Assignment 8
Problems
The Dot Product
One thing we used in our arguments building up the basics of spherical geometry was the fact that the dot product has a nice derivative rule.
Exercise 1 (Product Rule for Dot Product) Let \(f(t)=\langle f_1(t),f_2(t),f_3(t)\rangle\) and \(g(t)=\langle g_1(t),g_2(t),g_3(t)\rangle\) be two vector functions. Prove that the dot product satisifes the product rule: \[\frac{d}{dt}\left(f(t)\cdot g(t)\right)= f^\prime(t)\cdot g(t)+f(t)\cdot g^\prime(t)\]
Our use of the dot product overall is as a tool to give the sphere geometry it defines what we mean by infinitesimal length and by angle. Often we will use this just as a theoretical tool - but its good to get some hands-on practice at the beginning, measuring some actual angles.
Exercise 2 Consider the curves \(\alpha(t)=(\cos t,\sin t,0)\) (the equator of the sphere), and \(\beta(t)=(0,\sin(t),\cos(t))\) (a line of longitude). Prove that they
- Intersect each other at the \(t=\pi/2\)
- Form a right angle at their point of intersection.
Draw a picture of this situation in 3D on a sphere.
Isometries
Recall the definition of an isomery of \(\SS^2\) is a function \(\phi\colon\SS^2\to\SS^2\) which preserves the dot product (or equivalently, preserves infinitesimal lengths).
Exercise 3 A permutation matrix is a square matrix where every row and column has exactly one “1”, and the other entries are zero. Prove the following permuatation matrix \[A=\pmat{0&1&0\\1&0&0\\0&0&1}\]
can be used to define an isometry of \(\SS^2\) by the formula \[\phi(x,y,z)=A\pmat{x\\ y\\ z}\] directly from the definition of isometry.
In class we are working our way to prove some facts about isometries, mimicking what we did in the Euclidean plane. In particular, on Tuesday we will prove the following two important facts.
Theorem 1 (The Sphere is Homogeneous) Given any two points \(p\) and \(q\) on the sphere, there is an isometry taking \(p\) to \(q\):
Proposition 1 Let \(N\) be the north pole, and \(v\) be any unit vector in \(T_N\SS^2\). Then there exists an isometry \(\phi\) of the sphere which fixes \(N\) and takes \(\langle 1,0,0\rangle\in T_N\SS^2\) to \(v\).
The first is an analog of translations of the Euclidean plane: we can always find an isometry of the sphere that takes any point to any other. And the second is similar to when we proved that you could build rotations of \(\EE^2\) about the origin (we’ll actually prove it, using that exact theorem!)
Your goal in these next problems is to use these theorems to prove even more: first, prove that you can actually rotate the sphere fixing any point you wish (not just the north pole!)
Exercise 4 Use Proposition prp-sphere-rotate-north-pole and Theorem The Sphere is Homogeneous to show the sphere is isotropic: that given any point \(p\in\SS^2\) and any two unit vectors \(v,w\in T_p\SS^2\), there exists an isometry of \(\SS^2\) fixing \(p\) and taking \(v\) to \(w\).
(Hint: first show you can do this when \(p\) is the north pole! Then use homogenity and a conjugation. Be inspired by the Euclidean proof!)
Next, just like in Euclidean space we often find it useful to combine homogenity and isotropy into one condition: we can move any point and any tangent vector to any other point and tangent vector that we like!
Exercise 5 Let \(p,q\) be any two points on the sphere, and \(v\) a unit tangent vector at \(p\) and \(w\) a unit tangent vector at \(q\). Then there is an isomtery of \(\SS^2\) taking \((p,v)\) to \((q,w)\).
Hint: Look back at Ex 4 Homework 4!
Solutions
The Dot Product
The Product Rule
This problem is just an algebraic computation (but it was assigned since the result itself is very useful, and we used it to derive that the tangent space \(T_p\SS^2\) is orthogonal to \(p\) in \(\EE^3\).)
Given \(f(t)=\langle f_1(t),f_2(t),f_3(t)\rangle\) and \(g(t)=\langle g_1(t),g_2(t),g_3(t)\rangle\), we take their dot product and get \(f\cdot g = f_1g_1+f_2g_2+f_3g_3\). (I’ve left out the explicit time dependence \((t)\) in the notation since everything is a function of \(t\), we will just remember that all symbols in our expression are functions). Now we can take the derivative:
\[\begin{align*}(f\cdot g)^\prime &= (f_1g_1+f_2g_2+f_3g_3)^\prime\\ &= (f_1g_1)^\prime + (f_2g_2)^\prime+ (f_3g_3)^\prime \end{align*}\]
By the product rule, we have \((f_ig_i)^\prime = f_i^\prime g_i+f_i g_i^\prime\) for each \(i\in\{1,2,3\}\). Plugging this in, we can reorder the addition so that all terms with derivatives of \(f\) come first:
\[(f_1^\prime g_1+f_1g_1^\prime)+(f_2^\prime g_2+f_2g_2^\prime)+(f_3^\prime g_3+f_3g_3^\prime)\]
\[=(f_1^\prime g_1+f_2^\prime g_2+f_3^\prime g_3)+(f_1g_1^\prime+f_2g_2^\prime+f_3g_3^\prime)\]
Now by the definition of the dot product in \(\EE^3\) we see the terms in the first parentheses are just an unpacking of \(f^\prime \cdot g\) and in the second set are \(f\cdot g^\prime\). Thus putting it all together, we’ve shown whwat we needed to:
\[(f\cdot g)^\prime = f^\prime \cdot g + f\cdot g^\prime\]
Angles Beetween Curves
To prove \(\alpha(t)\) and \(\beta(t)\) intersect at \(t=\pi/2\), we just need to plug in this value of \(t\) to each curve and observe that we get the same point.
\[\alpha(\pi/2)=\left(\cos\frac\pi 2, \sin\frac\pi 2,0\right)=(0,1,0)\] \[\beta(\pi/2)=\left(0,\sin\frac\pi 2, \cos\frac\pi 2\right)=(0,1,0)\]
So, they meet the point \(p=(0,1,0)\). Our next goal is to show that they intersect orthogonally there. For this, we need to find their tangent vectors in \(T_{p}\SS^2\), which are the result of differentiation:
\[\alpha^\prime(t)=\langle -\sin t, \cos t, 0\rangle\] \[\beta^\prime(t)=\langle 0, \cos t, -\sin t\rangle\]
At \(t=\pi/2\) we then have
\[\alpha^\prime(\pi/2)=\langle -1,0,0\rangle\in T_p\SS^2\] \[\beta^\prime(\pi/2)=\langle 0,0,-1\rangle\in T_p\SS^2\]
Because these two vectors live in the same tangent space, they define an angle on the sphere. We can measure the angle using the sphere’s dot product: since they are unit vectors,
\[\begin{align*}\cos\theta &= \alpha^\prime(\pi/2)\cdot \beta^\prime(\pi/2)\\ &= \langle -1,0,0\rangle\cdot\langle 0,0,-1\rangle\\ &-=0 \end{align*}\]
Thus \(\cos\theta=0\), so \(\theta=\pi/2\), the vectors are indeed orthogonal as claimed.
Isometries
Permutation Matrix
Here we wish to show that the isometry \(\phi(x,y,z)=A(x,y,z)\) is an isometry, for the \(A\) the permutation matrix
\[A=\pmat{0&1&0\\1&0&0\\0&0&1}\]
Recall that we have two equivalent notions for what it means to be an isometry, we can either check that \(\phi\) preseves all infinitesimal lengths, or that \(\phi\) preserves the dot product. Neither one is easier or harder than the other, so I’ll do the latter.
This means I need to start with an arbitrary point \(p\in\SS^2\), and two arbitrary vectors \(v,w\in T_p\SS^2\), and show that applying \(\phi\) doesn’t change anything:
\[v\cdot w = (D\phi_p v)\cdot (D\phi_p w)\]
First, we note that since \(\phi\) is defined as a linear map, its easy to differentiate: its derivative is just itself! (This is what I call the cool lemma in class because its cool how fast it makes it to take the derivative of isometries)
Thus for any point \(p\), the map \(D\phi_p\) acts on vectors \(v=\langle v_1,v_2,v_3\rangle\) in \(T_p\SS^2\) as
\[D\phi_p v= \pmat{0&1&0\\1&0&0\\0&0&1}\pmat{v_1\\v_2\\v_3}=\pmat{v_2\\ v_1\\v_3}\]
Applying the same map to \(w\), we see that \(D\phi_p w= \langle w_2,w_1,w_3\rangle\).
Now its just a matter of algebra. We need to compute the dot products before and after applying \(\phi\) and see they’re equal:
\[\begin{align*} (D\phi_p v)\cdot (D\phi_p w)&=\langle v_2,v_1,v_3\rangle\cdot \langle w_2,w_1,w_3\rangle\\ &= v_2w_2+v_1w_1+v_3w_3\\ &= v_1w_1+v_2w_2+v_3w_3\\ &= v\cdot w \end{align*}\]
Thus, \(\phi\) preserves the dot product, so its an isometery!
The Sphere is Isotropic
Fix an arbitrary point \(p\in\SS^2\): we want to show that if we have any two unit vectors \(v,w\in T_p\SS^2\), we can build some isometry which takes \(v\) to \(w\).
Using what we know, let’s first take an isometry that moves \(p\) to the north pole. Call this isometry \(\phi\), so \(\phi(p)=N\). Applying this to our two vectors \(v\) and \(w\), we get two new vectors now based at the north pole. So that we can remember where they came from, I’ll name them using the capital letter vsion of their original name:
\[V=D\phi_p v\in T_N\SS^2\hspace{1cm} W=D\phi_p w\in T_N\SS^2\]
Now I will work to move \(V\) to \(W\). Unfortunately, we don’t have a pre-built tool for this! Instead, the tool we have (the theorem the problem statement asks us to use) just says that we can find an isomery fixing \(N\) which takes \(\langle 1,0,0\rangle\) to any vector in \(T_N\SS^2\). So, we’re going to have to get creative, and use this twice!
Let \(\alpha\) be the isometry this theroem guarantees which takes \(\langle 1,0,0\rangle\) to \(V\), and let \(\beta\) be the similar isometry taking \(\langle 1,0,0\rangle\) to \(W\). The inverse \(\alpha^{-1}\) is then an isometry which takes \(V\) to \(\langle 1,0,0\rangle\), and following this with \(\beta\) carries this to \(W\).
Thus, we have found an isometry \(\beta\circ\alpha^{-1}\), which fixes \(N\) and takes \(V\) to \(W\). Now there’s only one step left: we used \(\phi\) to get \(p\) up to \(N\), and now we need to use its inverse to get \(N\) back down to \(p\)! That means overall, our isometry is
\[\phi^{-1}\circ\beta\circ\alpha^{-1}\circ\phi\]
And now, we check that this does what we want. First, note it fixes \(p\): first \(p\) is moved to \(N\), then we fix \(N\), and finally we take \(N\) back to \(p\). Second, we need to see that this actually carries \(v\) to \(w\). But this is exactly how we designed it! First \(\phi\) takes \(v\) to \(V\), then \(\beta\alpha^{-1}\) takes \(V\) to \(W\), and finally \(\phi^{-1}\) takes \(W\) to \(w\), becuase we know \(\phi\) took \(w\) to \(W\) by definition!
Homogenity & Isotropy
Finally we wish to show that given any two points \(p,q\in\SS^2\) and any two unit vectors \(v\in T_p\SS^2\), \(w\in T_q\SS^2\), we can find a single isometry which simultaneously takes \(p\) to \(q\) and \(v\) to \(w\). We essentially did all the hard work in the last problem, so here we just assemble the pieces.
First, build any isometry we like that takes \(p\) to \(q\): we may remember one of these exists from class, but if not we can put one together quickly: if \(\phi\) takes \(p\) to \(N\) and \(\psi\) takes \(q\) to \(N\) we can to \(\phi\) then the inverse of \(\psi\) to get \(p\) to \(q\).
Now, we have some isometry \(f\) which takes \(p\) to \(q\), but unfortunately we have no idea what it does to our tangent vector \(v\in T_p\SS^2\). But that’s OK - we know it must take it to some vector in \(T_q\SS^2\), so let’s just call that vector \(V\). Now all we need to do is get \(V\) to \(w\), while holding \(q\) fixed. But this is exactly what the above problem taught us how to do! So, let \(g\) be such an isometry. Then the composition \(g\circ f\) is what we are after: this takes \(p\) to \(q\), and takes \(v\) to \(w\)!