Homology: counting holes in doughnuts and why balls and disks are radically different.

November 24, 2014 by Daniel Robert-Nicoud. 0 comments

There are some questions that are really easily posed, have an obvious answer, but are in fact really, really hard to answer in a mathematically satisfactory way. Two examples are:

  1. How many holes does a doughnut have?
  2. Are a ball and a disk “the same thing”? Meaning: can I deform the first to make it the second in a way that locally preserves its structure, i.e. without tearing, and without “squishing” things too much?

The intuitive (and actually correct) answers are: one, and no. However, in order to prove them true, we’ll have first to formalize the questions in mathematical language, and then to develop a theory that will allow us to work on them.

Let’s start with the first question. “A doughnut” can be interpreted in two ways: it could be a filled doughnut, that is the space \(S^1\times D^2\), where \(S^1\) denotes the circle, and \(D^2\) the \(2\)-dimensional disc , or it could be a hollow doughnut, corresponding to the torus \(T^2=S^1\times S^1\). Both of them are examples of topological spaces, so we will generalize the question to: given a topological space \(X\), how many holes does \(X\) have? If you don’t know what a topological space is, just take \(X\) to be one of the examples of doughnuts given above, unless otherwise specified.

The next question we have to answer is: what is a hole? We have many different examples of things we would like to consider holes. One is our hole in the doughnut, but we could also take the space, \(\mathbb{R}^3\), and take away a ball from it, or a infinitely long filled cylinder. Notice that the last two examples are fundamentally different, in the following way: if we take away an infinite cylinder from the space and draw a closed curve around it, we will never be able to deform it to a path not containing the cylinder, or to a single point, without “breaking” it, while we deform every closed curve to a single point in the very last example. However, if we were to put a sphere around the ball we removed in the last example, then we could not deform it to a point, while we could do it for every sphere in our space without a cylinder.

Noticing that a closed path looks very much like a circle, we can use this to distinguish between various kinds of holes. We will informally call an \(n\)-dimensional hole an \(n\)-sphere \(S^n\) that cannot be deformed to a single point without tearing it, where we define: $$S^n=\left\{x\in\mathbb{R}^{n+1}:\lvert x\rvert^2=1\right\}$$ Notice that \(S^1\) is the circle, and \(S^2\) is the sphere. Now, spheres look like really simple spaces, but are in fact quite difficult to work with. However, there’s something we can do to avoid working directly with spheres while keeping intact the essence of what we have said until now: we can cut up spheres in smaller “triangular” pieces. We make the following definition:

Definition: Let \(v_0,\ldots,v_n\in\mathbb{R}^{m}\), where \(m\ge n\). We define the affine \(n\)-dimensional singular simplex \([v_0,\ldots,v_n]\) as the closed convex hull of the points \(v_0,\ldots,v_n\), that is: $$[v_0,\ldots,v_n]=\left\{\sum_{k=0}^nt_kv_k:t_k\in[0,1]\forall k,\ \sum_{k=0}^nt_k=1\right\}$$ We also define the standard \(n\)-simplex as \(\Delta^n=[e_0,\ldots,e_n]\subset\mathbb{R}^{n+1}\), where \(e_i\in\mathbb{R}^{n+1}\) are the standard basis elements.

Notice that the standard \(2\)-simplex is in fact a triangle, and the standard \(3\)-simplex is a tetrahedron. So this definition gives a sensible generalization of what a triangle of dimension \(n\) should be. Let’s now cover the circle \(S^1\) with \(1\)-simplices, and the sphere \(S^2\) with \(2\)-simplices. We immediately notice that they have something special with respect to a random collection of simplices: they have no boundary, that is, the triangles composing them have the sides glued together in such a way that they cancel each other. This leads us to make the next definition:

Definition: The \(i\)-th face of an \(n\)-simplex \(a=[v_0,\ldots,v_n]\) is the \((n-1)\)-simplex \(a^{(i)}[v_0,\ldots,v_{i-1},v_{i+1},\ldots,v_n]\). The boundary of the simplex is the (formal) sum of \((n-1)\)-simplices: $$ \partial a=\sum_{k=0}^n(-1)^ka^{(k)}$$

Notice that the boundary of a simplex is a sum of simplices of lower dimension, so we would like to define some set of simplices where we are allowed to take sums in a sensible way. Also, we would like our simplices to live in our topological space \(X\), and not only in \(\mathbb{R}^m\). So we define the group of \(n\)-simplices in \(X\), denoted by \(S_n(X)\), as the free abelian group generated by continuous maps \(\sigma:\Delta^n\to X\). This simply means that the object (called chains) of \(S_n(x)\) are finite sums of simplices of dimension \(n\) in \(X\) (this is why we take maps from the standard simplex to \(X\)), and that we can sum two such objects in the obvious way, for example if we have two chains \(\sigma_1\) and \(\sigma_2\), then we have: $$(2\sigma_1+\sigma_2)+3\sigma_2=2\sigma_1+4\sigma_3$$ The boundary defines then a map (in fact, a group homomorphism) from \(S_n(x)\) to \(S_{n-1}(X)\), defined on the generators \(\sigma:\Delta^n\mapsto X\) as: $$\partial \sigma=\sum_{k=0}^n(-1)^k\sigma^{(k)}$$ where \(\sigma^{(k)}\) is simply the map \(\sigma\) restricted to the \(i\)-th face of \(\Delta^n\). The sequence of groups \(S_n(X)\) together with the boundary maps defines a sequence: $$\ldots\stackrel{\partial_{n+1}}{\longrightarrow} S_n(X)\stackrel{\partial_{n}}{\longrightarrow} S_{n-1}(x)\stackrel{\partial_{n-1}}{\longrightarrow}\ldots\stackrel{\partial_{2}}{\longrightarrow} S_1(X)\stackrel{\partial_{1}}{\longrightarrow} S_0(X)\to 0$$ where the composition of two consecutive arrows gives the zero map (this can be easily checked by writing down what happens to a generator and rearranging a couple of sums). This kind of sequence is really important in some areas of mathematics, and they have a special name: they are called chain complexes. As we have seen, of all the elements in the chain complex we want to consider those with no boundary, that is, elements in: $$Z_n=\ker(\partial_n)=\{c\in S_n(X)|\partial c=0\}\subseteq S_n(X)$$ We call these elements cycles. They represent in some sense “closed things”, like circles, spheres, but also for example tori (i.e. our hollow donuts) and similar stuff. Moreover, we would like to identify two cycles whenever they represent, for example, two closed paths that can be deformed in such a way that the first becomes equal to the second. Notice that if this is the case, then during the deformation the first curve will draw some kind of annulus, which we can cover with \(2\)-simplices and (as a chain) will have as boundary the first path minus the second one. This leads us to the idea of identifying two cycles whenever their difference is a boundary. Thus we define the homology groups of \(X\) by: $$H_n(X)=Z_n/B_n$$ where \(B_n=\partial_{n+1}(S_{n+1}(X))\) is the set of boundaries.

These groups have a great deal of nice properties. First of all, they are topological invariants. This means that if two spaces are “essentially the same” (the technical term is homeomorphic, meaning that there exist a continuous bijection with continuous inverse between the two), then their homology groups are equal. Another similar thing is that if \(Y\) is a subspace of \(X\), and we can deform \(X\) to \(Y\) without deforming \(Y\), then \(X\) and \(Y\) have the same homology groups. The other useful properties are mostly too technical to be stated here without making this post excessively long. Also, unfortunately, we don’t have enough tools to compute the homology groups of spaces more complicated than, say, finite unions of points, or balls, or \(\mathbb{R}^n\). So when needed I will just state the results, and if you are interested in the computations, you can try to consult one of the bibliographical references I will give at the end.

Whew! We’ve come a long way from the original question of counting the number of holes in a doughnut! But finally we can answer the question. From our definitions, the \(n\)-th homology group should more or less count the number of \(n\)-dimensional holes in the space. For example, for the filled doughnut we have: $$H_n(S^1\times D^2)=\begin{cases} \mathbb{Z}&\text{for $ n=0,2$} \\ 0 &\text{else } \end{cases}$$ The \(0\)-th group doesn’t really matters to us, in fact all it does is to count the number of “pieces” of which our space is made of. The \(1\)-st homology group, however, gives us the information we needed: up to equivalence, there is exactly one \(1\)-chain which is not a boundary (we get \(\mathbb{Z}\) because we can also count twice the same chain, or three times, etc.). This means that there is one circle (or something analogous) that cannot be deformed to be a point, and thus that there is exactly one \(1\)-dimensional hole. Similarly, for the hollow doughnut we have: $$H_n(S^1\times S^1)=\begin{cases} \mathbb{Z} &\text{for $n=0,2$}\\ \mathbb{Z}^2 &\text{for $n=1$}\\ 0 &\text{else}\end{cases}$$

Thus we have one \(2\)-dimensional hole (the whole hollow part of the doughnut), and two \(1\)-dimensional holes (the circle going around the central hole of the doughnut, and the circle going around the hole formed by the hollow part).

As a bonus, we can also use homology to answer our second question. As I have stated, two homeomorphic spaces (“essentially the same”, remember?) have the same homology groups. Now if a ball \(B^3\) in space were homeomorphic to a disk \(B^2\), then removing exactly one interior point in each of those spaces in a sensible way we would again get homeomorphic spaces. However a ball without a point deforms nicely to a sphere, and a disk without a point to a circle, and we have: \begin{gather}H_n(S^1)=\begin{cases}\mathbb{Z} &\text{if $n=0,1$}\\ 0 &\text{else}\end{cases},\\ H_n(S^2)=\begin{cases} \mathbb{Z} &\text{if $n=0,2$}\\ 0 &\text{else} \end{cases}\end{gather}

which is exactly what we would expect from our intuition about \(n\)-dimensional holes. Since these are not equal, the two spaces cannot be homeomorphic, and thus we are done.

Those two problems we just solved are two of the many applications of homology theory, and indeed of the larger framework, which is called algebraic topology. If you want to know more on the subject, here are three books you can try to read:

  • Allen Hatcher, Algebraic Topology
  • Glen E. Bredon, Topology and Geometry
  • Edwin H. Spanier, Algebraic Topology

Binary quadratic forms over the rational integers and class numbers of quadratic fields.

August 23, 2014 by Will Jagy. 3 comments

I wrote an article with Irving Kaplansky on indefinite binary quadratic forms, integral coefficients. At the time, I believe I used high-precision continued fractions or similar. It took me years to realize that the right way to solve Pell’s equation, or find out the “minimum” of an indefinite form (and other small primitively represented values), or the period of its continued fraction, was the method of “reduced” forms in cycles/chains, due to Lagrange, Legendre, Gauss. It is also the cheapest way to find the class number and group multiplication for ideals in real quadratic fields, this probably due to Dirichlet. For imaginary quadratic fields, we have easier “reduced” positive forms.

A binary quadratic form, with integer coefficients, is some $$ f(x,y) = A x^2 + B xy + C y^2. $$ The discriminant is $$ \Delta = B^2 – 4 A C. $$ We will abbreviate this by $$ \langle A,B,C \rangle. $$ It is primitive if \({\gcd(A,B,C)=1. }\) Standard fact, hard to discover but easy to check: $$ (A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$ where \({ X = x z – D yw, \; Y = A xw + C yz + B yw. }\) This gives us Dirichlet’s definition of “composition” of quadratic forms of the same discriminant, $$ \langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle. $$ In particular, if this \({D=1,}\) the result represents \({1}\) and is (\({SL_2 \mathbb Z}\)) equivalent to the “principal” form for this discriminant. Oh, duplication or squaring in the group; if \({\gcd(A,B)=1,}\) $$ \langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle. $$ This comes up with positive forms: \({ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle }\) is principal, the group identity. Probably should display some \({SL_2 \mathbb Z}\) equivalence rules, these are how we calculate when things are not quite right for Dirichlet’s rule: $$ \langle A,B,C \rangle \cong \langle C,-B,A \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B – 2 A, A – B +C \rangle. $$

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Zabreiko’s lemma and four fundamental theorems of functional analysis

June 25, 2014 by Norbert. 0 comments

There are no doubts that open mapping theorem, closed graph theorem, bounded inverse theorem, uniform boundedness principle are the fundamental theorems of functional analysis. All of them are similar in the sense that they explicitly or implicitly use the Baire category theorem, but in the details they are different. However, all these theorems can be formulated as continuity of certain seminorms defined on a normed spaces.

The proof of the lemma is very similar to the usual proof by Banach-Schauder of the open mapping theorem and is actually stronger than the usual consequences of the Baire category theorem in basic functional analysis. The lemma can be used to easily prove the aforementioned theorems.

I would like to thank the user t.b. for drawing our attention to this lemma and adaptation of the original proof from the general case of linear metric spaces, which he gave as an answer to the question Direct Approach to the Closed Graph Theorem on the main site.

Lemma. (Zabreiko, [1], [2]) Let \(X\) be a Banach space and let \(p: X \to [0,\infty)\) be a seminorm. If for all absolutely convergent series \(\sum_{n=1}^\infty x_n\) in \(X\) we have \[ p\left(\sum_{n=1}^\infty x_n\right) \leq \sum_{n=1}^\infty p(x_n) \in [0,\infty] \] then \(p\) is continuous. That is to say, there exists a constant \(C \geq 0\) such that \(p(x) \leq C\lvert x\rvert _{X}\) for all \(x \in X\).

Proof. Let \(A_n = p^{-1}([0,n])\) and \(F_n = \overline{A_n}\). Note that \(A_n\) and \(F_n\) are symmetric and convex because \(p\) is a seminorm. We have \(X = \bigcup_{n=1}^\infty F_n\) and Baire’s theorem implies that there is \(N\) such that the interior of \(F_N\) is nonempty.

Therefore, there are \(x_0 \in X\) and \(R \gt 0\) such that \(B_R(x_0) \subset F_N\). By symmetry of \(F_N\) we have \(B_{R}(-x_0) = -B_{R}(x_0) \subset F_n\), too. If \(\lvert x\rvert \lt R\) then \(x+x_0 \in B_{R}(x_0)\) and \(x-x_0 \in B_{R}(-x_0)\), so \(x \pm x_0 \in F_{N}\). By convexity of \(F_N\) it follows that \[ x = \frac{1}{2}(x-x_0) + \frac{1}{2}(x+x_0) \in F_N, \] so \(B_R(0) \subset F_N\). But, in fact, we can prove much more: \[ B_{R}(0) \subset A_N \] Suppose \(\lvert x\rvert \lt R\) and choose \(r\) such that \(\lvert x\rvert \lt r \lt R\). Fix \(0 \lt q \lt 1-\frac{r}{R}\), so \(\frac{1}{1-q} \frac{r}{R} \lt 1\). Then \(y = \frac{R}{r}x \in B_{R}(0) \subset F_N = \overline{A_N}\), so there is \(y_{0} \in A_N\) such that \(\lvert y-y_0\rvert \lt qR\), so \(q^{-1}(y-y_0) \in B_R\). Now choose \(y_1 \in A_N\) with \(\lvert q^{-1}(y-y_0) – y_1\rvert \lt q R\), so \(\lvert (y-y_0 – qy_1)\rvert \lt q^2 R\). By induction we obtain a sequence \( (y_k)\subset A_N\) such that \[ \left\lvert y – \sum_{k=0}^n q^k y_k\right\rvert \lt q^n R \quad \text{for all }n \geq 0, \] hence \(y = \sum_{k=0}^\infty q^k y_k\). Observe that by construction \(|y_k| \leq R + qR\) for all \(k\), so the series \(\sum_{k=0}^\infty q^k y_k\) is absolutely convergent. But then the countable subadditivity hypothesis in \(p\) implies that \[ p(y) = p\left(\sum_{k=0}^\infty q^k y_k\right) \leq \sum_{k=0}^\infty q^k p(y_k) \leq \frac{1}{1-q} N \] and thus \(p(x) \leq \frac{r}{R} \frac{1}{1-q} N \lt N\) which means \(x \in A_N\), as we wanted.

Finally, for any \(x \neq 0\), we have with \(\lambda = \frac{R}{\lvert x\rvert (1+\varepsilon)}\) that \(\lambda x \in B_{R}(0) \subset A_N\), so \(p(\lambda x) \leq N\) and thus \(p(x) \leq \frac{N(1+\varepsilon)}{R} \lvert x\rvert \), as desired. \(\blacksquare\)

Now the proof of the fundamental theorems becomes an easy exercise.

  1. The open mapping theorem. Hint: set \(p(y) = \inf_{x\in T^{-1}(y)}\lvert x\rvert \).
  2. The bounded inverse theorem. Hint: set \(p(x) = \lvert T^{-1}(x)\rvert \).
  3. The uniform boundedness principle. Hint: set \(p(x) = \sup_{i\in I}\lvert T_i (x)\rvert \).
  4. The closed graph theorem. Hint: set \(p(x) = \lvert T(x)\rvert \).

Detailed solutions can be found in theorems \(1.6.5\), \(1.6.6\), \(1.6.9\) and \(1.6.11\) in An Introduction to Banach Space Theory Graduate Texts in Mathematics. Robert E. Megginson.

[1] П. П. Забрейко, Об одной теореме для полуаддитивных функционалов, Функциональный анализ и его приложения, 3:1 (1969), 86–88

[2] P. P. Zabreiko, A theorem for semiadditive functionals, Functional analysis and its applications 3 (1), 1969, 70-72)