Infinite Swindles

 A puzzle I found very amusing when I was much younger is the following 'proof' that $1=0$: $1 = 1 + (−1 + 1) + (−1 + 1) + ... = 1 − 1 + 1 − 1 + ... = (1 − 1) + (1 − 1) + ... = 0$. Of course, this isn't valid because the middle object, known as the Grandi series, isn't well-defined at all, and serves as a warning to students dipping their toes into mathematics that they should be (especially) careful when infinite things appear. Nevertheless, this can be made rigorous in certain settings where the infinite sum is well-defined. The issue with the Grandi series is that the infinite object can only be made sense of relative to finite objects, but if the infinite object can be independently defined then one can legitimately use it to deduce properties of finite objects. This process is known as the Eilenberg-Mazur swindle. In this post we will discuss applications of infinite processes and infinite sums to topology and algebra.

Topology

Groups of manifolds

The (homeomorphism classes of) oriented $n$-manifolds have an addition operation given by connected sum, with $0$ the $n$-sphere. If $A + B$ is the $n-$sphere, then $A + B + A + B + \dots$ is Euclidean space (imagine removing the North pole of larger and larger spheres and then flattening it out to cover more and more of Euclidean space) so the Mazur swindle shows that the connected sum of $A$ and Euclidean space is Euclidean space, which shows that $A$ is the 1-point compactification of Euclidean space and therefore $A$ is homeomorphic to the $n-$sphere. Note however that there is no hope of making the processes involved smooth because there exist exotic spheres in higher dimensions. A problem that has received a great deal of interest is that of deciding how many diffeomorphism classes of exotic spheres there are in a given dimension.

The Whitehead manifold

While this isn't strictly speaking an infinite swindle, it is something perhaps slightly odd involving the infinite so I feel it is ok to include here. We will construct a contractible open $3-$manifold $W$, known as the Whitehead manifold, as an ascending union of non-simply connected manifolds.
Take a solid torus $T_1$ (in $S^3$) and take a loop generating the fundamental group. This is called a meridian curve. Homotope this loop into the interior of the torus and take a thickening, i.e. a tubular neighbourhood. Choose an embedded solid torus $T_2$ in $T_1$ so that $T_2$ and the tubular neighbourhood form a Whitehead link, displayed below with an image from Wikipedia. 

 

Note now that in $S^3 \backslash$ tubular neighbourhood $T_2$ is nullhomotopic because it has zero winding number around the tubular neighbourhood. This is what lets us conclude that the resulting manifold is simply connected.
We now iterate this process, finding $T_i$ inside $T_{i-1}$ so that it and the tubular neighbourhood of the meridian of $T_{i-1}$ form a Whitehead link. Define $C:= \cap_{i=1}^{\infty}T_i$, known as the Whitehead continuum. The promised manifold $W$ is $S^3 \backslash C$. By above this is simply connected, and one can compute further to show it has no homology groups in positive dimension, so the Whitehead theorem on homotopy equivalences shows that this is contractible.
This is all very well and good, but how do we know that we haven't produced $\mathbb{R}^3$ in a silly way?

A topological space $X$ is said to be simply connected at infinity if for any compact subset $C$ of $X$, there is a compact set $D$ in $X$ containing $C$ so that the induced map on fundamental groups

    \[\pi_ 1 ( X − D ) → \pi_ 1 ( X − C ) {\displaystyle \pi _{1}(X-D)\to \pi _{1}(X-C)} \] is zero. One should think of this condition as saying that all the $\pi_1$ is concentrating in a compact region, hence the name.
One checks that this is invariant under homeomorphism, that Euclidean space has this property, and that $W$ doesn't, to show that we have genuinely produced a different manifold. One can further show that $W \times \mathbb{R} \cong \mathbb{R}^4$, much like the dogbone space mentioned in a previous post. 

Infinite processes like this are an important tool for constructing examples in dimension 4, which is where the wild things are: this is the only dimension for which the smooth Poincare conjecture is open, and this is the only dimension where Euclidean space admits a different smooth structure (in fact uncountably many). Freedman's theorem that a homology 3-sphere topologically bounds a contractible 4-manifold is another important example of results proven by an infinite process. Other important objects constructed like the Whitehead manifold include the Alexander horned sphere, an embedding of the 2-sphere into $\mathbb{R}^3$ with non-simply connected complement (this can't happen for smooth embedings, which shows that even in dimension 3 one needs to be careful about what regularity one wants maps to have), and Antoine's necklace, an embedding of the Cantor set into $\mathbb{R}^3$ with non-simply connected complement. An exposition of the latter can be found here.

Algebra

We begin with four applications of the Eilenberg swindle in algebra.
 

• A module $A$ over a ring $R$ is said to be projective if there is some other $R-$module $Q$ and free module $F$ such that $P \oplus Q \cong F$. There are many equivalent definitions of projectivity which we won't get into. There are many reasons people care about projective modules: they are used to define homology, they have very nice module theoretic properties, they are the algebraic version of (topological) vector bundles etc. A little bit more on this later, but you don't need to understand all of the background to appreciate the following swindle: if $A$ is a projective module over a ring $R$, then there is a free module $F$ with $A \oplus F \cong F$. To see this, choose a module $B$ such that $A \oplus B$ is free, which can be done as $A$ is projective, and put \[ F = B \oplus A \oplus B \oplus A \oplus B \oplus \cdots, \] so that \[ A \oplus F = A \oplus (B \oplus A) \oplus (B \oplus A) \oplus \cdots = (A \oplus B) \oplus (A \oplus B) \oplus \cdots \cong F.\]
•  Finitely generated free modules over commutative rings $R$ have a well-defined natural number as their dimension which is additive under direct sums, and are isomorphic if and only if they have the same dimension. This is false for some noncommutative rings, and a counterexample can be constructed using the Eilenberg swindle as follows. Let $X$ be an abelian group such that $X \cong X \oplus X$ (for example the direct sum of an infinite number of copies of any nonzero abelian group), and let $R$ be the ring of endomorphisms of $X$. Then the left $R$-module $R$ is isomorphic to the left $R$-module $R \oplus R$.
• The Grothendieck group of an abelian category that admits infinite direct sums is trivial. (If you don't know what these words mean don't worry. This really is a direct application of the swindle so feel free to skip and move on.)
• If $A$ and $B$ are any groups then the Eilenberg swindle can be used to construct a ring $R$ such that the group rings $R[A]$ and $R[B]$ are isomorphic rings: take $R$ to be the group ring of the restricted direct product of infinitely many copies of $A \times B$. (This is mostly an exercise in figuring out how group rings of product groups works.) A consequence of this is that questions about whether the group ring determines the group must be asked for sensible coefficient rings, e.g. the integers or complex numbers or finite fields.
  
  

The first example leads to an important question: what if we insist the free module is finitely generated? This leads to the notion of being stably free: a module is said to be stably free if it becomes free after adding finitely generated free module. One can ask similar questions for other properties of modules. The motivation for this is in turn algebraic $K-$theory and algebraic topology, where the philosophy of the latter is that the 'stable' properties of spaces are the properties susceptible to algebraic methods, e.g. in topological $K-$theory the most important characteristic classes are stable under summing with a trivial line bundle, which is the topological analogue of a free module. We first show that being stably free is a notion in its own right by exhibiting modules which are stably free but aren't free.

Theorem Let $R$ be the ring $R[x, y, z]/(x^2 + y^2 + z^2 - 1)$. Let
\[
T = \{(f, g, h) \in {R}^3 : xf + yg + zh = 0 \text{ in } R\}.
\]
Then $R \oplus T \cong {R}^3$, but $T \ncong {R}^2$.


The module $T$ in this theorem is stably free (it is stably isomorphic to $R^2$), but it is not a free module. Indeed, if $T$ were free, then (since $T$ is finitely generated; the theorem shows it admits a surjection from $R^3$) for some $n$ we would have $T \cong R^n$, so $R \oplus R^n \cong R^3$. Since $R^a \cong R^b$ only if $a = b$ for nonzero commutative rings $R$, $1 + n = 3$, so $n = 2$. But this contradicts the non-isomorphism in the conclusion of the theorem.

It’s worth noting that the ranks in the theorem are as small as possible for a non-free stably free module. If $R$ is a commutative ring and $M$ is an $R$-module such that $R \oplus M \cong R$, then $M = 0$. If $R \oplus M \cong \mathbb{R}^2$, then $M \cong R$. The first time we could have $R \oplus M \cong \mathbb{R}^\ell$ with $M \ncong \mathbb{R}^{\ell-1}$ is $\ell = 3$, and the Theorem shows such an example occurs.
 

It will be easy to show $R \oplus T \cong {R}^3$, but the proof that $T \ncong {R}^2$ will require a theorem from topology about vector fields on the sphere. We denote the module as $T$ because it is related to tangent vectors on the sphere.

Proof: Since $R$ is a ring, on $R^3$ we can consider the dot product $R^3 \times R^3 \to R$. For example,
\[
(x, y, z) \cdot (x, y, z) = x^2 + y^2 + z^2 = 1.
\]
For all $v \in R^3$, let $r = v \cdot (x, y, z) \in R$. Then
\[
(v - r(x, y, z)) \cdot (x, y, z) = v \cdot (x, y, z) - r(x, y, z) \cdot (x, y, z) = r - r = 0,
\]
so $v - r(x, y, z) \in T$. That means $R^3 = R(x, y, z) + T$. This sum is direct since $R(x, y, z) \cap T = (0, 0, 0)$: if $r(x, y, z) \in T$, then dotting $r(x, y, z)$ with $(x, y, z)$ implies $r = 0$. So we have proved

\[R^3 = R(x, y, z) \oplus T.\]


Since $R \cong R(x, y, z)$ by $r \mapsto r(x, y, z)$, $R^3 \cong R \oplus T$. Thus $T$ is stably free.

Now we will show by contradiction that $T \ncong R^2$. Assume $T \cong R^2$, so $T$ has an $R$-basis of size 2, say $(f, g, h)$ and $(F, G, H)$. By
$R^3 = R(x, y, z) \oplus T$ the three vectors $(x, y, z)$, $(f, g, h)$, $(F, G, H)$ in $R^3$ form an $R$-basis, so the matrix
\[
\begin{pmatrix}
x & f & F \\
y & g & G \\
z & h & H
\end{pmatrix}
\]
in $M_3(R)$ must be invertible: it is the change-of-basis matrix between the standard basis of $R^3$ and the basis $(x, y, z)$, $(f, g, h)$, $(F, G, H)$. Therefore the determinant of this matrix is a unit in $R$:
\[
\det \begin{pmatrix}
x & f & F \\
y & g & G \\
z & h & H
\end{pmatrix} \in R^\times.
\]

It makes sense to evaluate elements of $R$ at points $(x_0, y_0, z_0)$ on the unit sphere $S^2$: polynomials in $R[x, y, z]$ that are congruent modulo $x^2 + y^2 + z^2 - 1$ take the same value at all $(x_0, y_0, z_0) \in S^2$ since $x_0^2 + y_0^2 + z_0^2 - 1 = 0$. A unit in $R$ takes nonzero values everywhere on the sphere: if $a(x, y, z)b(x, y, z) = 1$ in $\mathbb{R}$, then $a(x_0, y_0, z_0)b(x_0, y_0, z_0) = 1$ when $(x_0, y_0, z_0) \in S^2$.

In particular, at each point $v \in S^2$, the determinant above has a nonzero value, so $(f(v), g(v), h(v)) \in \mathbb{R}^3 \setminus \{0\}$. Thus $v \mapsto (f(v), g(v), h(v))$ is a nowhere vanishing vector field on $S^2$ with continuous components (since polynomial functions are continuous).

But this is impossible: the hairy ball theorem in topology says every continuous vector field on the sphere vanishes at least once. $\blacksquare$


Now we might ask when stably free modules must in fact be free. Rings with this property are known as Hermite rings. The Quillen-Suslin theorem, previously known as the Serre problem or Serre conjecture, states that for $R$ a polynomial ring, or possibly Laurent polynomial ring in some of the variables, over a PID, it is a Hermite ring. In algebraic geometry vector bundles are projective modules over the coordinate ring of the variety, so the question includes as a subquestion whether all algebraic vector bundles over affine space are trivial, which doesn't follow from the fact that they must be topologically or holomorphically trivial. The Hilbert-Serre theorem says that projective modules are stably free, and the gap from stably free to free was closed by Quillen and Suslin independently, who both developed impressive algebraic machinery. Later a short and completely elementary proof was found. A clean exposition of this can be found here. For an excellent and comprehensive discussion of the developments in this direction, including applications both inside and outside of mathematics, see this book.