## Popular Lecture series: ‘Applications of Analysis and Topology to Commutative Algebra’

The Department of Mathematics at IIT Bombay is running a Popular Lecture series this semester, and the first talk in the series was deliverd by Prof. Sudarshan Gurjar on 6th August. His talk was titled ‘Applications of Analysis and Topology to Commutative Algebra’. The talk was an hour long and also fairly elementary. He picked up two attractive examples of such applications, one each from Analysis and Topology.

## Topology in Algebra

Let $A$ be a commutative ring with identity and $I$ an ideal of $A$. Then, we have the natural projection map

$\displaystyle A \to A/I$ given by $\displaystyle a \mapsto a + I$

Similarly, one can talk about matrix groups over these rings, for example, the special linear groups $\mathrm{SL}_n(A)$ and $\mathrm{SL}_n(A/I)$. Then, the projection map from $A$ onto $A/I$ extends naturally to a map from $\mathrm{SL}_n(A)$ into $\mathrm{SL}_n(A/I)$.

Prof. Gurjar asked the question, “Does every element in $\mathrm{SL}_n(A/I)$ lift to some element in $\mathrm{SL}_n (A)$?” One does not expect this to be true in general. After all, if a matrix is in $\mathrm{SL}_n(A)$, then it is only the image of some matrix with entries in $A$ whose determinant is $1+r$ where $r \in I$. There is no reason to expect that we can always find such a matrix with $r = 0$.

And indeed, the question has a negative answer, and a proof for the case $n = 2$ uses a very neat application of topology.

Before proceeding to the proof, it is worth mentioning that there are important cases where it is true that one can always find such a lift. The example I know comes up in the study of modular forms, or more precisely, when working with discrete subgroups of $\mathrm{SL}_2(\mathbb{Z})$: the natural map from $\mathrm{SL}_2(\mathbb{Z})$ into the congruence subgroup $\Gamma(N) = \mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ is a surjective map. The interested reader can refer to Chapter 2 of Murty, Dewar & Graves’ “Problems in the Theory of Modular Forms”.

Returning to our claim, let $A = \mathbb{R}[X,Y]$ and $I = (X^2 + Y^2 - 1)$. We claim that the element

$\displaystyle \begin{pmatrix} \bar{X} & -\bar{Y} \\ \bar{Y} & \phantom{-}\bar{X} \end{pmatrix} \in \mathrm{SL}_2(A/I)$

has no lift to $\mathrm{SL}_2(A)$.

The proof is by contradiction. We will show that if there is a lift, then there is a retract of $\mathbb{R}^2$ onto $S^1$ (the circle group), which is known to not exist.

So, suppose that there is a lift to some element in $\mathrm{SL}_2(A)$. Then, that element would look like

$\displaystyle \begin{pmatrix} f_{11}(X,Y) & f_{12}(X,Y) \\ f_{21}(X,Y) & f_{22}(X,Y) \end{pmatrix}$,

where the polynomials $f_{ij}$ must satisfy $f_{11}(X,Y)f_{22}(X,Y) - f_{12}(X,Y)f_{21}(X,Y) = 1$ as well as the equations

\displaystyle \begin{aligned} f_{11}(X,Y) &= X + (X^2 + Y^2 - 1)h_{11}(X,Y)\\ f_{12}(X,Y) &= -Y + (X^2 + Y^2 - 1)h_{12}(X,Y)\\ f_{21}(X,Y) &= Y + (X^2 + Y^2 - 1)h_{21}(X,Y)\\ f_{22}(X,Y) &= X + (X^2 + Y^2 - 1)h_{22}(X,Y) \end{aligned}

for some polynomials $h_{ij} \in A$. Now, we have a map $\mathbb{R}^2 \to \mathrm{SL}_2(\mathbb{R})$ given by

$\displaystyle (a,b) \mapsto \begin{pmatrix} f_{11}(a,b) & f_{12}(a,b) \\ f_{21}(a,b) & f_{22}(a,b) \end{pmatrix}.$

We also have a map $\mathrm{SL}_2(\mathbb{R}) \to \mathrm{O}(2)$, the space of $2 \times 2$ orthogonal matrices, given by Gram-Schmidt orthogonalisation. But now observe that Gram-Schmidt orthogonalisation is a continuous map that does not change the determinant. Since $\mathrm{S}_2(\mathbb{R})$ is connected, this is actually a map into $\mathrm{SO}(2)$, the space of orthogonal matrices with determinant one. This space has a natural identification with $S^1$, using the map

$\displaystyle \theta \mapsto \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \phantom{-}\cos \theta \end{pmatrix}.$

Hence, we have a the following sequence of maps:

$\displaystyle S^1 \overset{i}{\hookrightarrow} \mathbb{R}^2 \overset{r}{\to} S^1$,

where $i$ is the canonical inclusion map. Observe that $r \circ i = \mathrm{id}_{S^1}$ because of the equations that $f_{ij}$ satisfy. So, $r$ is actually what one calls a retract of $\mathbb{R}^2$ onto $S^1$. However, it is a well-known result in topology that there is no retract of $\mathbb{R}^n$ onto $S^{n-1}$ for any $n \in \mathbb{N}$, and so we have arrived at a contradiction.

## Analysis in Algebra

Let $A = \mathbb{R}[X,Y] / (X^2 + Y^2 - 1)$. Consider the maximal ideal $M = (\bar{X} - \bar{1}, \bar{Y})$ of $A$. (Exercise: show that $M$ maximal!) It is an interesting question to ask whether $M$ is a principal ideal. We shall see that the answer is that it $M$ is not principal, and the proof neatly applies ideas from analysis.

It makes sense to view $A$ as the space of polynomials on $S^1$ modulo the equivalence relation

$\displaystyle f \sim g \iff f - g = 0$ on $\displaystyle S^1$.

An important idea that goes into the proof is localisation. Let $A_{\bar{X}-\bar{1}}$ and $M_{\bar{X}-\bar{1}}$ be the localisations of $A$ and $M$, respectively, at the element $\bar{X} - \bar{1}$. Then, $M_{\bar{X}-\bar{1}}$ is also an ideal of $A_{\bar{X}-\bar{1}}$. But, $\bar{X}-\bar{1}$ is a unit in $A_{\bar{X}-\bar{1}}$ and it is also one of the generators of $M$. So, there is a unit in $M_{\bar{X}-\bar{1}}$, which implies that $M_{\bar{X}-\bar{1}} = A_{\bar{X}-\bar{1}}$.

Similarly, consider the localisations of $A$ and $M$ at the element $\bar{X}+\bar{1}$. Since $(\bar{X}-\bar{1})(\bar{X}+\bar{1}) = \bar{Y}^2$, we see that $M_{\bar{X}+\bar{1}}$ is the principal ideal $(\bar{Y}$ in $A_{\bar{X}+\bar{1}}$.

Now, let us prove that $M$ is not maximal by contradiction. Suppose $M = (g(\bar{X},\bar{Y}))$. Then, $M_{\bar{X}-\bar{1}}$ and $M_{\bar{X}+\bar{1}}$ are also generated by the same element as ideals of $A_{\bar{X}-\bar{1}}$ and $A_{\bar{X}+\bar{1}}$, respectively. By the above examination of these localisations, we can say that there exist elements $h_1(\bar{X},\bar{Y}) \in A_{\bar{X}-\bar{1}}$ and $h_2(\bar{X},\bar{Y}) \in A_{\bar{X}+\bar{1}}$ such that

$\displaystyle 1 = g(\bar{X},\bar{Y}) h_1(\bar{X},\bar{Y})$ and $\displaystyle \bar{Y} = g(\bar{X},\bar{Y}) h_2(\bar{X},\bar{Y})$.

This implies that

$\displaystyle \bar{Y} = h_1^{-1}(\bar{X},\bar{Y}) h_2(\bar{X},\bar{Y}) = h_3(\bar{X},\bar{Y})h_2(\bar{X},\bar{Y})$,

for some $h_3(\bar{X},\bar{Y}) \in A_{\bar{X}-\bar{1}}$.

Now, the analysis kicks in. Observe that $h_3$ is a continuous function from $S^1 \setminus \{ (1,0)\}$ into $\mathbb{R}$. But $h_3$ is nonzero everywhere on this domain, and so it is actually a map into $\mathbb{R}^*$. So, $h_3$ has the same sign everywhere on this domain. Similarly, $h_2$ has the same sign everywhere on the domain $S^1 \setminus \{ (-1,0) \}$. Hence, the product $h_3(\bar{X},\bar{Y})h_2(\bar{X},\bar{Y})$ has the same sign everywhere on $S^1 \setminus \{(\pm 1, 0) \}$. But this product equals $\bar{Y}$ whose signs differ on the upper and lower arcs, which is a contradiction.

## Measuring subsets of the real line

Our interest is in finding a way to measure the “size” of any subset of $\mathbb{R}$. Before tackling the problem in so much generality, let us look at some special cases.

For an interval $I$ in $\mathbb{R}$ we define its length, $l(I)$, to be the difference of its endpoints. For example, $l(0,1) = 1$, $l[-2,2] = 4$, $l[0,\infty) = \infty$, etc. So, the length of any interval is a non-negative extended real number, that is, $l(I) \in [0,\infty) \cup \{ \infty \}$ for every interval $I$.

Note that if $A = I_1 \cup I_2$, where $I_1$ and $I_2$ are disjoint intervals, then it is natural to define the “size”, or measure, of $A$ to be

$\displaystyle m(A) = l(I_1) + l(I_2)$.

Now, any open set in $\mathbb{R}$ can be written as a union of a countable collection of pairwise disjoint open intervals. So, we can define its “size”, or measure, to be the sum of the lengths of these open intervals. That is, if $U \subseteq \mathbb{R}$ is open, and is expressed as

$\displaystyle U = \bigcup_{n=1}^\infty I_n$

for some countable collection $\{ I_n \}$ of pairwise disjoint open intervals, then we define $m(U)$, the measure of $U$, by

$\displaystyle m(U) = \sum_{n=1}^\infty l(I_n)$.

Note that the infinite sum on the right-hand side does not depend on the ordering of the intervals because the length is always non-negative and hence the series converges absolutely.

What about sets more general than these? Is there any way to measure the “size” of an arbitrary subset of $\mathbb{R}$?

The answer to this question is quite surprising.

But we’re getting ahead of ourselves. It is not obvious what one means by the “size” of an arbitrary subset. Intervals and open subsets have a natural description but apart these simple subsets it is not straightforward to describe how to measure the size of an arbitrary subset.

Instead, let us specify what properties we would like our ideal “size-measuring” function to satisfy, and then proceed to search for such a function.

For starters, we would like this function, say denoted by $m$, to input any subset of $\mathbb{R}$ and output its “size”. The “size” must be non-negative and can be infinite. So, we want a function

$\displaystyle m : \mathscr{P}(\mathbb{R}) \to [0,\infty) \cup \{ \infty \}$,

where $\mathscr{P}(\mathbb{R})$ denotes the power set of $\mathbb{R}$.

Next, we want that the “size” of an interval should be its length. Taking a hint from how we defined the size of an open set, we want the size of a disjoint union of sets to be the sum of the sizes of the sets in the disjoint union. That is, we want that

• $m(I) = l(I)$;
• If $\{ A_n \}$ is a countable collection of pairwise disjoint sets, then

$\displaystyle m\left (\bigcup_{n=1}^\infty A_n \right) = \sum_{n=1}^\infty m(A_n)$.

We would also like this function to be compatible with the additive structure of $\mathbb{R}$. Given $A \subseteq \mathbb{R}$ and $x \in \mathbb{R}$, the set $A + x := \{ a + x : a \in A \}$ is just the translation of $A$ to the right by $x$. Ideally, a translation of a set mustn’t change its size. So, we also want that

• For every $A \subseteq \mathbb{R}$ and $x \in \mathbb{R}$,

$\displaystyle m(A+x) = m(A)$.

That’s it! That’s all that we want our “size-measuring” function to satisfy. Now we can begin our search for such a function.

Except, there’s a small problem.

It turns out that it is impossible for any such function to exist.

.

.

.

This is such a disastrous and unexpected hurdle right at the beginning of our exploration! In a short while, I will prove that no such function as we have described can exist. But first, a few words on what this tells us: the structure of the real numbers is far, far more complicated than we imagined. In the proof, we will see that there are some pathological subsets of $\mathbb{R}$ which cannot be given a meaningful “size” without running into contradictions.

PROPOSITION: There is no set function $m : \mathscr{P}(\mathbb{R}) \to [0,\infty) \cup \{ \infty \}$ that satisfies all three properties listed above.

Proof: Suppose that there is such a function $m$. A clever bit of algebra will produce a set which produces a contradiction.

The idea is as follows. Consider the second bullet point: if we can find a set $A$ with $m(A) < \infty$, say $m(A) = 1$, and a decomposition of $A$ into a countable collection $\{ A_n \}$ of pairwise disjoint subsets such that $m(A_i) = m(A_j)$ for all $i$, $j$, then the equality in the second bullet point won’t hold for this collection.

To find such a collection of sets, a bit of group theory comes in handy. Let $A = [0,1)$. Under the operation of addition modulo $1$, $A$ becomes a group. The set $B$ of rationals in $A$, that is, $B = \mathbb{Q} \cap A$, is a subgroup of $A$. Hence, we can decompose $A$ into a disjoint union of cosets of $B$.

Construct a set $P \subseteq A$ by choosing one element from each coset of $B$. The translates (modulo $1$) of $P$ by the elements of $B$ are all pairwise disjoint. To see why, let $\{ q_n \}$ be an enumeration of the members of $B$ and let $P_n$ be the translate (modulo $1$) of $P$ by $q_n$. That is, let

$\displaystyle P_n = \{ p + q_n \mod 1 : p \in P \}$.

If $x \in P_n \cap P_m$, then $x = p + q_n = p' + q_m$ for some $p, p' \in P$. Hence, $p - p' = q_m - q_n \in B$ $\Rightarrow p$ and $p'$ are in the same coset of $B$. But only one element was chosen from each coset of $B$, so $p = p'$. Hence, $q_n = q_m \Rightarrow n = m$.

Observe that $A = \cup_{n=1}^\infty P_n$. This is because each $x \in A$ lies in the coset $x + B$, and if $p \in P$ is the element chosen from this coset, then $x - p \in B$, that is, $x - p = q_n$ for some $q_n \in B$. Hence, $x \in P_n$.

We now have all the elements to arrive at a contradiction. Since each $P_n$ is a translate of $P$, we must have that $m(P) = m(P_n)$ for all $n \in \mathbb{N}$. Since $A = \cup_{n=1}^\infty P_n$, we must have that

\displaystyle \begin{aligned} m(A) &= \sum_{n=1}^\infty m(P_n) \\ \Rightarrow 1 &= \sum_{n=1}^\infty m(P) \end{aligned}.

If $m(P) = 0$ then the sum on the right-hand side equals $0$. If $m(P) > 0$ (including $m(P) = \infty$), then the sum on the right-hand side equals $\infty$. In either case, we have a contradiction.

$\blacksquare$

Hence, the set $P$ that we have constructed throws a wrench into our efforts. There is no way to assign a value to $m(P)$ in a manner consistent with the properties we wish the function $m$ to have. Something needs to go. In the next post, we’ll see how to resolve this conundrum.

Before closing this post, a comment on the proof. $[0,1)$ is uncountable and the rationals are countable. So, there are uncountably many cosets of $B$. Therefore, choosing one element arbitrarily from each coset of $B$ requires the use of the Axiom of Choice. In fact, without the Axiom of Choice one cannot show the existence of any such pathological sets.