Topology bare facts: part 5

continued from previous blog (same reference text):

From the fact that we formulate the concept of a complex abstractly, its range of application is substantially enlarged. As long as one adheres to the elementary geometrical conception of a complex as a simplicial decomposition of a polyhedron, one cannot free oneself from the impression that there is something arbitrary which is connected with the choice of this concept as the basic concept of topology; why should this particular notion, simplicial decomposition of polyhedra, constitute the central point of all topology? The abstract conception of a complex as a finite scheme which is, a priori, suitable for describing different processes (for example, the structure of a finite system of sets) helps to overcome this skepticism. Here, precisely those abstract complexes which are defined as nerves of finite systems of sets play a decisive role: that is, it can be shown that the topoological investigation of an arbitrary closed set — therefore, the most general geometrical figure conceivable — can be completely reduced to the investigation of a sequence of complexes

(Equation 1) $K_{1}^{n}, K_{2}^{n}, \ldots, K_{h}^{n}, \ldots$

(n is the dimension of the set) related to one another by certain simplicial mappings. Expressed more exactly: for every closed set one can construct a sequence of complexes (10 and of simplicial mappings $f_{h}$ of $K_{h+1}$ into $K_{h}$ where ($h=1,2, \ldots$) , (which also satisfies certain secondary conditions which, for the moment, need not be considered). Such a sequence of complexes and simplicial mappings is called a projection spectrum. Conversely, every projection spectrum defines in a certain way, which we cannot describe here, a uniquely determined class of mutually homeomorphic closed sets; moreover, one can formulate exact necessary and sufficient conditions under which two different projections spectra define homeomorphic sets. In other words: the totality of all projection spectra falls into classes whose definition requires only the concepts “complex” and “simplicial mapping”, and which correspond in a one-to-one way to the classes of mutually homeomorphic closed sets. It turns out that the elements of a projection spectrum are none other than the nerves of increasingly finer coverings of the given closed sets. These nerves can be considered as approximating complexes for the closed set. (Note: until further notice, we are dealing only with geometrical complexes, that is, simplicial decompositions of (perhaps curved) polyhedra of a coordinate space).

A.

We now go over to a brief survey of the proof of the invariance of the Betti numbers of a complex promised earlier. Since we are only going to emphasize the principal idea of this proof, we shall forgo a proof of the fact that a geometrical complex has the same Betti numbers as any one of its subdivisions. We begin the proof with the following fundamental lemma:

Lebesgue’s lemma. For every covering

(Equation A1) $S=(F_{1}, F_{2}, \ldots, F_{s})$

of the closed set F., there is a number $\sigma = \sigma(S)$ — the Lebesgue number of the covering of S — with the following property: if there is a point a whose distance from certain numbers of the covering S — say $F_{i_{1}}, F_{i_{2}}, \ldots, F_{i_{k}}$ —- is less than $\sigma$, then the sets $F_{i_{1}}, F_{i_{2}}, \ldots, F_{i_{k}}$ have a non-empty intersection.

Proof:

Let us suppose that the assertion is false.

Then, there is a sequence of points

(Equation A2) $a_{1}, a_{2}, \ldots, a_{m}, \ldots$

and of sub-systems

(Equation A3) $S_{1}, S_{2}, \ldots, S_{m}, \ldots$

of the system of sets S such that $a_{m}$ has a disitance less than $\frac{1}{m}$ from all sets of the system $S_{m}$ while the intersection of the sets of the system $S_{m}$ is empty. Since there are only finitely many different sub-systems of the finite systems of sets S, there are, in particular among the $S_{i}$, only finitely many different systems of sets, so that at least one of them — say $S_{1}$— appears in the sequence $(A3)$ infinitely often. Consequently, after replacing (A2) by a subsequence if necessary, we have the following situation:

there is a fixed subsystem

$S_{1} = (F_{i_{1}}, F_{i_{2}}, \ldots, F_{i_{k}})$

of S and a convergent sequence of points

(Equation A4) $a_{1}, a_{2}, \ldots, a_{m}, \ldots$

with the property that the sets $F_{i_{k}}$ where $k=1,2, \ldots, k$ have an empty intersection, while, on the other hand, the distance from $a_{m}$ to $F_{i_{k}}$ is less than 1/m; however, this is impossible, because, under these circumstances the limit point $a_{m}$ of the convergent sequence (A4) must belong to all sets of the system, QED.

B.

For the second lemma, we make the following simple observation: Let P be a polyhedron, K a simplicial decomposition of P, and $K_{1}$ a subdivision of K. If we let each vertex b of $K_{1}$ correspond to the center of a barycentric star containing b, then (by a remark made in earlier blog article) the vertex b is mapped onto a vertex of the simplex of K containing b, so that, this procedure gives rise to a simplicial map f of $K_{1}$ into K. The mapping f — to which we give the same canonical displacement of $K_{1}$ with respect to K — satisfies the condition of the third conservation theorem, and hence, gives as the image of the complex $K_{1}$ the whole complex, K. (Note: The analogous assertion also holds with respect to every algebraic subcomplex of $K_{1}$ (or K); if C is an algebraic subcomplex of K, and $C_{1}$ a subdivision of C induced by $K_{1}$, then the condition of the third conservation theorem, are again fulfilled and we have $f(C_{1})=C$)

The same conclusion still holds if, instead f f, we consider the following modified canonical displacement $f^{'}$: first, we displace the vertex ba little — that is, less than $\epsilon = \epsilon(K)$ — and then define the centre of the barycentric star containing the image of the displacement as the image point $f^{'}(b)$. Then, by the previously mentioned assertion, it follows immediately that the condition of the third conservation theorem is also fulfilled for the mapping $f^{'}$, and consequently, $f^{'}(K_{1})=K$ (note: similarly, $f(C_{1}^{'}=C)$

C.

Now, that we have defined the concept of canonical displacement (and that of modified canonical displacement) for each subdivision of the complex K, we introduce the same concept for each sufficiently fine (curved) simplicial decomposition Q of the polyhedron P, where now Q is independent of the simplicial decomposition K except for the single condition that the diameter of the elements of Q must be smaller than the Lebesgue number of the barycentric covering of the polyhedron P corresponding to K. We consider the following mapping of the complex Q into the complex K; to each vertex b of Q we associate the centre of one of those barycentric stars of K which contains the point b. The barycentric stars which contain the different vertices of a simplex y of Q are all at a distance of less than the diameter of y from an arbitrarily chosen vertex of the simplex y; since this diameter is smaller than the Lebesgue number of the barycentric covering, the stars in question have a non-empty intersection, and their centres are thus vertices of one simplex of K. Our vertex correspondence thus actually defines a simplicial mapping g of Q into K; this mapping g we call a canonical displacement of Q with respect to K.

D.

Now we are in possession of all the lemmas which are necessary for a very short proof of the invariance theorem for the Betti numbers. Let P and $P^{'}$ be two homeomorphic polyhedra, and K and $K^{'}$ arbitrary simplicial decomposition of them. We wish to show that the r-dimensional Betti number p of K is equal to the r-dimensional Betti number $p^{'}$ of K. From symmetry considerations it suffices to prove that $p^{'} \geq p$.

For this purpose, we notice first of all, that under the topological mapping t of $P^{'}$ onto $P$, the complex $K^{'}$ and each subdivision $K_{1}^{'}$ of $K^{'}$ go over into curved simplicial decomposition of the polyhedron P. If we denote, for the moment, by $\sigma$ a positive number which is smaller than the Lebesgue number of the barycentric covering of K, and also smaller than the number $\epsilon(K)$ defined earlier, then one can choose the subdivision $K_{1}^{'}$ of $K^{'}$ so fine that under mapping t the simplexes and the barycentric stars of $K_{1}^{'}$ go over into point sets whose diameters are less than $\sigma$. These point sets form the curved simplexes and the barycentric stars of the simplicial decomposition Q of P into which t takes the complex $K_{1}^{'}$.

Recall definition of $\epsilon(K)$: One means by a barycentric star of K, the union of all simplexes of the barycentric subdivision $K_{1}$ of K which possess a fixed star a of K as their common (leading) vertex. The vertex a is called the centre of the star. It is easily shown that a point of a simplex $|x|$ of K can belong only to those barycentric stars whose centres are vertices of the simplex $|x|$. From this it follows that: (a) if certain barycentric stars $B_{1}, B_{2}, \ldots, B_{s}$ have a common point p, then their centers are vertices of one and the same simplex of K (namely, that simplex which contains the point p in its interior) (b)there is a positive number $\epsilon = \epsilon(K)$ with the property that all points of the polyhedron P (whose simplicial decomposition is K) which are at a distance of less than $\epsilon$ from a simplex x of K can belong only to those barycentric stars which have their centres at the vertices of x. (This follows simply from the fact that all other stars are disjoint from x, and consequently have a positive distance from this simplex). (end of definition of $\epsilon$).

Contd. Proof:

Now, let $K_{1}$ be a subdivision of K so fine that the simplexes of $K_{1}$ are smaller than the Lebesgue number of the barycentric covering of Q. Then, there exists (according to the following (recall section B above) ) a canonical displacement g of $K_{1}$ with respect to Q; furthermore, let f be a canonical displacement of Q with respect to K (this exists because the simplexes of Q are smaller than the Lebesgue number of the barycentric covering of K). Since, by means of g, every vortex of $K_{1}$ is moved to the centre of a barycentric star of Q containing it, and, therefore, is displaced by less than $\epsilon(K)$, the simplicial mapping $f(g(K_{1}))$ — written fg(K_{1}) for short — of the complex $K_{1}$ into the complex K is a modified canonical displacement of $K_{1}$ with respect to K, under which, by previous discussion(s),

$fg(K_{1})=K$.

Furthermore, if C is an algebraic subcomplex of K and $C_{1}$ its subdivision in $K_{1}$, then (as per the following note given earlier:)

$fg(C_{1})=C$

E.

Now let

$Z_{1}, Z_{2}, \ldots, Z_{p}$

be p linearly independent (in the sense of homology) r-dimensional cycles in K, and

$z_{1}, z_{2}, \ldots, z_{n}$

their subdivisions in $K_{1}$. The cycles

$g(z_{1}), g(z_{2}), \ldots, g(z_{p})$

are independent in Q, since if U is a subcomplex of Q bounded by a linear combination

$\sum_{i}c^{i}g(z_{i})$,

then $f(U)$ will be bounded by

$\sum_{i}c^{i}fg(z_{i})$

that is, $c^{i}Z^{i}$, which, according to the assumed independence of the $Z_{i}$ in K, implies the vanishing of the coefficients $c^{i}$.

Under the topological mapping t, the linearly independent cycles $g(z_{i})$ of the complex Q go over into linearly independent cycles of the complex $K_{1}^{'}$ (indeed, both complexes have the same combinatorial structure), so that there are at least p linearly independent r-dimensional cycles in $K_{1}^{'}$. Since we have assumed the equality of the Betti numbers of $K^{'}$ and $K_{1}^{'}$, it follows, therefore that $p^{'} \geq p$. QED.

With the same methods (and only slightly more complicated considerations), one could also prove the isomorphism of the Betti groups of K and $K^{'}$.

Cheers,

Nalin Pithwa.

(PS: same reference as previous blog articles on this topic)

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