Nothing better illustrates the difference between the intuitive understanding that something is true and being convinced by logic that something is true than the visual proof of the theorem egregium of Gauss found in “Visual Differential Geometry and Forms” by Tristan Needham and the 9 step algebraic proof in “The Geometry of Spacetime” by Jim Callahan.

Mathematicians attempt to tie down the Gulliver of our powerful appreciation of space with Lilliputian strands of logic.

First: some background on the neurology of vision and our perception of space and why it is so compelling to us.

In the old days, we neurologists figured out what the brain was doing by studying what was lost when parts of the brain were destroyed (usually by strokes, but sometimes by tumors or trauma). This wasn’t terribly logical, as pulling the plug on a lamp plunges you in darkness, but the plug has nothing to do with how the lightbulb or LED produces light. Even so, it was clear that the occipital lobe was important — destroy it on both sides and you are blind — https://en.wikipedia.org/wiki/Occipital_lobe but the occipital lobe accounts for only 10% of the gray matter of the cerebral cortex.

The information flowing into your brain from your eyes is enormous. The optic nerve connecting the eyeball to the brain has a million fibers, and they can fire ‘up to 500 times a second. If each firing (nerve impulse) is a bit, then that’s an information flow into your brain of a gigaBit/second. This information is highly processed by the neurons and receptors in the 10 layers of the retina. Over 30 retinal cell types in our retinas are known, each responding to a different aspect of the visual stimulus. For instance, there are cells responding to color, to movement in one direction, to a light stimulus turning on, to a light stimulus turning off, etc. etc.

So how does the relatively small occipital lobe deal with this?* It doesn’t.* At least half of your the brain responds to visual stimuli. How do we know? It’s complicated, but something called functional Magnetic Resonance Imaging (fMRI) is able to show us increased neuronal activity primarily by the increase in blood flow it causes.

Given that half of your brain is processing what you see, it makes sense to use it to ‘see’ what’s going on in Mathematics involving space. This is where Tristan Needham’s books come in.

I’ve written several posts about them.

and Here — https://luysii.wordpress.com/2022/03/07/visual-differential-geometry-and-forms-q-take-3/

OK, so what *is* the theorem egregium? Look at any object (say a banana). You can see how curved it is by just looking at its surface (e.g. how it looks in the 3 dimensional space of our existence). Gauss showed that you don’t

have to even look at an object in 3 space, just perform local measurements (using the distance between surface points, e.g. the metric e.g. the metric tensor) . Curvature is intrinsic to the surface itself, and you don’t have to get outside of the surface (as we are) to find it.

The idea (and mathematical machinery) has been extended to the 3 dimensional space we live in (something we *can’t* get outside of). Is our universe curved or not? To study the question is to determine its intrinsic curvature by extrapolating the tools Gauss gave us to higher dimensions and comparing the mathematical results with experimental observation. The elephant in the room is general relativity which would be impossible without this (which is why I’m studying the theorem egregium in the first place).

So how does Callahan phrase and prove the theorem egregium? He defines curvature as the ratio of the area on a (small) patch on the surface to the area of another patch on the unit sphere. If you took some vector calculus, you’ll know that the area spanned by two nonCollinear vectors is the numeric value of their cross product.

The vectors Callahan needs for the cross product are the normal vectors to the surface. Herein beginneth the algebra. Callahan parameterizes the surface in 3 space from a region in the plane, uses the metric of the surface to determine a formula for the normal vector to the surface at a point (which has 3 components x , y and z, each of which is the sum of 4 elements, each of which is the product of a second order derivative with a first order derivative of the metric). Forming the cross product of the normal vectors and writing it out is an algebraic nightmare. At this point you know you are describing something called curvature, but you have no clear conception of what curvature* is*. But you have a clear definition in terms of the ratio of areas, which soon disappears in a massive (but necessary) algebraic fandango.

On pages 258 – 262 Callahan breaks down the proof into 9 steps involving various mathematical functions of the metric and its derivatives such as Christoffel symbols, the Riemann curvature tensors etc. etc. It is logically complete, logically convincing, and shows that all this mathematical machinery arises from the metric (intrinsic to the surface) and its derivatives (some as high as third order).

For this we all owe Callahan a great debt. But unfortunately,** although I believe it, I don’t see it**. This certainly isn’t to denigrate Callahan, who has helped me through his book, and a guy who I consider a friend as I’ve drunk beer with him and his wife while listening to Irish music in a dive bar north of Amherst.

Callahan’s proof is the way Gauss himself did it and Callahan told me that Gauss didn’t have the notational tools we have today making the theorem even more outstanding (egregious).

Needham’s definition of curvature uses angular excess of a triangle. Angles are measured in radians, which is the ratio of the arc subtended by the angle to the radius of the circle (*not* the circumference as I thought I remembered). Since the circumference of a circle is 2*pi*radius, radian measure varies from 0 to 2*pi. So a right angle is pi/2 radians.

Here is a triangle with angular excess. Start with a sphere of radius R. Go to the north pol and drop a longitude down to the equator. It meets the equator at a right angle (pi/2). Go back to the north pole, form an angle of pi/2 with the first longitude, and drop another longitude at that angle which meets the equator at an angle of pi/2. The two points on the equator and the north pole form a triangle, with total internal angles of 3*(pi/2). In plane geometry we know that the total angles of a triangle is 2 (pi/2). (Interestingly this depends on the parallel postulate. See if you can figure out why). So the angular excess of our triangle is pi/2. Nothing complicated to understand (or visualize) here.

Needham defines the curvature of the triangle (and any closed area) as the ratio between the angular excess of the triangle to its area

What is the area of the triangle? Well, the volume of a sphere is (4/3) pi * r^3, and its area is the integral (4 pi * r^2). The area of the north hemisphere, is 2 pi *r^2, and the area of the triangle just made is 1/2 * Pi * r^2.

So the curvature of the triangle is (pi/2) / (1/2 * pi * r^2) = 1 / r^2. More to the point, this is the curvature of a sphere of radius r.

At this point you should have a geometric intuition of just what curvature is, and how to find it. So when you are embroiled in the algebra in higher dimensions trying to describe curvature there, you will have a mental image of what the algebra is attempting to describe, rather than just the symbols and machinations of the algebra itself (the Lilliputian strands of logic tying down the Gulliver of curvature).

The road from here to the Einstein gravitational field equations (p. 326 of Needham) and one I haven’t so far traversed, presently is about 50 pages.Just to get to this point however, you have been exposed to

*comprehensible geometrical expositions*, of geodesics, holonomy, parallel transport and vector fields, and you should have mental images of them all.Interested? Be prepared to work, and to reorient how you think about these things if you’ve met them before. The 3 links mentioned about will give you a glimpse of Needham’s style. You probably should read them next.