# Uniformly at Random

## Like a Shark

Richard Holmes begins his book, The Age of Wonder, with several quotations from writers and philosophers of the Romantic Period, including this one from Coleridge:

I shall attack Chemistry, like a Shark.

Reminds me of this comic from xkcd.

Written by uncudh

December 7, 2009 at 4:53 pm

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## Coleridge’s thoughts on Newton

Coleridge had the following to say (in a letter to Tom Poole) concerning Sir Isaac Newton (see Holmes, Coleridge: Early Visions):

My opinion is this—that deep Thinking is attainable only by a man of deep Feeling, and that all Truth is a species of Revelation.  The more I understand of Sir Isaac Newton’s works, the more boldly I dare to utter to my own mind & therefore to you, that I believe the Souls of 500 Sir Isaac Newtons would go to a making up of  a Shakespeare or a Milton . . . Newton was a mere materialist—Mind in his system is always passive—a Lazy Looker-on on an external World.  If the mind be not passive, if it be indeed made in God’s Image,  that too in the sublimest sense—the Image of the Creator—there is ground for suspicion, that any system build on the passiveness of the mind must be false, as a system.

Written by uncudh

November 29, 2009 at 5:58 pm

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## Concerning the machines of Archimedes

Below is an excerpt from Plutarch’s Life of Marcellus (the famous Dryden translation) concerning the mechanical work of Archimedes in devising war machines for the defense of Syracuse.  What is notable to me is the mention of Plato’s indignation at the application of mathematics to practical ends.

The manner of the death of Archimedes is quite famous, and is described by Plutarch (Marcellus):

But nothing afflicted Marcellus so much as the death of Archimedes, who was then, as fate would have it, intent upon working out some problem by a diagram, and having fixed his mind alike and his eyes upon the subject of his speculation, he never noticed the incursion of the Romans, nor that the city was taken. In this transport of study and contemplation, a soldier, unexpectedly coming up to him, commanded him to follow to Marcellus; which he declining to do before he had worked out his problem to a demonstration, the soldier, enraged, drew his sword and ran him through. Others write that a Roman soldier, running upon him with a drawn sword, offered to kill him; and that Archimedes, looking back, earnestly besought him to hold his hand a little while, that he might not leave what he was then at work upon inconclusive and imperfect; but the soldier, nothing moved by his entreaty, instantly killed him. Others again relate that, as Archimedes was carrying to Marcellus mathematical instruments, dials, spheres, and angles, by which the magnitude of the sun might be measured to the sight, some soldiers seeing him, and thinking that he carried gold in a vessel, slew him. Certain it is that his death was very afflicting to Marcellus; and that Marcellus ever after regarded him that killed him as a murderer; and that he sought for his kindred and honoured them with signal favours.

Written by uncudh

November 8, 2009 at 3:00 pm

Posted in history, math

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## Napkin thieves in ancient Rome

Apparently Romans got pretty upset when people would steal napkins from their dinner tables.  Catullus, at any rate, didn’t care for the practice.  He wrote, not one, but two poems berating someone for swiping napkins at dinner.  From Catullus 12 (Green’s translation):

Your left hand, friend Asinius, you provincial,
works its michief while we drink and gossip,
snitching napkins from distracted guests.  You
think this trick is smart?  So dumb, you can’t see
just how dirty your game is, how unlovely?
[…]
Either, then, you give me back my napkin,
or else you’ll get a scad of scathing verses.
It’s not so much the price that’s made me angry:
top-line real native hand-towels, that Fabullus—
and Veranius—sent me all the way from
Spain: so I must love them just as much as
sweet Veranius and my dear Fabullus.

Written by uncudh

October 25, 2009 at 8:40 pm

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## Sigurd as the “chosen one”

We have previously made some remarks concerning Tolkien’s recently published Lays of Sigurd and Gudrun.  One thing I neglected to mention was Tolkien’s one major departure from his sources: namely, his portrayal of Sigurd as the “chosen one” of the gods, whose presence at the Last Battle will determine its outcome and the future of the world to follow.  The Norse Poetic Edda begins with the famous poem Voluspa, which describes the prophecy of the Seeress concerning the Ragnarok, or the Doom of the gods.  Tolkien begins his Lay of Sigurd with his own version of the Voluspa; however, he inserts several stanzas describing the role of Sigurd in the Last Battle:

If in day of Doom
one deathless stands,
who death hath tasted
and dies no more,
the serpent-slayer,
seed of Odin,
then all shall not end,
nor Earth perish.

This conception of Sigurd is not at all present in any of the Norse texts.  One wonders why Tolkien felt it necessary to introduce such an element into the mythology.  Christopher Tolkien points out in his commentary to the Lay the connections to his father’s own imagined mythology, in particular to the tale of Turin Turambar:

This mysterious conception […] reappeared as a prophecy in the Silmarillion texts of the 1930s: so in the Quenta Noldorinwa, ‘it shall be the black sword of Turin that deals unto Melko [Morgoth] his death and final end; and so shall the children of Hurin and all Men be avenged.’

In general, the parallels between Sigurd and Turin are of course quite clear:  Turin as the Dragon-slayer, the wearer of the Dragon-helm, etc.

Written by uncudh

October 24, 2009 at 2:38 pm

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## The death of love

with one comment

Arguably the greatest long poem (mahakavya) in classical Sanskrit is Kalidasa’s Kumarasambhava (or “The Birth of Kumara”), which tells the tale of how the warrior god Skanda (the “Kumara” of the title) came to be born.  Once, the gods were suffering greatly from the attacks of the demon Taraka.  Unable to defeat Taraka, the gods approached the creator-god Brahma to ask for his help.  Specifically, the gods asked for a general who could lead them to victory against Taraka.  Brahma told the gods that they must find a way to convince Shiva the Destroyer to marry Parvati, the daughter of the mountain god.  The child of Shiva and Parvati would be the general that they were looking for.  However, Shiva was deeply absorbed in meditation and would not easily be tempted into marriage.  The gods approached Kamadeva, god of love (armed, like his Greco-Roman counterpart, with bow and arrow), and requested that he use his unique abilities to make Shiva fall in love with Parvati.  Kamadeva agreed, and went to the mountaintop where the god Shiva was engaged in meditation.  Shiva, however, sensed the intrusion of the love god.  Kalidasa describes what happened next (the Clay Sanskrit Library edition 3.69-72):

Then Three-eyed Shiva,
through his self-control
powerfully suppressing
the disturbance of his senses,
wished to see the cause
of his mind’s disturbance
and sent his gaze in all directions.

He saw Self-born Love ready to attack,
his lovely bow drawn right back
to form a circle,
his fist resting
at the corner of his right eye,
shoulder hunched,
left foot arched.

Enraged by the violation of his penance,
and from his third eye
a sparkling, blazing fire
suddenly flew forth.

hold back!”—
even as the cries of the wind-gods
crossed the sky,
that fire born from the eye of Shiva who is Being,
reduced to ashes Intoxicating Love.

His corporeal form having been disintegrated by the fire emanating from the mystical third eye of Shiva the Destroyer, Kamadeva, the god of love, is henceforth known as Ananga, the Bodiless God.

Written by uncudh

September 7, 2009 at 7:39 pm

## The General Burnside Problem

In 1902 William Burnside posed the following question concerning finitely generated groups:

Bounded Burnside Problem. If $G$ is a finitely generated group and there is an integer $n$ such that $g^n = 1$ for every $g\in G$, then must $G$ be finite?

The problem also has the following variant:

General Burnside Problem. If $G$ is a finitely generated group and every element of $G$ has finite order, then must $G$ be finite?

The answer to both questions was expected to be “yes”’; the solution to the General Burnside Problem was therefore anticipated to be somewhat harder than that of the Bounded Burnside Problem. However, the answer to both problems turned out to be “no”. A counterexample to the General Burnside problem was given by a beautiful and elegant construction of Golod and Shafarevich in 1964; and a counterexample to the Bounded Burnside Problem was given by Novikov and Adjan in 1968.

The Burnside Problem has the following ring-theoretic analogue:

Kurosh’s Problem. If $A$ is a finitely generated algebra over a field $F$ and every element of $A$ is nilpotent, then must $A$ be nilpotent?

(A element $a \in A$ is nilpotent if $a^n=0$ for some $n$; and $A$ is itself nilpotent if there is an $m$ such that $a_1a_2\cdots a_m = 0$ for all $a_1,a_2,\ldots,a_m \in A$.) The Golod-Shafarevich construction also provides a counter-example to Kurosh’s Problem. We sketch the construction below.

The Golod-Shafarevich Theorem

Let $F$ be a field and let $T = F\langle x_1,x_2,\ldots x_d \rangle$ be the free non-commutative algebra over $F$ generated by the variables $x_1,x_2,\ldots,x_d$. Let $T_n$ denote the subspace of $T$ consisting of all linear combinations of monomials of degree $n$. The elements of $T_n$ are the homogeneous elements of degree $n$. Let $I$ be a two-sided ideal of $T$ generated by a set $A$ of homogeneous elements, each of degree at least 2. Suppose that $A$ has at most $r_i$ elements of degree $i$ for $i \geq 2$. The following (which we do not prove here) is the main computational result of the Golod-Shafarevich construction:

Theorem. The quotient algebra $T/I$ is infinite dimensional over $F$ if the coefficients in the power series expansion of $(1 - dz + \sum_{i=2}^\infty r_i z^i)^{-1}$ are nonnegative.

Using this theorem, one constructs a counterexample to Kurosh’s Problem as follows.

Counterexample to Kurosh’s Problem

Let $T = F\langle x_1,x_2,x_3 \rangle$ be the free algebra over a countable field $F$. Let $T'$ be the ideal of $T$ consisting of all elements of $T$ without constant term. Enumerate the elements of $T'$ as $t_1, t_2, \ldots$. Choose an integer $m_1 \geq 2$ and write $t_1^{m_1} = t_{1,2} + t_{1,3} + \cdots + t_{1,k_1}$, where each $t_{1,j} \in T_j$. Choose another positive integer $m_2$ sufficiently large so that $t_2^{m_2} = t_{2,k_1+1} + t_{2,k_1+2} + \cdots + t_{2,k_2}$ for some $k_2 > k_1$. Continue in this way for sufficiently large powers of $t_3, t_4, \cdots$. Now let $I$ be the ideal generated by the $t_{i,j}$ defined in the process above. Consider the quotient $T'/I$. The construction of $I$ guarantees that each element of $T'/I$ is nilpotent; but the theorem above ensures that $T'/I$ is infinite dimensional over $F$ (and hence not nilpotent). Thus $T'/I$ is a counterexample to Kurosh’s Problem. From this construction, we can in turn build a counterexample to the General Burnside Problem.

Counterexample to the General Burnside Problem

Let us suppose now that $p$ is a prime number and $F$ is the field with $p$ elements. Let $T$ and $I$ be as defined above. Let $a_1, a_2, a_3$ be the elements $x_1 + I, x_2 + I, x_3 + I$ of the quotient $T/I$ respectively. Let $G$ be the multiplicative semigroup in $T/I$ generated by the elements $1+a_1$, $1+a_2$, and $1+ a_3$. An element of $G$ has the form $1+a$ for some $a \in T'/I$. The element $a$ is nilpotent by the construction of $T'/I$, and so for sufficiently large $n$, we have $a^{p^n} = 0$. Since we are in characteristic $p$ we have $(1+a)^{p^n} = 1 + a^{p^n} = 1$. It follows that $1+a$ has an inverse, whence $G$ is a group. Moreover every element $1+a$ of $G$ has finite order (indeed order a power of $p$). Thus $G$ satisfies the conditions of the General Burnside Problem. It remains to show that $G$ is infinite. If $G$ were finite, then the linear combinations of its elements would form a finite dimensional algebra $B$ over $F$. Moreover, since both 1 and $1+a_i$ are in $G$, the linear combination $(1+a_i) - 1 = a_i$ is in $B$. Thus, $1, a_1, a_2, a_3$ are all in $B$. But $1, a_1, a_2, a_3$ suffice to generate $T/I$, which was previously shown to be infinite dimensional. The algebra $B$ is therefore also infinite dimensional, a contradiction. Thus $G$ is infinite, as required.

Written by uncudh

September 4, 2009 at 12:20 am