Now suppose we are given two oriented hyperbolic lines, say $L_i = (a_i, b_i)$, for $i = 1, 2$. We could define the cross ratio of $L_1$ and $L_2$ to be the cross-ratio of the $4$ corresponding ideal points, namely

$$ \operatorname{CR}(L_1, L_2) = \frac{(a_1 – b_2) (a_2 – b_1)}{(a_1 – b_1)(a_2 – b_2)},$$

where we are identifying the sphere at infinity $S^2_\infty$ with the Riemann sphere via stereographic projection from the “North” pole $(0, 0, 1)^T$. Thus, each of the $a_i$, $b_i$, for $i = 1, 2$, is a complex number or possibly $\infty$, with the usual conventions for $\infty$. When in doubt, the reader should use homogeneous coordinates on $\mathbb{P}^1_\mathbb{C}$.

Let $\delta$ be the hyperbolic distance between lines $L_1$ and $L_2$: it is the hyperbolic distance between the points $p_1 \in L_1$ and $p_2 \in L_2$, now thinking of each $L_i$ as an oriented hyperbolic line, rather than a pair of ideal points, that are closest to each other. If $L_1$ and $L_2$ intersect at a finite point in hyperbolic $3$-space, then their hyperbolic distance $\delta$ vanishes. Let $v_1$ be the unit vector based at $p_1$ which is tangent to $L_1$ and oriented along the orientation of $L_1$. Let $v’_1$ be the parallel transport of $v_1$ along the geodesic line segment $p_1 p_2$; in particular, $v’_1$ is based at $p_2$. Let $\alpha$ be the angle, using the right-hand rule (thinking of the unit tangent vector to $p_1 p_2$ at $p_2$, as a kind of normal vector), between $v’_1$ and $v_2$. Note that $\alpha$ is well defined up to a multiple of $2 \pi$.

One can also check that, even if we permute $L_1$ and $L_2$, both $\delta$ and the angle $\alpha$ remain the same. We then define the complex distance between $L_1$ and $L_2$ to be

$$ d_\mathbb{C}(L_1, L_2) = \delta + i \frac{\alpha}{2} \in \mathbb{C}/(i \pi \mathbb{Z}).$$

In turns out, and it can be proved, that

$$ \operatorname{CR}(L_1, L_2) = \operatorname{cosh}^2(d_\mathbb{C}(L_1, L_2)). $$

Note: I have independently rediscovered this formula, but, even though not very well known, this formula is not new. Indeed, as Sam Nead mentioned on MathOverflow (and I paraphrase him here), the complex distance between two hyperbolic lines appears on p. 355 in Marden’s book Outer circles: an introduction to hyperbolic 3-manifolds. In the second edition of the book, with the title, Hyperbolic manifolds: an introduction in 2 and 3 dimensions, the formula appears on page 432.

However, I am not sure whether Sam Nead’s reference, i.e. Marden’s book, either editions, mentions the formula for the cross-ratio in terms of the complex distance. I have no doubt it is somewhere in the literature, though. If someone knows of some references containing such formulas, then please email me to inform me about them.

]]>$$ F = *d_{\omega} \Phi $$

where $F$ is the curvature of $\omega$, i.e. $F = d\omega$ (since the Lie group $U(1)$ is abelian), $*$ is the Hodge star with respect to the Euclidean metric on $B$ and $d_{\omega}$ is the exterior differential twisted with the connection $\omega$. In this special case, the Lie group $G = U(1)$ is abelian, so $d_{\omega} \Phi$ is just $d \Phi$.

There is also a converse to that construction, so that any gravitational instanton with $U(1)$ symmetry can be, under some assumptions, obtained from this construction.

When learning about the Gibbons-Hawking ansatz, it is natural to ask the question: how can one view the simplest $4$-dimensional gravitational instanton with $U(1)$ symmetry, $M = \mathbb{H}$, with metric $g = dq d\bar{q}$, where $\mathbb{H}$ denotes the quaternions, $q$ is a quaternionic coordinate on $M$, from the point of view of the Gibbons-Hawking ansatz?

I struggled a bit with the calculation and authors tend to completely omit the calculation, so I thought I would post about it here, hoping it will be useful to somebody else, or possibly myself at some point in the future.

Write $q = a e^{i \frac{\psi}{2}}$ where $a$ is a pure imaginary quaternion and $\psi \in [0, 4\pi)$. Differentiating, we get

$$ dq = (da + \frac{1}{2} d\psi a i) e^{i \frac{\psi}{2}} $$

Therefore

\begin{aligned} g &= (da + \frac{1}{2} d\psi a i) (d\bar{a} – \frac{1}{2} i \bar{a} d \psi) \\ &= da d\bar{a} + \frac{1}{2}( ai d\bar{a} – da i \bar{a}) d\psi + \frac{r}{4} d\psi^2 \\ &= da d\bar{a} + \frac{r}{4} \left( d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a}) \right)^2 – \frac{1}{4r} (a i d\bar{a} – da i \bar{a})^2 \end{aligned}

Let $\mathbf{r} = q i \bar{q} = a i \bar{a}$. We then obtain that

$$ da d\bar{a} – \frac{1}{4r} (a i d\bar{a} – da i \bar{a})^2 = \frac{1}{4r} d \mathbf{r}^2 $$

so that

$$ g = \frac{1}{4r} d \mathbf{r}^2 + \frac{r}{4} \left( d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a}) \right)^2. $$

Let $\omega = d\psi + \frac{1}{r}(ai d\bar{a} – da i \bar{a})$. I claim that $d \omega = * d(1/r)$. Note that $\Phi = 1/r$ is (up to a factor) the fundamental solution of the Laplacian on $\mathbb{R}^3$.

We differentiate $\omega$ and obtain

$$ d \omega = – \frac{dr}{r^2} \wedge (ai d\bar{a} – da i \bar{a}) + \frac{2}{r} da i \wedge d\bar{a} .$$

But $r^2 = – \mathbf{r} \mathbf{r}$, which yields, upon differentiation

$$ 2 r dr = -(d\mathbf{r} \mathbf{r} + \mathbf{r} d \mathbf{r}). $$

Hence

\begin{aligned} d \omega &= \frac{1}{2 r^3} d \mathbf{r} \mathbf{r} \wedge ( ai d\bar{a} – da i \bar{a}) – \frac{1}{2 r^3} ( ai d\bar{a} – da i \bar{a}) \mathbf{r} \wedge d \mathbf{r} – \frac{2}{r^3} da i \bar{a} \mathbf{r} a i d\bar{a} \\ &= – \frac{1}{r^3} a i d\bar{a} \mathbf{r} \wedge da i \bar{a} – \frac{1}{r^3} da i \bar{a} \mathbf{r} \wedge a i d\bar{a} \\ &= – \frac{1}{2r^3} d\mathbf{r} \mathbf{r} \wedge d \mathbf{r} + \frac{1}{2 r^3} (a i d\bar{a} – da i \bar{a}) \mathbf{r} \wedge (a i d\bar{a} – da i \bar{a}) \end{aligned}

But the last term on the last line vanishes, so we get

$$ d \omega = – \frac{1}{2r^3} d\mathbf{r} \mathbf{r} \wedge d \mathbf{r}, $$

which can be checked to be nothing but $*d ( 1 / r)$, as claimed.

Many authors sweep this calculation under the rug, so I thought I’d write a post about it!

]]>- I have studied Electrical Engineering during my undergraduate studies.
- I have then studied Mathematics at the graduate level.
- I have worked as a Mathematics Professor for about 7 years.
- I am now a professional Python programmer.

Q: How did *that* happen?

A: Python started out as being a hobby of mine, maybe since 2014 or 2015. I first started using it because I was curious about it. My first programming language was C++. From my own humble point of view, I liked the fact that if you wanted to do some scientific programming, you had many official Python libraries to help you accomplish your work, such as Numpy, Scipy and Matplotlib.

Note: recently, I should mention that I really enjoyed the way the Julia programming language was doing things for scientific programming and I think that Julia may have a big impact on the scientific programming community in the near future. Time will tell.

In addition to the above Python libraries, I have played around a bit with Panda, for data manipulation (a very useful library!), some libraries for automating tasks (which I have used to shuffle problems and answers for multiple choice exams) and I have played around a bit with Django (a Python library which enables you to interact with a database on a server, create Content Management Systems, restful APIs etc.).

I have recently started working for a company as a Python programmer. I am now using virtual Python environments, a professional IDE for Python called PyCharm which is really helpful (I wonder why I was not using serious IDEs before!), pushing code onto a server using the magic of git and learning a lot in the process.

This is my story. Does it mean I will stop doing research in Mathematics? No. I mean it will be more difficult to find time for it though. I am enjoying programming though, so it is good!

]]>When CAMS (Center for Advanced Mathematical Sciences) was first founded circa 1999, I was an Electrical Engineering student at the American University of Beirut (AUB) where CAMS is physically located. I was curious and wanted to attend a talk by Sir Michael Atiyah, whom I will refer to as Sir Michael for the rest of the post. It took place in Hotel Al-Bustan actually. I think it may have been at some point in December 1999, or perhaps in January 2000. The talk was about an “elementary” problem in geometry, which is the problem on configurations of points. This was also the first time I had see Sir Michael in person, and also the first time I had learned about stereographic projection (remember, I was an Electrical Engineering student back then, not a Mathematics student).

I was really interested in the problem, and of course in Mathematics in general, particularly geometry (broadly understood). I decided to email Sir Michael, and asked him about careers in Mathematics. He answered me that there were many career opportunities in Mathematics nowadays. That email was key in making me decide to study Mathematics later on.

I emailed the poor man several times with ideas and sometimes wrong proofs. He used to reply to me with short emails, explaining to me why it does not work etc. Sometimes he would give me some advice too: try proving this, or that. Once also, Sir Michael told me to “gry my hands” at the planar case of the problem.

Let us move forward in time to October 29, 2016. I emailed Sir Michael about an idea for how to generalize his problem to a Lie-theoretic setting. On November 1, 2016, he replied to me, and the following is an excerpt from his email (which was also sent to Paul Sutcliffe and Roger Bielawski):

“I like your idea, which ties up closely with mine but is more algebraic. Please send me a fuller version (which need not be perfect). Do not rush to publish as yet. I will advise and help publication and see how it fits with my own work. It may be that that we can write a joint paper.”

I was really happy of course. My favorite Mathematician, who is also one of the best Mathematicians in the World, liked one of my ideas. This was a turning point.

I then visited him twice in Edinburgh, once in February 2017 and once in July 2018. Both times, we discussed many mathematical subjects (including of course the problem on configurations of points), with mostly him doing the talking. I must say it was really impressive to listen to his thoughts.

As many people may know, Sir Michael came a lot under attack in the past couple of years of his life, especially after uploading a series of preprints on various subjects: an argument for the lack of complex structures on $S^6$, a fixed point idea for simplifying the proof of the Feit-Thompson theorem, an explanation for the value of the fine structure constant and an argument for the Riemann hypothesis.

While flaws and/or inaccuracies were found in these preprints, I would like to say a few remarks.

Sir Michael’s idea for the Feit-Thompson theorem is, in my humble opinion, quite plausible. I find it possible that someone may manage to apply a fixed point argument to show the statement that any finite group of odd order is solvable. Perhaps the details in that preprint were incorrect, but the general idea itself is in my opinion quite plausible, and certainly worthy of further study. Alain Connes took up this idea and pursued it a bit further, his way, but so far the approach has not been successful. Anyway, I do think it is worthy of further study.

Regarding his preprint on the fine structure constant, I think that there are some interesting ideas in that preprint, regardless of whether or not they can applied to the fine structure constant. They may possibly have other applications. I did not understand all the Math behind this article, because, as of now, I am not familiar with Von Neumann algebras.

I have to say that, even when he was facing all that opposition to his ideas in the last couple of years of his life, this did not stop him from expressing himself. Some people thought this was sad. I thought it was remarkable. Moreover, his brilliant intuition never suffered a single bit. I think he was less concerned with details so much, but wanted to gain a deeper understanding. As he used to say, there are those who like to think, and those who like to understand. Sir Michael was definitely among those who wanted to understand.

]]>I will explain in this post, how to define the Atiyah-Sutcliffe normalized determinant function, which is a smooth complex-valued function on $C_n(\mathbb{R}^3)$.

Given a configuration $\mathbf{x} = (\mathbf{x}_1, \ldots, \mathbf{x}_n) \in C_n(\mathbb{R}^3)$ of $n$ distinct points in $\mathbb{R}^3$, I have explained in my previous post how to associate to $\mathbf{x}$ complex polynomials $p_i(\zeta)$, for $i = 1, \ldots, n$ of degree (at most) $n-1$ using the Hopf map. I will paraphrase this construction (due to Atiyah and Sutcliffe) here.

Given any $i,j \in \{1,\ldots,n\}$, with $i \neq j$, we define $\zeta_{ij}$ to be the complex number (possibly $\infty$) corresponding to the normalized direction from $\mathbf{x}_i$ to $\mathbf{x}_j$, after identifying the unit sphere $S^2$ with the Riemann sphere $\hat{S}$ (see my previous post). Let $p_{ij}(\zeta)$ be the polynomial of degree (at most) $1$ having $\zeta_{ij}$ as its unique root. Note that $p_{ij}(\zeta)$ is only defined up to scaling by a nonzero complex number. For instance, if $\zeta_{ij}$ is “finite”, one may for instance take $p_{ij}(\zeta) = \zeta – \zeta_{ij}$, while if $\zeta_{ij} = \infty$, one may take for example $p_{ij}(\zeta) = 1$. This last convention may seem odd at first, but it makes sense when one thinks of the Riemann sphere as $P^1(\mathbb{C})$ and uses a pair of homogeneous coordinates to describe the location of a point on it.

Then one may define, for $1 \leq i \leq n$:

$$p_i(\zeta) = \prod_{j \neq i} p_{ij}(\zeta).$$

Denote by $\det(p_1, \ldots, p_n)$ the determinant of the complex $n$ by $n$ matrix containing the coefficients of $p_j$ (ordered say by increasing powers of $\zeta$) as its $j$-th column. Note that, due to the scaling ambiguity of each polynomial $p_{ij}(\zeta)$, which therefore results in a scaling ambiguity of each of the $p_j(\zeta)$, the expression $\det(p_1,\ldots, p_n)$ is ill-defined. We may however divide it by an expression with exactly the same kind of scaling ambiguity and then the resulting quotient will be well defined. More precisely, Atiyah and Sutcliffe defined

$$ D(\mathbf{x}) = \frac{ \det(p_1, \ldots, p_n) } { \prod_{1 \leq i < j \leq n} \det(p_{ij}, p_{ji}) }. $$

The Atiyah-Sutcliffe conjecture $1$ states that for any $\mathbf{x} \in C_n(\mathbb{R}^3)$, the $n$ polynomials $p_i(\zeta)$, for $i = 1, \ldots, n$, are linearly independent over $\mathbb{C}$. This conjecture is also equivalent to $D$ being non-vanishing on $C_n(\mathbb{R}^3)$.

The Atiyah-Sutcliffe conjecture $2$ states that for any $\mathbf{x} \in C_n(\mathbb{R}^3)$, $| D(\mathbf{x} | \geq 1$. It is clear that the Atiyah-Sutcliffe conjecture $2$ is stronger than conjecture $1$.

As of the time of writing, both conjectures are open for a general $n > 4$. The case $n = 3$ was proved by Atiyah and Sutcliffe. Conjecture 1 for $n = 4$ was proved by Eastwood and Norbury, using symbolic manipulations in Maple, shortly after the conjectures appeared. They also came close to proving conjecture 2. More precisely, they showed that for any $\mathbf{x} \in C_4(\mathbb{R}^3)$, $|D(\mathbf{x})| \geq \frac{15}{16}$, so their lower bound is just a little below $1$.

Many years later, a proof using linear programming was done by Bou Khuzam and Johnson, published in SIGMA. They even proved a stronger conjecture, called the Atiyah-Sutcliffe conjecture 3 (for $n = 4$), which is even stronger than conjecture 2. Roughly at the same time and independently, alternative proofs of conjectures 2 and 3 for $n = 4$ were also made and presented by D. Svrtan. There are currently no other general results for $n > 4$. There are some special cases (meaning for some special classes of configurations) for which some of the conjectures are known (see the works of Svrtan and Urbiha for instance).

I mention also that Sir Michael Atiyah used to say that he would offer a bottle of liquor to whoever solves the conjectures (a bottle of Vodka if they are Russian, a bottle of Arak if they are Lebanese and so on). I had the great honor to meet with him a few times and discuss with him this problem and other mathematical topics. I will probably write at some point about that in a later post.

]]>

Let $C_n(\mathbb{R}^3)$ denote the configuration space of $n$ distinct points in $\mathbb{R}^3$. Given $\mathbf{x} = (\mathbf{x}_1,\ldots,\mathbf{x}_n) \in C_n(\mathbb{R}^3)$, so that each $\mathbf{x}_i \in \mathbb{R}^3$, for $i = 1, \ldots, n$ and the $\mathbf{x}_i$ are distinct, we will associate to the configuration $\mathbf{x}$ $n$ complex polynomials $p_i(\zeta)$, for $i = 1, \ldots, n$ of degree (at most) $n-1$, each of them only defined up to multiplication by a nonzero complex factor.

Let $i \in \{1,\ldots,n\}$. For each $j \neq i$, $1 \leq j \leq n$, we form the normalized vector

$$v_{ij} = \frac{\mathbf{x}_j-\mathbf{x}_i}{\lVert \mathbf{x}_j-\mathbf{x}_i \rVert} \in S^2.$$

Using stereographic projection, there is a natural complex structure on $S^2$, which turns it into the Riemann sphere $\hat{S}$ and a natural complex coordinate $\zeta$ defined on $S^2 \setminus \{N\}$, where $N = (0,0,1)^T$ is the “North” pole, so to speak.

So $v_{ij}$ corresponds to a complex number $\zeta_{ij}$ if $v_{ij} \neq N$ and to the point at infinity $\infty$ on the Riemann sphere $\hat{S}$ if $v_{ij} = N$.

Let $p_i(\zeta)$ be the complex polynomial of degree (at most) $n-1$, having the $\zeta_{ij}$, for $1 \leq j \leq n$ and $j \neq i$, as its set of roots, taking multiplicity into account. Note that each polynomial $p_i(\zeta)$ is only defined up to a nonzero complex factor.

Atiyah conjectured that given any $\mathbf{x} \in C_n(\mathbb{R}^3)$, the corresponding polynomials $p_i(\zeta)$, for $1 \leq i \leq n$, are always linearly independent over $\mathbb{C}$. This is also known as the Atiyah-Sutcliffe conjecture $1$. If it is true, then it would imply that the map

$$\mathbf{x} \mapsto (p_1(\zeta), \ldots, p_n(\zeta))$$

gives, after orthogonalization, a smooth solution of the Berry-Robbins problem.

This is a problem I have been working on for a very long time. It is still open for general $n$, though it was proved to be true by Atiyah for $n = 3$ and by Eastwood and Norbury for $n = 4$.

There is more to the story. Indeed, Atiyah and Sutcliffe have defined a very interesting normalized determinant function $D$

$$D: C_n(\mathbb{R}^3) \to \mathbb{C}$$

whose non-vanishing is equivalent to the linear independence conjecture above (the Atiyah-Sutcliffe conjecture 1). However, Atiyah and Sutcliffe made a stronger conjecture, known as the Atiyah-Sutcliffe conjecture 2, which states that

$$|D(\mathbf{x})| \geq 1 \text{ for any $\mathbf{x} \in C_n(\mathbb{R}^3)$.}$$

However, I should tell the reader first how to define $D$. This will be the subject of a future post.

]]>Let $C_n(\mathbb{R}^3)$ denote the configuration space of $n$ distinct particles. Then the symmetric group $S_n$ acts on $C_n(\mathbb{R}^3)$ by permuting the components of any configuration. Moreover, let $U(n)$ denote the group of unitary $n \times n$ matrices and let $T^n$ denote the subgroup of diagonal unitary matrices. Then $S_n$ acts on $U(n)/T^n$ by permuting the columns of a representive of any left coset $gT^n$ (where $g \in U(n)$). The Berry-Robbins asks if there is a continuous mapping from $C_n(\mathbb{R}^3)$ into $U(n)$, which is $S_n$ equivariant.

It turns out that the Berry-Robbins problem does indeed have a solution. Even more, Atiyah and Bielawski in ^{2} show the existence of a *smooth* solution of the Berry-Robbins. This then shows that the Berry and Robbins’ argument does carry through for $n$ particles. While Atiyah and Bielawski’s article is quite a nice article, involving Nahm’s equations, Lie algebras and even proposing a conjectural link with the work of Kazhdan and Lusztig, yet there is another attempt at solving the Berry-Robbins problem, which I personally find quite interesting. It is what is known as the Atiyah problem on configurations of points, or sometimes the Atiyah-Sutcliffe conjectures.

In ^{3} and ^{4} , Atiyah, and then Atiyah and Sutcliffe, propose an alternative approach to the Berry-Robbins problem, which is very explicit and elementary. Indeed, to state the problem only involves the Hopf map. But for the approach to succeed in providing a solution to the Berry-Robbins problem, a linear independence conjecture has to hold. This is the so-called Atiyah-Sutcliffe conjecture 1.

I am planning to introduce this problem in a subsequent post.

- 1.Berry MV, Robbins JM. Indistinguishability for quantum particles: spin, statistics and the geometric phase.
*Proc R Soc Lond A*. Published online August 8, 1997:1771-1790. doi:10.1098/rspa.1997.0096 - 2.Atiyah M, Bielawski R. Nahm’s equations, configuration spaces and flag manifolds.
*Bull Braz Math Soc*. Published online July 2002:157-176. doi:10.1007/s005740200007 - 3.Atiyah M. Configurations of points. Arnold VI, Bruce JW, Moffatt HK, Pelz RB, eds.
*Philosophical Transactions of the Royal Society of London Series A: Mathematical, Physical and Engineering Sciences*. Published online July 15, 2001:1375-1387. doi:10.1098/rsta.2001.0840 - 4.Atiyah M, Sutcliffe P. The geometry of point particles.
*Proc R Soc Lond A*. Published online May 8, 2002:1089-1115. doi:10.1098/rspa.2001.0913