
#1
Mar110, 01:10 PM

Sci Advisor
P: 2,341

This thread is similar to but more elementary than
http://www.physicsforums.com/showthread.php?t=379608 I plan to
Lets begin with a very natural question: It helps to know from the outset that AT falls into two topics (which eventually turn out to be related):
And it helps to know from the outset that the origins of homology/cohomology lie with some natural approaches to some basic problems:
(I hope to eventually explain some of the terms I didn't define above, either in this BRS thread or the Cellular Homology BRS thread. In a later post in the BRS thread cited above, "Cellular Homology with Macaulay2", I hope to say more about Schubert calculus note that another Macaulay2 package allows us to compute with the cohomology ring of Grassmannian manifolds. I hope to eventually discuss Caratheodory's ideas in another BRS thread on the common ground shared by electrical networks and some formally identical problems involving vibrations, etc.) It also helps to know from the outset that one of the most important developments in AT was the recognition that all varieties of homology/cohomology share a common ground in a more fundamental subject, commutative algebra and homological algebra, which deals with sequences of Rmodule homomorphisms between certain Rmodules (some fixed ring R), forming a "chain complex". Here, the key idea is that as soon as you have a sequence of maps which obey the key property "boundary of a boundary vanishes" [itex]\partial^2 = 0[/itex], the image of each map in the sequence is a submodule of the kernel of the next map, both being submodules of a chain group, so one can define the quotient modules as the homology modules. Then one can define a graded Rmodule which is the direct sum of all the homology modules, so that we need only refer to "the homology", instead of a sequence of homology modules. In algebraic topology, for a given topological space X, it turns out that the homology graded module H(X), but not the chain graded module C(X), is invariant under homeomorphisms, and thus a topological invariant of X. In fact, H(X) is invariant under a much more general equivalence relation which arises in homotopy theory, called homotopic equivalence. Because this relation is weaker than homeomorphism, homology cannot be a complete topological invariant; there are spaces which are not homeomorphic but which homology cannot distinguish. (I plan to give and sketch some examples in a later post in this thread.) It turns out that the abelianization of the fundamental group of X is just the homology in dimension one, [itex]H_1(X)[/itex]. And very roughly, the role of the various competing theories goes something like this:




#2
Mar310, 12:12 AM

Sci Advisor
P: 2,341

Let's look at a small simplicial complex consisting entirely of
We want to consider our complex as representing an entire equivalence class of topological spaces under homeomorphism, call it X. If you have some experience with homotopy theory, you can probably see that the edge labeled [itex]e_1[/itex] can be homotopically deformed by pushing it into the point b, and then the remainder can be homotopically deformed into a configuration which we recognize as the wedge sum of three circles, so in our example [itex]X = \vee_{j=1}^3 S^1[/tex] In other words, we have here three loops sharing a single common point, and nothing else of topological significance or more precisely, nothing else of homological significance. One basic idea of (integer) homology is to replace paths with formal (integer coefficient) linear combinations of edges, called onechains. Similarly we replace twodimensional topological surfaces with formal linear combinations of twosimplices (triangles), which we call twochains. These chains form abelian groups, the chain groups [itex]C_0(X), \; C_1(X), \; C_2(X), \ldots[/itex]. But the really essential idea is to define an equivalence relation on qchains by means of certain boundary maps [tex] C_{q1}(X) \stackrel{\partial_q}{\leftarrow} C_q(X) \stackrel{\partial_{q+1}}{\leftarrow} C_{q+1}(X) [/tex] Then we can say that two qchains are homologous if their formal difference is a qbound, i.e. lies in the image of the boundary map. Of course, the key to making this work is to discover the correct definition of the boundary map. This definition turns out to be [tex] \partial (p_0 \, p_1 \, \ldots p_q) = (1)^q \; \sum_{j=0}^q (1)^j (p_0 \, p_1 \, \ldots \hat{p_j} \, \dots p_q) [/tex] where the points have been placed in order and where the caret signifies "omit me". For example: [tex] \partial \, (ab) = a  b, \; \; \partial \, (bcd) = cd + bd  bc [/tex] So why is this the right definition? Well, we want the boundary map to make sense when applied to formal linear combinations (we want it to be a homomorphism of abelian groups), and we have for example [tex] \partial (ab + bc) = a  b + b  c = a  c [/tex] which is just what we would expect for the boundary of the onechain ab+bc. Even better, the path around the loop containing the points a,b,c is as a formal linear combination ab+bcac (note the sign!), and then [tex] \partial(ab + bc  ac) = a  b + b  c  (a  c) = 0 [/tex] In other words, the boundary map [itex]\partial_1[/itex] vanishes on loops. So we can take [itex]Z_1(X) = \ker \partial_1[/itex], the group of onecycles, to be the group generated by the onechains on which the boundary operator vanishes, and in our example, we expect it to be generated by three fundamental loops. Here is how we would represent our first example of a simplical complex using the package SimplicalComplexes package in Macaulay2:
Compare the figure below with the boundary map produced by Macaulay2 [tex] \mathbb{Z}^8 \stackrel{\partial_1}{\leftarrow} \mathbb{Z}^{10} [/tex] which is [tex] \partial_1 = \left[ \begin{array}{rrrr rrr rrrl} e_1 & e_2 & e_3 & e_4 & e_5 & e_6 & e_7 & e_8 & e_9 & e_{10} & \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & b\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & c \\ 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & d \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & e \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & f \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & g \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & h \end{array} \right] [/tex] (Ignore the boundary map [itex]\partial_0[/itex], which uses a silly trick to compute the reduced homology rather than the homology proper.) Note well that we have ten edges (onesimplices) and eight points (zerosimplices) in our chain complex; as our remark about homotopical deformation suggests, these are certainly not topological invariants! But using linear algebra you can show that [itex]\ker \partial_1= Z_1(X)[/itex] is generated by three fundamental cycles, so [itex]Z_1(X) = \mathbb{Z}^3[/itex]. Because our simplicial complex consists only of at most one dimensional simplices, this is actually a topological invariant. Indeed, the command
[tex] 0 \stackrel{\epsilon}{\leftarrow} \tilde{H}_0(X) \stackrel{\partial_1}{\leftarrow} H_1(X) \leftarrow 0 [/tex] is [tex] 0 \stackrel{\epsilon}{\leftarrow} 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z}^3 \leftarrow 0 [/tex] But we are not quite there yet; in general, simply by considering the groups of cycles, we don't yet have a topological invariant. It turns out that the final step is to define the aforementioned equivalence relation on cycles: two qcycles are cohomologous if their difference is the boundary of a (q+1)chain. To understand how this works for 1cycles, we need to introduce some 2chains and to look at their boundaries. So let's add a twosimplex to our first example:
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z}^2 \stackrel{\partial_2}{\leftarrow} 0 \leftarrow 0 [/tex] This means that by adding triangles (just one, in this case) so that one of our loops becomes the boundary of a topological disk, we can shrink that loop to a point, so we have eliminated one of the cycles, without adding twodimensional structure of any topological significance. So what would be "topologically significant" twodimensional structure? To find out, lets consider a third simplicial complex, formed by four triangles making the "skin" of a tetrahedron:
[tex] \begin{array}{rcl} \partial_1 & = & \left[ \begin{array}{rrr rrrl} ab & ac & ad & bc & bd & cd & \\ \hline 1 & 1 & 1 & 0 & 0 & 0 & a \\ 1 & 0 & 0 & 1 & 1 & 0 & b\\ 0 & 1 & 0 & 1 & 0 & 1 & c \\ 0 & 0 & 1 & 0 & 1 & 1 & d \end{array} \right], \\ &&\\ \partial_2 & = & \left[ \begin{array}{rrrrl} abc & abd & acd & bcd & \\ \hline 1 & 1 & 0 & 0 & ab \\ 1 & 0 & 1 & 0 & ac \\ 0 & 1 & 1 & 0 & ad \\ 1 & 0 & 0 & 1 & bc \\ 0 & 1 & 0 & 1 & bd \\ 0 & 0 & 1 & 1 & cd \end{array} \right] [/tex] So for example (see the figure below) [tex] \partial \, abc = ab + ac  bc, \; \; \partial \, bcd = bc + bd cd [/tex] so that [itex]\partial \, \partial \, abcd = 0[/itex] (add the boundaries of the four triangles). Notice that the orientations of the four faces cancels along common edges. Now we can define [itex]B_1(X) = \operatorname{im} \partial_2[/itex] to be the group of onebounds. We have two nested subgroups of an abelian group [tex] B_1(X) < Z_1(X) < C_1(X)[/tex] so we can form the quotient group [tex]H_1(X) = Z_1(X)/B_1(X)[/tex] and similarly for other dimensions. The resulting homology groups (or Rmodules, if we are using a ring other than the ring of integers) are shown, in algebraic topology textbooks, to be topological invariants. And the elements of the qth homology groups are precisely the equivalence classes of qcycles modulo boundaries. In our example, [itex]H_2(X) = \mathbb{Z}[/itex], showing we have a nontrivial 2cycle (not homologous to any 2bound). If we add [itex]abcd[/itex] to our simplicial complex (if we "fill in the interior of the tetrahedron"), this 2cycle is now the boundary of the interior, so now [itex]H_2(X) = 0[/itex]. So "homologically significant structure" arises whenever we have nontrivial qcycles, and by "filling in the interiors" we can systematically destroy such structure. The crucial property of the boundary maps which ensures that the qbounds always form a subgroup of the qcycles is "the boundary of a boundary vanishes", and you can check using matrix multiplication that in our "skin of the tetrahedron" example [itex]\partial_1 \; \partial_2 = 0[/itex] (composing from right to left, or multiplying matrices from left to right). It is not very hard to prove that this general property we need is true: [itex]\partial_q \, \partial_{q+1} = 0[/tex]. Figures:




#3
Mar310, 12:43 PM

Sci Advisor
P: 2,341

Example Four: let's represent a two dimensional disk by three triangles (think of the faces of a tetrahedron, with one face removed), and just for the heck of it, lets use rational coefficients instead of integer coefficients:
Here, recall from "BRS: Cellular Homology with Macaulay2" that reduced homology is the same as homology, except that the rank of the module [itex]H_0(X)[/itex] is reduced by one. In integer homology, [itex]H_0(X) = \mathbb{Z}^r[/itex] where X has r path components; in reduced homology, [itex]\tilde{H}_0(X) = \mathbb{Z}^{r1}[/itex]. (It is traditional to indicate reduced homology groups by tildes, although this is redundant except in dimension q=0.) Example Five: if we add the fourth face to our complex, we obtain "the skin of a tetrahedron", and now the reduced homology indicates the presence of nontrivial twodimensional structure (a twocell, in the lingo of cellular homology):
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} 0 \stackrel{\partial_2}{\leftarrow} \mathbb{Q} \leftarrow 0 [/tex] Example Six: if we add the interior of the tetrahedron, we are back to trivial reduced homology:
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z} \stackrel{\partial_2}{\leftarrow} \mathbb{Z} \leftarrow 0 [/tex] showing the presence of a onecell and a twocell (in the language of cellular homology), and nothing else of homological significance. Example Nine: a Moebius band (see the figure below)
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z} \stackrel{\partial_2}{\leftarrow} 0 \leftarrow 0 [/tex] The circle is a deformation retract of the Moebius band, so we should expect that this result, which agrees with the reduced homology of a circle. Example Ten: a torus (see figure below):
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z}^2 \stackrel{\partial_2}{\leftarrow} \mathbb{Z} \leftarrow 0 [/tex] showing the presence of a twocell and two onecells. Example Eleven: Klein bottle (see figure below):
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z} \oplus \mathbb{Z}/2 \stackrel{\partial_2}{\leftarrow} 0 \leftarrow 0 [/tex] showing two onecycles, one of which has a contractible "square", and no other structure of any homological significance. Contractible square: one of the two onecycles here has the property that if you follow it twice, the resulting loop is contractible to a point. Example Twelve: the real projective plane (see figure below):
[tex] 0 \leftarrow 0 \stackrel{\partial_1}{\leftarrow} \mathbb{Z}/2 \stackrel{\partial_2}{\leftarrow} 0 \leftarrow 0 [/tex] You can experiment with different triangulations of these surfaces; of course, you should get the same results irrespective of which triangulation you choose! Figures: left to right:



Register to reply 
Related Discussions  
BRS: Cellular homology with Macaulay2  Quantum Physics  8  
pushout of simplicial sets  Differential Geometry  0  
simplicial homology  Differential Geometry  2  
Non planar simplicial homology?  General Math  3  
is a homomorphism sending an element  General Math  5 