# Topology…bare facts…part 3

II. Algebraic Complexes

A. There is something artificial about considering a manifold as a polyhedron : the general idea of the manifold as a homogeneous structure of n-fold extent, an idea which goes back to Riemann, has nothing intrinsically to do with the simplicial decompositions which were used to introduce polyhedra. Poincare, who undertook the first systematic topological study of manifolds, and thus changed topology from a collection of mathematical curiosities into an independent and significant branch of geometry, originally defined manifolds analytically with the aid of systems of equations. However, within only four years after the appearance of the first pioneering work he took the point of view which today is known as combinatorial topology, and essentially amounts to the consideration of manifolds as polyhedra. The advantage of this viewpoint lies in the fact that with its help the difficult — partially purely geometric, partially set-theoretic — considerations to which the study of manifolds leads are replaced by the investigation of a finite combinatorial model — namely the system of simplexes of a simplicial decomposition of the polyhedron (that is, the geometrical complex) — which opens the way to the application of algebraic methods.

Thus, it turns out that the definition of a manifold which we use here is currently the most convenient, although it represents nothing more than a deliberate compromise between the set-theoretic concept of topological space and the methods of combinatorial topology, a compromise which, at present, can scarcely be called an organic blending of these two directions. The most important of the difficult problems connected with the concept of manifold are by no means solved by the definition which we have adopted.

B. We shall now turn to briefly discuss the algebraic methods of the topology of manifolds (and general polyhedra). The basic concepts in algebraic topology are those of oriented simplex, algebraic complex and boundary of an algebraic complex.

An oriented one-dimensional simplex is a directed line segment $(a_{0}a_{1})$, that is, a line which is traversed from the vertex $a_{0}$ to the vertex $a_{1}$. One can also say: an oriented one-dimensional simplex is one with a particular ordering of its endpoints. If we denote the oriented line $(a_{0}a_{1})$ by $x^{1}$ (where the superscript 1 gives the dimension), the oppositely oriented simplex $(a_{1}a_{0})$ will be denoted by $-x^{1}$. This same line considered without orientation we denote by

$|x^{1}|= |a_{0}a_{1}| = |a_{1}a_{0}|$

An oriented two dimensional simplex — an oriented triangle — is a triangle with a particular sense of rotation or with a particular ordering of its vertices; at the same time, no distinction is made between orderings which differ from another by an even permutation, so that $(a_{0}a_{1}a_{2})$, $(a_{1}a_{2}a_{0})$ and $(a_{2}a_{0}a_{1})$ represent one permutation, and $(a_{0}a_{2}a_{1}), (a_{1}a_{0}a_{2}), (a_{2}a_{1}a_{0})$, the other is called $-x^{2}$. The triangle considered without orientation will again be denoted by $|x^{2}|$. The essential thing here is that in an oriented triangle the boundary is also to be understood as an oriented (directed) polygon. The boundary of an oriented triangle $(a_{0}a_{1}a_{2})$ is the collection of oriented lines $(a_{0}a_{1})$, $(a_{1}a_{2})$ and $(a_{2}a_{0})$. If one denotes the boundary of $x^{2}$ by $\dot{x}^{2}$, then our last statement is expressed by the formula

$(1^{'})$ $\dot{x}^{2} = (a_{0}a_{1}) + (a_{1}a_{2}) + (a_{2}a_{0})$

or equivalently

$(1)$ $\dot{x}^{2} = (a_{1}a_{2}) - (a_{0}a_{2}) + (a_{0}a_{1})$

We may also say that in the boundary of $x^{2}$, the sides $(a_{1}a_{2})$ and $(a_{0}a_{1})$ appear with the coefficient +1, and the side $(a_{0}a_{2})$ with the coefficient -1.

C. Consider now any decomposition into triangles (or, triangulation) of a two-dimensional polyhedron $P^{2}$. The system comprised of the triangles together with their edges and vertices forms what we called earlier a two-dimensional geometrical complex $K^{2}$. Now, we choose a particular (but completely arbitrary) orientation $x_{i}^{2}$ of any one of the triangles $|x_{i}^{2}|$, $1 \leq i \leq \alpha_{2}$, of our complex; in a similar way, we can choose any (Note: where $\alpha_{0}, \alpha_{1}, \alpha_{2}$) denote-, the number of two-, one-, or zero-dimensional elements of the geometrical complex.) (Note: if one imagines $x^{2} = (a_{0}a_{1}a_{2})$ as a symbolic product of three “variables”, $a_{0}, a_{1}, a_{2}$, one may write $\dot{x}^{2} = \sum_{i=0}^{2} (-1)^{i} \frac{\partial {x^{2}}} {\partial {a_{i}}}$) particular orientation $x_{j}^{1}$ of one of the sides $|x_{j}^{1}|$, $1 \leq j \leq \alpha_{1}$. The system of all $x_{i}^{2}$ we call an oriented two-dimensional complex $C^{2}$, that is, an orientation of the geometrical complex $K^{2}$. For the oriented complex $C^{2}$ we use the notation

$C^{2} = \sum_{i=1}^{\alpha_{2}} x_{i}^{2}$

In order to indicate that $C^{2}$ is the result of orienting the complex $K^{2}$, we shall sometimes write $|C^{2}|=K^{2}$.

The boundary of each oriented triangle $x_{i}^{2}$ can now be represented by a linear form

(2) $\dot{x}_{i}^{2} = \sum_{j=1}^{\alpha_{i}}t_{i}^{j}x_{j}^{1}$

where $t^{j}=+1, -1, 0$ according to whether the oriented line $x_{j}^{1}$ occurs in the boundary of the oriented triangle $x_{i}^{2}$ with the coefficient +1, -1 or not at all.

If one sums equation (2) over all i, $1 \leq i \leq \alpha_{2}$, one obtains

$\sum_{i=1}^{\alpha_{2}} \dot{x}_{i}^{2} = \sum_{i=1}^{\alpha_{2}} \sum_{j=1}^{\alpha_{1}}t_{i}^{j}x_{j}^{1} = \sum_{j=1}^{\alpha_{1}} u^{j}x_{j}^{1}$, where $u^{j} = \sum_{i=1}^{\alpha_{2}}t_{i}^{j}$

The above expression

$\sum_{j=1}^{\alpha_{1}} u^{j}x_{j}^{1}$

is called the boundary of the oriented complex $C^{2}$ and is denoted by $\dot{C}^{2}$.

Examples.

Eg.1. Let K be the system composed of the four triangular faces of a tetrahedron; let the orientation of each of the faces be as indicated by the direction of arrows. (I am not able to produce the diagram in LaTeX here, you will have to do reverse engineering sort of thing and produce the diagram from the system of equations given):

The boundary of the resulting oriented complex is:

$C^{2}=x_{1}^{2} + x_{2}^{2} + x_{3}^{2}+x_{4}^{2}$

equals zero, because each edge of the tetrahedron appears in the two triangles of which it is a side with different signs. In formulae:

$x_{2}^{2} = (a_{0}a_{1}a_{2})$

$x_{2}^{2} = (a_{1}a_{0}a_{3})$

$x_{3}^{2} = (a_{1}a_{3}a_{2})$

$x_{4}^{2} = (a_{0}a_{2}a_{3})$

and

$x_{1}^{1} = (a_{0}a_{1})$

$x_{2}^{1} = (a_{0}a_{2})$

$x_{3}^{1} = (a_{0}a_{3})$

$x_{4}^{1} = (a_{1}a_{2})$

$x_{5}^{1} = (a_{1}a_{3})$

$x_{6}^{1} = (a_{2}a_{3})$

(PS: from the above, you will be able to reconstruct the diagram of the planar tetrahedron)

Thus,

$\dot{x}_{1}^{2} = + x_{1}^{1} - x_{2}^{1} + x_{4}^{1}$

$\dot{x}_{2}^{2}= - x_{1}^{1} + x_{3}^{1} - x_{5}^{1}$

$\dot{x}_{3}^{2} = - x_{4}^{1} + x_{5}^{1} - x_{4}^{1}$

$\dot{x}_{4}^{2} = + x_{2}^{1} - x_{3}^{1} + x_{6}^{1}$

Adding all the above, we get $\dot{C}^{2} = \sum_{i=1}^{4}\dot{x}_{i}^{2}=0$.

Eg. 2:

If one orients the ten triangles of the triangulation of the projective plane shown (again I am not able to produce in LaTeX) as indicated by the arrows, and puts:

$C^{2} = \sum_{i=1}^{10} x_{i}^{2}$

then

$\dot{C}^{2} = 2x_{1}^{1}+2x_{2}^{1}+2x_{3}^{1}$….equation (3)

The boundary of the oriented complex consists, therefore, of the projective line AX (composed of the three segments $x_{1}^{1}, x_{2}^{1}, x_{3}^{1}$) counted twice. With another choice of orientations $x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, \ldots, x_{10}^{2}$ of the ten triangles of this triangulation one would obtain another oriented complex and its boundary would be different from equation (3). Hence, it is meaningless to speak of the “boundary of the projective plane”; one must speak only of the boundaries of the various oriented complexes arising from different triangulations of the projective plane.

One can easily prove that no matter how one orients the ten triangles of this figure, the boundary of the resulting complex

$C^{2} = \sum_{i=1}^{10}x_{i}^{2}$

is never zero. In fact, the following general result (which can be taken as the definition of orientability of a closed surface) holds:

A closed surface is orientable if and only if one can orient the triangles of any of its triangulations in such a way that the oriented complex thus arising has boundary zero.

Eg. 3.

In the triangulation and orientation for the Mobius band given in the figure (again not shown but can be reverse engineered) we have:

$\dot{C}^{2} = 2x_{1}^{1}+x_{2}^{1} + x_{3}^{1} + x_{4}^{1} + x_{5}^{1}$

C. Oriented complexes and their boundaries serve also as examples of so-called algebraic complexes. A two-dimensional oriented complex, that is, a system of oriented simplexes taken from a simplicial decomposition of a polyhedron, was written by us as a linear form, $x_{i}^{2}$; furthermore, as the boundary of the oriented complex $C^{2}=\sum x_{i}^{2}$, there appeared a linear form $\sum u^{i}x_{j}^{1}$ whose coefficients are, in general, taken as arbitrary integers. Such linear forms are called algebraic complexes. The same considerations hold in the n-dimensional case, if we make the general definition:

Definition I.

An oriented r-dimensional simplex $x^{r}$ is an r-dimensional simplex with an arbitrarily chosen ordering of its vertices,

$x^{r} = (a_{0}a_{1}\ldots a_{r})$

where orderings which arise from one another by even permutations of the vertices determine the same orientation (the same oriented simplex), so that each simplex $|x^{r}|$ possesses two orientations, $x^{r}$ and $-x^{r}$.

Note: A zero dimensional simplex has only one orientation, and thus it makes no sense to distinguish between $x^{0}$ and $|x^{0}|$.

Remark:

Let $x^{r}$ be an oriented simplex. Through the r+1 vertices of $x^{r}$ passes a unique r-dimensional hyperplane $R^{r}$ (the $R^{r}$ in which $x^{r}$ lies ), and to each r-dimensional simplex $|y^{r}|$ of $R^{r}$ there exists a unique orientation $y^{r}$ such that one can map $R^{r}$ onto itself by an affine mapping with a positive determinant in such a way that under this mapping the oriented simplex $x^{r}$ goes over into the simplex $y^{r}$. Thus, the orientation $x^{r}$ of $|x^{r}|$ induces a completely determined orientation $y^{r}$ for each simplex $|y^{r}|$ which lies in the hyperplane $R^{r}$ containing $x^{r}$. Under these circumstances, one says that the simplexes $x^{r}$ and $y^{r}$ are equivalently — or consistently — oriented simplexes of $R^{r}$. One says also that the whole coordinate space $R^{r}$ is oriented by $x^{r}$, which means precisely that from the oriented simplex $x^{r}$ all r-dimensional simplexes of $R^{r}$ acquire a fixed orientation. In particular, one can orient each r-dimensional simplex lying in $R^{r}$ so that it has an orientation equivalent to that of $x^{r}$.

Definition II. A linear form with integral coefficients $C^{r} = \sum t^{i} x_{i}^{r}$ whose indeterminants $x_{i}$ are oriented r-dimensional simplexes, is called an r-dimensional algebraic complex.

(the above definition also has meaning in the case $r=0$. A zero dimensional algebraic complex is a finite system of points with which some particular (positive, negative or vanishing) integers are associated as coefficients ; in modern terminology, $C^{r}$ would be called an (integral) r-dimensional chain)).

Expressed otherwise: an algebraic complex is a system of oriented simplexes, each of which is to be counted with a certain multiplicity (that is, each one is provided with an integral coefficient). Here it will generally be assumed only that these simplexes lie in one and the same coordinate definite simplicial decomposition of a polyhedron (that is, a geometrical complex). On the contrary, the simplexes of an algebraic complex may, in general, intersect one another arbitrarily. In case, the simplexes of an algebraic complex $C^{r}$ belong to a geometrical complex (that is, re obtained by orienting certain elements of a simplicial decomposition of a polyhedron ), $C^{r}$ is called an alagebraic subcomplex of the geometrical complex (of the given simplicial decomposition) in question; here, self-intersections of simplexes, of course, cannot occur. This case is to be considered as the most important.

D. Algebraic complexes are to be considered as a higher dimensional generalization of ordinary directed polygonal paths; here, however, the concept of polygonal path is taken from the outset in the most general sense: the individual lines may intersect themselves, and there may also exist lines which are traversed many times; moreover, one should not forget that the whole thing is to be considered algebraically, and a line which is traversed twice in opposite directions no longer counts at all. Furthermore, polygonal paths may consist of several pieces (thus, no requirement of connectedness). Thus, the two figures 8 and 9 represent polygonal paths which, considered as algebraic complexes, have the same structure. (that is, represent the same linear form).

Since the r-dimensional algebraic complexes of $R^{n}$ may as linear forms, be added and subtracted according to the usual rules of calculating with such symbols, they form an Abelian group $L_{r}(R^{n})$. One can also consider, instead of the whole of $R^{n}$, a subspace G of $R^{n}$, for example, the r-dimensional algebraic complexes lying in it then form the group $L_{r}(G)$ — a subgroup of $L_{r}(R^{n})$.

Also, the r-dimensional algebraic subcomplexes of a geometric complex K form a group — the group $L_{r}(K)$; it is the starting point for almost all further considerations. Before we continue with these considerations however, I would like to direct the attention of the reader to the fact that the concepts “polyhedron,” “geometrical complex,” and “algebraic complex” belong to entirely different logical categories: a polyhedron is a point set, thus a set whose elements are ordinary points of $R^{n}$ ; a geometrical complex is a finite set whose elements are simplexes, and, indeed, simplexes in the naive geometrical sense, that is, without orientation. An algebraic complex is not a set at all; it would be false to say that an algebraic complex is a set of oriented simplexes, since the essential thing about an algebraic complex is that the simplexes which appear in it are provided with coefficients and therefore, in general, are to be counted with a certain multiplicity. This distinction between the three concepts, which often appear side by side, reflects the essential difference between the set theoretic and the algebraic approaches to topology.

E. The boundary $C^{r}$ of the algebraic complex $C^{r} = \sum t^{i} x_{i}^{r}$ is defined to be the algebraic sum of the boundaries of the oriented simplexes $\sum t^{i} x_{i}^{r}$ where the boundary of the oriented simplex $x^{r} = (a_{0}a_{1}\ldots a_{r})$ is the $(r-1)$-dimensional algebraic complex

(Equation 4) $\dot{x}^{r} = \sigma_{i=0}^{r}(-1)^{i}(a_{0}\ldots \hat{a}_{i}\ldots a_{r} )$

where $\hat{a}_{i}$ means that the vertex $a_{i}$ is to be omitted. In case the boundary of $C^{r}$ is zero, then $C^{r}$ is called a cycle. Thus, in the group $L_{r}(R^{n})$, and analogously in $L_{r}(K)$ and $L_{r}(G)$, the subgroup of all r-dimensional cycles $Z_{r}(R^{n})$, or, respectively, $Z_{r}(K)$ and $Z_{r}(G)$, is defined.

We can now say : a closed surface is orientable if and only if one can arrange, by a suitably chosen orientation of any simplicial (that is, in this case, triangular) decomposition of the surface, that the oriented complex given by this orientation is a cycle. Without change, this definition holds for the case of a closed manifold of arbitrary dimension. Let us remark immediately: orientability, which we have just defined as a property of a definite simplicial decomposition of a manifold, actually expresses a property of the manifold itself, since it can be shown that if one simplicial decomposition of a manifold satisfies the condition of orientability, the same holds true for every simplicial decomposition of this manifold.

Remark: If $x^{n}$ and $y^{n}$ are two equivalently oriented simplexes of $R^{n}$ which have the common face $|x^{n-1}|$, then the face $x^{n-1}$ (with some orientation) appears in $x^{n}$ and $y^{n}$ with the same or different signs according to whether the simplexes $|x^{n}|$ and $|y^{n}|$ lie on the same side or or different sides of the hyperplane $R^{n-1}$ containing $|x^{n-1}|$. The proof of this assertion is left to the reader as an exercise.

F. As is easily verified, the boundary of a simplex is a cycle. But from this it follows that the boundary of an arbitrary algebraic complex is also a cycle. On the other hand, it is easy to show that for each cycle $Z^{r}$, with $r>0$, in $R^{n}$ there is an algebraic complex lying in this $R^{n}$ which is bounded by $Z^{r}$ (Note: on the other hand, a zero-dimensional cycle in $R^{n}$ bounds if and only if the sum of its coefficients equals zero (the proof is by induction on the number of sides of the bounded polygon)) ; indeed, it suffices to choose a point O of the space different from all the vertices of the cycle $Z^{r}$ and to consider the pyramid erected above the given cycle (with the apex at O). In other words, if

$Z^{r} =\sum_{(i)}c^{i}x_{i}^{r}$ and $x_{i}^{r} = (a_{0}^{i}, a_{1}^{i}, \ldots, a_{r}^{i})$ then one defines the $(r+1)-$ dimensional oriented simplex $x_{i}^{r+1}$ as

$x_{i}^{r+1} = (O, a_{0}^{1}, \ldots, a_{r}^{i})$

and considers the algebraic complex

$C^{r+1} = \sum_{(i)}c^{i}x_{i}^{r+1}$.

The boundary of $C^{r+1}$ is $Z^{r}$, since everything else cancels out.

If we consider, however, instead of the whole of $R^{n}$, some region G in $R^{n}$ (or more generally an arbitrary open set in $R^{n}$), then the situation is no longer so simple: a cycle of $R^{n}$ lying in G need not bound in G. Indeed, if the region G is a plane annulus, then it is easy to convince oneself that there are cycles which do not bound in G (in this case closed polygons which encircle the center hole). Similarly, in a geometrical complex, there are generally some cycles which do not bound in the complex.

Consequently, one distinguishes the subgroups $B_{r}(G)$ of $Z_{r}(G)$, and $B_{r}(K)$ of $Z_{r}(K)$ of bounding cycles: the elements of $B_{r}(G)$ or $B_{r}(K)$ are cycles which bound some $(r+1)-$ dimensional algebraic complex in G, or respectively K.

In the example of the triangulation given (of the projective plane; again figure not shown here; but can be reconstructed or reverse engineered) we see that it can happen that a cycle z does not bound in K, while a certain fixed integral multiple of it (that is, a cycle of the form tx where t is an integer different from zero) does bound some algebraic subcomplex of K. We have, in fact, seen that the cycle $2x_{1}^{1}+2x_{2}^{1}+2x_{3}^{1}=2z^{1}$ (the doubly counted projections line) in the triangulation of the bounds of the figure, while in the same triangulation there is no algebraic complex which has the cycle $z^{1} = (x_{1}+x_{2}+x_{3})$ as its boundary. It is thus suitable to designate as boundary divisors all of those cycles $z^{r}$ of K (of G) for which there exists a non-zero integer t such that tz bounds in K (in G). Since t may have the value 1, the true boundaries (that is, bounding cycles) are included among the boundary divisors. The boundary divisors form, as is easily seen, a subgroup of the group $Z_{r}(K)$, which we denote by $B_{r}^{'}(K)$ or $B_{r}^{'}(K)$; (obviously), the group $B_{r(K)}$ is contained in the group $B_{r}^{'}(K)$.

G. If $z^{r}$ bounds in K (in G) we also say that $z^{r}$ is strongly homologous to zero in K (in G), and we write $z^{r} \sim 0$ (in K or in G); if $x^{r}$ is a boundary divisor of K (of G), we say that z is weakly homologous to zero and write $z \approx 0$ (in K or in G).

If two cycles of a geometrical complex K (or of region G) have the property that the cycle $z_{1}^{r} - z_{2}^{r}$ is homologous to zero, one says that the cycles $z_{1}^{r}$ and $z_{2}^{r}$ are homologous to one another; the definition is valid for strong as well as for weak homology, so that one has the relations $z_{1}^{r} \sim z_{z}^{r}$ and $z_{1}^{r} \approx z_{2}^{r}$. Examples of these relations are given in Fig 12 ($z_{1} \sim z_{2}$) and in the following figures.

In the Figures 15 and 16, G is the region of three-dimensional space which is complementary to the closed Jordan curve S or, respectively, to the lemniscate A.

H. Thus, the group $Z_{r}(K)$ falls into an so-called homology classes, that is, into classes of cycles which are homologous to one another; there are in general both weak and strong homology classes, according to whether the concept of homology is meant to be weak or strong. If one again takes for K the geometrical complex of Figure 6, then there are two strong homology classes of dimension one, for every one-dimensional cycle of K is either homologous to zero (that is, belongs to the zero-class) or homologous to the projective line(that is, say, the circle $x_{1}+x_{2}+x_{3}$). Since every one-dimensional cycle of K in our case is a boundary divisor, there is only one weak homology class — the zero class.

As for the one-dimensional homology classes of the complexes in Figures 12 and 13, they may be completely enumerated if one notices that in Figure 12 every one-dimensional cycle satisfies a homology of the form $z \sim tz_{1}$, and, in Figure 13, a homology of the form $z \sim uz_{1} + vz_{2}$, where t, u, and v are integers; furthermore, the strong homology classes coincide with the weak in both complexes (for there are no boundary divisors which are not at the same time boundaries).

If $\zeta_{1}$ and $\zeta_{2}$ are two homology classes and $z_{1}$ and $z_{2}$ are arbitrarily chosen cycles in $\zeta_{1}$ and $\zeta_{2}$ respectively, then one denotes by $\zeta_{1}+\zeta_{2}$ the homology class to which $z_{1}+z_{2}$ belongs. This definition for the sum of two homology classes is valid because, as one may easily convince oneself, the homology class designated by $\zeta_{1}+\zeta_{2}$ does not depend on the particular choice of the cycles $z_{1}$ and $z_{2}$ in $\zeta_{1}$ and $\zeta_{2}$.

The r-dimensional topology classes of K therefore form a group — the so-called factor group of $Z_{r}(K)$ modulo $B_{r}(K)$, or modulo $B_{r}^{'}(K)$; it is called the r-dimensional Betti group of K. Moreover, one differentiates between the full and the free (or reduced) Betti groups —- the first corresponds to the strong homology concept (it is, therefore, the factor group $Z_{r}(K)$ modulo $B_{r}(K)$, denoted $H_{r}(K)$), while the second is the group of the weak homology classes (the factor group $Z_{r}(K) modulo$latex B_{r}^{‘}(K), denoted $F_{r}(K)$. (Note : In fact, one may write $H_{r}(K)=F_{r}(K) \bigoplus T_{r}(K)$ where $T_{r}(K)$ is the subgroup of $H_{r}(K)$ consisting of the elements of finite order; the so-called torsion subgroup of $H_{r}(K)$. In current usage, the group $H_{r}(K)$ is more often referred to as the r-dimensional homology group rather than the r-dimensional Betti group).

From the above discussion, it follows that the full one-dimensional Betti group of the triangulation of the projective plane given in Figure 6 is a finite group of order two; on the other hand, the free (one-dimensional) Betti group of the same complex is the zero group. The one dimensional Betti group of the complex K is the infinite cyclic group, while in Figure 13 the group of all linear forms $u\zeta_{1}+v\zeta_{2}$ (with integral u and v) is the one-dimensional Betti group. In the latter two cases, the full and reduced Betti groups coincide.

From simple group-theoretic theorems it follows that the full and the reduced Betti groups (of any given dimension r) have the same rank, that is, the maximal number of linearly independent elements which can be chosen from each group is the same. This common rank is called the r-dimensional Betti number (you can easily prove that the zero-dimensional Betti number of a complex K equals the number of its components (that is, the number of disjoint pieces of which the corresponding polyhedron is composed))of the complex K. The one dimensional Betti number for the projective plane is zero; for Figures 12 and 13 it is, respectively, 1 and 2.

I. The same definitions are valid for arbitrary regions G contained in $R^{n}$. It is especially important to remember that, while in the case of a geometrical complex all of the groups considered had a finite number of generators, this is by no means necessarily the case for regions of $R^{n}$. Indeed, the region complementary to that consisting of infinitely many circles converging to a point (Figure 18) has, as one may easily see, an infinite one-dimensional Betti number (consequently, the one-dimensional Betti group does not have finite rank, therefore, certainly not a finite number of generators).

J. The presentation of the basic concepts of the so-called algebraic topology which have given is based on the concept of the oriented simplex. In many questions, however, one does not need to consider the orientation of the simplex at all — and can still use the algebraic methods extensively. In such cases, moreover, all considerations are much simpler, because the problems of sign (which often leads to rather tedious calculations) disappears. The elimination of orientation throughout, wherever it is actually possible, leads to the so-called “modulo-2” theory in which all coefficients of the linear forms that we have previously considered are replaced by their residue classes modulo 2. Thus, one puts the digit 0 in place of any even number, the digit 1 in place of any odd number, and calculates with them in the following way:

$0+0=0$,

$0+1=1+0=1$

$0-1=1-0=1$

$0-0=1-1=0$

$1+1=0$

In particular, an algebraic complex mod 2 is a linear form whose indeterminants are simplexes considered without orientation and with coefficients 0 and 1. (Note: One can consider geometrical complexes as a special case of the algebraic complexes modulo 2, if one interprets the coefficient 1 as signifying the occurence, and the coefficient 0 as signifying the non-occurence, of a simplex in a complex. This remark allows us to apply to geometrical complexes theorems which are proved for algebraic complexes).

The boundary of a simplex $x^{n}$ appears in the theory mod 2 as a complex mod 2 which consists of all $(n-1)-$ dimensional faces of the simplex $x^{n}$. Hence, the boundary mod 2 of an arbitrary complex $C^{n}$ is defined as the sum (always mod 2) of the boundaries of the individual $n-$ dimensional simplexes of $C^{n}$. One can also say that the boundary mod 2 of $C^{n}$ consists of those and only those $(n-1)-$ dimensional simplexes of $C^{n}$ with which an odd number of n-dimensional simplexes are incident. You may easily construct examples which illustrate what has been said.

The concepts of cycle, homology, and Betti group mod 2 can be introduced exactly as in the “oriented” case. It should be especially noticed that all of our groups $L_{r}(K), Z_{r}(K), H_{r}(K)$ and so on, are now finite groups (which we shall denote by $L_{r}(K;Z_{2}), Z_{r}(K;Z_{2}), H_{r}(K; Z_{2})$ etc. where $Z_{2}$ is the group of residue classes modulo 2), because we now have, throughout, linear forms in finitely many indeterminants whose coefficients take only the two values 0 and 1. The triangulation of the projective plane given in Figure 6 can serve as an example of a two-dimensional cycle mod 2, for — considered as an algebraic complex mod 2 — it obviously has vanishing boundary. In the case of an n-dimensional complex $K^{n}$ (that is, a complex consisting of simplexes of dimension less than or equal to n), just as $Z_{n}(K^{n})$ is isomorphic to $H_{n}(K^{n}; Z_{2})$; therefore, the two-dimensional Betti group modulo 2 for the projective plane is different from zero (its order is 2); the one-dimensional Betti group modulo 2 in the case of the projective plane is also of order 2.

Finally, one can also introduce the concept of the r-dimensional Betti number modulo 2; this is the rank mod 2 of the group $H_{r}(K; Z_{2})$, that is, the greatest number of elements $u_{1}, u_{2}, \ldots, u_{s}$ of this group such that a relation of the form

$t_{1}u_{1}+t_{2}u_{2}+ \ldots + t_{s}u_{s}=0$

is satisfied only if all $t_{i}$ vanish (where the $t_{i}$ take only the values 0 and 1).

The zero-, one-, and two-dimensional Betti numbers of the projective plane modulo 2 all have the same value 1.

(The theory modulo 2 is due to Veblen and Alexander; it plays a very important role in modern topology, and has also prepared the way for the most general formulation of the concept of “algebraic complex”: If J is any commutative ring with identity, we mean by an algebraic complex of coefficient domain J is linear form whose indeterminants are oriented simplexes whose coefficients are elements of the ring J. Then, one defines boundaries, cycles, homology, etc. exactly as before but with respect to the ring J; in particular, the coefficient 1 or -1 is now to be interpreted as an element of the ring (which, indeed, according to the hypothesis contains an identity). If J is the ring of residue classes modulo m we speak of algebraic complexes modulo m. These complexes are gaining more and more significance in topology. Of greater importance as a coefficient domain is the set R of rational numbers; in particular, the cycles which we have called boundary divisors are nothing else but the cycles with integer coefficients which bound in K with respect to R (but not necessarily with respect to the ring of integers).

K. We close our algebraic-combinatorial considerations with the concept of subdivision. If one decomposes each simplex of a geometrical complex K into (“smaller”) simplexes such that the totality of all simplexes thus obtained again forms a geometrical complex $K_{1}$, then $K_{1}$ is called a subdivision of K. If K consists of a single simplex, then the elements of the subdivision which lie on the boundary of the simplex form a subdivision of the boundary. From this it follows that if $K^{n}$ is a geometrical complex, $K_{1}^{n}$ its subdivision, and $K^{r}$ the complex consisting of all the r-dimensional elements of $K^{n}$ (together with all of their faces) and $r \leq n$, the totality of those elements of $K_{1}^{n}$ which lie on simplexes of $K^{r}$ forms a subdivision of $K^{r}$.

One can speak of subdivisions of algebraic complexes; we shall do this for the most important special case, in which the algebraic complex $C^{n} = \sum t^{i}x_{i}^{n}$ is an algebraic subcomplex of a geometrical complex. Then, it is also true that the totality of all simplexes (considered without coefficients or orientation) of $C^{n}$ forms a geometrical complex $K^{n}$. Let $K_{1}^{n}$ be a subdivision of $K^{n}$, and $|y^{n}|$ some simplex of $C^{n}$; then $|y^{n}|$ lies on some particular simplex $|x_{i}^{n}|$ of $C^{n}$. We now orient the simplex $|y^{n}|$ the same as $x_{i}^{n}$ amd give it the coefficient $t^{i}$. In this way. we obtain an algebraic complex which is called a subdivision of the algebraic complex $C^{n}$. One can easily see that the boundary of the subdivision $C_{1}^{n}$ of $C^{n}$ is a subdivision of the boundary of $C^{n}$. (Considered modulo 2, the process gives nothing beyond the subdivisions of a geometrical complex).

Cheers,

Nalin Pithwa.

I have an advice worth two cents. If you have read the previous blog and this blog, you might like to build up more your grasp of topology with the following two articles from Quanta Magazine:

1.

https://www.quantamagazine.org/topology-101-how-mathematicians-study-holes-20210126/

2.

https://www.quantamagazine.org/how-mathematicians-use-homology-to-make-sense-of-topology-20210511/

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