Thedi vanda lakshmi

Posted November 9, 2009 by N
Categories: click, humour, movies

Our very own charade , just a decade later but in black and white.

lakshmijaishankar

vs

audreygrant

Grumpy ?

Posted November 6, 2009 by N
Categories: click, daily news

Posted October 31, 2009 by N
Categories: click, humour, music

It (even has raravenu bits incorporated)

Nostalgia

Posted October 21, 2009 by N
Categories: music, poetry

..
valarndhida
thaneerai vaarthavan
verai , paripadhazhago ?


Posted October 18, 2009 by N
Categories: click, crazes

This is why everyone should know how to skate.

The problem with ..

Posted October 11, 2009 by N
Categories: humour

Ashwin’s one line solution to this:

To prove that if ( \sum a_i g_i) (\sum b_j h_j) = 0 then either all a_i or all b_j are zero, consider the subgroup H generated by g_i and h_j. H is finitely generated and torsion free and hence free.

The problem with generalizations

Posted October 8, 2009 by N
Categories: thoughts

My office mate stumbled upon this innocuous statement in a paper,

Note that because P is an abelian multiplicative group with no torsion elements , the group ring ZP has no zero divisors

Breaking down the fancy terminology, apparently \mathbb{Z}P is nothing but the ring of all formal integral linear combinations of elements of P with the condition that any set of elements of P are \mathbb{Z } linearly independent in this ring. More simply,

any element of \mathbb{Z}P looks like \sum_{i=1}^{n}a_ip_i where a_i\in\mathbb{Z} and p_i\in P and this element can be 0 if and only of all the a_i’s are 0.

After trying in vain to come up with counter examples, we decided to prove it. Note that a nice example is \mathbb{Z}[x,\frac{1}{x}], the ring of laurent polynomials in one variable, which sort of convinces one why this must be true. And what else to appeal to first, but induction ? So here goes.

Let us induct on the number of  basis elements (k) needed to express a divisor.

k=1

That is, the zero divisor looks like rp where r is a non-zero integer and p is an element of P. Let

rp(a_1p_1+a_2p_2+\ldots a_np_n)=0,

where p_1,p_2, ..p_n are distinct elements of P and a_is, non zero integers. So you have

\sum_{i=1}^{n}ra_ipp_i=0 ,

and as p_is are distinct, so are pp_is, and you get ra_i=0\forall i, which gives you r=0. Cheap, yes. And we don’t use the fact about torsionlessness (er) at all, but then, you hit case number 2,

k=2

Let (ap+bq)(a_1p_1+a_2p_2+\ldots + a_np_n)=0 where a,b,a_i\neq 0 are integers and p\neq q, p_i\neq p_j\forall i,j are elements of P. Expanding, you get

\sum_{i=1}^{n}aa_ipp_i + \sum_{i=1}^{n}ba_iqp_i = 0.

Note that pp_i\neq pp_j for any i\neq j and similarly for qq_is too. But to get 0, they must cancel out! So given any i, there exists a j such that pp_i=qp_j. Also note that this j is unique, as pp_i=qp_j=qp_s would imply, p_j=p_s. And also, no two i, i’ can have the same j (By that I mean pp_i=qp_j, pp_{i'}=qp_j is not possible if i\neq i')

So all we are finally saying is that, there exists a permutation \sigma of the set {1,2,..n} such that

pp_i=qp_{\sigma(i)} ———————————————————–(*).

Now, for the trick. Take product of  (*) over all i\leq n. We get,

\prod_{i}pp_i=\prod_iqp_{\sigma(i)}, which simplifies to p^n=q^n or (\frac{p}{q})^n=1. Since P is torsion free (aha!), this would mean p=q, a contradiction to our initial assumption.

This looks more like it (or so we thought). However hopes for an inductive proof were lost at this point of time. But then, whatever works here must generalize, right ? Nope. I mean, I don’t know if it does or doesn’t :P. After two days of conscientiously avoiding googling the problem, I gave up and found this on wikipedia.

A long-standing conjecture of Kaplansky (~1940) says that if G is a torsion-free group, and K is a field, then the group ring K[G] has no non-trivial zero divisors. In fact, the condition that K is a field can be relaxed to any ring that can be embedded into an integral domain.The conjecture remains open in full generality, however some special cases of torsion-free groups have been shown to satisfy the zero divisor conjecture.

So much for that. It just seemed so doable.

However, as a consolation, for any finitely generated abelian group P, torsion free would mean that P\cong\mathbb{Z}^n by structure theorem, and so the ring ZP would just be the ring of laurent polynomials in finitely many variables, (ie)

ZP=\mathbb{Z}[x_1^{\pm 1},x_2^{\pm 2},\ldots x_n^{\pm 1}],

which is a domain. The proof  I expect follows on the similar lines of proving that a polynomial ring of  several variables is a domain, which incidentally I always have trouble proving. Can someone give me a proper proof which is also not very messy :P ?

Ask rajini

Posted October 6, 2009 by N
Categories: click

The shop around the corner

Posted October 3, 2009 by N
Categories: movies

After wondering of what use film reviews are, especially if they are just passages exuberating one’s favourite films and scenes, as this one, I have come to the conclusion that they are meant only for the eyes of kindred souls, who have seen the movie and are equally in love with it as the writer. However, any hobby or habit which decides to go under the name of a passion, forces the person to become a minor evangelist, especially if they own a blog. That’s why this article exists, for human nature lives on constant hope that one can convert people, hopeless though the case has been proven to be

shoparoundcorner

The old old story about the man and the woman (nowadays, it is the high school dude and the class nerd) falling in love over letters (emails) but hating each other in real life has been wrung dry to the point that the normal procedure to follow when you realize that you have been trapped into watching a film with such a premise is to either cringe in your seat and fidget like a toothy overactive bunny or to make yourself comfortable and doze off into a pleasant stupor and occasionally wake up to see if either of the lead pair have realized who their letter writing partner is.

Of course, when Lubitsch directs it, it becomes a masterpiece. The film opens with a quiet title card announcing that , about to follow is a story about Matuschek and Co, that shop around the corner in Budapest, Hungary. You can comfortably predict the story two minutes into the movie, the key word here being comfortably. I think the film’s greatest achievement lies in the fact that you(er, the saner among you) would still want to watch it after the said two minutes of the film.

We have the Lubitsch regulars playing their bit (or not so bit) roles. Remember the wanna-be Shylock from “To be or not to be” ? He is Ferencz Vadas here, the man who flees everytime Mr Matuschek wants an honest opinion from his employees and rightfully thinks that good friends are those who visit one after dinner is done.

“Well, we started talking about love.. culturally, of course” “What else can you do over letters?”

Margaret Sullavan as Klara is passable, but James Stewart takes the cake. Somehow I can only recollect one constant expression of his in all the films I have seen him in, but he seems to fit whatever role he is playing to a T, if you will allow the cliched usage. One of the priceless scenes in the film according to me, is when Stewart with Vadas stands outside the cafe, finding out who his angelic letter writer is. Especially if you compare this with its rip off – “You’ve got mail”.

The difference, I think, lies in the fact that Shop around the corner, is just what it claims to be, a story about Matuschek and co. And since I’ve admitted this is an evangelistic post, let me parade

Doctor: Pardon me Mr.Katona? Precisely what position do you hold with Matuschek and Company?
Pepi Katona: Well, I would describe myself as a contact man. I keep contact between Matuschek and the customers… on a bicycle.
Doctor: Do you mean, an errand boy?
Pepi Katona: Doctor, do I call you a pill-peddler?

quotes from IMDB. Shameless, yes, but if you’re still reading this, you clearly must know it already.

Photographs and memories

Posted September 16, 2009 by N
Categories: music

Jim Croce

Photographs and memories,
Christmas cards you sent to me
All that I have are these
To remember you

Memories that come at night
That take me to another time
Back to a happier day
When I called you mine

But we sure had a good time
When we started way back then
Morning walks and bedroom talks
Oh, how I loved you then

Summer skies and lullabies
Nights we couldn’t say goodbye
And all of the things that we knew
Not a dream survives

Photographs and memories
All the love you gave to me
Somehow it just can’t be true
That’s all I have left of you