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I came across a simple statement in finite group theory that I’m almost upset no one told me earlier. The source is Serre’s book ‘Linear Representations of Finite Groups’. Serre used this statement below to define / prove Brauer’s theorem on induced representations of finite groups, using which one proves the meromorphicity of Artin-L-functions. Here it goes.

$G$ is a finite group and $p$ is a fixed prime. An element $x$ of $G$ is called $p$unipotent if $x$ has order a power of $p$ and $p$regular if it’s order is prime to $p$.

Cool result:  Every element $x$ in $G$ can be uniquely written as

$x = x_u x_r;$

where

• $x_u$ is $p$-unipotent, $x_r$ is $p$-regular,
• $x_u$ and $x_r$ commute and
• they are both powers of $x$.

The proof is really easy. Just replace $G$ by the (finite) cyclic group generated by $x$!

I call it the Jordan decomposition because we have a similar decomposition for endomorphisms (among other things).

Let $V$ be a finite dimensional vector space over an algebraically closed field of characteristic zero (just in case!). Each $x \in \text{End }(V)$ can be uniquely written as

$x = x_s x_u;$

where

• $x_s$ is semisimple (diagonalizable), $x_u$ is nilpotent,
• they both commute and
• they are polynomials in $x$ without a constant term.

Pretty cool, huh!

Sohei YASUDA gave me a proof of a lemma whose proof is instructive and interesting, using many little facts from group theory and topology. I thought I should blog it.

Lemma: G is a compact topological group and H is an open subgroup. Then H is also closed (!) and of finite index.

Proof. G can be written as a union of cosets, each of which is open in G thanks to continuity of the group product. Thus the complement of H is a union of open (co)sets so is open. H is thus closed.

Consider now the quotient map from G to G/H. (Here, I’m not assuming H is normal in G but rather thinking of G/H as the quotient topological group with it’s topology coming from the quotient map). By definition of the quotient map, H cl-open in G implies it’s image, the identity coset of G/H too is cl-open.

Compactness of G inherits to the topology on G/H so the identity element is discrete on a compact set. The set G/H better be finite now, thus proving the second assertion. $\blacksquare$

In their book, Singer & Thorpe say, “At the present time, the average undergraduate Mathematics major finds math heavily compartmentalized.” One learns many things but does not see the connections between seemingly different things. Indeed, as the great Poincare says, Mathematics is the art of giving the same name to different things. In this and the subsequent post, we shall see the connections in algebra and topology with respect the Galois theory.

Galois theory in Algebra

This topic is covered in most standard algebra texts. It deals with studying the roots of polynomials and their relations. Given a field F and an irreducible polynomial p(x) with coefficients in F, we look at the smallest field K containing F and which has all roots of p(x). The set of all permutations of the roots of p  correspond to automorphisms of K which fix F element-wise. These automorphisms form a group known as the Galois group. There is a beautiful correspondence between subgroups of this group and subfields of K fixing F.

The 19-th century mathematicians Galois and Abel studied this group and Galois came up with useful characterization on this group as to when its roots could be expressed in terms of the coefficients. Of course the theory has now been much more generalized, abstracted and is used indispensably in many parts of mathematics – number theory, algebraic geometry and more.

Galois theory in Topology

(Even this can be found in any standard text on algebraic topology. But here I am talking about Riemann surfaces – and the connections between the two Galois theories — the too-fascinating-to-be-true connection, although not very difficult, its not found in lower-level texts. I will explain that in a follow-up post to this soon).

Given a point P on a topological space S, one talks about the equivalence class (modulo continuous deformations) of paths starting and ending at P. They form a group, the fundamental group. Also, given two (path)-connected spaces R and S, one says that R is a covering of S if there is a continuous surjective map from R to S such that every point of S has a neighbourhood U whose inverse image is a disjoint union of copies each resembling U. One may imagine the real line to be a covering of the unit circle via the map t being mapped to (cos t, sin t).

Now comes the Galois correspondence. For a `nice’ topological space, there is a natural one-to-one correspondence between subgroups of the fundamental group of that space and its covering spaces (rather, isomorphism classes of covering spaces, to be pedantic). Further, the fundamental groups of these spaces have fundamental group isomorphic to the subgroup we started with!

More analogy

The analogy between Galois groups of algebraic objects and fundamental groups and covering spaces of topological spaces goes beyond just one-to-one correspondence. A field extension is normal if it has enough automorphisms. A covering map too is normal (or Galois) if it has enough automorphisms!

Normal field extension $\longleftrightarrow$ Normal subgroups of the Galois group

Normal covering $\longleftrightarrow$ Normal subgroup of the fundamental group

In the next post, we shall see a deeper connection between the four objects above. Namely, we shall take a polynomial, construct its Galois group, get a covering map for this field extension and see that the two groups are the same!

Galois group = fundamental group.

Amazing stuff!

For the entire post, assume that $\mathbb N$ is any countable set. (One may assume it to be the set of natural numbers. ) Further, let

$\mathcal S = \{ f_1 \dots : \mathbb N \to \mathbb N, f \text{ is a bijection} \}$

It is a familiar proof (due to Cantor) that the set of all functions from $\mathbb N$ to $\mathbb N$ is uncountable. But can the same be said about bijections of $\mathbb N$? Here are three proofs:

• We mimic the Cantor’s diagonal argument. Suppose that $\mathcal S$ is countable and enumerate its elements as $f_1, f_2, \dots$. We produce a bijection $f$ different from the $f_i$‘s. Partition $\mathbb N$ into two (disjoint) infinite sets $X$ and $Y$. Define $f(2)$ to be any element of $X$ other than $f_1(2)$. Having defined $f(2), f(4), \dots, f(2n-2)$, define $f(2n)$ to be any element of

$X \backslash \{ f(2), f(4), \dots, f(2n-2), f_n(2n) \}$.

We need to fill in the odd places of $f$ so that it is a bijection. Let $Z = \{ X\backslash \{ f(2n) : n \in \mathbb N \} \}$. Notice that $Z$ is infinite since $Y\subseteq Z$. Now map $Z$ bijectively with odd places of $f$.

This is a contradiction to the countability of $\mathcal S$.

• The second proof needs some facts from higher algebra. $\mathbb Q$ is countable. So is $\bar{\mathbb Q}$, its algebraic closure. Hence, there is a bijection between $\mathbb N$ and $\bar{\mathbb Q}$. Now, the automorphisms of $\bar{\mathbb Q}$ that fix $\mathbb Q$ are, in fact, bijections of $\bar{\mathbb Q}$, or to bijections of $\mathbb N$. In fact,

$\text{Gal }(\bar{\mathbb Q})/\mathbb Q)$ is a subgroup of $\mathcal S$.

That the former is uncountable is a (deep but) standard result in Algebraic Number Theory.

• For the third proof, I quote a result (due to Riemann) on the rearrangements of a conditionally-convergent series, from Rudin’s Principles of Mathematical Analysis, weakened as required in this context.

Theorem: Let $\sum a_n$ be a series of real numbers which converges but not absolutely. Suppose that

$-\infty \leq \alpha \leq \infty$.

Then there exists a rearrangement $\sum a_n'$ with partial sums $s_n'$ such that

$\displaystyle\lim_{n\to\infty} s_n' = \alpha$.

Take $\displaystyle a_n = \frac{(-1)^n}{n}$; it is conditionally convergent. For every (extended) real number, we have a rearrangement of the $a_n$‘s. Further the uniqueness of a limit in the complete metric space $\mathbb R$ suggests that the map (given by the theorem) from $\mathbb R \to \mathcal S$ is injective. Q.E.D.

As a quick revision, let me recollect the way we define the addition of countably many numbers:
Suppose $a_0, a_1, \dots$ are non-negative real numbers. Then the series $a_1 + a_2 + \dots$ is said to converge to $L$ if the sequence $s_n$ defined by $s_n = a_1 + a_2 + \dots + a_n$ of partial sums of $a_n$ converges to $L$.

In other words, the tail of the sequence must add up to atmost as small a number as one desires.

Now, in the quest of adding up uncountably many non-negative numbers, let us index the set of these numbers by $I$. Suppose $f : I \to [0, \infty)$ is a function that associates every element of the index set $I$ to a non-negative number. Our mission is to define $\sum_{x \in I} f(x)$.

Turn $I$ into a discrete measure space by defining the measure of each point in $I$ to be 1. Clearly, now $I$, with the discrete measure $\mu : \mathcal P (I) \to {1}$ is a measure on the measure space $I$.

Now define

$\sum_{ x \in I} f(x)= \int_I f(t) d\mu (t)$

In other words, the sum of all elements indexed by $I$, is the integral of f with respect to the discrete measure $\mu$ taken over the entire index set $I$.

I then wondered if the sum of uncountably many positive numbers could be finite. (Note that a countable sum of positive numbers can be finite; the reciprocals of positive powers of two add up to unity.) The following proposition helps (cf. [Rud]) :

Proposition: Let $(I, \mathcal P (I), \mu : I \to \{1\} )$ be a measure space. If $f : I \to [0,\infty)$ is such that $\int_I f(t) d\mu(t)<\infty$, then $f\equiv 0$ except perhaps on a countable subset of I.

Proof:
Decompose $I$ as:
$I = \{ x | f(x) = 0 \} \bigcup_{n \in N} \{ x | f(x) > 1/n \}$
Call the right set as $A_0$ and the further right ones as $A_n$. Suppose that some of the $A_n$ is infinite. Then
$\int_I f(t) d\mu(t) > \int_{A_n} \frac{1}{n} d\mu(t) = \infty$
contrary to our assumption that the integral is finite. Hence, every $A_n$ is finite and so its union is countable and the proposition is proved.

Reference:
[Rud] Walter Rudin, “Real and Complex Analysis”, 3rd ed. TMH

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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