1
$\begingroup$

A few questions based on the definition of homology from Wikipedia (at the bottom):

  1. What the chain complex elements are called (e.g. if they're actually called p-chains).
  2. If chain complex elements can only be groups/modules. Or can they be any mathematical structure.
  3. Why the composition of any two maps is the zero map in chain complexes. Why it is defined to be zero. Not sure I'm following how the maps are supposed to work. To me that says that the chain will always resolve to $0$ when you apply more than one boundary operator.
  4. The use case behind $H_{n}(X):=\ker(\partial _{n})/\mathrm {im} (\partial _{n+1})=Z_{n}(X)/B_{n}(X)$. I understand what a kernel, image, and quotient group are at a basic level, just not what this is trying to accomplish. The explanation using holes doesn't make sense to me outside of circles/spheres. Wondering what you can learn about a topology by using homology groups / homologies. Perhaps it would help to know the reason for the names "cycle" and "boundary", not sure.

Intuitively it seems a chain complex is simply a bunch of groups connected together by homomorphisms. Maybe it can be arbitrary mathematical structures connected by homomorphisms which I'm wondering. Then a homomorphism is a very simple chain between two things. This resource sort of described homology groups, saying they are used to understand the topology of a graph using groups. But I don't yet grasp the meaning / purpose of the homology group equation. It sounds like homologies are defined in order to learn about the structure of the topology, but not quite sure how/when to apply it / when I need to learn about the structure. Wondering what I can learn.


An image is:

  1. The subset of a function's codomain.

A kernel $f:G \to H$ is:

  1. A measure of the degree to which a homomorphism fails to be injective.
  2. For the two groups $G$ and $H$ (if $e_H$ is the identity element of $H$): $$\operatorname {ker} f=\{g\in G:f(g)=e_{H}\}{\mbox{.}}$$
  3. More generally: $$\operatorname {ker} f=\{(a,a')\in A\times A:f(a)=f(a')\}{\mbox{.}}$$

A quotient group is:

  1. Obtained by aggregated elements of another group using an equivalent relation.
  2. And preserving the original group structure.

A chain complex $(C_{\bullet },\partial_{\bullet })$ is:

  1. A sequence of abelian groups or modules $\dotsc, C_0, C_1, C_2, \dotsc$, called p-chains.
  2. Connected by homomorphisms (called boundary operators or differentials) $\partial_n : C_n\to C_{n−1}$.
  3. The composition of any two consecutive maps is the zero map, $\partial_n \circ \partial_{n+1} = 0$.
  4. The chain complex can be written as: $$\cdots {\xleftarrow {\partial_{0}}}C_{0}{\xleftarrow {\partial_{1}}}C_{1}{\xleftarrow {\partial_{2}}}C_{2}{\xleftarrow {\partial_{3}}}C_{3}{\xleftarrow {\partial_{4}}}C_{4}{\xleftarrow {\partial_{5}}}\cdots$$

A cochain complex $(C^{\bullet },\partial^{\bullet })$ is:

  1. The dual of the chain complex.
  2. A sequence of abelian groups or modules $\dotsc, C^0, C^1, C^2, \dotsc$
  3. Connected by homomorphisms $\partial^n : C^n\to C^{n+1}$ satisfying $\partial^{n+1} \circ \partial^n = 0$.
  4. The cochain complex can be written as: $$\cdots {\xrightarrow {\partial^{-1}}}C^{0}{\xrightarrow {\partial^{0}}}C^{1}{\xrightarrow {\partial^{1}}}C^{2}{\xrightarrow {\partial^{2}}}C^{3}{\xrightarrow {\partial^{3}}}C^{4}{\xrightarrow {\partial^{4}}}\cdots$$

A homology group is:

  1. Defined over a topological space $X$ with a corresponding chain complex $C(X)$ encoding info about $X$.
  2. "That the boundary of a boundary is trivial implies $\mathrm {im} (\partial _{n+1})\subseteq \ker(\partial _{n})$, where $\mathrm {im} (\partial _{n+1})$ denotes the image of the boundary operator and $\ker(\partial _{n})$ its kernel." (I don't understand this wiki quote).
  3. Elements of $B_{n}(X)=\mathrm {im} (\partial _{n+1})$ are called boundaries.
  4. Elements of $Z_{n}(X)=\ker(\partial _{n})$ are called cycles.
  5. Then one can create the quotient group called the nth homology group of $X$: $$H_{n}(X):=\ker(\partial _{n})/\mathrm {im} (\partial _{n+1})=Z_{n}(X)/B_{n}(X)$$
  6. Elements of $H_n(X)$ are called homology classes.
  7. Each homology class is an equivalence class over cycles and two cycles in the same homology class are said to be homologous.
  8. The homology groups of $X$ measure "how far" the chain complex associated to $X$ is from being exact.
$\endgroup$
  • $\begingroup$ One can certainly have chain complexes taking values in Abelian categories, notably categories of sheaves. I suppose one can look at even more general categories than Abelian categories. If I had a "complex" where the compositions of successive maps were not all zero, I wouldn't call it a chain complex. $\endgroup$ – Lord Shark the Unknown May 5 '18 at 2:30
2
$\begingroup$

To address each of your questions:

  1. Yes, the elements of the $p^\mathrm{th}$ chain group $C_p$ are commonly called "$p$-chains". Or in cohomology, elements of $C^p$ are called "cochains".

  2. I don't know about any mathematical structure, but typically one can talk about chain (or cochain) complexes in any abelian category.

  3. You sort of answered your own question lower down in your post, in your point #2 under "A homology group is:". The purpose of having $d^{n+1} \circ d^n = 0$ is so that it makes sense to take the quotient of the kernel of one map by the image of the previous.

  4. In the case of the simplicial/CW/cellular homology of a topological space, you "triangulate" your space into a simplicial complex, and the chain groups $C_p$ are generated by the $p$-dimensional cells/simplices in this decomposition. The boundary maps then literally represent the boundary of these cells. For example, if you have a 1-simplex $\sigma \in C_1$ (i.e. a line segment) with endpoints $v_0$ and $v_1$, then $\partial (\sigma) = v_1 - v_0$. Then for $\sigma$ to be in the kernel of $\partial$, this means that $v_0 = v_1$, and so its endpoints are the same. This is most likely the intuition behind the term "cycle" for elements of the kernel.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.