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Unformatted text preview: 1 3. The Sphere l. Fundamental Concepts The Sphere
1.1 Deﬁnition: The (unit) sphere is deﬁned to be the set
S2 = {u E Rallul = 1}. Given u E 52, the point ~u is called the antipodal point of u. The Cross Product in R3 1.2 Deﬁnition: Recall the following deﬁnitions. Let u and v be vectors
in R3. Recall that the dot product of u and v is deﬁned to be uv = ulvl +U2’02 +u3v3. If u and v are non—zero, the angle between u and v in R3 is the angle 0(u,v)=cos_1<u'v) E[0,7r]. IUI lvl The cross product of u and v is deﬁned to be where {61, 62, 63} is the standard basis for R3. 1.3 Theorem: Recall the following facts about the cross product. 1) u x v = —v X u. 2) u X 1; = 0 if and only ifu and v are linearly dependent. 3) IfuxvaéOthenuxviu,uxvl_vanduxvpointsin the
direction of the thumb of the right hand if the ﬁngers point
from u to 1) (so that {u,v,w} is a positively oriented basis for
R3). 4) u x v = [U] v sin0(u,v), Which is equal to the area of the
parallelogram on the vectors u and v. 1.4 Deﬁnition: The tri ﬁle product of three vectors u,v,w E R3 is
deﬁned to be u,v,w = (u x v)w. 1.5 Theorem: Recall the following facts about the trippéle product. M x \T
L , \ M
lb”: “Obi“ N “A
V“
’jl‘r~\ x .
7 (”6 [Vrf‘clll‘il (ch (kl ("\ ﬁgure fl’. 3 . 3 ,
on m CMA \I “1 v1 wi 1) lu,v,wl= 112 v2 wz .
U3 ’03 W3
2) u,v,wl = lv,w,ul = lw,u,v = —v,u,w = —lu,w,v =
—w,v,ul.
3) luauavl : lu’v)ul = lu,v,vl : 0' 4) lu,v,w = 0 if and only if {u,v,w} is linearly dependent. If
lu, v, wl > 0 then {u,v,w} is a positively oriented basis for R3,
and if lu,v, wl < 0 then {u,v,w} is a negatively oriented basis
for R3. 5) llu,v,wl is equal to the volume of the parallelipiped on the
vectors u, v and w. 1.6 Theorem: For u,v,w,x E R3 we have
1) (uxv)xw=(u w)v—(v w)u.
2) (11 X 1)) (w x x): (u w)(v z)— (:2  w)(u . 3:). Proof: The proof 15 left as an exercise. D Distance 1.7 Deﬁnition: Given two points 11 and v in 52, we deﬁne the distance
between 11 and v in 52 to be equal to the angle between 11 and ’U in R3, that is
d(u,v) = cos—1(uv). Notice that d(u,v) S 7r with equality if and only if u _—_ ~12. 1.8 Theorem: The distance d(u,v) between two points it and v in 32
determines and is determined by the Euclidean distance [11 — vl between
11 and v in R3. Proof: For 11,1) 6 S2, we have 0 _<_ d(u,v) _<_ 7r and 0 S lu — vl S 2.
By the law of cosines, we have Iu — 1)]2 = 2 — 2cos 0(11, 1)) , hence we see that
lu —vl = W and
d(u,v) = cos 1(1 — —lu—v2).
El 1.9 Theorem: The distance function is a metric on $21 In other words,
for all 11,12,112 E S2 we have 1) d(u, v): d(v, u), 2) d_(u, u)=0andifu19év thend(u, v)>0, and MMJ),LMIKJ\P "2”.) Proof: Parts 1) and 2) should be clear. To prove part 3), we have
cos(6(u, '0) + 0(1), w)) = cos 0(u, '0) cos 9(1), w) — sin 0(u, '0) sin 0(1), w)
:(uv)(v~w)—lu xvllv xwl
§(u~v)(vw)—(uxv)(vxw)
= (u  v)<v  w) — ((u  v)(v  w) — (u  wxv  v»
= (u  w)
= cos 0(u, 111). Since c030 is decreasing, this shows that 0(u,w) S 0(u,v) + 0(v,w). 1:! Lines lbev Km gillel Q: i on Ux/V amoi W 3 1.10 Deﬁnition: Given a point n E S2, we deﬁne the line Ln, with
pole n, to be Ln={uESZun=0}, (or equivalently, Ln = 52 ﬂ ker 72‘). The line L.n can also be described
by a parametric equation as follows: let u be any point on the line, and
let u = n x u (so that v is another point on the line, and {u,v} is an
orthonormal basis for ker n‘), and we have Ln = {f(t)0 S t _<_ 27r},
Where f(t) = ucost+vsint. The map f(t) deﬁnes a curve which starts at f(0) = u and traces out
the line Ln in the direction of v, with f(%) = v. The line is traced
anti—clockwise if the vector u points towards us, or in other words, the
line is traced in the direction in which the ﬁngers on our right hand
would point if our thumb were in the direction of n. We deﬁne a line in 32 to be any set of the form Ln for some n 6 S2,
(or equivalently, any set of the form 52 0V for some 2—dimensional vector
space V C R3). Notice that Ln = Lm if and only if n = ::m. The
points 72 and —n are called the poles of the line Ln. Two lines are said
to be perpendicular in 52 if their poles are perpendicular in R3, (or
equivalently, if their planes are perpendicular in R3). 1.11 Theorem: The following statements should be intuitively clear. 1) There is a unique line With a given pole. 2) Given u and v in 32 with u  v = 0, there is a unique line which
passes through u in the direction of v. 3) There is a unique line through any two nonantipodal points
( on the other hand, every line through a point it also passes
through the antipodal point —u). 4) Every two lines have exactly two points of intersection, these
points being antipodal. 5) Given a point u and a line L at L", there is a unique line
through u perpendicular to L (on the other hand, every line
through u is perpendicular to Lu). 6) There is a unique line perpendicular to any two given lines. Proof: Rather than proving these statements, we shall give a couple of
illustrative examples. [I 1.12 Example: Find a point 77. such that the line Ln passes through it = %(0,2, 1) and v = $0,171). Solution: If L7, is the required line, then since u 6 L, we must have
u  n = 0, and since 1) 6 Ln we must have 71 ~ 72 = 0. Thus n must be
perpendicular to both u and v, or in other words, n must be a multiple
of u x 1). Since Inl = 1, we have UX’U n=:l: .
uxv In this particular example, we have u x v = ﬁb—B, 1:2) so that we
must choose n = :tﬁﬁ, —1,2). 1.13 Example: Let n = 91—;(1, 1,0), and m = i“), 1, 1). Find the two
points of intersection of the lines Ln and Lm. 4 Solution: A point u lies on both the lines Ln and Lm if and only if
u  n = u ' m = 0, which is true if and only if u is perpendicular to the
span of n and m. So u must lie in the span of n X m. Since 1u:1, we have
n x m u=:l: .
n><m In this particular example, we have n = %(1, 1,0) and m = %(0,1,1),
and so n x m = %(1,—1,1). Thus u = i%(1,—1,1). Line Segments 1.14 Deﬁnition: There is only one line between any two nonantipodal
points u and 12, but (unlike R2) there are two line segments between u
and '0; one is of length d(u,v), which is less than 7r, and the other is
of length 27r — d(u, 1)). We let [11,12] denote the line segment between
u and v of length d(u,v). If we let n = (u X v)/u >< v and then let
w = n X u (so that {u, w} is an orthonormal basis for the plane spanned
by u and v ) then we have [um] = {u cost + wsint'O S t S d(u,v)}. 1.15 Example: Verify that if f(t) = ucost+ to sin t, With w as above,
then we do in fact have f(d(u,v)) = 1). Verify also that the length of
the curve f(t), with 0 S t S d(u, v), is equal to d(u, v). Solution: We have f(d(u, 11)) = u cos d(u,v) + wsin d(u, v)
=(uv)u+u xvw
u X 7) =(uv)u+uxv[ ><u luxvl
:(uv)u+(uxv)xu
:(uv)u+(uu)v—(vu)u To ﬁnd the length of f(t), we note that f’(t) = ——u sint + w cost so we
have f'(t)2 = (—u sint + wcos t)  (—u sint — wcos t)
= (u  u)sin2t + (w w)coszt — 2(u  w)sintcost
= sin2t+ coszt :1. Thus lf’(t) = l and so the length of the curve is d(u,v) d(u,v)
L=/ If’(t)dt=/ 1dt=d(u,v).
0 0 Circles 5 1.16 Deﬁnition: Given a point n E S2 and a real number 7' with
0 S r 3 7r, we deﬁne the circle about n of radius r to be the set C(n,r) = {u E Szld(n,u) = 7'}. Notice that a line on 52 is also a circle, in fact Ln = C(n, g). Notice
also that C(n,r) = C(—n, 7r—r), so the centre of a circle is not uniquely
determined. We deﬁne the solid circle (or the disc) about n of radius
7" to be the set D(n,r) = {u E Szld(n,u) S 7‘}. 1.17 Theorem: Every circle is of the form 52 {1 P for some plane P in
R3 whose distance (in R3) from the origin is less than or equal to 1. Proof: We have C(n,r) = {u E 52d(n,u) = r}
= {u 6 S2] cos d(n,u) = cos r}
= {u E S2lnu : cosr}
= S2 nP, where P is the plane in R3 perpendicular to n which passes through the point (cos 7") n. Note that the distance (in R3) between the plane P
and the origin is d(0,P) = cosr. Cl 1.18 Example: Find a formula for the circumference C of C(n, r). Solution: From the picture, we see that the circle C(n, 7‘) on 82 may be
thought of as a circle of radius sinr in the plane P. So its circumference
is equal to C = 27r sinr . 1.19 Example: Find a formula for the area A of the disc D(n,r). Solution: From the same picture, we see that the disc is a cross section \M
of the sphere of width A1: = 1  cos 7", so from the formula for the area
of a cross section of a sphere, we obtain A = 27r(1 — cosr). 1.20 Remark: The sphere S2 has radius 1. If we were working on a
sphere of radius R, then we would obtain the formulas C = 27rR£sin ﬁ and
A = 27rR (1 — cos ﬁ) for the circumference and area of C(n, r) and D(n, 7‘). Notice that as ry—rO. Thus we see that , y _ R — lﬂm ,. , Tangent Vectors and Angles 6 1.21 Deﬁnition: A curve on 52 is a continuous map f : I —> S2,
where I is an interval in R. We can write f(t) : (:c(t),y(t),z(t)) with
f(t)2 = $(t)2 + y(t)2 + z(t)2 = 1. The curve f is called smooth if it
is differentiable, that is if the derivative fl(t) : (371(0) yl(t)a 2’0» exists for all t. 1.22 Deﬁnition: Given a point u E Sz, a tangent vector at u is
deﬁned to be any vector v E R3 such that u  'u = 0. Notice that the set
of all tangent vectors at u is a subspace of R3 (it is the kernel of the
linear map ut). 1.23 Theorem: Iff : I —+ 52 is a smooth curve then, for all t E I,
f'(t) is a tangent vector at f(t). Proof: We have f(t)[2 = 1. Taking the derivative yields 0 = (W) f(t))'
= (13(02 + W)2 + 2(t)2)'
= 2(I(t)$'(t) + y(t)y'(t) + 2(t)2'(t))
= 2(f(t) f'(t)) Thus f(t)  f’(t) : 0, which means that f’(t) is a tangent vector at f(t). 1.24 Example: Recall that, for u,v E 52 with u  v = 0, the line
through u which is initially in the direction of '0 is given by the equation
f(t) = ucost + vsint. This is a smooth curve, and we have f’(t) =
——u sint + vcos t. Note that f(0) = u and f’(0) = 1). Let a E R. The map g(t) = cos(at)u+sin(at)v is also a curve which
traces out the line through u = 9(0) in the direction of 1), but it moves
at a different speed; in fact we have g’(t) = —asin(at)u + acos(at)v so
that g’(0) = av. This shows that every tangent vector at u is of the
form g’ (t) for some smooth curve g. 1.25 Deﬁnition: Given two tangent vectors u and v at a point n 6 S2, we deﬁne the angle between u and v to be the angle between u and v
in R3, that is the angle _1 U'U [6 u,v I: cos .
( ) lul lvl Notice that [6(u, v)! E [0, 7r]. As in R2, and unlike the situation in R3,
we can also think of an angle as having an orientation. We deﬁne the oriented angle 9(u,v) from u to v at n as follows. Let 61 = 11 = Iii—I’ and then let 62 = n x (31 so that {81, 62} is an orthonormal basis for the
space of tangent vectors at n. This basis looks like the standard basis
for R2 if we hold the sphere with n pointing towards us. We can write
0 (uniquely) as v = wlcl + 10262, with wl = v  61 and wz = v  62. Then
we deﬁne 9(u,v) to be the angle in R2 from (1,0) to (101,102), that is,
the angle of w = (w1,w2) E R2. 1.26 Example: Find a formula for sin 0(u,v) and cos 0(u, v). (i Swill”! ) ale Wx Pm (‘8 a)" . :7
W8. if QM ant/lb 1W} (t0 M l’k L 2'1 l\ Q N K VK / (\o l’ (3's 7 Solution: Notice ﬁrst that, since el and 62 are orthonormal, we have v2 = v v =(w161 + 7.0262) (w161+ 1.0262) : w12 + 1022 = Iw2
so that v = w[. Thus we have
wl 61'1} ﬁv uvv
cos6(u v) = c030 w = = — =
’ ( ) le lvl Iv! Iul Ivl 11—2 _ ezv _ (n x d)sv [‘n,u,v]I
M M M I“! I”! ’ sin0(u,v) = sin9(w) = where In,u, vl denotes the triple product n,u,v = (nxu)v = n(uxv).
We remark also that, since n is in the same direction as u X 2), we have
ln,u,v = ilu x v. This means that luxvl sin0(u,v) = = sin 10(u,v). IUI lvl 1.27 Deﬁnition: Given two non—antipodal points u and v in 82, let u”
denote the unit tangent vector at u which points in the direction of 1)
along the line segment [u, v]. To be explicit, UX’U uv=nxu ,where n: .
luxvl Given three points u, v and w with u 7e iv and u 5£ :lzw, we deﬁne the
oriented angle Avuw by Zvuw = 9(uv,uw) where uv and um are the unit tangent vectors at u in the directions of
v and w. 1.28 Example: Let a = Zvuw. Find a formula for sina and for cos a. Solution: We have :(uxv)xu_v—(uv)u up and luxvl _ uxv
_(uxw)xu_w—(uw)w uw— quwI _ luxwl Using our formulas from example 1.26, we obtain cosa = cos 0(uv,uw) = uv uw =(v—(uv)u><w—<uw>u>
quvHuxw
=<vw>—<uv><uw)
luvauxwl
=(uXv)(u><w)
uvauXw sina = sin0(uv,uw) = u,u,,,uw
u. (v—(uv)u) x(w—(uw)u))
Iu >< vHu x ml
u~(va—(uw)vxu—(u~v)uxw)
[u X vu X w] lu,v,w In x U] u X to! I
1.29 Example: Let u = (0,—1,0), v = —%(1,1,0) and w : %(1,1,1)
Find the vectors u,, and uw, and ﬁnd the angle a = Zvuw. Solution: A short calculation shows that uxv uxw
= (0,0,1) and [u x w = —12—(—1,0,1). uxv Another calculation then yields u,, = (u—l::—)l—;l(—u = (1,0,0) , and
_(u><w)xu 1
“w — W = ﬁ(1,0,1). Using the formulas from example 1.26, we have cos9(uv,uw) = uv ~uw = i , and sin9(uv,uw) = u,uv,uw[ = %. Thus a = 9(uv,uw) = 5 4.
1.30 Remark: Actually, the above problem can be solved without
doing any calculations. If lines are drawn on a cube, as shown, and
then the cube is blown up like a balloon until it becomes a sphere, then
the sphere will have lines on it as shown (try drawing these lines on an
orange or on a ping pong ball). All the triangles will be congruent, with 1r 7? interior angles 3’ 3 and 325. The angle a can be seen from the picture. For example, from the picture (or better yet, from the ping pong ball) we can see that Zwvu = g and Zuwv = 27". Triangles 1.31 Deﬁnition: A triangle in 52 is given by three noncolinear points
in 82, called the vertices of the triangle. We can think of a triangle
in several different ways. We could think of a triangle simply as a set
of three points {u, v, w} or, if we wish to keep track of the order of the
points, as an ordered triple (u, v, w) of points. Since u, v and w are non—colinear, no two of them are antipodal
and so the edges [v,w], [w,u] and [u,v] are uniquely determined. A
triangle can be thought of as the union of its three edges. Alternatively, we can think of a triangle as including its interior
points, in which case we denote the triangle by [u,v,w}. Note that 9 three points are noncolinear in S2 if and only if they do not all lie in a
single 2—dimensional vector space in R3, which is true if and only if they
are linearly independent in R3. Thus every point in R3 (and certainly
every point in 52) can be written uniquely in the form Tu + 31) + tw for
some r,s,t E R. The solid triangle is the set [u,v,w] = {ru+sv+tu E S2r,s,t Z 0}. 1.32 Deﬁnition: Given a triangle [u,v,w], we shall let a, b and c
denote the lengths of the edges [11,10], [w,u] and [u, u], so a = d(v,w), b = d(w,u) and c = d(u,v),
and we shall let (1, ﬂ and 7 denote the oriented angles
a = Avuw, ,8 = éwvu and 7 = Zuwv. The angles la], (,6 and 7 are called the interior angles of the triangle
(recall that lal = a when a E [0,7r] and Ial = —a when a E [—7r,0], so
that we always have la] 6 [0,7r]). 1.33 Deﬁnition: If {u, v, w} is a positively oriented basis for R3 (recall
that this means that Iu, v, w] > 0), then we say that the triangle [u, v, w]
is positively oriented. Otherwise, we say that it is negatively ori ented. Since sina = u,v,w/(u x v] [u X wl), we have sina > 0 if
and only if u,v,w > 0. Similarly, sinﬂ > 0 (E) lv,w,u[ > 0 and
sin7 > 0 (=> lw,u,v > 0. But u,v,w] = v,w,u = w,u,v], so we see that if [u,v,w] is positively oriented, then sin a, sinﬂ and sin7 are
all positive, while if [u, v, w] is negatively oriented then the sines of the
angles are all negative. 1.34 Theorem: The area of the triangle [u, v, w] is equal to [al + ﬂ +
7  7r Proof: Let Hu be the hemisphere containing u with the line MD as its boundary,
11,, = {ru+sv+tw E Szlr Z 0}, and similarly for H” and Hw. Let
W0=HvﬂHw , WngwﬂHu ,and W7=HuﬂHv. From the picture, we can see that the area of the wedge Wu is lal/7r times the area of the whole sphere, that is , area(Wa) = @471 : 4a,
and similarly for the areas of Wﬁ and W7, so we have area(Wa) = 4104 , area(Wﬁ) = 4M3] , and area(W7) = 47I. Since 1) G Wu <==> —p G Wu, we know that, for any hemisphere H, we
have area(Wa H H) = area(Wa O H) = %area(Wa). Now, let us divide the hemisphere Hu into the four regions A, B,
C and T as shown, so that T = [u,v,w] anchéaamWHE“U' Wm. Then we have WaﬂHu = AUT, WpﬂHu = BUT and
WVHHu =CUT, and so
2004 + Iﬁl + M) = %area(Wa) + %area(Wﬂ) + iarea(W7)
= area(Wa n H”) + area(Wg O H“) + area(W., n H")
= area(A U BU C U T) + 2area(T)
= areaHu +2areaT
= 21r+2areaT. Thus area[u,v,w] = [a] + [,5] + [7 — 7r. D 10 1.35 Deﬁnition: Given a triangle [u,v,w], let u’, v’ and w’ denote the
poles of the directed edges [v,w], [w,u] and [um], that is , vxw , wxu , uxv u:
vaw] The triangle [u',v',w’] is called the polar triangle of triangle [u,v,w]. 1.36 Theorem: Let [u, v,w] be apositively oriented triangle with polar
triangle [u’, v’, w’]. Then 1) [u”,v",w”] = [u,v,w]. 2) [u’,v’,w’] is positive] oriented. 3) a’=7r—a,b’=7r—ﬂandc’=7r—7. 4) a'=7r—a,ﬂ’=7r—band7'=7r—c. Proof: We have v,xw,=(w><u)x(uxv)
wauHuxvl _ w(uxv)u—u(uxv)w
.— [wxulluxvl
lu,v,wu _ lwquuxvh
This is a positive multiple of u, and so I I
,, va u U . 2W: Similarly we have u” = v and w” = w. This shows that [u”,v",w"] =
[11,7510].
Now, let us ﬁnd the orientation of [u’, v’,w’]. We have lu',v',w'[ = u'  ('0' x w')
_ 1) X w u,v,wu _vxw'wxuuxv _ lumwl2
— luxvvxwwxu which is positive. So [u’,v’,w’] is positively oriented.
Next, let us ﬁnd the lengths of the edges of the polar triangle. We have
2,1,0, = W1) 2 “com
Iw x u] {u X "0
This means that a’ = 71' i a. Since [u,v,w] is positively oriented, we
have a E [0,7r] and so we must have a’ = 7r — a. Similarly, b’ = 7r — ﬂ
and c’ = 7r — 7.
Finally, let us calculate the angles of the polar triangle. I I . I I
cowl: MM =w~.(_,~)=_w.,=_.cosa_
Iu’ Xv’Hu’ xw’l This means that a’ = 7r i (1. But since [u’,v’,w'] is positively oriented,
we must have a = 7r — a. Similarly, ﬂ’ = 7r — b and 7’ : 7r — c. I3 11 1.37 Theorem: (The Sine Law) For any triangle [u,v,w] on 82 we have . _ .
sm a mm b sm c sina sinﬂ _ sin7'
Proof: We have
sina: u7v)wl : Iu,v,w '
quvlluxw] sincsinb
A similar formula holds for sinﬂ and sin 7, so we obtain
u,v,w = sinasinbsinc = sin a sin ﬂ sinc = sinasinbsin'y.
The sine law follows by dividing these equations by one another. D 1.38 Theorem: (The First Law of Cosines) For any triangle [u, v, m],
we have cos a — cosb cosc ll cos a , sin 6 sin c
cos b  cos a cosc ll cos ,3 , and sin a sin 0 cos c — cosa cos b
cos'y = ——T—.—
s1na s1nb Proof: We have
(u x v)  (u x w)
[u X 1)] Iu >< w
_ (v.w)_(u.w)(v.u)
_ Iu X vl Iu X wl COSCY = cos a —— cos 6 cos 6 sin c sin b
The other two formulas may be proved similarly. I] 1.39 Corollary:
(SSS) If we know the lengths of the three sides ofa triangle then we
can ﬁnd the three interior angles.
(SAS) If we know the lengths of two of the sides and the angle at the
common vertex, then we can find the length of the third side
(and hence also the other two angles). 1.40 Theorem: (The Second Law of Cosines) For any triangle [u, v, w]
in 82 we have cosoz + cosﬂ cos'y
cosa = — sinﬂsin’y
cosﬂ+cos7 cosa
cosb: —..—
sm'y sma
cosy+cosa cosﬂ
cosc = ——————————. sin a sin a
Proof: Suppose that [u,v,w] is positively oriented, and let [u’,v’,w’]
be its polar triangle. Then we have a’ = 7r — a and 01’ = 7r — (1, etc. The
second law of cosines is found by applying the ﬁrst law of cosines to the
polar triangle: , —cos a’ +cos b’ cos c’ cosa+cosﬂcos7
cosa: —cosoz = ———,——,—.I— 2 ~——.——.———
smb s1nc smﬂsm’y The case in which [u,v,w] is negatively oriented is left as an exercise.
Cl 12 1.41 Corollary: (AAA) If we know the three interior angles of a triangle, then we can
ﬁnd the lengths of its three sides. (ASA) If we know the length of one side of a triangle and the angles
at the side’s two vertices, then we can ﬁnd the third angle (and
hence also the lengths of the other two sides). Isometries 1.42 Deﬁnition: Recall that an isometry on 82 is a bijective map 5 : S2 + 82 such that S (and hence also its inverse) preserve distance
in S2, in other words d(Su,Sv) = d(u,v) for all u,v 6 S2. 1.43 Theorem: Every orthogonal map on R3 restricts to an isometry of
$2, and conversely every isometry on S2 extends to a unique orthogonal
map on R3. Thus the group of isometries on S2 may be identiﬁed with
the group 0(3) of orthogonal maps on R3. Proof: First note that, since the spherical distance d(u,v) determines
and is determined by the Euclidean distance u — v[, a bijective map
S : 52 —) 32 will be an isometry if and only if it preserves Euclidean
distance, that is if and only if ISu — $1)! = Iu — 1;] for all u,v 6 S2. Recall that an orthogonal map A is an isometrty on R3 , so it
preserves Euclidean distance. For u E 52 we have lAu = IAu — A0} =
[u — OI = lul = 1 and hence Au 6 S2. Thus A restricts to a bijection
S : 82 —> 52 which preserves Euclidean distance. By the above remark,
we know that S is an isometry on 52. Now suppose that we have been given an isometry S : S2 —> 52.
Then we have seen that S preserves Euclidean distance. The polariza
tion identity then shows that 5' also preserves the dot product, in fact
for u,v E 52 we have 2(Su  .912) 2 [SUV + [SUI2 — lS'u — 5le
= 1+ 1 — lu — 012
= IUI + IUI — I“ v2
= 2 (u  1)).
In particular, if we let {81, 62,63} be the standard basis for R3, then we
have Sei  Sej = 6M so that the set {$61,562,5e3} is an orthonormal
basis for R3. Given any element 1) of 52, we can write 1) uniquely in the form 1) = ZciSug. Then we have u  Sej = Zci(Sei  Sej) : Cj. Thus
3
v = 2(1)  Seg)S'e,.
i=1
Replacing v by Su where u : Eaiei E S2 we obtain a formula for S:
a 3 3
Su = 2(511 ' 56056,” = 2(1).  6056; = E aiSei.
i=1 i=1 i=1 But this formula deﬁnes an orthogonal map if we allow u to be any vector
in R3. Thus 5' : 82 —> 82 extends to the orthogonal map A : R3 —> R3
deﬁned by the formula A(Z age.) = Z agSei. El 13 From the above theorem, and from our knowlege of orthogonal maps and isometries in R3, we obtain the following examples and classiﬁcation
theorem for isometries on 32. 1.44 Example: Let L be the directed line in the direction of the unit
vector u E 52. Then the rotation RL,9 is an orthogonal map, so its
restriction to S2 is an isometry on 52, which we call the rotation about
u by 9 and which we denote by Riga. Let U be any vector perpendicular to u, let w = u X 1), and let A be the matrix with columns u, v and 11;. Then
1 0 0 Ru,‘9=At 0 cos0 —sin0 A. 0 sin0 cos 0 Note that Rum, = RWs if and only if either u = v and a = ﬂ (modulo
27r) or u = —v and a = —ﬂ (modulo 27r) (or a = ﬂ = 0 (modulo 27r) in
which case the rotation is the identity). 1.45 Example: Let P be the plane through the origin with unit normal
vector 72 E SZ. Then the reﬂection Fp is an orthogonal map, so its
restriction to S2 is an isometry on 52, which we call the reﬂection in
the line Ln and which we denote by FL" or by F". For any u E 82 we
have Fn(u) = u — 2(u  n)n or alternatively we can express F" as the orthogonal matrix a2 ab ac
Fn = I— 2 ab b2 be
ac bc c2 We have FL“ = FL," if and only if Ln : Lm which is true if and only
if n = :l:m. 1.46 Example: The antipodal map A0 about the origin is an isometry
on S2. It is given by Ao(u) = —u, or as a matrix we have Ao=—I. 1.47 Example: The composite of two rotations is equal to a single
rotation, and R1219 = Ru,—o = R_u,9. The composite of two reﬂections
is a rotation, and F;1 = F" = F_n. Conversely, every rotation is
the composite of two reﬂections. The composite of a rotation about u
with the reflection in LH is called a rotaryreﬂection. The composite
of a rotation with the antipodal map is called a rotaryinversion. If
a+7r=ﬂthe we have Ru,aFu : FuRu,alpha : AORu,ﬂ : Ru,ﬂA0 so in particular a rotary—reflection is the same thing as a rotary—inversion 1.48 Theorem: Every isometry on S2 is either a rotation or a rotary
inversion (the identity being the trivial rotation). ...
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 Linear Algebra, Algebra

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