User paul-olivier dehaye - MathOverflow

Answer by Paul-Olivier Dehaye for Location of Archimedes' grave in Syracuse (math/archaelogy trivia)

2012-10-20T15:33:36Z

I think the book http://books.google.ch/books?id=6TBY0R4ZPVYC&lpg=PP1&hl=fr&pg=PA598#v=onepage&q&f=false corroborates much of what you say about the connection Necropolis del Fusco-Archimedes. This book was written in 1904. If you search for "Orsi" in the book, you see that a professor Paolo Orsi was an "antiquarian" who contributed extensively to the book. Interesting terms to search for: "Archimedes", "Fusco", "Agragian". With all this you can patch together a topographic description of the area.

So in short, this professor (indirectly) confirms the link Fusco-Archimedes. It is possible the shopping center is sitting right on top of the grave, but the area of the "west of the necropolis" is still pretty big, so it might be a bit much to deduce that.

Product on representations of an integer by a quadratic form?

2012-08-25T07:06:18Z

Define the quadratic form $$Q(z_1,z_2,z_3,z_4) = 13 + \sum_{i=1}^4 (10+i)z_i +5 \sum_{1 \le i \le j \le 4} z_iz_j.$$ Then, $r_Q(n) := \left|{(z_1,z_2,z_3,z_4) \in \mathbb{Z}^4 : Q(z_1,z_2,z_3,z_4) = n }\right|$ is weakly multiplicative. I can prove this by using the generating function $\sum r_Q(n) q^n$ which is in the Eisenstein subspace of $\mathcal{M}_2(\chi_5)$, with $\chi_5$ the Legendre symbol.

Because $r_Q(n)$ counts something, a likely alternative explanation is that there exists some product on solutions to $Q(\vec{z})=n$ that would explain this multiplicativity directly, hence the question:

Does there exist a product $\times$ on solutions to $Q(\vec{z})=n$ such that, whenever $(m,n)=1$,

$Q(\vec{x})=m$ and $Q(\vec{y})=n$ imply $Q(\vec{x}\times \vec{y})=mn$;

$Q(\vec{z})=mn$ implies we can find unique $\vec{x}, \vec{y}$ with $Q(\vec{x})=m$, $Q(\vec{y})=n$ and $\vec{x} \times \vec{y} = mn$?

I am interested in this because results of Garvan, Kim and Stanton give a bijection between the representations of $n$ by $Q$ and the number of $5-$core partitions of size $n-1$, which would lead to a product on 5-cores that I would like to understand. This multiplicativity has been used at 5 only by GKS to show combinatorially that $p(5n+4) \equiv 0 \mod5$.

Note 1: After the change of variables $v_i := 5z_i+i$ and the introduction of the fifth variable $v_0 := -v_1-v_2-v_3-v_4$, one can also define $Q$ by the more symmetric and homogeneous $$ Q(v_0,v_1,v_2,v_3,v_4) = \frac{1}{10} \sum_{i=0}^4 v_i^2, $$ but we are not looking at all solutions then: we need $\sum_{i=0}^4 v_i = 0$ and $v_i \equiv i \mod 5$.

Note 2: For the multiplicativity, 5 is special at the moment, but could conceivably be replaced by 7 and 11 later, judging from the theory of Garvan, Kim and Stanton. However I am hoping that a combinatorial construction for the product on 5-cores could be generalized more widely.

UPDATE: I am sure the $r_Q(n)$ is not completely multiplicative. In fact, here is a list of the first 50 values, starting at 1: 1, 1, 2, 3, 5, 2, 6, 5, 7, 5, 12, 6, 12, 6, 10, 11, 16, 7, 20, 15, 12, 12, 22, 10, 25, 12, 20, 18, 30, 10, 32, 21, 24, 16, 30, 21, 36, 20, 24, 25, 42, 12, 42, 36, 35, 22, 46, 22, 43, 25,... ( it's at http://oeis.org/A053723 ). There are actually closed forms there for $r_Q(p^e)$, but I don't see how they help for my question.

In particular, we have $r_Q(2) = 1$, and $r_Q(4) = 3$, corresondign respectively to the solutions $(0, -1, 0, -1)$ and $(-1, 0, -1, 0), (0, -1, -1, -1), (0, 0, 0, -1)$.

Answer by Paul-Olivier Dehaye for Ingenuity in mathematics

2010-09-28T17:08:26Z

Certainly Cantor diagonalization? You have Russell's paradox, which is perfectly understandable to the lay person, going to the uncountability of reals, going to Godel's incompleteness. I think each incarnation passes Andrew Stacey's tests...

Answer by Paul-Olivier Dehaye for Is there a generalisation of the "sunflower spiral" to higher dimensions?

2010-09-27T20:15:20Z

I think a good reference for this is "The discrepancy method" by Bernard Chazelle, page 59 and following. It's available at http://books.google.com/books?id=dmOPmEh6LdYC&printsec=frontcover&dq=the+discrepancy+method&hl=en&ei=WvqgTO7xLIXKOLzCxa4M&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false to read. Chapter title is "An orbital construction for points on a spehere", and the method is compared to $\phi$ for the circle.

Answer by Paul-Olivier Dehaye for A sum involving irreducible characters of the symmetric group

2010-09-27T12:58:22Z

Hi Thomas,

This is very similar to things I have been working on or thinking about. I would be very interested to know the source of this problem. My own source for similar questions was in Random Matrix Theory. What do you want to know about those sums?

First a remark. In your definition of H(n,L), you could see this as a sum over all partitions of $2n$. The term $ \chi^{Y_{i,j,w}}(\tau)$ ensures only the diagrams you describe give non-zero contribution.

You might know this, but the $-p+q$ is the content of the cell $(p,q)$.

You can compute $\chi^{Y_{i,j,w}}([2^n])$ via the Murnaghan-Nakayama rule indeed: it is counting all the different ways to express $Y_{i,j,w}$ as an incremental union of 2-ribbons (with signs!). Another way uses the so-called rim hook lattice. If the 2-core $Y_{i,j,w}$ is non-empty, this character value is 0. If it is empty, take the two 2-quotients of $Y_{i,j,w}$ and compute the dimension of each. Combining this with a binomial sum, you can obtain the character value you are looking for. [*]

Finally, I would send you to Stanley's recent paper (look at the version on his homepage http://math.mit.edu/~rstan/papers.html#hooks ) called ''Some combinatorial properties of hook lengths, contents, and parts of partitions'' There he has sums over partitions of functions of the contents (like you do), but weighted by the Plancherel measure. I think you can make that measure aappear out of your sum. His result would typically take the form of a polynomiality in $n$ of the $L^i$ coefficient of $H(n,L)$.

[Edit], concerning part [*]: This is essentially what's done in the accepted answer to this question: http://mathoverflow.net/questions/8776/statistics-of-irreps-of-s-n-that-can-be-read-off-the-young-diagram-and-consequen (I was describing the Fomin-Lulov method, while that answer to the MO question describes their result).

How to recover partition from its multiset of hook lengths?

2010-09-14T11:17:25Z

One of the invariants associated to a partition is its multiset of hook lengths. For instance, as shown here, the partition (5,4,1) has hook lengths {1,1,1,2,3,3,4,5,5,7}. Is there a good way to go backwards (up to conjugation, of course)? An actual algorithm that does not involve brute forcing through a bunch of possibilities, and would give me one (or all) of the possible partitions?

What about the decision problem "Is this multiset the multiset of hook lengths of a partition?" For instance, for {1, 1, 1, 2, 3, 3, 4, 4, 5, 7, 8, 10} the answer is no, but I can only come up with very ad hoc ways to show that.

Answer by Paul-Olivier Dehaye for How to recover partition from its multiset of hook lengths?

2010-09-15T16:20:45Z

One condition on hooks that hasn't been mentioned yet is that $$ h_{(1,1)} + h_{(i,j)} = h_{(i,1)} + h_{(1,j)}. $$ (and similarly, based at other points).

When $i<>1$ and $j<>1$, this means that the cells $(1,1)$ and $(i,j)$ in that property are corners of a (unique) square, whose other two corners are the other two points.

Since 10+4 is not expressible as the sum of any two of the other given hook lengths in my negative example, we know that 4 and 4 are on the same line as 10 (not that suprising given the large size of first hook length 10 compared to the size 12, but could be useful for bigger partitions).

Answer by Paul-Olivier Dehaye for Explicit computations using the Haar measure

2010-08-23T22:23:53Z

Couldn't resist to point this out: actually, $$ \lim_{N \rightarrow \infty} \frac{1}{N^{k^2}} \mathbb{E}_{U\in U(N)} |Z_U|^{2k} = \frac{G(k+1)^2}{G(2k+1)}, $$ with $G$ the Barnes $G$ function, $Z_U$ the characteristic polynomial of the matrix $U$ in $U(N)$, taken according to Haar measure and evaluated somewhere along the unit circle in $\mathbb{C}$, say 1 (where on the circle is irrelevant as Haar measure is rotationally invariant).

When $k=3$, this actually evaluates to $$ \frac{\mathbf{42}}{9!}, $$ and accounts for the 3rd (and first new) case in the Keating-Snaith discovery of the interest of random matrices for quantitative formulations of analytic number theory conjectures, as explained in their paper "Random Matrix Theory and $\zeta (1/2 + i t )$" or less formally in http://seedmagazine.com/content/article/prime_numbers_get_hitched/ .

Incidentally, the connections between random matrix theory and number theory indeed lead to many practical computations for Haar-random matrices in classical compact groups. Some are easier to understand (for instance via the Weyl integration formula or reformulations in terms of Selberg integrals), while some are much less clear (for instance, subsitute in the above statement $|Z_U'|$ instead of $|Z_U|$ and study again the behaviour for large k, or analytic continuation in k of the RHS. The renormalization in that case, however, is know, and would be $\frac{1}{N^{k^2+2k}}$).

Since this is a reference request, look also at papers by Conrey or Hughes for examples of such explicit computations.

Comment by Paul-Olivier Dehaye

2012-10-20T15:16:31Z

Britannica says a tiny bit more about the location (And various other sources too): books.google.ch/… It looks like the necropolis was inside the walls, which included the Fusco gorge, a defensive weakness.

Comment by Paul-Olivier Dehaye

2012-10-20T15:01:56Z

books.google.ch/… Footnote 39. It was the line from Syracuse to Noto. This is fun on a lazy Saturday.

Comment by Paul-Olivier Dehaye

2012-10-20T14:43:31Z

I am not sure "rotabile" translate to "railway". In fact, there is no railway leading to Floridia (they go either north or south, if you follow them on Google maps, not west to Floridia), but there is a road in the area leading in the right direction (Strada Statale 124). About railways, it says somewhere that Orsi's work were prompted by the construction of a railway. So this would certainly be consistent with putting the necropolis south and east of the "via Necropolis del Fusco" (note that it goes on after crossing the railway), leaving a chance the tomb would be in an open field.

Comment by Paul-Olivier Dehaye

2012-08-27T08:48:13Z

Yes, a potential way to get the product is by showing that $Q$ behaves like a norm on an order of a particular quaternion algebra. I am hoping invariants of the form can help narrow down where to look.

Comment by Paul-Olivier Dehaye

2012-08-26T03:39:39Z

It is not completely multiplicative. I think the two conditions you wrote are equivalent. Anyways, it fails $r_Q(4) = r_Q(2)^2$, as shown in my update.

Comment by Paul-Olivier Dehaye

2010-10-07T09:17:55Z

This is also what I would look for, but it won't work here. To work, you need for every single term to be able to "cancel poles" and give a meaning to that term. But consider $$ \alpha =0 , \beta_i =0, \delta_i =0 , \epsilon=0 $$ and $n=k$. Then $T$ will have one $(-n)!$, with all other factors involving only factorials of positive integers...

Comment by Paul-Olivier Dehaye

2010-10-04T13:13:50Z

You could collect seven or eight million, not six or seven (depending on Yang Mills): the P=NP prize is the only one formulated in such a way that either a positive or negative answer is guaranteed 1 million (in contrast, if one found a zero of zeta off the line, a committee would have to decide whether this was really a significant achievement worthy of that million). So if you prove inconsistency, you (or your lawyer) can claim P=NP and P<>NP...

Comment by Paul-Olivier Dehaye

2010-10-01T13:26:42Z

There is also the work of Sylvester, but I haven't seen that presented anywhere in detail.

Comment by Paul-Olivier Dehaye

2010-10-01T13:23:34Z

There is also a summary of that paper of Diamond and Erdos in Diamond's "Elementary methods in the study of the distribution of prime numbers" in the Bulletin AMS (sections 3 and 9). It exposes the ideas in several steps: first $$\frac{(2n)!}{n!n!}$$, then $$\frac{(30 n)! n!}{(15 n)! (10 n)! (6 n)!}$$ (Chebyshev's approach) and finally a (related) general form, due to Erdos-Diamond, which uses a function $\mu_T$ close to the Moebius $\mu$ function. The closest analogue of Chebyshev's is then $$ \frac{(30 n)!^5(5n)!}{(15 n)!^5(10 n)!^5(6 n)!^5}. $$

Comment by Paul-Olivier Dehaye

2010-09-28T19:12:38Z

I plead guilty to being a nerd, in any case.

Comment by Paul-Olivier Dehaye

2010-09-28T19:11:40Z

What I had in mind when I wrote "Russell's paradox" is the barber paradox version. I think it can be understood by a painter. If it helps, you can even turn it into a painter paradox version, with the painter drawing portraits! The second part, about Cantor or Godel, can be put into language very similar to the barber paradox. While I don't expect a painter to understand all the words that fit in these statements, it should be possible to explain the similarity with the more tangible barber paradox. And you can confidently say that this idea has revolutionized math twice in the past century.

Comment by Paul-Olivier Dehaye

2010-09-28T08:31:05Z

Thanks, I learned a new "word".

Comment by Paul-Olivier Dehaye

2010-09-17T08:44:33Z

There is also Selberg, Atle, Chowla, S, "On Epstein's zeta-function." J. Reine Angew. Math. 227 1967 86--110. (Selberg's only joint paper)

Comment by Paul-Olivier Dehaye

2010-09-15T16:24:08Z

For the record, another ad hoc way I have to exclude {1, 1, 1, 2, 3, 3, 4, 4, 5, 7, 8, 10} is as follows. Only 2 cells are not in the first hook and they thus have hook lengths 1 and 2. 8 has to be next to 10, either aligned with the 2,1 we just placed or not. Both choices quickly lead to contradiction, as we now know exactly how long the arm based at 8 is.

Comment by Paul-Olivier Dehaye

2010-09-15T16:03:28Z

Yes. this is the "very ad hoc" ways I was referring in my original post. Actually, the argument does not require branching if you take k=4, since you then have the multiset {1,1,2} which is supposed to be the union of 4 partitions. Since I know what those are (empty, empty, [1], ([2] or $[2]^t$)) and I know the core as well (empty), I "only" have 2 * 4! /2 possiblities to check, corresponding to the different orderings of the quotients. The hooks have to satisfy tremendous divisibility conditions, and it is precisely to clarify this that I am asking the question...