 Having introduced the singular, simplicial chain complex for a simplicial complex as a sub complex of, of the singular chain complex of mod k. We shall now go for another simplification, a further simplification. This time we are not going to treat it as a sub complex, but as a quotient complex which is the natural way of doing it. So let this sigma, capital sigma n plus 1 denote the permutation group on n plus 1 letters. To each alpha inside sigma n plus 1, that means a permutation of n plus 1 letters, we will get a simplicial homeomorphism from delta n to delta n. Because any man, any satiric map from vertex set to vertex set is a simplicial map in delta n. In particular, if you have a permutation of vertices that can be extended linearly to mod delta n to mod delta n, which will be automatically a homeomorphism. So that we denote by g alpha. So this element represents an action of sigma n plus 1 on a sin of delta n by merely composing on the right, namely take a lambda which is a simplicial map, take lambda composite g alpha. That means you take first g alpha and then composite with lambda. So that gives you right action of sigma n plus 1 on sigma, on set of all sin of delta n. Because g n's are all simplicial, the composite will be also a simplicial map. That is all that we have to write. Once you have it on simplicial, a simplicial con, simplicial, you know, one single simplex, then you can extend it linearly for all the elements of S n by summation nj lambda j of alpha going to nj lambda j of alpha, summed up. This notation is standard notation in the many literatures about the action, writing it on the top. Similarly it comes from group theory wherein conjugation by an element was simplified by this writing like this. And most often it is the conjugation of the group on itself which is studied, that action is studied very thoroughly. So that must be the motivation for writing like this. So I am following that notation that is all here. A simplicial map delta n to any simplicial complex k is called degenerate and simplex. If the dimension of lambda of delta n is strictly less than n, it is the same thing as saying that the vertex function lambda from vertices of delta n to vertices of k is not injective. If it is an injective and it is a simplicial map, then more delta n would have been embedded as a n simplex, so the dimension would have been the same. So such things are called degenerate and anything which is not degenerate would be non-degenerate, that means when lambda is injective. So let us look at all those elements in Sn of k which are generated by subgroups of certain things namely degenerate simplexes, all degenerate simplexes and also of the form lambda minus the signature of alpha times lambda alpha. So all these elements as well as degenerate elements we are taking together, they may not form a subgroup but take the subgroup generated by them, this abelian group. So they said that you generate a abelian subgroup of that. The idea is that throw away all these things, go model all these things, kill them because the simplest thing that we want to say is if u comma u as a geometric object it will be a single element though it is representing a edge just like a constant loop or just like anything which is contractible loop etc we have taken them away in the homotopy theory. So this is a very obvious thing that we have to do namely all the degenerate simplexes they do not count similarly if you have a lambda and then you interchange the two vertices that should be minus lambda so that is the kind of relation that we want to introduce here. Alpha is a transfer position, signature of alpha is minus 1. So lambda plus the interchanger thing that should be killed which is same thing as lambda will be equal to minus of lambda alpha where alpha is a transfer position. More generally when alpha is any permutation I have to put the signature there. So this is what we want to identify. So I am taking this as a subgroup to generate these subgroups and then throw them away. One by one I could have done but I can do them together and simultaneously namely all degenerate simplexes and all lambda minus this one they will generate a subgroup. That subgroup is S0 of k. Obviously it is a chain subgroup. In other words boundary of this S0 of n will be contained inside S0 of n minus 1 and so on and if we verify that then this will be a sub chain complex. Whenever you have a sub chain complex you can take the quotient and that will give you another chain complex. So this is what we have to verify here boundary of S0 n is contained in the S0 n minus 1. So here for the generators if I show it then the whole thing will follow because lambda, daba is also a linear map. So daba is also a linear map. daba of a plus b is daba a plus daba b. daba of n times a is n times daba a. So it is enough to verify this for generator. First we shall first check that the boundary of a degenerate simplex is inside here. Boundary of degenerate simplex may not be a degenerate simplex. It may be a chain or it may have this kind of elements also. You see that is the whole idea of putting them together. So let us start with a lambda from delta into k such that dimension of lambda n is less than n. Let us call it as m. If this m is actually less than n minus 1 that means there are at most n minus 2 elements out of n plus 1 element that means either 3 elements have gone to the same element or there must be pairs of 2 pairs of elements say u and u2 and u3, u4 like that which have gone to the same element. This is purely synthetic argument. Number of elements from n plus 1 if you have gone to n minus 1 then that is what must have happened. If 3 elements have gone to the same element when you take the boundary what happens you are deleting only 1 at a time and then putting minus sign and then doing that in the boundary formula. In each of them still there will be 2 different elements going to the same element. Therefore it is a sum of degenerate simplex its boundary of that one. So that is the whole idea here. So once since boundary of lambda is a sum of simplex each obtained by deleting only one vertex vector it follows that boundary of lambda is a sum of degenerate n minus 1 simplex. So this is happens when both the cases if you if you have 3 of them going to same element you delete one of them mainly we do you are deleting one at a time which one you do not know if it happens to be one of the 3 still the other 2 will be going to same element. So it is a degenerate. If 2 pairs of things are going to same elements then when you are deleting one of them from the one of the pair the other pair is still there other pair 2 of them are going to same element. Therefore in either case this should be sum of degenerate simplex the boundary. So that is done. The second case the second case is what m e could n minus 1 itself. That means that means only a pair of elements have been identified and all other things are injectively going different distinct elements. So let us say ith and jth vertices of the same of the say are mapped to the same vertices for some 0 less than i less than j less than less than equal to n we can assume. So in this case boundary of daba consists of a sum of a number of degenerate n minus 1 simplex is wherein when you are dropping some element it is neither ith element nor jth element. Okay so all that in i and jth both f them are going to they are present and they are going to the same element so it is a degenerate. Okay so there will be two terms namely when you drop ith element whatever you get is a non-degenerate simplex. Similarly when you delete jth element that will be also non-degenerate. So we need to check what happens in this case. So this is the interesting case. So what happens is all other terms have been degenerate we can forget about it. What are the two terms remaining here? This is in the expression of boundary of lambda. Okay minus 1 to the i vi hat minus 1 to the j vj hat all other things are there. Is that clear? But vi vi hat and vj hat the images are the same right. So this vj and vi hat is sorry vi it is vi and vj are images are same. So this v0 vj minus 1 is there. This vi is now whatever let us say vi is some u. It is in the j minus 1 place whereas here it is in the ith place. So you have to bring that to an element into the ith element and then see what happens. Pi by transforming by changing the order. When you change that order what you get is this element is equal to minus of that element with the appropriate sign okay. So when you do that performing i plus j minus 1 transpositions you can bring it to this place okay. Then the form if you write the first one as tau the second one will be minus signature of alpha times tau alpha where alpha is the transposition is the permutation required to make the second one look like the first one. So boundary of lambda is always in s0 of k minus 1. The next case is you start with a lambda minus signature of alpha times lambda alpha. What is the boundary of this? This should also belong to s k minus 1 no that also you have to show okay. By writing it as a sum of a number of writing alpha as a product of transpositions okay and then adding and starting adding and deleting several times what you can do is you can bring this to verifying lambda plus alpha lambda less to alpha where alpha is a permutation alpha is a transposition. When alpha is a transposition signature of alpha is minus so this becomes lambda plus lambda put a power off okay. So boundary of this how does it look like? If you if you show that these things are here then an element of this form is sum of such elements finitely sum of so that will be also there. So the case is reduced to the case when alpha is just a transposition okay. So we will leave this to you as an exercise. You have to just write the boundary of alpha in each of them case lambda could be just like here you have to do that this we will leave it to you as an exercise okay. Unless you toil that much you won't know what these symbols are what this what is the meaning of boundary and so on okay. So this part we will leave it as an exercise. Now we can define whatever we are promised here namely a further simplification of the chain complex. So what we do we take this subgroup S naught go module of that from SNK. Remember SNK was all singular simplicial maps okay. Simplicial chains to go module of those which are degenerate or which correspond to under permutation okay lambda minus signature of alpha times lambda power alpha okay go module of that this will become now C dot will become a chain group okay a chain complex SN of K to SN minus 1 of K the same daba induces a homomorphism of degree minus 1 such that daba square is 0 on CK CN of K to CN minus 1 of K. So this is chain complex this chain complex is called the simplicial chain complex of K. The homology groups of this group this chain complex are called simplicial homology groups. So this is the ultimate simplification that we solved. This CK as given is not a sub chain complex of the singular chain complex of mod K it is the quotient of a sub. So this is called a sub quotient it is a quotient of a sub chain complex okay. Nevertheless this is our life this is this brings a lot of simplification in the simplicial in the in the homology of a mod K of a of a simplicial complex. So let us take a closer look at it how things look like okay in the very special case namely K is just delta n itself if you do not understand that we we cannot do much so let us understand what is C of delta n okay. To begin with there is no confusion about the zero chains they should not be but that can be just now as pointed out they are linearly linear combinations of vertices of delta n okay therefore C naught of K is just that for n plus 1 because there are n plus 1 vertices in delta n okay. Whether degenerate or non-degenerate they are all same there is one one point map so one point to one point so it is all injective so there is nothing nothing special about that. Now let us look at the one chains we know that S1 of delta n is center S2 n plus 1 square okay because it is set of all maps from two elements set into n plus 1 element set so that is why it is like this but any simplest given by a non-injective map and two of them are going to same thing they have to be thrown away okay. Non-injective map is a degenerate simplex so it goes into S naught right similarly so at that level already we can only look at the simplexes which are given by injective mapping namely proper edges now u comma v where u and v are distinct that will be an edge because it is a simple shell map okay from delta 1 to k alright even now if u v and v u will be two different things we are counting so this is the next step we want to throw away to v u and u v differ by what by transposition and what is the sign of the transposition signature is minus 1 so u v plus v u will go into the subgroup which is same thing as saying that u v and is equal to minus v u which is how we are bringing the geometry into this namely an edge trace in the other way direction is stated as a minus of that edge minus of the simplex okay so that is the next step that you have to do it just means that now once you have u v you do not have to take v u okay so u v plus v u belong to the same thing this is the one simplex this is a story of S naught this just means that we need not count an edge in delta n we need to count as delta n only once in the other direction we do not count what is the meaning of that given the number of all pairs of u comma v they form a they will form a simplex they will form an edge inside delta n therefore it is the number of choices that we can make out of n plus one we have to make two of them two of them you have to choose so that is n plus one choose two so there are exactly n plus one choose two non degenerate simplex is up to order order means what only one of them we are taking ordered pairs sorry unordered pairs we are taking subsets once you have taken u v there is no chance to take v u so that forms a basis okay so this leaves us with n plus one choose two edges in delta n which will form a basis for 700 it follows that c 1 of delta n is addressed to n plus one choose two the boundary operator is obviously boundary of u v is v minus u okay remember that boundary operator comes here you drop out u you get v when you drop out v you have to put a sign here right so this is this much understanding is there from c 1 to c 0 okay likewise the same argument will give you the q chains delta n will be generated by n plus one choose q plus one such element namely q plus one subsets of n plus one set and it is generated over that freely so this is z power n plus one choose q plus now boundary of of this map c q to c q minus one is given by this formula same boundary map okay q q naught q n minus remember all these when I write q and q I am thinking them as now elements of c q which is a quotient space which is actually equivalence class okay if I change the order here I will have to put a minus sign if I change the order twice it does not matter it is the same element for example u naught u 1 u 2 is equal to minus u 1 u naught u 2 but it is equal to plus u 1 u 2 u naught so that is the meaning of this in this group that is the way we have to think of elements okay in any case this formula comes from directly from the formula for s s itself you have to operate you have to take this and operate the the phase operator on this side then this is how it is and put a minus one extra and take the summation just the entire chain looks like starting with one choose n plus one which is zero simplex says to choose an n plus one choose two one simplex and so on this will be there is only one n plus one there is only one n simplex namely you have to take the entire set okay if you if you change it by a permutation which is odd then it will be a negative generator if it is even permutation it is the same generator one single this is just a set and after that all of them are zero there are no chains left out here is that clear because you cannot choose n plus two subset out of n plus one set those are empty sets so they saw all zero chains zero groups here so this is the biggest simplification which is coming here which was not possible in the case of s s s double s of k also right it was going on infinitely this stops at the n plus one and nth stage itself so this is cn it is z okay then z power n plus one z power n plus one choose two and so on z power n okay if you put one more z here by augmentation this looks perfectly symmetric just like binomial expansion and for that beauty at least it is better to consider the augmented chain complexes and doing combinatorics that is much simpler object than this one so you better put that one more element in the augmentation by extending it by of epsilon with augmentation to z to zero after that okay given a simple shell complex okay cnk is purely an algebraic object there is no continuity no topology nothing you have to discuss okay once the simple shell complex is given which is an abstract algebraic way of combinatorial way of discriminating cnk is purely an algebraic object okay obviously the boundary maps themselves will depend on the incident relations which simplex is sitting where i mean what are the vertices of an edge what are the vertices of a triangle which triangle as which of the edges are there this is only incidence is incidence relation okay so that is the only thing that matters and that is like a combinatorial information that has the full control of topology is what what we we have seen already studying simple shell complex to some extent so it comes again here therefore one may expect that you know because just purely algebra here it may have very little to say topology however the following results come as a pleasant surprise so what is that surprise these are the two statements here the inclusion map from the full singular simple shell chain complex to the singular chain complex of mod k and model here they are this part is purely topological this is purely purely combinatorial this inclusion map is a chain homotopy equivalent exactly same way the quotient from this smaller thing to ckl this is a chain homotopy equivalent you already know that chain homotopy equivalence is induce isomorphisms on the homology groups combining these two what you get there is a homology of this topological space is isomorphism homology of this object is isomorphism homology of this object so that is the big result that we get so I am going to state that one for pairs of simple shell complexes there are canonical isomorphisms of singular homology h star of s dot mod k model the singular simple shell homology h star of s dot of k l and the simple shell homology h star of ckl so what are they first one is inclusion induced map okay if the h star of s we are we are not writing this s at all we are taking it for granted we are writing h star of k l okay but here I have to write it because this is something different from this one similarly phi star from s star of k to c of k l okay so that gives you i star from p star inverse so you take the inverse of this one and follow by that one from ckl to h star of k l so from here go here and come back here so that is a canonical isomorphism that is the statement we are going to postpone the proofs of these isomorphism theorem here the homology of c dot k l will be referred to as a simple shell homology of the simple shell pair when l is empty it is purely c dot of k okay simple shell homology of k of a simple shell complex okay as such it depends upon the actual simple shell complex that much is obvious but however once we have proved the above theorem it means that it will depend only on the topological space underlying topological space that means if I change the triangle take a simple shell complex take its underlying space I may be able to put another simple shell complex structure on that and then compute the homology there no problem the homology will be the same so that is the it does not depend upon what triangulation you take provided you have a triangulation then it is the same thing it will give you the homology of the underlying topological space okay therefore this freedom is very important this was a very very deep result active very strong result with the the singular chain complex I know was able to attain so the similarity I see here you know there are problems about Riemann integration to solve that you had to invent you know Lebesgue had to invent another integration theory namely Lebesgue it is so called Lebesgue integration theory come out of the Riemann integration then only you could prove the theorem of Riemann integration namely the characterization of of functions which are Riemann integrable in terms of the number of continuous number of points of discontinuities of the function it is similar to that to prove that it is invariant of the underlying you know topological space so you have to invent another homology just reframing the simple homology there was no way in fact some way which was thought by Poincare was wrong here he he had summed up and no way to complete that into a proof okay so we will stop here and continue this study next time