Tag Archives: Theorem Egregium

A visual proof of the the theorem egregium of Gauss

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).

Well now,  onto Needham’s geometrical proof.  Disabuse yourself of the notion that it won’t involve much intellectual work on your part even though it uses the geometric intuition you were born with (the green glasses of Immanuel Kant — http://onemillionpoints.blogspot.com/2009/07/kant-for-dummies-green-sunglasses.html)

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.

The Reimann curvature tensor

I have harpooned the great white whale of mathematics (for me at least) the Reimann curvature tensor.  Even better, I understand what curvature is, and how the Reimann curvature tensor expresses it.  Below you’ll see the nightmare of notation by which it is expressed.

Start with curvature.  We all know that a sphere (e.g. the earth) is curved.  But that’s when you look at it from space.  Gauss showed that you could prove a surface was curved just be performing measurements entirely within the surface itself, not looking at it from the outside (theorem egregium).

Start with the earth, assuming that it is a perfect sphere (it isn’t because its rotation fattens its middle).  We’ve got longitude running from pole to pole and the equator around the middle.  Perfect sphere means that all points are the same distance from the center — e.g. the radius.  Call the radius 1.

Now think of a line from the north pole to the plane formed by the equator (radius 1).  Take the midpoint of that line and inscribe a circle on the sphere, parallel to the plane of the equator.  Its radius is the half the square root of 3 (or 1.73). This comes from the right angle triangle just built with hypotenuse is 1 and  one side 1/2.   The circumference of the equator is 2*pi (remember the sphere’s radius is 1).  The circumference of the newly inscribed circle is 1.73 * pi.

Now pick a point on the smaller circle and follow a longitude down to the equator.  Call this point down1.  Move in one direction by 1/4 of the circumference of the sphere (pi/2).  Call that point on the equator down then across

Now go back to the smaller circle at the first point you picked and move in the same direction as you did on the equator by absolute distance pi/2 (not by pi/2 radians).  Then follow the longitude down to the equator.  Call that point across then down.  The two will not be the same.  Across then down is farther from down 1 than down then across.

The difference occurs because the surface of the sphere is curved, and the difference in endpoints of the two paths is exactly what the Reimann curvature tensor measures.

Here is the way the Riemann curvature tensor is notated.  Hideous isn’t it?

If you’re going to have any hope of understanding general relativity (not special relativity) you need to understand curvature.

I used paths in the example, Riemann uses the slope of the paths (e.g derivatives) which makes things much more complicated.  Which is where triangles (dels), and the capital gammas (Γ) come in.

To really understand the actual notation, you need to understand what a covariant derivative actually is, which is much more complicated, but knowing what you know now, you’ll see where you are going when enmeshed in thickets of notation.

What the Riemann curvature tensor is actually saying is that the order of taking covariant derivatives (which is the same thing as the order of taking paths)  is NOT commutative.

The simplest functions we grow up with are commutative.  2 + 3 is the same as 3 + 2, and 5*3 = 3*5.  The order of the terms doesn’t matter.

Although we weren’t taught to think of it that way, subtraction is not.  5 – 3 is not the same as 3 – 5.  There is all sorts of nonCommutativity in math.  The Lie bracket is one such, the Poisson bracket  another, and most groups are nonCommutative.  But that’s enough.  I wish I’d known this when I started studying general relativity.

Two math tips

Two of the most important theorems in differential geometry are Gauss’s Theorem egregium and the Inverse function theorem. Basically the theorem egregium says that you don’t need to look at the shape of a two dimensional surface (say the surface of a walnut) from outside (e.g. from the way it sits in 3 dimensional space) to understand its shape. All the information is contained in the surface itself.

The inverse function theorem (InFT) is used over and over. If you have a continuous function from Euclidean space U of finite dimension n to Euclidean space V of the same dimension, and certain properties of its derivative are present at a point x of U, then there exists a another function to get you back from space V to U.

Even better, once you’ve proved the inverse function theorem, proof of another important theorem (the implicit function theorem aka the ImFT) is quite simple. The ImFT lets you know if given f(x, y, .. .) –> R (e.g. a real valued function) if you can express one variable (say x) in terms of the others. Again sometimes it’s difficult to solve such an equation for x in terms of y — consider arctan(e^(x + y^2) * sin(xy) + ln x). What is important to know in this case, is whether it’s even possible.

The proofs of both are tricky. In particular, the proof of the inverse function theorem is an existence proof. You may not be able to write down the function from V to U even though you’ve just proved that it exists. So using the InFT to prove the implicit function theory is also nonconstructive.

At some point in your mathematical adolescence, you should sit down and follow these proofs. They aren’t easy and they aren’t short.

Here’s where to go. Both can be found in books by James J. Callahan, emeritus professor of Mathematics at Smith College in Northampton Mass. The proof of the InVT is to be found on pages 169 – 174 of his “Advanced Calculus, A Geometric View”, which is geometric, with lots of pictures. What’s good about this proof is that it’s broken down into some 13 steps. Be prepared to meet a lot of functions and variables.

Just the statement of InVT involves functions f, f^-1, df, df^-1, spaces U^n, R^n, variables a, q, B

The proof of InVT involves functions g, phi, dphi, h, dh, N, most of which are vector valued (N is real valued)

Then there are the geometric objects U^n, R^n, Wa, Wfa, Br, Br/2

Vectors a, x, u, delta x, delta u, delta v, delta w

Real number r

That’s just to get you through step 8 of the 13 step proof, which proves the existence of the inverse function (aka f^-1). The rest involves proving properties of f^-1 such as continuity and differentiability. I must confess that just proving existence of f^-1 was enough for me.

The proof of the implicit function theorem for two variables — e.g. f(x, y) = k takes less than a page (190).

The proof of the Theorem Egregium is to be found in his book “The Geometry of Spacetime” pp. 258 – 262 in 9 steps. Be prepared for fewer functions, but many more symbols.

As to why I’m doing this please see https://luysii.wordpress.com/2011/12/31/some-new-years-resolutions/