Tony’s Take October 2025

This month’s topics:

Geometry: trying the Socratic method on ChatGPT.

Can ChatGPT generate knowledge, or merely recollect it? To investigate this question, Nadav Marco (Hebrew University of Jerusalem) and Andreas Stylianides (University of Cambridge) turn to an ancient discussion about human learning. In Plato’s dialogue Meno, Meno asks: “Yes, Socrates; but what do you mean by saying that we do not learn, and that what we call learning is only a process of recollection?” In the passage that ensues, usually referred to as “Socrates and the slave-boy,” the great man uses leading questions to “remind” the youngster that he already knows how to duplicate a square. (Full translation of the passage is here, thanks to R.A.G. Seely.)

In a recent article in the International Journal of Mathematical Education in Science and Technology, Marco and Stylianides present Socrates’ example to ChatGPT, and examine Chat’s responses for evidence of “knowledge as a recollection from memory” versus “knowledge as ongoingly generated from experiences.” Live Science covered the work in “Scientists asked ChatGPT to solve a math problem from more than 2,000 years ago — how it answered it surprised them.”

Left: A square cut into four squares, with the lower left square highlighted. Right: The same image, now with a red square whose sides are the diagonals of the four small squares.
The “doubling the square” problem. Given a $2\times 2$ blue square (left), how to construct a red square with exactly double the area? Figure 1 from
Marco and Stylianides, used under a CC BY 4.0 License.

The authors begin by repeating, with Chat, Socrates’s prompts: “Do you know what a square is?”, etc. (The complete text of the interaction is recorded here.) Presented with the doubling problem, Chat suggests multiplying the square’s side lengths by $\sqrt{2}$. This answer is presumably recalled from information acquired during training. The authors ask for a method that is more exact, since the number $\sqrt{2}$ can only be known approximately. Chat answers that “reflecting on your challenge with fresh perspective” it has found an exact method (that shown in the illustration above) that it “should have emphasized” earlier. So another answer recalled from memory, even though the recollection required prompting.

To explore if Chat has gained new knowledge during this exercise, the authors then ask it to double the area of a $2\times 3$ rectangle while preserving the exact proportions of its sides. Chat initially gives the algebraic answer (scale both sides by $\sqrt{2}$). When asked if “your beautiful solution with the diagonal” could be adapted to this context, Chat incorrectly interprets diagonal as necessarily referring to the diagonal of the rectangle (which doesn’t work) and seems to insist that algebraic scaling is the only “practical approach.” The authors remark that the response is more like generation of knowledge than recollection, since it is unlikely that this incorrect idea could have been retrieved from memory. It seems to me that Chat’s inability to find a geometric solution to the problem in its training data is another (negative) example of knowledge as recollection.

A blue rectangle with squares extended from its left and top edges. The diagonals of the squares are shown. By shifting one diagonal to meet the other at a right angle, we get a new rectangle.
A geometric construction for doubling the area of a rectangle. Figure 7 from Marco and Stylianides, used under a CC BY 4.0 License.

The authors then teach Chat the geometric solution, as in the “proof without words” diagram above. Chat responds with rapturous praise (“Your insight to leverage the properties of squares to address a challenge with rectangles shows deep understanding and ingenuity”) and the conversation goes on to other doubling problems. When prompted for a geometric solution, Chat explains how to double the area of an equilateral triangle using the same technique the prompter had used for the rectangle. This is clearly an example of “knowledge generated from experience” since Chat initially did not know this tactic but has learned it; the authors have shown, as they state in the abstract, that “the Chat’s responses reflected both types of knowledge.”

I’m more intrigued by their observation that Chat displays “overgeneralization”: it found a diagonal construction that worked for the square, and tried to repeat the strategy for a rectangle. Overgeneralization is familiar to us from children’s grammar errors like “bringed.” Is there a connection? The possibility that investigations with AI may lead to progress in understanding human thought processes was brought up recently by James Somers in The New Yorker. He quotes neuroscientist Doris Tsao (University of California, Berkeley), who claims that the deepest insight she has gained from ChatGPT “is that I think it radically demystifies thinking.”

Walking through rotation space, and coming home.

With doubling and scaling, almost any continuous string of rotations can be turned into a loop. That’s what Jean-Pierre Eckmann (University of Geneva) and Tsvi Tlusty (Ulsan National Institute of Science and Technology in South Korea) prove in “Walks in Rotation Spaces Return Home when Doubled and Scaled,” published October 1 in Physical Review Letters. New Scientist covered the research on October 16.

Here, I’ll focus on special cases where these walks correspond to a solid body rolling on the $xy$-plane. Let’s assume we have a sphere of radius 1 which starts at the origin, and begins rolling in the positive $x$ direction. It rolls, without sliding or swiveling, a distance $\pi/2$ in that direction, and then rolls a distance $\pi/4$ in the negative $y$ direction. Paths like this correspond to paths in the space of rotations of the sphere: In this case, first, rotate the sphere by $\pi/2$ around the $y$-axis, then down by $\pi/4$ around the $x$-axis. (For a more general curve, the axis of rotation would vary from point to point.)

A path in the plane starting at $\mathrm{O}$ gives a path in the space of rotations of the sphere. Here the sphere is first rotated through an angle $\pi/2$ around an axis parallel to the $y$-axis, then through an angle $-\pi/4$ around an axis parallel to the $x$-axis. Note that the points of tangency trace out a lifted curve up on the sphere. Image credit: Tony Phillips.

Can a given path in the space of rotations be rescaled so that the sphere ends up back in its original orientation? This is what the authors call “returning home.”

The point of tangency traces out another path, this one on the sphere. In terms of this lifted path, returning home is governed by two parameters. First, let $p$ be the point on the sphere that touches the plane at the start of the walk. Second, let ${\bf v}$ be a tangent vector to the sphere at $p$, corresponding to the direction in which the sphere begins to roll. (In our example above, ${\bf v}$ initially points along the $x$-axis.) If the walk successfully returns home, then at the end the sphere once again touches the plane at $p$, with ${\bf v}$ parallel to its initial direction.

The authors assert that while just rescaling a path in rotation space can almost never meet the two requirements, almost any path can be doubled and rescaled so as to meet them. To explain this, they rely on the geometry of parallel transport, a way of generalizing parallelism to curved surfaces.

Parallel transport. In the plane, we know how to determine if two vectors at different locations $p$ and $q$ are parallel: We just slide them together and check if they match. But what about vectors at two different points on a curved surface? The question, partly motivated by physics, was settled in the early 20th century by the Italian geometer Tullio Levi-Civita. The solution was parallel transport along a curve. Given any piecewise smooth curve $c$ from $p$ to $q$, and any tangent vector ${\bf v}$ at $p$, there was a differential equation which would produce a new, same-length vector at $q$, “the parallel transport of ${\bf v}$ along $c$”.

Parallel transport has intuitive appeal. In the plane, the parallel transport of ${\bf v}$ along any curve is parallel to ${\bf v}$ in the usual sense. Also, if $c$ is the shortest curve (geodesic) between $p$ and $q$, then the angle between ${\bf v}$ and $c$ stays constant during transport; in two dimensions, this determines parallel transport along geodesics completely.

It turns out (a recent reference is here) that for surfaces there is a simple way to realize parallel transport. Here’s how it works. Say we have a surface $S$, a curve $c$ on $S$ starting at $p$ and ending at $q$, and a vector ${\bf v}$ tangent to $S$ at $p$. We are going to transport ${\bf v}$ along $c$.

Set the surface down so that it is tangent at $p$ to a plane (a non-convex surface may have to intersect the plane; this is not a problem). Then ${\bf v}$ coincides with a vector in the plane; let’s call it ${\bf v’}$. Now roll the surface, without slipping or swiveling, so that the point of tangency follows $c$. At the end of the process, the point of tangency will be $q$, and vectors in the tangent space to $S$ at $q$ will coincide again with vectors in the plane. The vector parallel (in the plane) to ${\bf v’}$ will be the parallel transport of ${\bf v}$ along $c$. (An example is in the next figure.)

This formulation of parallel transport echoes the process of Eckmann, Tlusty et al. relating initial and final orientations of the sphere with respect to the plane. It means that the authors can use a striking relation between parallel transport and curvature: If you transport a vector counterclockwise around a loop, the final result will be rotated by an angle equal to the total (Gaussian) curvature enclosed by loop. Thus the change in orientation while rolling around a curve is equal to the total curvature enclosed by that curve.

Here is an example on the unit sphere, using the curve that borders a spherical octant—one-eighth of the sphere’s surface, a quarter of a hemisphere. Parallel transport around the boundary of an octant is especially simple since its three sides, arcs of great circles, are geodesics. Furthermore, the unit sphere has Gaussian curvature $1$ and area $4\pi$, so total curvature $4\pi$, and the total curvature enclosed by the octant, an eighth of that, is $\pi/2$.

A sphere with an octant whose corners are p1, p2, and p_3. On the plane, the boundary of the octant forms three sides of a rectangle, with corners q1, q2, q3, and q4.
The unit sphere rolls along the path $\mathrm{O}q_2q_3q_4$. That path lifts to the spherical triangle $p_1p_2p_3$, an octant. At the end, the orientation relative to the plane has rotated by by $\pi/2$. In terms of parallel transport, the vector ${\bf v}$, transported around $p_1p_2p_3$, is rotated by $\pi/2$. And $\pi/2$ is precisely the total curvature enclosed by the octant. Image credit: Tony Phillips.

Back in rotation space. This means that if a curve encloses half the sphere, then the parallel transport of $\bf{v}$ around the curve rotates $\bf{v}$ by $2\pi$. That is, $\bf{v}$ ends up in the same position and orientation as it started. So if a path can be doubled and rescaled so as to form a loop enclosing half the sphere, it can always return home. This is how Eckmann and Tlusty proceed. They manipulate the lifted path (up on the sphere), roll the new path out into the plane and interpret it, as above, as a path in rotation space.

Doubling (schematic). If the endpoints of the lifted curve are different, the authors draw a geodesic (great-circle arc) on the sphere between them. They identify the midpoint of that arc and rotate a copy of the curve $180^{\circ}$ about that point to form a closed curve. Image credit: Tony Phillips.
Rescaling (schematic). When the sphere orientations at start and finish do not match, the curve is rescaled to make the enclosed area exactly half of that of the sphere. Then the geometry of parallel transport implies that the orientation vector will have rotated by exactly $2\pi$, i.e. the beginning and ending orientations will coincide. Image credit: Tony Phillips.
End of procedure (schematic). The doubled and rescaled path on the sphere is unrolled onto the plane. Note that the sphere’s initial and final tangency points are the same, with parallel orientations: We have “returned home.” Image credit: Tony Phillips.

The authors’ construction answers a question about solid bodies rolling along a periodic path. A simple example is the hexasphericon. The cylindrical and conical parts of its surface line up enough so that positioned properly it will roll down an inclined plane in an entertaining way. (For an earlier example, the sphericon, see Ian Stewart’s column in the October 1999 issue of Scientific American).

a. Construction of hexasphericon. A hexagon is spun around its axis and split in two. The two halves are joined after one has been rotated by $60^{\circ}$. b. A hexasphericon will roll down an inclined plane following a periodic path. In blue, one period of the corresponding curve in rotation space. c. Lifted up to the sphere, the curve is closed and splits the surface into two equal areas. Image credit: Tony Phillips.

We can abstract the hexasphericon’s cylindrical/conical surface structure (which determines its rolling motion) to a curve on a round sphere; in the plane, the rolling tangent point traces out a curve and its translates (dashed line in the image above). This is an indefinitely extended periodic path. In an earlier publication, the authors had asked which other paths in the plane could be generated, and extended periodically, by a rolling solid. Their work implies that, after doubling and rescaling, almost any path in the plane works.

—Tony Phillips, Stony Brook University