Inversive Geometry
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Transcript of Inversive Geometry
OPEN UNIVERSITY
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Intro to Inversive Gerometry Shirleen Stibbe
M203 Pure Mathematics Summerschool
NB OA = |A| = , |A'| = k|A| 22 yx +
Inversion in a circle, centre 0, radius r
Can't cope - yet. So get rid of it in the meantime.
Punctured plane
points outside points inside
points on the circle fixed
t: A' A , and t-1 = t (self inverse)
what happens to the centre????
→
⇔
Example: Invert A = (3, 4) in the unit circle (r = 1)
r2 = 1 = |(3, 4)| x |(3k, 4k)| = 5 x 5k = 25k
So k = 1/25 and A' = (3/25, 4/25)
t: A A' OA.OA' = r2
→r
O A = (x, y)
A' = (X, Y) = (kx, ky),
0k ≠
2
Inversion in the unit circle C (centre 0, radius 1)
1r)yx(k yxk yxAOOA 2222222 ==+=+×+=ʹ′⋅
A = (x, y) A' = (X, Y) = (kx, ky), k > 0
To find the image of: a circle: (x - a)2 + (y - b)2 - r = 0 a line: ax + by = 0
under inversion in the unit circle:
1) Replace x by x / (x2 + y2) y by y / (x2 + y2) in the equation
2) Simplify
No! Get a life!
Strategy: HB p62
22 yx1k+
= ⎟⎟⎠
⎞⎜⎜⎝
⎛
++= 2222 yx
y,yx
x)Y,X(and so
3
line through 0
line through 0
c = 0 d = 0
line not through 0
circle through 0
c = 0 d 0
circle through 0
line not through 0
c 0 d = 0
circle not through 0
circle not through 0
c 0 d 0
Inverts to Curve Equation
≠
≠
≠
≠
Inversion in C - the easy way
Images
Write the equation of the curve C in the form: d(x2 + y2) + ax + by + c = 0 (*)
C is a circle if d ≠ 0, a line if d = 0
C passes through the origin if and only if c = 0
To invert C in the unit circle: interchange c and d in (*)
That is: t(C) is the curve with equation c(x2 + y2) + ax + by + d = 0
!
4
Oops - we're not doing very well
Can't cope with the origin
Doesn't preserve circles
Doesn't preserve lines
Doesn't preserve size
Some good news - at last
Angle theorem:
Inversion preserves size, reverses direction
of angles between curves - just as reflection does
5
Examples: Invert i) L, the line x = 1/2, and
ii) C, the circle centre (1, 0), radius 2
in the unit circle C
0 -1 1 2 3
L
C
C
Draw a picture
Equation for L: 0(x2 + y2) - x + 1/2 = 0
Equation for t(L): 1/2(x2 + y2) -x = 0 Simplifies to: (x-1)2 + y2 = 1
Equation for C: 1(x2 + y2) -2x - 3 = 0
Equation for t(C): -3(x2 + y2) -2x +1 = 0 Simplifies to (x+1/3)2 + y2 = 4/9
Do some algebra:
6
Even easier way:
-1 1 2 3
L
C
C
L'
C' -1/3
L: line not through 0 → circle through 0 symmetric about x axis (right angles preserved) points on the circle fixed (1/2, 0) (nearest to 0) → (2, 0) (furthest from 0) Image: circle centre (1, 0), radius 1
C: circle not through 0 → circle not through 0 symmetric about x axis (right angles preserved) (-1, 0) fixed (3, 0) (furthest from 0) → (1/3, 0) (nearest to 0) Image: circle centre (-1/3, 0), radius 2/3
7
Now for the really cunning part
Use complex numbers
⎟⎟⎠
⎞⎜⎜⎝
⎛
++→+= 2222 yx
y,yx
x)y,x(~iyxz
z1
iyx1
)iyx)(iyx(iyx
yxiyx~ 22 =
−=
+−+
=+
+
compositions of reflections
Translation: t(z) = z + c, c = a + ib
Reflection in x-axis: t(z) =
Rotation about 0: t(z) = az a = Cosθ + iSinθ
z
z/1z→Inversion in C is a doddle:
Scaling: t(z) = kz, k R, k > 0 ∈
z z+c
a b
z
z
z
az θ
Euclidean transformations in C
8
Rotate through angle θ
= reflect in line p, then reflect in line q
Translate through vector c
= reflect in line p, then reflect in line q
q
|c|
½|c|
p
z
t(z)
q
θ/2
p
z t(z)
9
Inversion in C, centre c = a +ib, radius r
Discovery! tC and t are conjugates
ccz
r)z(ttt)z(t2
122C +
−== −
ccz
rczr
rczz
2ttt 2
12 +
−⎯→⎯
−⎯→⎯
−⎯⎯→⎯−
tC:
Convert C to C: translate through -c, scale by 1/r
Invert in C:
Convert C to C: scale by r, translate through +c
z1z:t →
Note: t1 = t2-1 crzz:t2 +→
We know how to invert in the unit circle C, so …
Abuse the plane to convert C to C
Do the inversion in C
Conceal the evidence; convert C back to C
,rczz:t1
−→
10
Introducing the very lovely
extend the plane: C Ĉ
solves (nearly) all our problems
under inversion
define a line as a (generalised) circle with infinite radius
generalised circles generalised circles
we're back in business
inversion (nearly) back in line with reflection
A late arrival at the Inversion Ball
Generalised Circle:
Ordinary circle if it doesn't contain ∞ Extended line if it contains ∞
0 and 0 →∞∞→
→
=∞∪ }{
∞Point at Infinity
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Inversive Group
Elements: set of inversive transformations
Operations: composition of functions
Properties: a) map gen circles to gen circles
b) preserve magnitude of angles
Inversion in Ĉ
C a generalised circle in Ĉ
t = inversion in C
If C is a 'proper' circle, centre 0, radius r:
t(0) =
t( ) = 0
t(z) = normal inversion,
If C is an extended line L ∪ { }:
t( ) =
t(z) = reflection in L,
∞
∞
∞ ∞
∞
∞≠≠ z,0z
∞≠z
12
Möbius Functions
M(z) : direct - preserves orientation of angles
composite of even number of inversions
indirect - reverses orientation of angles
composite of odd number of inversions
:)z(M
∞=−⎩⎨⎧
≠
=∞=∞ )c/d(M
0c if c/a0c if
)(M
Workingform
⎟⎠
⎞⎜⎝
⎛β−α−
=zzK)z(M
c/d ,a/b ,c/aK −=β−=α=
Example1: Express as a Möbius transformation ccz
r2+
−
dzcbza)z(M
dczbaz)z(M
++
=++
=M : C Ĉ →
czzc
cz)ccr(zc
cz)cz(cr c
czr 222
−α−
=−−+
=−−+
=+−
c/rc 2−=αwhere
a, b, c, d ∈ C, ad – bc ≠ 0
13
Example2:
i) calculate M(2), M(i) and M( )
ii) show that M-1M(z) = z
i)
2ziz)z(M
−−
=
∞
Associated matrix:
Composition of transformations = matrix multiplication
⎟⎠
⎞⎜⎝
⎛
−
−=⎟
⎠
⎞⎜⎝
⎛= −
acbd
A,dcba
A 1
ii) Matrix is
dczbaz)z(M
++
=Möbius and Matrices
0i2bcad21i1
A ≠+−=−⎟⎠
⎞⎜⎝
⎛
−
−=
⎟⎠
⎞⎜⎝
⎛
+−
+−=⎟
⎠
⎞⎜⎝
⎛
−
−⎟⎠
⎞⎜⎝
⎛
−
−=⎟
⎠
⎞⎜⎝
⎛
−
−= −−
1200i2
21i1
11i2
AAso 11i2
A 11
1 z/21z/i1lim )(M
0)i(M ,)2(M
|z|=
−−
=∞
=∞=
∞→
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Fundamental Theorem
Given any 3 points in C, there exists a Möbius transformation that maps these to any other three given points.
Note:
Two points do not fix a generalised circle:
But three do:
∞ ∞
15
Putting Möbius to work
To determine whether four points, a, b, c, d, all lie on the same generalised circle:
1 The first 3 points (a, b and c) determine a generalised circle, C.
2 Find a Möbius transformation M(z) that takes a, b and c to the real axis
[ Easiest way: send them to 0, 1 and ∞ ]
3 The M you found maps the C, the whole C and
nothing but the C to the real axis, and M-1 maps the real axis back to the generalised circle, C.
4 Apply M to the 4th point, d.
If M(d) is real, say M(d) = x ∈ R, then M-1(x) = d lies on the generalised circle C.
5 If not, then it doesn't.
6 If M(∞) is real, then ∞ lies on C, so C is an extended line. Otherwise, C is a 'proper' circle.
16
17
Example
1 Find a Möbius transformation that maps 1 - i to 4 + 4i to 0, and 5 to 1.
∞
⎟⎠
⎞⎜⎝
⎛−−+−
=⇒∞=−
=+
−−
=)i1(z)i44(zK)z(M
)i1(M0)i44(M
,bzazK )z(M
1i4i41K
)i1(5)i44(5K)5(M =⎟
⎠⎞
⎜⎝⎛
+−
=⎟⎠
⎞⎜⎝
⎛−−+−
=
i i41)i41(i
i41i4 K So =
−−
=−+
=
i1zi44iz
)i1(z)i44(zi)z(MTherefore
+−−+
=⎟⎠
⎞⎜⎝
⎛−−+−
=
2 Do the points 1 - i, 4 + 4i, 5 and 3i all lie on the same generalised circle, and if so, what kind of circle is it?
1 - i, 4 + 4i, 5 lie on a generalised circle, C, which is mapped by M to the real axis.
1 i41
i41 i1i3
i443 )i3(M −=+−−
=+−−+−
=
which is real, so 3i must lie on C
M(∞) = K = i, which is not on the real axis, so ∞ is not on C
So C is a 'proper' circle and not an extended line.