=0,\dots,\dim(X)$ consisting of $k$-cells $e^k_i$ and their attaching maps to the $(k-1)$-skeleton.

Cellular homology

Definition: The long exact sequence in relative homology for the pair $X_k,X_{k-1}$ shares terms with the long exact sequence for the pair $X_{k+1},X_k$, as well as $X_{k-1},X_{k-2}$. By letting $d_k$ be the composition of maps in different long exact sequences, for $k>1$, that make the diagram

commute, we get a complex of equivalence classes of chains
\[
\cdots \to H_{k+1}(X_{k+1},X_k) \tov{d_{k+1}} H_k(X_k,X_{k-1})\tov{d_k} H_{k-1}(X_{k-1},X_{k-2})\to \cdots \to H_1(X_1,X_0)\tov{d_1} H_0(X_0) \tov{d_0} 0,
\]
whose homology $H_k^{CW}(X) = \ker(d_k)/\text{im}(d_{k-1})$ is called the cellular homology of $X$. The map $d_1$ is the connecting map in the long exact sequence of the pair $X_1,X_0$, and $d_0=0$.

This seems quite a roundabout way of defining homology groups, but it turns out to be very useful. Note that for $k=1$, the map $d_1$ is the same as for a simplicial complex, hence

Theorem: In the context above,

1. for $k\>0$, $H^{CW}_k(X)\cong H_k(X)$;
2. for $k\>1$, $H_k(X_k,X_{k-1})=\Z^\ell$, where $\ell$ is the number of $k$-cells in $X$; and
3. for $k\>2$, $d_k(e^k_i) = \displaystyle\sum_j\deg(\underbrace{\dy e^k_i}_{S^{k-1}}\tov{f_{k,i}} X_{k-1}\tov{\pi} \underbrace{X_{k-1}/X_{k-1}-e^{k-1}_j}_{S^{k-1}})e^{k-1}_j$.

Example: Real projective space $\R\P^n$ has a cell decomposition with one cell in each dimension, and 2-to-1 attaching maps $\dy(e_k) =2X_{k-1}$ for $k>1$. This gives us a construction
\[
X_0 = e_0,
\hspace{1cm}
X_1 = e_1 \bigsqcup_{\dy(e_1)=e_0} X_0,
\hspace{1cm}
X_2 = e_2 \bigsqcup_{\dy(e_2)=2e_1} X_1,
\hspace{1cm}
X_3 = e_3 \bigsqcup_{\dy(e_3)=2e_2} X_2, \dots
\]It is immediate that $d_0=d_1=0$, and for higher degrees, we have
\[
d_k(e^k) = \deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1})e^{k-1}.\]
Since this is a map between spheres, we may apply local degree calculations. The first part is the 2-to-1 cover, where every point in $\R\P^{k-1}$ is covered by two points from $S^{k-1}$, one in each hemisphere. One covers it via the identity, the other via the antipodal map. As long as we choose a point not in $\R\P^{k-2}\subset \R\P^{k-1}$, the second step doesn't affect these degree calculations. The antipodal map $S^{k-1}\to S^{k-1}$ has degree $(-1)^k$, hence for $a$ the antipodal map, the composition has degree
\[
\deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1}) = \deg(\id_{S^{k-1}}) + \deg(a_{S^{k-1}}) = 1+(-1)^k = \begin{cases} 2 & k\text{ even}, \\ 0 & k \text{ odd.}\end{cases}
\]

Products in (co)homology

Recall that an $n$-chain on $X$ is a map $\sigma:\Delta^n\to X$, where $\Delta^n=[v_0,\dots,v_n]$ is an $n$-simplex. These form the group $C_n$ of $n$-chains. An $n$-cochain is an element of $C^n = \Hom(C_n,\Z)$, though the coefficient group does not need to be $\Z$ necessarily.

Definition: The diagonal map $X\to X\times X$ induces a map on cohomology $H^*(X\times X)\to H^*(X)$, and by Kunneth, this gives a map $H^*(X)\otimes H^*(X)\to H^*(X)$, and is called the cup product.

For $a\in H^p(X)$ and $b\in H^q(X)$, representatives of the class $a$ are in $\Hom(C_p,\Z)$ and representatives of the class $b$ are in $\Hom(C_q,\Z)$, though we will conflate the notation for the class with that of a representative. Hence for a $(p+q)$-chain $\sigma$ the cup product of $a$ and $b$ acts as
\[
(a\smile b)\sigma = a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot b\left(\sigma|_{[v_p,\dots,v_{p+q}]}\right).
\]
Definition: The cap product combines $p$-cochains with $q$-chains to give $(q-p)$-chains, by
\[
\begin{array}{r c l}
\frown\ :\ H^p(X) \times H_q(X) & \to & H_{q-p}(X), \\\
(a, \sigma) & \mapsto & a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot \sigma|_{[v_p,\dots,v_q]}.
\end{array}\]
The cap product with the orientation form of an orientable manifold $X$ gives the isomorphism of Poincare duality.

Remark: Given a map $f:X\to Y$, the cup and cap products satisfy certain identities via the induced map on cohomology groups. Let $a,b\in H^*(Y)$ and $c\in H_*(X)$ be cochain and chain classes, for which
\[f^*(a\smile b) = f^*(a)\smile f^*(b),
\hspace{1cm}
a\frown f_*c = f_*(f^*a\frown c).\]
The first identity asserts that $f^*$ is a ring homomorphism and the second describes the commutativity of an appropriate diagram. The cup and cap products are related by the equation
\[a(b\frown \sigma) = (a\smile b)\sigma,\]for $a\in H^p$, $b\in H^q$ and $\sigma\in C_{p+q}$.

References: Hatcher (Algebraic topology, Chapter 2.2), Prasolov (Elements of homology theory, Chapter 2)

Degree and orientation

 Preliminary exam prep

Topology

 Recall that a topological manifold is a Hausdroff space in which every point has a neighborhood homeomorphic to $\R^n$ for some $n$. An orientation on $M$ is a choice of basis of $\R^n$ in each neighborhood such that every path in $M$ keeps the same orientation in each neighborhood. Every manifold $M\owns x$ appears in a long exact sequence (via relative homology) with three terms
\[
H_n(M-\{x\}) \tov{f} H_n(M) \tov{g} H_n(M,M-\{x\}).
\]
The first term is 0, because removing a point from an $n$-dimensional space leaves only its $(n-1)$-skeleton, which is at most $(n-1)$-dimensional. For $U$ a neighborhood of $x$ in $M$, the last term (via excision) is
\[
H_n(M-U^c, M-\{x\}-U^c) = H_n(U, U-\{x\})\cong H_n(\R^n, \R^n-\{x\}) \cong H_n(\R^n, S^{n-1}),
\]
which in turn fits into a long exact sequence whose interesting part is
\[
H_n(\R^n)\to H_n(\R^n,S^{n-1}) \to H_{n-1}(S^{n-1})\to H_{n-1}(\R^n),
\]
and since the first and last terms are zero, $H_n(M,M-\{x\})=\Z$. Since $f$ is zero, $g$ into $\Z$ must be injective, meaning that $H_n(M)=\Z$ or 0.

Theorem: Let $M$ be a connected compact (without boundary) $n$-manifold. Then

1. if $M$ is orientable, $g$ is an isomorphism for all $x\in M$, and
2. if $M$ is not orientable, $g=0$.

Definition: Let $f:M\to N$ be a map of connected, oriented $n$-manifolds. Since $H_n(M)=H_n(N)$ is infinite cyclic, the induced homomorphism $f_*:H_n(M)\to H_n(N)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$.

In the special case when we are computing the degree for a map $f:S^n\to S^n$, by excision we get
\[
\deg(f) = \sum_{x_i\in f^{-1}(y)} \deg\left(H_n(U_i,U_i-x_i)\tov{f_*} H_n(V,V-y)\right),
\]for any $y\in Y$, some neighborhood $V$ of $y$, and preimages $U_i$ of $V$. This is called the local degree of $f$.

Geometry

Let $M$ be a smooth $n$-manifold. Recall $\Omega_M^r$ is the space of $r$-forms on $M$ and $d^r:\Omega^r_M\to \Omega^{r+1}_M$ is the differential map. Also recall the de Rham cohomology groups $H^r(M) = \text{ker}(d^r)/\text{im}(d^{r-1})$.

Definition: An $n$-manifold $M$ is orientable if it has a nowhere-zero $n$-form $\omega\in \Omega^n_M$. A choice of $\omega$ is called an orientation of $M$.

We also have a map $H^n(M)\to \R$, given by $\alpha\mapsto \int_M\alpha$, where the integral is normalized by the volume of $M$, so that integrating 1 across $M$ gives back 1. It is immediate that this doesn't make sense when $M$ is not compact, but when $M$ is compact and orientable, we get that $H^n(M)\neq0$. Indeed, if $\eta\in \Omega^{n-1}_M$ with $d\eta =\omega$, by Stokes' theorem we have
\[
\int_M\omega = \int_Md\eta = \int_{\dy M}\eta = \int_\emptyset\eta = 0,
\]
as $M$ has no boundary (since it is a manifold). But $\omega$ is nowhere-zero, meaning the first expression on the left cannot be zero. Hence $\omega$ is not exact and is a non-trivial element of $H^n(M)$.

Theorem: Let $M$ be a smooth, compact, orientable manifold of dimension $n$. Then $H^n(M)$ is one-dimensional.

Proof: The above discussion demonstrates that $\dim(H^n(M))\>1$. We can get an upper bound on the dimension by noting that the space of $n$-forms on $M$, given by $\Omega^n_M = \bigwedge^n(TM)^*$, has elements described by $dx_{i_1}\wedge \cdots \wedge dx_{i_n}$, with $\{i_1,\dots,i_n\}\subset \{1,\dots,n\}$. By rearranging the order of the $dx_{i_j}$, every element looks like $\alpha dx_1\wedge \cdots\wedge dx_n$ for some real number $\alpha$. Hence $\dim(\Omega^n_M) \leqslant 1$, so $\dim(H^n(M))$ is either 0 or 1. Therefore $\dim(H^n(M))=1$. $\square$

Definition: Let $f:M\to N$ be a map of smooth, compact, oriented manifolds of dimension $n$. Since $H^n(M)$ and $H^n(N)$ are 1-dimensional, the induced map $f^*:H^n(N)\to H^n(M)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$. Equivalently, for any $\omega\in \Omega^n_N$,
\[
\int_M f^*\omega = d\int_N\omega
\]

References: Hatcher (Algebraic Topology, Chapters 2, 3.3), Lee (Introduction to Smooth Manifolds, Chapter 17)