## Saturday, October 16, 2010

### Whence 4?

Time for some more math! At my (new! yay!) supervisor's behest I've been trying like a primordial lungfish to breath the fresh air called spectral sequences --- how they ever managed to acclimate to such thin and rarefied reference to anything tangible is beyond me; but then, the calculations themselves are thick as water to my more-terrestrial brain.

Today is not about spectral sequences, but back to geometric measure theory. Some time ago I got as far as to outline the proportion of surface areas vs. angles
$$S(\triangle qrs):S(\mathbb{S})::\angle qrs + \angle rsq + \angle sqr - \pi:4\pi$$
for a spherical triangle $qrs$ --- although for reasons of presentation, that tale refered to the hemispheres $A,B,C$ described by the arcs $qr,$ $rs,$ $sq$ and containing the triangle $\triangle qrs$. (Do you see some ambiguity creeping into the tale? Don't worry: make your choices and then show that they aren't important!)

So, our task today is to improve the proportionality by establishing, for a unit sphere $\mathbb{S}$, an equality $S(\triangle qrs) = \angle qrs + \angle rsq + \angle sqr - \pi$, which by the complementary proportion will give $S(\mathbb{S})=4\pi$.

For simplicity, we'll assume the angle at $s$ --- that is, $\angle rsq$ is a right angle, also known as $\frac{\pi}{2}$, or half of $\pi$. (Not, btw, half a pie... half a pie is a whole $\pi$, and we're also trying to show today that half a unit pea has rounded area $2\pi$!). This means reducing our problem to the study of $\angle qrs+\angle sqr-\frac{\pi}{2}$, a form which should resonate in the mind of anyone who has thought seriously about right triangles in the more extraordinary, flat world.

A logician I once knew exulted in declaring that Man's office of naming things is also the mathematician's first tool: it does wonders to our reasoning to give a short name to otherwise clunky perilocutions as "the angle subtended at the spherical center by the arc $sq$" --- this angle I will call $\sigma$. The other arc's angle, that of $rs$ will be $\tau$. For similar reasons, the angles at $q$ and $r$ will become $\phi$ and $\theta$. One more angle needs a name, that of the altitude from $s$ to the hypotenuse, which I shall call $\zeta$. Its sole purpose is to simplify the formulae following our sketch of the situation thus far:
with this picture, we have the two equations
$$\sin \tau \cos\theta = \cos\tau \sin\zeta$$
$$\sin \sigma \cos\phi = \cos\sigma \sin\zeta$$
Incidentally, it's quickest to derive this formula if you already know ---as we do--- that great circles lie in planes through the origin, each of which is a set of points $(x,y,z)$ satisfying an equation like $x\cos\phi + y\cos\theta = z\sin\zeta$. By far the simplest situation to deal with is when you have some relation like $\cos^2\phi + \cos^2\theta + \sin^2 \zeta = 1$ --- and as I've described the angles $\phi,\theta,\zeta$ above, that's exactly the case here. Why? Hmm... it's more linear algebra. There ought to be auxiliary diagrams that clear it all up...

Anyways, The Trick, at this point, is to study the area of our triangle, and compare it to other things that are similar. For instance, when $\sigma<1$ we have

$$\sigma - \frac{\sigma^3}{6} < \sin\sigma < \sigma < \tan\sigma < \frac{\sigma}{1-\sigma^2/2}$$

This means, for instance, that the area of a rectangle with sides $\tan\tau$ and $\tan\sigma$ is very close to $\tau\sigma$, which itself looks very close to twice the area of our spherical triangle --- how close? Well, in this case, it means if we fix $\theta$, and then imagine $\zeta$ decreasing to zero, then $\tau,\sigma$ also tend to zero, so that $1<\frac{\tan\sigma}{\sigma} < \frac{1}{1-\sigma^2/2}$ squeezes the middle ratio to $1$, sandwich-wise. The use of the accidental simplicity $\cos^2\theta+\cos^2\phi + \sin^2\zeta=1$ is that it can equally be written $\sin^2\zeta = \sin^2\theta - \cos^2\phi$, which if you remember your double-angle formulas, and your heterodyne equation, is
$$\sin^2\zeta = - \frac{\cos 2\theta + \cos 2\phi}{2} = -\cos(\theta+\phi)\cos(\theta-\phi)$$
What good is that? Well, it's handy that we can (again) just as well write
$\sin(\theta+\phi-\frac{\pi}{2}) = - cos(\theta+\phi)$...
oh dear, that's getting clunky, again... let $\alpha = \theta+\phi-\frac{\pi}{2}$. Note that this *is* the angle excess which we know to be exactly proportional to the area of our triangle.
Then $\sin^2\zeta = \sin(\alpha) \cos(\theta-\phi)$.
This is nifty, because on the one side we actually have $\tan\sigma\tan\tau \cos\phi\cos\theta$, which is proportional to our imagined rectangle, and the proportion only depends on $\theta$ and $\phi$; the other side is proportional to $\sin\alpha$ --- which is nearly $\alpha$ when $\alpha$ gets small, as it will --- so it seems we're left with comparing
$$\cos\phi\cos\theta$$ vs. $$\cos(\theta-\phi).$$
Now, you can't see --- because I skipped the step --- but what I've been writing $\cos(\theta-\phi)$ once used to be $\cos\phi\cos\theta + \sin\phi\sin\theta$. Oh! hooray! there's a $\cos\phi\cos\theta$! Oh! drat! there's another term! What is that other term?
Well, with the new name ($\alpha$) at our disposal, it's easily rewritten
$$\cos(\theta-\alpha)\cos(\phi-\alpha)$$.

Compiling our results so far, we're at the point where we can say
$$\frac{\tan\sigma\tan\tau}{\sin\alpha} = 1 + \frac{\cos(\theta-\alpha)}{\cos\theta}\frac{\cos(\phi-\alpha)}{\cos\phi}$$
In other words, the ratio of rectangle to angle-excess is $1$ plus that other thing --- that the other thing tends to $1$ as $\zeta$ tends to $0$ is yet more of these angle-sum and -difference rules, but as you can probably see, we're not far off.

The rough rectangle to roughly the angle excess is roughly 2, so the triangle to the angle excess is exactly 1.

We're done.