Noncommutative Analysis

Category: Fun stuff

The perfect Nullstellensatz just got more perfect

After giving a talk about the perfect Nullstellensatz (the commutative free Nullstellensatz) at the Technion Math department’s pizza and beer seminar, I had a revelation: I think it holds over other fields as well, not just over the complex numbers! (And in particular, contrary to what I thought before, it holds over the reals. It seems to hold over other fields as well). 

To explain, I will need some notation. 

Let k be a field. We write A = k[z_, \ldots, z_d] – the algebra of all polynomials in d (commuting) variables over the field k

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Polya’s three rules of style

In G. Polya‘s book “How to Solve It”, one of the shortest sections is called “Rules of style”. This section contains Polya’s three rules of style, which are worth repeating.

“The first rule of style”, writes Polya, “is to have something to say”.

“The second rule of style is to control yourself when, by chance, you have two things to say; say first one, then the other, not both at the same time”.

Polya’s third rule of style is: “Don’t say what does not need to be said” or maybe “Don’t say the obvious”. I am not sure of the exact formulation, because Polya doesn’t write the third rule down – that would be a violation of the rule!

Polya’s three rules are excellent and one is advised to follow them if one strives for good style when writing mathematics. However, style is not the only criterion by which we measure mathematical writing. There is a tradeoff between succinct and elegant style, on the one hand, and clarity and precision, on the other.

“Don’t say the obvious” – sure! But what is obvious? And to whom? A careful writer leaving a well placed exercise in a textbook is one thing. An author of a long and technical paper that leaves an exercise to the poor, overworked referee, is something different. And, of course, a mathematician leaving cryptic notes to his four-months-older self, is the most annoying of them all.

“Don’t say the obvious” – sure, sure! But is it even true? I think that all the mistakes that I am responsible for publishing have originated by an omission of an “obvious” argument. I won’t speak about actual mistakes made by others, but I do have the feeling that some people have gotten away with not explaining something non-trivial, and were lucky that things turned out to be as their intuition suggested (granted, having the correct intuition is also a non-trivial achievement).

I disagree with Polya’s third rule of style. And you see, to reject it, I had to formulate it. QED.

The perfect Nullstellensatz

Question: to what extent can we recover a polynomial from its zeros?

Our goal in this post is to give several answers to this question and its generalisations. In order to obtain elegant answers, we work over the complex field \mathbb{C} (e.g., there are many polynomials, such as x^{2n} + 1, that have no real zeros; the fact that they don’t have real zeros tells us something about these polynomials, but there is no way to “recover” these polynomials from their non-existing zeros). We will write \mathbb{C}[z] for the algebra of polynomials in one complex variable with complex coefficients, and consider a polynomial as a function of the complex variable z \in \mathbb{C}. We will also write \mathbb{C}[z_1, \ldots, z_d] for the algebra of polynomials in d (commuting) variables, and think of polynomials in \mathbb{C}[z_1, \ldots, z_d] – at least initially – as a functions of the variable z = (z_1, \ldots, z_d) \in \mathbb{C}^d

[Update June 24, 2019: contrary to what I thought, the main theorem presented below holds over arbitrary fields, not just over the complex numbers, very much by the same proof. See this post.]

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Journal of Xenomathematics

I am happy to advertise the existence of a new electronic journal/forum/website: Journal of Xenomathematics. Don’t worry, it’s not another new research journal. The editor is John E. McCarthy. The purpose is to discuss mathematics that is out of this world. Aren’t you curious?

A proof of Holder’s inequality

One of the parts of this blog that I am most proud of is my series of “Souvenirs” post, where I report about my favorite new finds in conferences. In July I went to a big conference (IWOTA 2016 in St. Louis) that I was looking forward to going to for a long time, but I did not write anything after I returned. It’s not that there was nothing to report – there was a lot, it was great conference. I was just too busy with other things.

Why am I so busy? Besides being the father of seven people (most of them “kids”) and preparing for next year, I am in the last stages of writing a book, partly based on the lecture notes on “Advanced Analysis” that appeared in this blog, and on lecture notes that evolved from that. (When it will be ready I will tell you about it, you can be sure). I want to share here and now one small excerpt from it (thanks to Guy Salomon for helping me finesse it!)

Working on the final touches to the book, I decided to include a proof of Holder’s inequality in it, but I did not want to copy a proof from somewhere.  So I came up with the following proof, which I think is new (and out of curiosity I am asking you to please tell me if you have seen it before). The lazy idea of the proof is to use the fact that we already know – thanks to Cauchy-Schwarz – that the inequality holds in the p =2 case, and to try to show how the general case follows from that.

In other words, instead of bringing you fancy souvenirs from St. Louis, I got you this little snack from the nearby mall (really, the proof crystallized in my head when my daughter, my dog and I were sitting and waiting on a bench in the mall until other members of our family finish shopping).

Definition. Two extended real numbers p,q \in [1, \infty] are said to be conjugate exponents if

\frac{1}{p} + \frac{1}{q} = 1.

If p=1 then we understand this to mean that q = \infty, and vice versa.

For any (finite or infinite) sequence x_1, x_2, x_3, \ldots, and and any p \in [1,\infty], we denote

\|x\|_p =\big(\sum |x_k|^p \big)^{1/p}.

Theorem (Holder’s inequality): Let p,q \in [1, \infty] be conjugate exponents.
Then for any two (finite or infinite) sequences x_1, x_2, \ldots and y_1, y_2, \ldots

\sum_k |x_k y_k| \leq \|x\|_p \|y\|_q.

Proof. The heart of the matter is to prove the inequality for finite sequences. Pushing the result to infinite sequences does not require any clever idea, and is left to the reader (no offense).
Therefore, we need to prove that for every x = (x_k)_{k=1}^n and y = (y_k)_{k=1}^n in \mathbb{C}^n,

(HI)   \sum |x_ky_k| \leq \big(\sum |x_k|^p \big)^{1/p} \big( \sum |y_k|^q \big)^{1/q}.

The case p=1 (or p=\infty) is immediate. The right hand side of (HI) is continuous in p when x and y are held fixed, so it enough to verify the inequality for a dense set of values of p in (1,\infty).


S = \Big\{\frac{1}{p} \in (0,1) \Big| p satisfies  (HI)  for all x,y \in \mathbb{C}^n \Big\}.

Now our task reduces to that of showing that S is dense in (0,1). By the Cauchy-Schwarz inequality, we know that \frac{1}{2} \in S. Also, the roles of p and q are interchangeable, so \frac{1}{p} \in S if and only if 1 - \frac{1}{p} \in S.

Set a = \frac{q}{2p+q} (a is chosen to be the solution to 2ap = (1-a)q, we will use this soon). Now, if \frac{1}{p} \in S, we apply (HI) to the sequences (|x_k| |y_k|^{a})_k and (|y_k|^{1-a})_k, and then we use the Cauchy-Schwarz inequality, to obtain

\sum |x_k y_k| = \sum|x_k||y_k|^a |y_k|^{1-a}

\leq \Big(\sum |x_k|^p |y_k|^{ap} \Big)^{1/p}\Big(\sum |y_k|^{(1-a)q} \Big)^{1/q}

\leq \Big((\sum |x_k|^{2p})^{1/2} (\sum |y_k|^{2ap})^{1/2} \Big)^{1/p}\Big(\sum |y_k|^{(1-a)q} \Big)^{1/q}

= \Big(\sum |x_k|^{p'} \Big)^{1/p'} \Big(\sum|y_k|^{q'} \Big)^{1/q'}

where \frac{1}{p'} = \frac{1}{2p} and \frac{1}{q'} = \frac{1}{2p} + \frac{1}{q}. Therefore, if s = \frac{1}{p} \in S, then \frac{s}{2} = \frac{1}{2p} \in S; and if s = \frac{1}{q} \in S, then \frac{s+1}{2} = \frac{1}{2}\frac{1}{q}+\frac{1}{2} = \frac{1}{q} + \frac{1}{2}\frac{1}{p} is also in S.

Since \frac{1}{2} is known to be in S, it follows that \frac{1}{4} and \frac{3}{4} are also in S, and continuing by induction we see that for every n \in \mathbb{N} and m \in \{1,2, \ldots, 2^n-1\}, the fraction \frac{m}{2^n} is in S. Hence S is dense in (0,1), and the proof is complete.