Alexander Grothendieck

This morning I read a Quanta Magazine article about a mathematician named Alexander Grothendieck titled How Alexander Grothendieck Revolutionized 20th Century Mathematics. I'd never heard of him. The article focuses on his mathematical contributions but says very little about his eccentric personal life. I asked Gemini the following:

Grothenddieck was a famous 20th century mathematician. Find out all you can about his personal life as opposed to his contributions to Mathematics. In particular what were his reasons for living as a recluse in the final years of his life. He fathered five children so presumably he turned his back on his family. 

Figure 1

Figure 1, taken from the Quanta Magazine article carries the following caption:

Grothendieck, seen here in 1954, was fascinated by hidden geometric structure. “If there is one thing in mathematics which fascinates me more than any other (and undoubtedly always has), it is neither ‘number’ nor ‘size,’ but invariably shape,” he wrote.

Here was Gemini's  response to my query:

You are entirely correct in your presumption. Despite fathering five children, Alexander Grothendieck was notoriously detached from his family and eventually abandoned those relationships completely. His colleague and biographer Pierre Cartier once noted that Grothendieck was "as absent a father to them as his own father was to him."

A Chaotic Family Life

Grothendieck's personal life was highly unstable, and his five children were born to three different women:

• Serge: Born to his landlady in Nancy during his early university years. Grothendieck later unsuccessfully sued for custody.
  
  
• Johanna (1959), Alexander (1961), and Mathieu (1965): Born to his wife, Mireille Dufour.
  
  
• A fifth child: Born to Justine Skalba, with whom he lived in a radical pacifist commune in the early 1970s.

Ultimately, his intense, manic obsession with his work—and later his extreme political and mystical beliefs—left almost no room for traditional familial bonds. He eventually severed all ties with his children and his past.

Early Trauma and a Stateless Youth

To understand his eventual rejection of society, it is crucial to look at his deeply traumatic upbringing.

Born in Berlin in 1928, his parents were radical anarchists. His father, a Russian Jew, was handed over to the Nazis by the French Vichy government and murdered in Auschwitz. Alexander and his mother were interned in French refugee and concentration camps (such as Rieucros) during World War II, and he spent years hiding in the woods from authorities.

Grothendieck was technically stateless for most of his life, only taking French citizenship in 1971. This early exposure to the brutality of the state instilled in him a permanent, uncompromising distrust of institutions, authority, and eventually, the scientific establishment itself.

Grothendieck lived the final decades of his life
as a recluse in the French countryside.
This 2013 photo was taken a year before his death.

Reasons for His Final Retreat

Grothendieck did not become a recluse overnight. His withdrawal was a gradual, decades-long process fueled by profound moral convictions, deep disillusionment, and an increasingly fragile psychological state.

• The 1970 Fracture: The major turning point occurred in 1970 when he discovered that his prestigious research home, the Institut des Hautes Études Scientifiques (IHÉS), received a small fraction of its funding from the French military. A strict pacifist, Grothendieck resigned immediately.
  
  
• Radical Pacifism: He shifted his immense intellectual focus from mathematics to environmentalism and anti-militarism, founding a group called Survivre et Vivre (Survive and Live). He became convinced that mainstream science was inextricably linked to militarism and global destruction.
  
  
• Rejection of the Establishment: By the 1980s, he felt completely alienated from his peers. In 1988, he famously refused the highly prestigious Crafoord Prize (and its $250,000 award). He wrote a scathing public letter to the Swedish Academy declaring that the scientific world had become ethically compromised and spiritually dead.
  
  
• Descent into Mysticism: In his later years, his writings became highly focused on spirituality, the apocalypse, and mystical visions. He wrote thousands of pages of personal, philosophical reflections—most notably the massive Récoltes et Semailles (Reaping and Sowing)—which blended brilliant self-analysis with bitter grievances against his former friends and colleagues.

In 1991, he completely vanished from public life. He moved to Lasserre, a tiny village in the French Pyrenees, and legally demanded that all of his unpublished works be destroyed. For the last 23 years of his life, he refused visitors, ignored letters, and lived a fiercely ascetic, solitary existence until his death in 2014 at the age of 86.

The MacTutor has a quite detailed biography about Grothendieck. Within this biography, there are links to even more detailed information about him with this link being especially informative.

A Mathematical Look At 2024

Well, 2024 is almost upon us and so it's time to look at some of the mathematical properties of that number. First and foremost is its factorisation which is:$$2024 = 2^3 \times 11 \times 23$$It can be noted that this factorisation involves only the digits 1, 2 and 3. The final day of 2023 can be written in MM-DD-YY format as 12-31-23 or 123123 which also contains only the digits 1, 2 and 3.

FIRST FUN FACT

The first entry in the OEIS is for A000292:

A000292

Tetrahedral (or triangular pyramidal) numbers:$$\text{a}(n) = \text{C}(n+2,3) = \frac{n \times (n+1) \times (n+2)}{6}$$

Figure 1 illustrates the triangular pyramidal numbers as a sum of triangular numbers stacked upon each other. In the case of 2024, \(n=22\) and this number represents the number of balls in the triangular pyramid in which each edge contains 22 balls. The sequence progresses as follows:

0, 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, 1140, 1330, 1540, 1771, 2024, 2300, 2600, 2925, 3276, 3654, 4060, 4495, 4960, 5456, 5984, 6545, 7140, 7770, 8436, 9139, 9880, 10660, 11480, 12341, 13244, 14190, 15180

Figure 1: source

SECOND FUN FACT

The tetrahedron is one of the Platonic Solids and therefore a shape of great significance.

However, 2024 is also connected with the dodecahedron because it is a member of OEIS A006566 with \(n=8\):

A006566

Dodecahedral numbers: $$ \text{a}(n) = \text{C}(3n,3)  =\frac{n \times (3n - 1) \times (3n - 2}{2} $$

The initial members of the sequence are:

0, 1, 20, 84, 220, 455, 816, 1330, 2024, 2925, 4060, 5456, 7140, 9139, 11480, 14190, 17296, 20825, 24804, 29260, 34220, 39711

So 2024 represents the number of balls in the triangular pyramid in which each edge contains 8 balls. The following video shows how to construct a dodecahedron from nanodots:

The connection between the tetrahedron and the dodecahedron is visible in the GIF below:

Dodecahedron with five tetrahedra inside (source)

THIRD FUN FACT

 The next sequence for 2024 listed in the OEIS is A003242:

A0032425

Number of compositions of \(n\) such that no two adjacent parts are equal (Carlitz compositions).

In the case of 2024, \(n=15\) and here a few examples of the 2024 possible Carlitz compositions (permalink):

• 1, 2, 1, 2, 1, 2, 1, 2, 1, 2
• 1, 4, 2, 1, 4, 3
• 2, 1, 6, 1, 2, 1, 2
• 3, 2, 1, 4, 5
• 6, 2, 1, 2, 3, 1

FOURTH FUN FACT

Watching this video on YouTube, I learned that 2024 is also the sum of consecutive cubes beginning with \(2^3\) and ending with \(9^3\). Thus we have:$$2024=2^3+3^3 + \dots + 8^3+9^3$$This means that next year, 2025, can be represented as:$$2025=1^3+2^3 + \dots +8^3+9^3$$

The March of Time

I've written about what I term AD and BC numbers in a post titled, quite sensibly, AD and BC Numbers. I was reminded of them because my diurnal age today, 27068, converts to 69BC in hexadecimal. For some time now, these BC numbers have been occurring every 256 days.

24764 --> 60bc

25020 --> 61bc

25276 --> 62bc

25532 --> 63bc

25788 --> 64bc

26044 --> 65bc

26300 --> 66bc

26556 --> 67bc

26812 --> 68bc

27068 --> 69bc

However, this regular march of time is now at an end because if 256 is added to 27068, the resultant number (27324) is 6ABC. The decimal equivalent of 70AD is 28845, representing a jump of 1792 or 7 x 256 days.

Similarly, I recently turned 27053 days old which is 69AD in hexadecimal. Notice the difference of 14 days between 69AD and 69BC. This number was also the end of a run of numbers differing by 256 days.

 

24749 --> 60ad

25005 --> 61ad

25261 --> 62ad

25517 --> 63ad

25773 --> 64ad

26029 --> 65ad

26285 --> 66ad

26541 --> 67ad

26797 --> 68ad

27053 --> 69ad

The decimal equivalent of 70AD is 28845, again a jump of 1792 or 7 x 256 days. Taken over a long enough time period, this represents an average advance of a little over 395.6 days, about a month longer than the solar year. Both AD and BC dates are advancing at this average rate.

However, it will almost five years before I encounter another hexadecimal AD and BC number so I thought it important to mark the fact in this post. While exercises like this may seem frivolous, they nonetheless provide an opportunity to work with hexadecimal numbers and convert from decimal to hexadecimal and vice versa. I'm always thinking in terms of the former mathematics teacher who I once was.

How can hexadecimal numbers be made interesting for students? These AD and BC numbers are a way of doing this. The question could be asked of students:

Find out your diurnal age and determine when you will next have a connection to AD or BC year (via decimal to hexadecimal conversion). What was significant about that year.

See my post titled 69BC for details on what was significant about this year in history. Students could be shown how to determine their diurnal age using Wolfram Alpha and they could also use it to convert between decimal and hexadecimal. Overall, a useful and interesting exercise.

General Solution to Cubic Equation

In a recent post titled All Cubic Polynomials Are Point Symmetric, the key point was that the \(x\) coordinate of the point of symmetry is given by: $$ x=\frac{-b}{3a}$$ and this fact also plays a crucial role in devising a solution to the general cubic equation \( ax^3+bx^2+cx+d=0\). It enables the point of symmetry in the graph of the polynomial to be translated horizontally so it lies on the \(y\) axis and the new equation is much easier to solve because the \(x^2\) term is removed.

Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. Source

The technique finds a single real root and the other two roots (real or complex) can be found by polynomial division and the use of the quadratic formula. The first step in the solution is to horizontally translate the point of symmetry so it lies on the \(y\) axis. This is achieved by means of the substitution:$$x=y-\frac{b}{3a}$$This produces:$$a(y-\frac{b}{3a})^3+b(y-\frac{b}{3a})^2+c(y-\frac{b}{3a})+d=0$$Multiplying this out gives$$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^2}{27a^2}-\frac{bc}{3a})=0$$After division by \(a\), this equation is of the general form \(y^3+Ay=B\) and was solved by Scipione del Ferro (1465-1526).

Consider a specific cubic equation e.g. \(x^3-15x^2+81x-175=0\).

Here the substitution will be \(x=y+5\) because \(a=1\) and \(b=-15\) leading to \(y^3+6y-20=0\). The situation is shown in the graph below, where it can be seen that the root of the original equation is 7 and that of the translated equation is 2 (although algebraically we don't know that yet):

We need to find \(s\) and \(t\) such that \(3st=A\) and \(s^3-t^3=B\).

It can be shown that \(y=s-t\) is then a solution of \(y^3+Ay=B\):$$(s-t)^3+3st(s-t)=s^3-t^3$$ $$ (s^3-3s^2t+3st^2-t^3)+(3s^2t-3st^2)=s^3-t^3$$Let's solve \(3st=A\) and \(s^3-t^3=B\) for the simplified cubic \(y^3+6y=20\).

\(3st=6\) and so \(s=2/t\).

Substituting into \(s^3-t^3=20\) gives \((2/t)^3-t^3=20\).

Multiplying by \(t^3\) gives \(t^6+20t^3-8=0\)

Taking the positive root \(t^3=-10+\sqrt{108}\) and \(t=\sqrt[3]{-10+\sqrt{108}}\)

Now \(s^3=20+t^3\) and so \(s^3=20+(-10+\sqrt{108})=10+\sqrt{108}\)

This means that \(s=\sqrt[3]{10+\sqrt{108}}\) and then, because \(y=s-t\), we have:

\(y=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}\)

But \(x=y+5\) and so \(x=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}+5\)

Remarkably, this formidable root (the only real root) evaluates to 7!

So that means \( \sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}=2\). Why is it so?

Well, here is how it all comes about:

\(\sqrt[3]{10 + \sqrt{108}} - \sqrt[3]{\sqrt{108} - 10}\)

\(=\sqrt[3]{10 + 6 \sqrt{3}} - \sqrt[3]{6 \sqrt{3} - 10}\)

\(=\sqrt[3]{1^3+3\sqrt{3}+3(\sqrt{3})^2+(\sqrt{3})^3}-\sqrt[3]{-1^3+3\sqrt{3}-3(\sqrt{3})^2+(\sqrt{3})^3}\)

\(=\sqrt[3]{(1+\sqrt{3})^3}-\sqrt[3]{(-1+\sqrt{3})^3}\)

\(=1+\sqrt{3}+1-\sqrt{3}\)

\(=2\)

Reference: The "Cubic Formula"