Download Homework 1. Solutions 1 a) Let x 2 + y2 = R2 be a circle in E2. Write

Homework 1. Solutions
1 a) Let x + y = R be a circle in E2 . Write down explicitly formulae for stereographic projections with respect to the North pole (the point (0, 1)) and South pole (the
point (0, −1)). b) Do the same exercise for the sphere x2 + y 2 + z 2 = R2 in E3 . (North
pole (the point (0, 0, 1)) and South pole (the point (0, 0, −1)).
a) Let N = (0, R) be North Pole and A be a point on the circle. Let chorde N A
intersects the line y = 0 at the point P . Let B be a point on the axis OY such that BP
is parallel to the axis OX. Consider similar triangles △N AB and △N P O, where O is
origin. Let A = (x, y) and P = (0, t). We have:
2
2
2
|N B|
|AB|
R−y
x
=
i.e.
=
|N O|
|OP |
R
t
Solving this equations we come to t =
t2 (R−y)2
R2
Rx
R−y .
Respectively: x =
t(R−y)
,
R
Using x2 + y 2 =
+ y 2 = R2 we come to
t2 (R − y)
=R+y ⇒
R2
(
x=
y=
2
t(R−y)
= R2tR
2 +t2
R
t2 −R2
t2 +R2 R
.
Analogously for South Pole: Let S = (0, −R) be South Pole and A′ a point on the circle.
Let chorde SA intersects the line y = 0 at the point P ′ . Consider similar triangles △SAB
and △SP ′ O. We have:
|AB|
R+y
x
|SB|
=
i.e.
= ′
′
|SO|
|OP |
R
t
Solving this equations we come to t′ =
(
Rx
R+y .
x=
y=
Respectively:
2t′ R2
R2 +t′2
R2 −t′2
R2 +t′2 R
.
b) For two dimensional sphere the considerations are analogous.
Let N = (0, 0, R) be North Pole and A a point on the sphere with coordinates (x, y, z)
Let chorde N A intersects the line y = 0 at the point P . Let B be a point the axis OY such
that BP is parallel to the axis OX. Let (u, v) be coordinates of the point P . Consider
similar triangles, △N AB and △N P O, where O is origin. Let A = (x, y) and P = (0, t).
We have:
|N B|
|AB|
R−z
x
y
=
i.e.
= = .
|N O|
|OP |
R
u
v
(
Rx
u = R−z
u(R−z)
, y = v(R−z)
,
Solving this equations we come to
Ry . Respectively: x =
R
R
v = R−z
Using
2 2
u(R − z)
u2 (R − z) v 2 (R − z)
v(R − z)
2
2
2
x +y +z =
+
+ z 2 = R2 ⇒
+
= R−z
R
R
R2
R2
1
we come to


x =
y=


z=
2uR2
u2 +v 2 +R2
2vR2
u2 +v 2 +R2
u2 +v 2 −R2
u2 +v 2 +R2 R
Analogously for stereographic projection with respect to South Pole.
Let S = (0, 0, −R) be South Pole and A a point on the sphere with coordinates (x, y, z)
Let chorde N A intersects the line y = 0 at the point P ′ with coordinates (u′ , v ′ ). Then we
come to

2uR2
(
x = u2 +v

2 +R2

Rx
u = R−z
2
2vR
y = u2 +v2 +R2 .
Ry ,

v = R−z
2
2

−R2
z = uu2 +v
+v 2 +R2 R
2 Consider in En the transformation:
ϕ(r) =
r
|r|2
(1)
inversion with the centre at origin. (Strictly speaking this transformation is defined on
En \0).
Analyze its geometrical meaning.
Show that this trasformation is an involution: ϕ(ϕ(r)) = r.
Let a be an arbitrary vector attached at the arbitrary point r of En (r 6= 0). Let a′ be
a vector attached at the point r′ = ϕ(r), such that a′ = ϕ∗ a is the image of vector a under
inversion (1), Show that
ar2 − 2r(r, a)
a′ =
,
(2)
|r|4
where ( , ) is the scalar product in Euclidean space En .
Using this equation show that inversion preserves angles between vectors.
Find the image of the hyperplane xn = a under the inversion (1). ((x1 , . . . , xn ) are
standard Cartesian coordinates.) 1) .
For simplicity you may consider just cases n = 2, 3
Solution
The involution property follows from geometrical picture. Points r, r′ = ϕ(r) and
r̃ = ϕ(r′ ) = ϕ(ϕ((r)) all belong to the same ray. On the other hand |r′ ||r| = |r̃||r′ | = 1,
hence |r̃| = |r|. Hence r̃ = r.
1)
Show that this is a sphere of radius
1
n
0, x = 2a
.
1
2|a|
with the centre at the point x1 = . . . = xn−1 =
2
If a point r with coordinates xi tansformes under the inversion to the point r′ with
coordinates y i = y i (x) then the tangent vector a at this point transforms to the tangent
vector a′ at the point r′ such that
(a′ )i =
For inversion (1) we have y i (x) =
xi
r2
=
∂y i (x) m
a
∂xm
xi
(x1 )2 +...+(xn )2
hence
i
i
∂y i
δm
2xi xm
r 2 δm
− 2xi xm
=
−
=
,
∂xm
r2
r4
r4
hence
∂y i (x) m
(a ) =
a =
∂xm
′ i
i
r 2 δm
− 2xi xm
r4
am =
r2 ai − 2xi (x1 a1 + . . . + xn an
,
r4
i.e.
r2 a − 2r(a, r)
a =
. where r2 = |r|2 = (r, r) .
4
r
To show that inversion does not change the angles between tangent vectors, calculate the
scalar products. Let a, b be two arbitrary vectors attached at arbitrary point r (r 6= 0).
Let a′ , b′ be vectors attached at the point r′ = ϕ(r), such that vector a′ is the image of the
vector a under inversion ϕ, and respectively vector b is the image of the vector b under
inversion ϕ. Then according to previous calculations we have
2
r4 (a, b) − 4r2 (a, r)(b, r) + 4r2 (a, r)(b, r)
a, b
r a − 2r(a, r) r2 b − 2r(b, r)
′
′
=
,
= 4
(a , b ) =
4
4
8
r
r
r
r
′
We see that inversion does not change the scalar product of vectors up to a factor depending
on the point.
Now find the image of hyperplane. Its image has to be sphere which passes via origin,
centre of inversion. Show it. Let A = (u1 , u2 , . . . , un−1 , a) be an arbitrary point of the
plane xn = a. Image of this point will be the point A′ with coordinates (v 1 , . . . , v n , b) such
that
a
ui
,b = 2
, where r2 = (u1 )2 + . . . + (un )2 .
vi = 2
2
a +r
a + r2
Perform calculations which check that point (v 1 , . . . , v n−1 , b) belongs to the sphere
P i i
P i i
2
X
1
1
a2
1
1
1
i i
2 b
iu u
iu u
v v + b−
= 2
+
+b
−
=
+
− 2
+ 2 = 2.
2
2
2
2
2
2
2
2
2
2a
(a + r )
a 4a
(a + r ) a + r a + r 4a
4a
i
3∗ Consider in E3 the transformation:
ϕN (r) = N +
Cr
,
|r − N|2
3
C 6= 0 ,
(2)
where N is an arbitrary vector.
(Stricly speaking this transformation is defined on En \N .
Analyze geometrical meaning of this transformation: in particular analyze the relation
of this transformation with the transformation (2).
Show that this transformation preserves angles between vectors.
Find a transformation inverse to transformation (2).
Find an image of a plane under the transformation (2).
The transformation (2) is compostions of four transformations
i) translation r → r − N
ii) inversion r → rr2
iii) dilation r Cr
iy) translation r + N
All these transformations preserve angles between tangent vectors. For translations,
dilations this is evident, and for inversion it was proved in the previous exercie. Hence the
composition of these transformations preserve also angles.
We see that transformation (2) is the composition
TN ◦ DC ◦ inversion ◦ T−N ,
where Ta : r → r + a is a translation on the vector a, and DC : r → Cr is dilation.
Using the fact that inversion is involution, and the decomposition abive, we see that
inverse of (2) is a trasformation
TN ◦ D C1 ◦ inversion ◦ T−N ,
Indeed:
TN ◦ inversion ◦ D C1 ◦ T−N ◦(TN ◦ DC ◦ inversion ◦ T−N , ) = TN ◦inversion◦D C1 ◦T−N ◦TN ◦DC ◦inve
TN ◦inversion◦D C1 ◦DC ◦inversion◦T−N = TN ◦inversion◦inversion◦T−N = TN ◦T−N = id .
We proved that the transformation inverse to transformation (2) is the trasnformation
TN ◦ inversion ◦ D C1 ◦ T−N :
r−N
C(r − N)
+N=
r′ = C 2 + N =
|r − N|2
r−N C
if C > 0. We see that transformation (2) is an involution if C > 0, and it changes the sign
in the case if C < 0. Sure this can be checked straightforwardly.
Transformation (2) is an inversion with centre at the point N. the radius of the sphere
√
√
of inversion is equal to C, (C > 0)—th points which are on the distance C of the centre
stay intact under the inversion.
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4∗ Let ϕ be stereographic projection of unit sphere x2 + y 2 + z 2 = 1 in E3 on the plane
z = 0 with respect to the North pole N = (0, 0, 1). Find a transformation (2) of E3 such
that its restriction on the sphere is this stereographic projection.
Consider the inversion (2) with centre at the point N = (0, 0, 1) (North pole of the
√
sphere) with the radius of inversion 2:

x 7→ x′ = x2 +y22x

+(z−1)2

2y
′
y 7→ y = x2 +y2 +(z−1)2


z 7→ z ′ = 1 + x2 +y2(z−1)
2 +(z−1)2
ne can see that this is inversion (2), the points of the equator x2 + y 2 = 1, z = 0 of the
sphere S 2 remian fixed, the North pole of the sphere ‘goes to infinity’. Hence the image of
this sphere under the inversion is the plane z = 0. It follows from the formula above that
if x2 + y 2 + z 2 = 1 (point is on the sphere) then its image has coordinates
x 7→ u =
y 7→ v =
2x
x2 +y 2 +(z−1)2
2y
x2 +y 2 +(z−1)2
z 7→ 1 +
2x
x2 +y 2 +z 2 −2z+1
2y
= x2 +y2 +z
2 −2z+1
2(z−1)
x2 +y 2 +(z−1)2 = 0
=
=
=
x
1−z
y
1−z
To find inverse formulae (from plane (u, v) to the sphere) we just use the fact that inversion
(2) is the involution (see the previous exercise):

′

2x
′

x′ 7→ x = x′2 +y′22x
x
→
7
x
=
′ −1)2
2 +y 2 +(z−1)2


+(z
x


′
2y
′
2y
′
y 7→ y = x2 +y2 +(z−1)2
y 7→ y = x′2 +y′2 +(z′ −1)2
⇔




2(z−1)
2(z ′ −1)
 z ′ 7→ z = 1 +
z 7→ z ′ = 1 + 2 2
2
x +y +(z−1)
x′2 +y ′2 +(z ′ −1)2
In particular if u = x′ , v = y ′ are stereographic coordinates then z ′ = 0 and we come to

′
2u

u = x′ 7→ x = x′2 +y′22x

+(z ′ −1)2 = u2 +v 2 +1

′
2v
y ′ 7→ y = x′2 +y′22y
′ −1)2 = 1+u2 +v 2
+(z

2
2

2(z ′ −1)
 z ′ = 0 7→ z = 1 +
= u +v −1
= 1 + −2
x′2 +y ′2 +(z ′ −1)2
u2 +v 2 +1
u2 +v 2 +1
(Compare these calculations with calculations performed in the exercise 1)
5∗ Using the result of previous exercise explain why stereographic projection of unit
sphere x2 + y 2 + z 2 = 1 establsihes bijection between points with rational coordinates on
the unit sphere with points with rational coordinates on the plane z = 0.
THis is because the inversion which leads to the stereographic projection evidently is
birational map (see above)
6 Show that the condition of non-degeneracy for a symmetric matrix ||gik || follows
from the condition that this matrix is positive-definite.
5
Solution Suppose det g = 0, i.e. g is degenerate matrix (rows and columns of the
matrix are linear dependent). Then there exists non-zero vector x = (x1 , x2 ) such that
gik xk = 0, hence gik xi xk = 0 for x 6= 0. Contradiction to the condition of positivedefiniteness.
7 Let (u, v) be local coordinates on 2-dimensional Riemannian manifold M . Let Riemannian metric be given in these local coordinates by the matrix
||gik || =
A(u, v)
C(u, v)
B(u, v)
D(u, v)
,
(2)
where A(u, v), B(u, v), C(u, v), D(u, v) are smooth functions. Show that the following conditions are fulfilled:
a) B(u, v) = C(u, v),
b) A(u, v)D(u, v) − B(u, v)C(u, v) 6= 0,
c) A(u, v) > 0,
d∗ ) A(u, v)D(u, v) − B(u, v)C(u, v) > 0.
e)∗ Show that conditions a), c) and d) are necessary and sufficient conditions for
matrix ||gik || to define locally a Riemannian metric.
Solution
Consider Riemannian scalar product G(X, Y) = gik X i Y k .
a) The condition that G(X, Y) = G(Y, X) means that gik = gki , i.e. B(u, v =
C(u, v)).
b) det G = A(u, v)D(u, v) − B(u, v)C(u, v) = AD − B 2 6= 0 since it is non-degenerate
(see the solution of exercise 1)
c) Consider quadratic form G(x, x) = gik xi xk = Ax2 +2Bxy+Dy 2 . (We already know
that B = C) Positive -definiteness means that G(x, x) > 0 for all x 6= 0. In particular if
we put x = (1, 0) we come to G(x, x) = A > 0. Thus A > 0.
d) Consider quadratic form G(x, x) = gik xi xk = Ax2 + 2Bxy + Dy 2 . We have an
identity
G(x, x) = gik xi xk = Ax2 + 2Bxy + Dy 2 =
(Ax + By)2 + (AD − B 2 )y 2
.
A
(1)
We already know that A > 0 (take x = (x, 0)). Now take x = (x, y): Ax + By = 0 (e.g.
2
)y 2
> 0. Hence (AD − B 2 ) = det G > 0 ∗ .
x=(-B,A)) we come to G(x, x) = (AD−B
A
∗
This special trick works good for dimension is n = 2. We could notice that A and
AD − B 2 are principal main minors of the matrix G. In the general case (if G is n × n
symmetric matrix) using triangular transformations one can show that quadratic form
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e) it follows from condition a) that matrix (1) is symmetric. It follows from conditionis
(c) and (d) and equation (2) that G(x, x) > 0 for any non-zero vector x.
8 Consider two-dimensional Riemannian manifold with Euclidean metric
G = dx2 +
x = ax′ + by ′ + e
dy 2 . How this metric will transform under arbitrary linear transformation
?
y = cx′ + dy ′ + f
Solution: Perform straightforward calculations: dx = adx′ + bdy ′ and dy = cdx′ + dy ′ .
Hence
G = dx2 +dy 2 = (adx′ +bdy ′ )2 +(cdx′ +dy ′ )2 = (a2 +c2 )(dx′ )2 +2(ab+cd)dx′ dy ′ +(b2 +d2 )(dy ′ )2 .
In coordinates (x, y) ||gik || =
1
0
0
1
′
′
and in coordinates (x , y )
′
||gik
||
=
a2 + c 2
ab + cd
ab + cd
.
b2 + d 2
9 Consider domain in two-dimensional Riemannian manifold with Riemannian metric
G = du2 + 2bdudv + dv 2 in local coordinates u, v, where b is a constant.
Show that b2 < 1
1 b
is positive definite, hence detg = 1−b2 > 0,
Solution: Matrix of the metric G
b 1
i.e. b2 < 1.
Another solution: for any non-zero vector x, G(x, x) > 0. Consider x = (t, 1). Then
for an arbitrary t (t, 1) 6= 0 and G(x, x) = t2 + 2bt + 1 > 0. Hence polynomial t2 + 2bt + 1
has no real roots, i.e. b2 < 1.
One can see that the condition b2 < 1 is not only necessary but it is sufficient condition
for G to be a metric.
10 ∗ Show that G = dx2 + dy 2 + cdz 2 in R3 defines Riemannian metric iff c > 0.
∗
Find null-vectors of pesudo-Riemannian metric G if c < 0.
It is symmetric matrix which is positive definite iff c > 0. If c < 0 condition of positivedefiniteness is failed, but matrix G is still non-degenerate. Null-vector X = (x, y, z):
Gik X i X k = x2 + y 2 + cz 2 = 0, i.e. vector X belongs to the cone x2 + y 2 − cz 2 = 0.
A(X, X) = aik xi xk (and respectively) is positive-definite if and only if all the leading principal minors ∆k are positive (leading Principal minor ∆k of the matrix A is a determinant
of the matrix formed by first k columns and first k rows of the matrix A.). In this case
matrix Gik of bilinear form is transformed to unity matrix.
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