### DARPA Challenge

Submissions that merely promise incremental improvements over the existing state of the art will be deemed unresponsive.I feel yet another funding proposal coming on ... HAPPY NEW YEAR!

occasional meanderings in physics' brave new world

At first I thought the problem list was a mildly amusing, handwaving bit of entertainment, but it turns out that the U.S. DARPA Mathematical Challenge has funding opportunities, open also to foreigners! And the 3 page announcement is the coolest I've ever seen, including the words

Submissions that merely promise incremental improvements over the existing state of the art will be deemed unresponsive.I feel yet another funding proposal coming on ... HAPPY NEW YEAR!

Registration for Neutrino 08 here at UC is open, so make sure you consider heading up this way!

I know it's a bit late for this year, but I found the perfect cheap present for a budding M theorist: the Sudokube! Of course, some basic knowledge of magic squares makes it too easy to solve, but it would look good on the shelf. And if you don't mind me saying so, Santa, I was a bit disappointed with The Golden Compass. Why were all the physicists male? And that extended arm double ice axe arrest was just plain ridiculous.

Putting together the Hoffman and Castro expressions, for real $s$ and $t$ in the critical interval with $|s+t| < 1$, we obtain

$\sum_{m,n} s^m t^n \zeta (x^m y^n) = t [ \frac{\zeta (s)}{\zeta (1-s)} \frac{\zeta (1+t)}{\zeta (-t)} \frac{\zeta (1-s-t)}{\zeta (s+t)}]$

where the left hand side is the expression

$\sum_{m} \frac{s^m}{m!} \sum_{k_1,k_2,\cdots,k_m} \frac{1}{k_1 k_2 \cdots k_m} \sum_{n} \frac{t^n}{(k_1 + k_2 + \cdots + k_m)^{n}}$

Specific values of the zeta function include, for the choice $t = 0.5$, $\zeta (1.5) = 2.612$ and, using the functional equation,

$\zeta (- \frac{1}{2}) = \frac{1}{\sqrt{2}} \pi^{\frac{-3}{2}} \Gamma (\frac{3}{2}) \zeta (\frac{3}{2})$

so that the centre ratio in the first equation above becomes

$\sqrt{2} \pi^{\frac{3}{2}} \frac{2}{\sqrt{\pi}} = 2 \sqrt{2} \pi$

giving a particularly interesting relation for the parameter $s < \frac{1}{2}$ involving the expression

$\frac{\zeta (0.5 - s)}{\zeta (0.5 + s)} \frac{\zeta (s)}{\zeta (1-s)}$

It would be nice to extend this to complex values of the parameters, because zeroes of the zeta function occur in conjugate pairs and the finite positivity of an MZV could then rule out zeroes lying in this region.

$\sum_{m,n} s^m t^n \zeta (x^m y^n) = t [ \frac{\zeta (s)}{\zeta (1-s)} \frac{\zeta (1+t)}{\zeta (-t)} \frac{\zeta (1-s-t)}{\zeta (s+t)}]$

where the left hand side is the expression

$\sum_{m} \frac{s^m}{m!} \sum_{k_1,k_2,\cdots,k_m} \frac{1}{k_1 k_2 \cdots k_m} \sum_{n} \frac{t^n}{(k_1 + k_2 + \cdots + k_m)^{n}}$

Specific values of the zeta function include, for the choice $t = 0.5$, $\zeta (1.5) = 2.612$ and, using the functional equation,

$\zeta (- \frac{1}{2}) = \frac{1}{\sqrt{2}} \pi^{\frac{-3}{2}} \Gamma (\frac{3}{2}) \zeta (\frac{3}{2})$

so that the centre ratio in the first equation above becomes

$\sqrt{2} \pi^{\frac{3}{2}} \frac{2}{\sqrt{\pi}} = 2 \sqrt{2} \pi$

giving a particularly interesting relation for the parameter $s < \frac{1}{2}$ involving the expression

$\frac{\zeta (0.5 - s)}{\zeta (0.5 + s)} \frac{\zeta (s)}{\zeta (1-s)}$

It would be nice to extend this to complex values of the parameters, because zeroes of the zeta function occur in conjugate pairs and the finite positivity of an MZV could then rule out zeroes lying in this region.

The original Hoffman post mentioned an expression in $\Gamma$ functions, similar to that appearing in the relation

$B(a,b)= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} + \frac{\Gamma(a)\Gamma(c)}{\Gamma(a + c)} + \frac{\Gamma (1 - a - b) \Gamma (b)}{\Gamma (1 - a)}$

$ = \frac{\zeta (1 - a)}{\zeta (a) } \frac{\zeta (1 - b)}{\zeta (b) } \frac{\zeta (a + b)}{\zeta (1 - a - b) }$

which appears in Castro's discussion of the zeroes of the Riemann zeta function. The $B$ function is the familiar 4-point amplitude of Veneziano, which we have been expressing in terms of chorded polygons; in this case a square with two diagonals representing the 1 dimensional associahedron, the interval.

$B(a,b)= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} + \frac{\Gamma(a)\Gamma(c)}{\Gamma(a + c)} + \frac{\Gamma (1 - a - b) \Gamma (b)}{\Gamma (1 - a)}$

$ = \frac{\zeta (1 - a)}{\zeta (a) } \frac{\zeta (1 - b)}{\zeta (b) } \frac{\zeta (a + b)}{\zeta (1 - a - b) }$

which appears in Castro's discussion of the zeroes of the Riemann zeta function. The $B$ function is the familiar 4-point amplitude of Veneziano, which we have been expressing in terms of chorded polygons; in this case a square with two diagonals representing the 1 dimensional associahedron, the interval.

Hoffman's 1997 paper begins with this example of an MZV relation:

$\zeta (2) \zeta (2,1) = 2 \zeta (2,2,1) + \zeta (2,1,2) + \zeta (4,1) + \zeta (2,3)$

which M theorists can try to draw in a number of ways, such as the 2-ordinal picture This suggests that zeta relations are in some sense functorial, or categorified, and arise from relations amongst arguments. In the last post, for instance, the argument of the Riemann zeta function was given by a complex cosmic time coordinate, which is often substituted in M theory for a value of $\hbar$ or $N$th root on the unit circle.

$\zeta (2) \zeta (2,1) = 2 \zeta (2,2,1) + \zeta (2,1,2) + \zeta (4,1) + \zeta (2,3)$

which M theorists can try to draw in a number of ways, such as the 2-ordinal picture This suggests that zeta relations are in some sense functorial, or categorified, and arise from relations amongst arguments. In the last post, for instance, the argument of the Riemann zeta function was given by a complex cosmic time coordinate, which is often substituted in M theory for a value of $\hbar$ or $N$th root on the unit circle.

By now we've all heard about the relation between the Riemann zeta function and Hermitian operators associated to matrix models. With CFT/AdS in the air now, it is not surprising to find this paper by McGuigan, which discusses brane partition functions.

Somehow, according to McGuigan, on the gravity side we are supposed to end up with modular functions like those appearing in the already notorious Witten paper on 2+1D gravity. In fact, the so-called cosmological constant (just think extra time coordinates) appears as the variable $z$ in a function whose zeroes must lie on the real axis, namely

$\Theta (z) = \zeta (iz + \frac{1}{2}) \Gamma (\frac{z}{2} + \frac{1}{4}) \pi^{- \frac{1}{4} - \frac{iz}{2}} (- \frac{z^2}{2} - \frac{1}{8})$

Who would have thought such stuff could get published on the arxiv?

Somehow, according to McGuigan, on the gravity side we are supposed to end up with modular functions like those appearing in the already notorious Witten paper on 2+1D gravity. In fact, the so-called cosmological constant (just think extra time coordinates) appears as the variable $z$ in a function whose zeroes must lie on the real axis, namely

$\Theta (z) = \zeta (iz + \frac{1}{2}) \Gamma (\frac{z}{2} + \frac{1}{4}) \pi^{- \frac{1}{4} - \frac{iz}{2}} (- \frac{z^2}{2} - \frac{1}{8})$

Who would have thought such stuff could get published on the arxiv?

Following up on the GRG18 news, the LIGO collaboration have posted a paper on the non-observation of gravitational waves from the bright electromagnetic event GRB 070201. Of course, this has been reported in a number of places already.

Here are some carols for the celebration of Newton's birthday. I quite like We Three Quarks, which begins

We three quarks fine particles are.

Bearing charm we travel afar.

Fields and forces, spin of course is

Multiplied by h-bar.

Oh, Quarks are wondrous, quarks are light.

Quarks have colors, clear and bright.

Still misleading, still exceeding

All the physicists' insight.

In an increasingly fascinating series of blogposts, the great mathematician Lieven Le Bruyn has finally reached the stringy topic of superpotentials. Apparently Grothendieck's children's drawings are dimers for Dedekind tesselations. Here is the recommended paper by Stienstra.

Aside: The very colourful graduation went well, on a stunning day. There were bagpipes, trumpet fanfares, Maori greetings, a Brahms sonata, singing in Maori, English and Latin, and the town hall organ was played. I would like to check my UC mail, but unfortunately somebody has managed to crash the system on the first day of the holidays, as usual, so I may have to wait until the New Year.

Aside: The very colourful graduation went well, on a stunning day. There were bagpipes, trumpet fanfares, Maori greetings, a Brahms sonata, singing in Maori, English and Latin, and the town hall organ was played. I would like to check my UC mail, but unfortunately somebody has managed to crash the system on the first day of the holidays, as usual, so I may have to wait until the New Year.

An intriguing paper [1] by Kalman Gyory discusses the equation

$m(m+1) \cdots (m + i - 1) = b k^{l}$

For $b = 1$ Erdös and Selfridge proved in 1975 [2] that this equation has no non-trivial solutions in the positive integers. The $(i,l,b) = (3,2,24)$ case can be seen to correspond to the cannonball problem under the substitution $n \mapsto \frac{m}{2}$. In general this suggests that the sequence of switchback expressions

$P_i \frac{\textrm{sum of squares}}{in + T_i}$

may hardly ever be expressed in the form $b k^{l}$ for $k \geq 2$, where $T_i$ is the triangular number $\sum_{j=1}^{i} {j}$, even though it is certainly a positive integer. This is an interesting fact about the cardinality of these faces of the permutohedra, and for some mysterious reason the proof for $b=1$ seems to involve the mathematics of Fermat's last theorem. Note also the similarity between the denominator above and terms in the associahedra sequences $F_{n}(i)$.

[1] K. Gyory, Acta Arith. 83 (1998) 87-92

[2] P. Erdös and J.L. Selfridge, Illinois J. Math. 19 (1975) 292-301

$m(m+1) \cdots (m + i - 1) = b k^{l}$

For $b = 1$ Erdös and Selfridge proved in 1975 [2] that this equation has no non-trivial solutions in the positive integers. The $(i,l,b) = (3,2,24)$ case can be seen to correspond to the cannonball problem under the substitution $n \mapsto \frac{m}{2}$. In general this suggests that the sequence of switchback expressions

$P_i \frac{\textrm{sum of squares}}{in + T_i}$

may hardly ever be expressed in the form $b k^{l}$ for $k \geq 2$, where $T_i$ is the triangular number $\sum_{j=1}^{i} {j}$, even though it is certainly a positive integer. This is an interesting fact about the cardinality of these faces of the permutohedra, and for some mysterious reason the proof for $b=1$ seems to involve the mathematics of Fermat's last theorem. Note also the similarity between the denominator above and terms in the associahedra sequences $F_{n}(i)$.

[1] K. Gyory, Acta Arith. 83 (1998) 87-92

[2] P. Erdös and J.L. Selfridge, Illinois J. Math. 19 (1975) 292-301

As reported in New Scientist, one of my esteemed colleagues from Mt Cook Village has expired after eating too much chocolate. Seriously folks, what are you doing throwing chocolate into the garbage can in a National Park? The kea is now officially endangered.

Courtesy of a commenter at God Plays Dice we have this nice link about the fact that there are no solutions to the sum of squares problem for $n > 24$. This was proved in 1918 by G. N. Watson, in the paper The problem of the square pyramid. In fact, the only solutions are $n = 1$ and $n = 24$. The equation

$\frac{1}{6} n (n+1) (2n + 1) = k^{2}$

originally described a pile of cannonballs, built from a base layer of $k \times k$ balls into a square pyramid of height $n$. So it's really a sphere packing problem.

$\frac{1}{6} n (n+1) (2n + 1) = k^{2}$

originally described a pile of cannonballs, built from a base layer of $k \times k$ balls into a square pyramid of height $n$. So it's really a sphere packing problem.

The 2-ordinal polytopes associated with the symmetric groups are the permutohedra. The number of codimension $k$ faces of the $n$th permutohedron is given in sequence A019538.

The second diagonal has some nice properties. For example, Alexander Povolotsky observed that these numbers $P_{i}$ arise as the right hand coefficients for the following sequence of expressions, indexed by $i$.

$n(n+1)[n + (n+1)] = 6(1 + 4 + 9 + \cdots + n^2 )$

$n(n+1)(n+2)[n + (n+1) + (n+2)] = 36(1 + (1+4) + (1+4+9) + \cdots + (1 + 4 + \cdots + n^2 ))$

This brings to mind the Leech sequence

$1 + 2^2 + 3^2 + \cdots + 24^{2} = 70^{2}$

for $n=24$, for which the first element of the list is expressed

$\frac{1}{3}(\sum_{i=1}^{n} i ) (2n + 1) = n^{2}$

If the squares of integers up to $n > 24$ cannot be summed to a square, it follows that the left hand side can never be a square.

The second diagonal has some nice properties. For example, Alexander Povolotsky observed that these numbers $P_{i}$ arise as the right hand coefficients for the following sequence of expressions, indexed by $i$.

$n(n+1)[n + (n+1)] = 6(1 + 4 + 9 + \cdots + n^2 )$

$n(n+1)(n+2)[n + (n+1) + (n+2)] = 36(1 + (1+4) + (1+4+9) + \cdots + (1 + 4 + \cdots + n^2 ))$

This brings to mind the Leech sequence

$1 + 2^2 + 3^2 + \cdots + 24^{2} = 70^{2}$

for $n=24$, for which the first element of the list is expressed

$\frac{1}{3}(\sum_{i=1}^{n} i ) (2n + 1) = n^{2}$

If the squares of integers up to $n > 24$ cannot be summed to a square, it follows that the left hand side can never be a square.

Continuing with the wonders of table A033282, the $k = 3$ recursion results in the relation

$F_{n+2}(3) F_{n+1}(1) = \frac{1}{2} F_{n+2}(2) [F_{n+1}(2) + \frac{1}{3} n F_{n+1}(1)]$

For example, considering the codimension 3 edges of the 4d polytope we obtain the relation

$84 \times 9 = \frac{1}{2} \times 56 \times (21 + \frac{2}{3} \times 9)$

Isn't it wonderful how the combinatorics of the associahedra gives us so many relations between integers? One might be forgiven for guessing that operads can tell us something about factorization of an integer into primes.

$F_{n+2}(3) F_{n+1}(1) = \frac{1}{2} F_{n+2}(2) [F_{n+1}(2) + \frac{1}{3} n F_{n+1}(1)]$

For example, considering the codimension 3 edges of the 4d polytope we obtain the relation

$84 \times 9 = \frac{1}{2} \times 56 \times (21 + \frac{2}{3} \times 9)$

Isn't it wonderful how the combinatorics of the associahedra gives us so many relations between integers? One might be forgiven for guessing that operads can tell us something about factorization of an integer into primes.

The expression $F_{n}$ for the codimension 1 faces of the associahedron is

$F_{n}(1) = \frac{1}{2} n (n + 3) = \frac{1}{2} n (n + 1) + n = (\sum_{i=1}^{n} i) + n$

This corresponds to rewriting the sequence in the form

2

2 + 2 + 1

2 + 2 + 1 + 3 + 1

2 + 2 + 1 + 3 + 1 + 4 + 1

After eliminating the bold $n$ terms this becomes

2

1 + 2

1 + 2 + (1 + 2)

1 + 2 + 3 + (1 + 1 + 2)

and so on. The previous post also separated the set of $F_{n}(1)$ Young diagrams into a set of $n$ yellow tiled diagrams and another set of $1 + 2 + \cdots + n$ diagrams with at most two purple tiles in a row.

Now consider the $k = 2$ sequence $F_{n}(2)$, which counts the number of divisions of an $(n+3)$-gon into three pieces by two diagonals. The terms of this sequence are given by the formula

$F_{n}(2) = \frac{1}{3} B(n + 4, 2) B(n,2) = \frac{1}{12} n(n-1)(n+3)(n+4) = \frac{1}{6} (n-1)(n+4) F_{n}(1)$

What set of diagrams counts this sequence? The $F_{n}(1)$ factor says that before cutting an $n$-gon into three pieces we must cut it into two. Note also that $F_{2}(2) = F_{2}(1) = 5$ since having cut a pentagon into two pieces, there is only one way to cut it into three. It is therefore more natural to write

$F_{n+1}(2) = \frac{1}{6} n(n+5) F_{n+1}(1) = [\frac{1}{3} (\sum_{i=1}^{n} i) + \frac{2}{3} n] F_{n+1}(1) $

$= [ \frac{1}{3} F_{n}(1) + \frac{1}{3} n] F_{n+1}(1) = \frac{1}{3} F_{n+1}(1)(n + F_{n}(1))$

This says that once we have chopped the $(n+4)$-gon into two pieces, either we have a triangle and an $(n+3)$-gon to chop up, or we have at least an $\frac{1}{2}(n+6)$-gon (ignoring odd $n$ for now) for which we can choose $n$ diagonals meeting the existing diagonal. For example, the $21$ edges of the 3d Stasheff polytope arise from the relation

$21 = \frac{1}{3} 9 (2 + 5)$

which says that once a hexagon is cut into two, in one of $9$ ways, either there is a pentagon to chop up, in one of $5$ ways, or the hexagon is split into two squares, one of which may be cut up in two ways. The full set of $9$ diagrams for $n=3$ appears, along with the pentagon subset and the $n$ yellow pieces. Note that either $F_{n+1}(1)$ is divisible by $3$, or $F_{n}(1) + n$ is divisible by $3$. For example, $14 + 4 = 18$ and $35 + 7 = 42$. The overcount factor of $3$ comes from the familiar cyclic symmetry of a central triangle in the hexagon, the three bisections of which mark the three possible choices for a square face on the 3d polytope. Similarly, the factor of $2$ in the $F_{n}(1)$ sequence came from the two diagonals of a square, which obey an $S_2$ symmetry under rotation.

Young diagrams are usually used to label irreducible representations of the symmetric group $S_n$. If certain collections of Young diagrams are used to label associahedra (which may be obtained from the permutohedra with vertices the elements of $S_n$) then there is a close connection between the collection of groups $\{ S_n \}$ and its representations. However, the whole heirarchy in all dimensions needs to be considered if we want to understand this correspondence between a theory and its models.

$F_{n}(1) = \frac{1}{2} n (n + 3) = \frac{1}{2} n (n + 1) + n = (\sum_{i=1}^{n} i) + n$

This corresponds to rewriting the sequence in the form

2

2 + 2 + 1

2 + 2 + 1 + 3 + 1

2 + 2 + 1 + 3 + 1 + 4 + 1

After eliminating the bold $n$ terms this becomes

2

1 + 2

1 + 2 + (1 + 2)

1 + 2 + 3 + (1 + 1 + 2)

and so on. The previous post also separated the set of $F_{n}(1)$ Young diagrams into a set of $n$ yellow tiled diagrams and another set of $1 + 2 + \cdots + n$ diagrams with at most two purple tiles in a row.

Now consider the $k = 2$ sequence $F_{n}(2)$, which counts the number of divisions of an $(n+3)$-gon into three pieces by two diagonals. The terms of this sequence are given by the formula

$F_{n}(2) = \frac{1}{3} B(n + 4, 2) B(n,2) = \frac{1}{12} n(n-1)(n+3)(n+4) = \frac{1}{6} (n-1)(n+4) F_{n}(1)$

What set of diagrams counts this sequence? The $F_{n}(1)$ factor says that before cutting an $n$-gon into three pieces we must cut it into two. Note also that $F_{2}(2) = F_{2}(1) = 5$ since having cut a pentagon into two pieces, there is only one way to cut it into three. It is therefore more natural to write

$F_{n+1}(2) = \frac{1}{6} n(n+5) F_{n+1}(1) = [\frac{1}{3} (\sum_{i=1}^{n} i) + \frac{2}{3} n] F_{n+1}(1) $

$= [ \frac{1}{3} F_{n}(1) + \frac{1}{3} n] F_{n+1}(1) = \frac{1}{3} F_{n+1}(1)(n + F_{n}(1))$

This says that once we have chopped the $(n+4)$-gon into two pieces, either we have a triangle and an $(n+3)$-gon to chop up, or we have at least an $\frac{1}{2}(n+6)$-gon (ignoring odd $n$ for now) for which we can choose $n$ diagonals meeting the existing diagonal. For example, the $21$ edges of the 3d Stasheff polytope arise from the relation

$21 = \frac{1}{3} 9 (2 + 5)$

which says that once a hexagon is cut into two, in one of $9$ ways, either there is a pentagon to chop up, in one of $5$ ways, or the hexagon is split into two squares, one of which may be cut up in two ways. The full set of $9$ diagrams for $n=3$ appears, along with the pentagon subset and the $n$ yellow pieces. Note that either $F_{n+1}(1)$ is divisible by $3$, or $F_{n}(1) + n$ is divisible by $3$. For example, $14 + 4 = 18$ and $35 + 7 = 42$. The overcount factor of $3$ comes from the familiar cyclic symmetry of a central triangle in the hexagon, the three bisections of which mark the three possible choices for a square face on the 3d polytope. Similarly, the factor of $2$ in the $F_{n}(1)$ sequence came from the two diagonals of a square, which obey an $S_2$ symmetry under rotation.

Young diagrams are usually used to label irreducible representations of the symmetric group $S_n$. If certain collections of Young diagrams are used to label associahedra (which may be obtained from the permutohedra with vertices the elements of $S_n$) then there is a close connection between the collection of groups $\{ S_n \}$ and its representations. However, the whole heirarchy in all dimensions needs to be considered if we want to understand this correspondence between a theory and its models.

We have seen how the Catalan numbers come up in many places. For example, they count the number of vertices of the $n$th associahedron, which is fully labelled by chorded $n+3$-gons. The full list of the number of codimension $k$ elements of the $n$th associahedron is given by sequence A033282 at the integer database. For instance, the left diagonal row

$2, 5, 9, 14, 20, 27, \cdots$

counts two ends to an interval, five edges on a pentagon, nine faces of the 3d Stasheff polytope, and so on. These codimension 1 faces are labelled by an $n$-gon with a single diagonal, splitting the $n$-gon into two parts.

Another way of viewing the particular row above is in terms of Young diagrams with only two rows. For example, the 3d and 4d polytopes have faces labelled by the following Young diagrams. Observe how the recursion is visible in the complements of the white tiles. For $n = 3$, by omitting both the full white diagram and the yellow diagrams one obtains five remaining purple diagrams corresponding to the Young pictures for the $n = 2$ pentagon edges. The recursion relation for codimension 1 faces must therefore be

$F_{n} = F_{n - 1} + n + 1 = \frac{n(n+3)}{2}$

Recall that codimension 1 elements are also labelled by trees with two internal vertices. For example, the whole pentagon is labelled by a single vertex tree with four leaves, but the pentagon edges are labelled by four leaved trees with two internal vertices. The homework problem is to figure out how these trees correspond to the Young diagrams.

The full list of left diagonals for A033282, with the above row as the $k = 1$ entry, gives the number of codimension $k$ faces. The general formula for these numbers takes the form

$F_{n} (k) = \frac{1}{k + 1} B(n + k + 2,k) B(n,k)$

for binomial coefficients $B(n,m)$. The sequence A126216 is a mirror image of A033282, which counts the number of Schroeder paths. These arose in the construction of Abel sums and interesting maps between sets of trees.

Update: It turns out that R. P. Stanley worked out a relation between trees and Young tableaux (numbered Young diagrams) in the short 1996 paper Polygon dissections and standard Young tableaux in J. Comb. Theory A76, pages 175-177.

$2, 5, 9, 14, 20, 27, \cdots$

counts two ends to an interval, five edges on a pentagon, nine faces of the 3d Stasheff polytope, and so on. These codimension 1 faces are labelled by an $n$-gon with a single diagonal, splitting the $n$-gon into two parts.

Another way of viewing the particular row above is in terms of Young diagrams with only two rows. For example, the 3d and 4d polytopes have faces labelled by the following Young diagrams. Observe how the recursion is visible in the complements of the white tiles. For $n = 3$, by omitting both the full white diagram and the yellow diagrams one obtains five remaining purple diagrams corresponding to the Young pictures for the $n = 2$ pentagon edges. The recursion relation for codimension 1 faces must therefore be

$F_{n} = F_{n - 1} + n + 1 = \frac{n(n+3)}{2}$

Recall that codimension 1 elements are also labelled by trees with two internal vertices. For example, the whole pentagon is labelled by a single vertex tree with four leaves, but the pentagon edges are labelled by four leaved trees with two internal vertices. The homework problem is to figure out how these trees correspond to the Young diagrams.

The full list of left diagonals for A033282, with the above row as the $k = 1$ entry, gives the number of codimension $k$ faces. The general formula for these numbers takes the form

$F_{n} (k) = \frac{1}{k + 1} B(n + k + 2,k) B(n,k)$

for binomial coefficients $B(n,m)$. The sequence A126216 is a mirror image of A033282, which counts the number of Schroeder paths. These arose in the construction of Abel sums and interesting maps between sets of trees.

Update: It turns out that R. P. Stanley worked out a relation between trees and Young tableaux (numbered Young diagrams) in the short 1996 paper Polygon dissections and standard Young tableaux in J. Comb. Theory A76, pages 175-177.

Having blogged a bit already about Geometric Representation Theory, I was going to leave it to people to watch this lecture series for themselves. But lecture 8 is way too cool to ignore!

Here we see how paths in a $q$ deformed Pascal triangle can be counted. First note that in M Theory we usually draw the triangle as a quadrant in a plane, on which we consider paths for the noncommutative Fourier transform. A step to the right picks up multiples of a power of $q$, whereas a step up simply multiplies the entry below by 1. In this way one obtains polynomials in $q$ with integer coefficients. Even though $q$ starts out labelling the number of elements in a finite field, we recall counting trees in a similar fashion, but ending up with complex roots of unity. Physicists may smell a sneaky Wick rotation in the shrubbery.

Here we see how paths in a $q$ deformed Pascal triangle can be counted. First note that in M Theory we usually draw the triangle as a quadrant in a plane, on which we consider paths for the noncommutative Fourier transform. A step to the right picks up multiples of a power of $q$, whereas a step up simply multiplies the entry below by 1. In this way one obtains polynomials in $q$ with integer coefficients. Even though $q$ starts out labelling the number of elements in a finite field, we recall counting trees in a similar fashion, but ending up with complex roots of unity. Physicists may smell a sneaky Wick rotation in the shrubbery.

A quick note: check out John Baez's TWF 259 for more details on the **absolute point**! Clearly I'm not the only crazy one who sees ghosts in the machine. To quote the summary:

In short, a mathematical phantom is gradually taking solid form before our very eyes! In the process, a grand generalization of algebraic geometry is emerging ...

James Dolan speaks of categorification and decategorification, and of information and entropy.

In logos land, these processes have a dimension raising or lowering aspect. It is often said that categorification is ill defined, in comparison to decategorification, but with dual processes it should not be so. Therefore, categorification itself must be defined in some canonical way that generalises the turning of natural numbers into sets or spaces. One way to do this would be to put the heirarchy on a loop, such as the loop labelled by the $q$ parameters at roots of unity. There would be $n$-categories for $n \in \mathbb{N}$ and $r$-categories for $r \in \mathbb{Q}$, and $n \rightarrow \infty$ would look like the limit $q \rightarrow 1$ again, where spaces begin to look like sets.

After all, projective geometry has its horizons, and the cohomology of motives would move left and right, like the mass interaction, or Stokes' theorem, or the Riemann zeta function.

In logos land, these processes have a dimension raising or lowering aspect. It is often said that categorification is ill defined, in comparison to decategorification, but with dual processes it should not be so. Therefore, categorification itself must be defined in some canonical way that generalises the turning of natural numbers into sets or spaces. One way to do this would be to put the heirarchy on a loop, such as the loop labelled by the $q$ parameters at roots of unity. There would be $n$-categories for $n \in \mathbb{N}$ and $r$-categories for $r \in \mathbb{Q}$, and $n \rightarrow \infty$ would look like the limit $q \rightarrow 1$ again, where spaces begin to look like sets.

After all, projective geometry has its horizons, and the cohomology of motives would move left and right, like the mass interaction, or Stokes' theorem, or the Riemann zeta function.

Recall that Kapranov and Smirnov have also been thinking about the field with one element. They say an affine line over $F_1$ should be zero along with all the roots of unity.

Looking at polynomials with $F_1$ coefficients, the group $GL(n, F_{1}[x])$ is just the braid group on $n$ strands. For example, $3 \times 3$ matrices are associated with the three strand braid group, as often discussed. Then the map $B_n \rightarrow S_n$ is thought of as the $q \rightarrow 1$ limit, since the symmetric group acts on sets as vector spaces.

The field $F_{1}(n)$ extends $F_1$ by containing zero and the set of all nth roots of unity. A vector space over this field is a pointed set (marked by zero) with an action by the roots of unity. Direct sum and smash product become the operations on such spaces.

Note that Weber and others have considered the category of pointed sets as a 2-categorical Cat analogue of subobject classifier, and the category Set plays the role here of a one element set. There seem to be a number of ways in which the field $F_1$ introduces new topos theoretic arrows.

Looking at polynomials with $F_1$ coefficients, the group $GL(n, F_{1}[x])$ is just the braid group on $n$ strands. For example, $3 \times 3$ matrices are associated with the three strand braid group, as often discussed. Then the map $B_n \rightarrow S_n$ is thought of as the $q \rightarrow 1$ limit, since the symmetric group acts on sets as vector spaces.

The field $F_{1}(n)$ extends $F_1$ by containing zero and the set of all nth roots of unity. A vector space over this field is a pointed set (marked by zero) with an action by the roots of unity. Direct sum and smash product become the operations on such spaces.

Note that Weber and others have considered the category of pointed sets as a 2-categorical Cat analogue of subobject classifier, and the category Set plays the role here of a one element set. There seem to be a number of ways in which the field $F_1$ introduces new topos theoretic arrows.

Whoa! I wasn't expecting it that soon. Motives appear already, at least conjecturally, in John Baez's lecture 6! In the diagram the vertical arrows are the decategorification of either sets or projective spaces. An isomorphism class of $n$ element sets is mapped to the number $n$. Natural numbers become $q$-numbers in the case of spaces, which is to say rational functions in the parameter $q$ corresponding to the number of elements in the finite field, or secretly really polynomials with integer coefficients. But what replaces the category of finite sets? There is more structure to the projective spaces, and we also want to understand the bottom arrow, which considers a set as a space over a one element field.

This is a rather delicate mathematical question. Baez mentions a recent paper by Durov (with lots of stuff on monads) about the idea of a one element field. When we understand this properly, do we find motives? Now, that is the question.

The plane of a (finite) field $F_{q}^{2}$ is the $q$-analogue of a two element set, which plays an important role in the Boolean topos Set, namely as the subobject classifier. The vector space version of this is commonly known as a qubit. Somehow the reason that a one element field $F_1$ doesn't usually make sense is because the logical 0 and 1 are not distinct. Since $q$ is, in the first instance, just a natural number (= $p^{k}$ for some prime $p$), we can ask ourselves first what it means to collapse a finite plane to a one element field. For the topos Set, this would amount to turning the whole category into the trivial category 1, since there is no way to distinguish a subset of a set $S$ from its complement and all sets have effectively only one element. Now this one element set is like a basis for spaces over $F_1$. But the map that takes a basis to a space is just the functor 1 $\rightarrow$ Set, which picks out a set!

But this doesn't sound right. Maybe what we need here is not the trivial 1-category, or a one point 0-category (set), but rather a -1-category. This idea always lurks in the operad heirarchy, where the left hand side of the table starts with the single leaf tree, despite the fact that a point is actually a two leaf tree.

Anyway, think of a finite set $S$ lying at the endpoints of the unit vectors in a vector space. The empty set at the origin is the smallest piece of the power set of $S$, and the one element subsets are the next smallest pieces. The power set fills out a cube of dimension $|S|$. Since the field in question is $F_1$ there is no extent to the axes. Only the elements of $S$ really exist.

This is a rather delicate mathematical question. Baez mentions a recent paper by Durov (with lots of stuff on monads) about the idea of a one element field. When we understand this properly, do we find motives? Now, that is the question.

The plane of a (finite) field $F_{q}^{2}$ is the $q$-analogue of a two element set, which plays an important role in the Boolean topos Set, namely as the subobject classifier. The vector space version of this is commonly known as a qubit. Somehow the reason that a one element field $F_1$ doesn't usually make sense is because the logical 0 and 1 are not distinct. Since $q$ is, in the first instance, just a natural number (= $p^{k}$ for some prime $p$), we can ask ourselves first what it means to collapse a finite plane to a one element field. For the topos Set, this would amount to turning the whole category into the trivial category 1, since there is no way to distinguish a subset of a set $S$ from its complement and all sets have effectively only one element. Now this one element set is like a basis for spaces over $F_1$. But the map that takes a basis to a space is just the functor 1 $\rightarrow$ Set, which picks out a set!

But this doesn't sound right. Maybe what we need here is not the trivial 1-category, or a one point 0-category (set), but rather a -1-category. This idea always lurks in the operad heirarchy, where the left hand side of the table starts with the single leaf tree, despite the fact that a point is actually a two leaf tree.

Anyway, think of a finite set $S$ lying at the endpoints of the unit vectors in a vector space. The empty set at the origin is the smallest piece of the power set of $S$, and the one element subsets are the next smallest pieces. The power set fills out a cube of dimension $|S|$. Since the field in question is $F_1$ there is no extent to the axes. Only the elements of $S$ really exist.

Planar Young diagrams represent partitions of a natural number $n = n_1 + n_2 + n_3 + \cdots + n_k$. The $k$ rows are the pieces $n_i$ of the partition. But categorified numbers $n_i$ are actually sets with $n_i$ elements, or perhaps vector spaces of dimension $n_i$, or projective spaces of dimension $n_i - 1$. In this setting the expression $n = n_1 + n_2 + n_3 + \cdots + n_k$ is about the decomposition of a space into subspaces.

We have seen something like this before. Let $n_i$ instead represent $V^{\otimes n_{i}}$ for a fixed finite dimensional vector space $V$. Then the $O(n_{i})$ piece of the operad is the space of linear maps from $n_i$ to $V$ in Vect. The operad rules come from compositions $n_1 \otimes n_2 \otimes n_3 \cdots \otimes n_k \rightarrow n$ of these maps. Maybe instead of categorification of $\mathbb{N}$ we can look at operadification.

One then wonders what happens for higher dimensional ordinals. Actually, partitions are a lot like 2-tree ordinals. The simplest generalisation would allow each $n_{i}$ of a planar partition to be itself replaced by a partition of $m$ in a third direction. The first permutation group to fill a three dimensional diagram would be $S_4$, with four box partitions. This is the only additional diagram to the planar labels for the $S_4$ barycentrically divided tetrahedron. Similarly, $S_3$ is the first group to fill a truly planar Young diagram. This pattern continues for all $n$.

We have seen something like this before. Let $n_i$ instead represent $V^{\otimes n_{i}}$ for a fixed finite dimensional vector space $V$. Then the $O(n_{i})$ piece of the operad is the space of linear maps from $n_i$ to $V$ in Vect. The operad rules come from compositions $n_1 \otimes n_2 \otimes n_3 \cdots \otimes n_k \rightarrow n$ of these maps. Maybe instead of categorification of $\mathbb{N}$ we can look at operadification.

One then wonders what happens for higher dimensional ordinals. Actually, partitions are a lot like 2-tree ordinals. The simplest generalisation would allow each $n_{i}$ of a planar partition to be itself replaced by a partition of $m$ in a third direction. The first permutation group to fill a three dimensional diagram would be $S_4$, with four box partitions. This is the only additional diagram to the planar labels for the $S_4$ barycentrically divided tetrahedron. Similarly, $S_3$ is the first group to fill a truly planar Young diagram. This pattern continues for all $n$.

Lecture 3 by James Dolan starts with diagrams for the subgroups of $S_3$. For example, the cyclic group produces a diagram where all three green vertices have been identified by rotations of the triangle, as have the blue vertices. Now we see the need for the Young diagram boxes, which can represent indeterminate elements of a set. This example splits the six elements of the group into two sets of three permutations, represented by the two triangles. One such set appeared as a basis for the mass matrix Fourier transform. M theory is like a child's game of connect the dots.

In Geometric Representation Theory lecture 2 James Dolan draws a 2-simplex with a barycentric subdivision, looking like a hexagon with a central point, or rather a cube with a missing hidden vertex. This picture is labelled with Young diagrams associated to the group $S_3$ of permutations on three letters. Diagrams of height 1 are associated with vertices, diagrams of height 2 with edges and diagrams of height 3 with faces. Note that horizontal lists are unordered. So the six faces represent the elements of the group. Dolan wants to think of these diagrams as axiomatic theories in a categorical sense. Such diagrams, and their subdiagrams, are associated with sequences of subspaces of the three element set, in analogy with flag spaces for vector spaces. Sets are just a classical kind of vector space. This is clear when counting vectors in vector spaces over finite fields, as discussed by Baez in lecture 1. For vector spaces over $F$ the permutation groups would naturally be replaced by the example of $GL(n,F)$.

The example above generalises as expected. Before we know it, we'll probably be doing motivic cohomology and Langlands geometry using Hecke pictures. Of course, Kontsevich has already been thinking about such things.

The example above generalises as expected. Before we know it, we'll probably be doing motivic cohomology and Langlands geometry using Hecke pictures. Of course, Kontsevich has already been thinking about such things.

S duality in the guise of the Langlands program is a truly awe inspiring component of stringy triality.

Recall that the complex number form of S duality has a modular group symmetry. This group appears all over the place in M theory. For example, we looked at the Banach-Tarski paradox in terms of a ternary tiling of hyperbolic space as a Poincare disc. This tiling marks the boundary of the circle with a nice triple pattern of accumulating points.

Alain Connes and Matilde Marcolli say that the Riemann zeta function is related to the problem of mass, which in turn we have seen is related to three stranded braids and M theory triality, of which S duality is a piece. It appears that no part of mathematics is left untouched by gravity.

Update: a new paper by Witten et al on 3D gravity, prominently featuring the modular group, has appeared on the arxiv. See the picture on page 49, and the J invariant on page 52. The paper shows that the partition function cannot be given a conventional Hilbert space interpretation. Holomorphic factorisation is suggested as a possible mechanism for extending the degrees of freedom. For instance, the complexified Einstein equations are considered. They say:

Recall that the complex number form of S duality has a modular group symmetry. This group appears all over the place in M theory. For example, we looked at the Banach-Tarski paradox in terms of a ternary tiling of hyperbolic space as a Poincare disc. This tiling marks the boundary of the circle with a nice triple pattern of accumulating points.

Alain Connes and Matilde Marcolli say that the Riemann zeta function is related to the problem of mass, which in turn we have seen is related to three stranded braids and M theory triality, of which S duality is a piece. It appears that no part of mathematics is left untouched by gravity.

Update: a new paper by Witten et al on 3D gravity, prominently featuring the modular group, has appeared on the arxiv. See the picture on page 49, and the J invariant on page 52. The paper shows that the partition function cannot be given a conventional Hilbert space interpretation. Holomorphic factorisation is suggested as a possible mechanism for extending the degrees of freedom. For instance, the complexified Einstein equations are considered. They say:

we think another possibility is that the non-perturbative framework of quantum gravity really involves a sum not over ordinary geometries in the usual sense, but over some more abstract structures that can be defined independently for holomorphic and antiholomorphic variables. Only when the two structures coincide can the result be interpreted in terms of a classical geometry.I confess to finding this statement a little ill-phrased, since some more abstract structure presumably does not begin with traditional complex analysis.

This month's conference on Langlands and QFT has naturally drawn some attention. Matti Pitkanen has some comments. Louise Riofrio and David Ben-Zvi (one of the speakers) have been taking notes. Louise points out that Witten is interested in four dimensions, but according to Ben-Zvi's notes from his talk, Witten said it was natural to look at six dimensions (think twistors), or at least four, but for the talk he focused on two.

Admittedly, I can't make that much out of Ben-Zvi's notes, and I'm sure that's not his fault! Here's a link to Frenkel's helpful paper.

Admittedly, I can't make that much out of Ben-Zvi's notes, and I'm sure that's not his fault! Here's a link to Frenkel's helpful paper.

Dr Motl reports on a new astro-ph paper on the lack of a Dark Force (cosmological constant). The abstract suggests

a new concordance model with 90% dark matter, 10% baryons, no dark energy and 14.8 Gyr as the age of the universe.This sounds familiar. It includes a reference to the now published paper by D. Wiltshire. On Wednesday, Dr Motl reported on the difficulties that Fermi initially faced having his ideas accepted. Meanwhile, Louise Riofrio brings us reports from New York, and a small town nearby.