Geometry
Lecture 1 - 06-02-2024
Course info
Differential geometry bit:
Ch7- Curves (envelopes and evolutes)
Ch8- Surfaces (first and second fundamental form and gaussian curvature)
Algebraic geometry bit:
Ch1- Affine geometry
Ch2,3,4- fast recollection/introduction to Euclidean geometry.
Ch5- Projective geometry
Ch6- Quadrics. Ignore this part.
If you want to do self-study: Coxeter- Introduction to Geometry. Mainly relevant as far as the geometry
of surfaces is concerned. It has a proper discussion of geodescics that Audin misses. Might get exercises
on homeworks.
Teaching team:
• Nikolay Martynchuck (lecturer, 4th floor, integrable systems)
• Tobias V. Henriksen (TA)
• Leendert Los (TA)
Exercises require mostly time. Hints will be distributed. Solutions will be posted at some point.
Affine geometry
Let Kn = {(x1 , . . . , xn ) |xi ∈ K} with K a given field.
Definition Let (E, K) be a vector space over a field K. Let E be a set and
f :E ×E →E
−−→
f (A, B) 7→ AB
such that
1. ∀ A ∈ R, f (A, ·) : E → E is a bijection;
2. ∀ A, B, C ∈ E, f (A, B) + f (B, C) = f (A, C)
Then E is called an affine space.
−−→
Vectors are not points. If you have a pair of points, then f (A, B) = AB connects these points.
Using the vector notation, we have that
−−→
1. b ∈ E 7→ AB, where A is fixed, is a bijection.
−−→ −−→ −→
2. AB + BC = AC
A bit more on affine spaces.
−−→ −−→
Theorem (paralellogram rule) If A, B, C, D are 4 points in an affine space such that AD = BC, then
−−→ −−→
AB = DC
→
−
Definition (Addition of points and vectors in an affine space) If A ∈ E, V ∈ E, then C =
−→
A+→ −
v ∈ E such that AC = →
−
v.
1
,Examples
• Rn as an affine space or Cn as an affine space. You can also have them as vector spaces.
Remark Affine spaces are vector spaces without the origin
Curves in Rn or Cn are given by parametric equations
γ − (x1 (t), . . . , xn (t)) : R → Rn
Chain rule: Let f : Rn → Rm which is differentiable at a point x0 ∈ Rn . Let G : Rm → Rk which is
differentiable at y0 = f (x0 ). Then G ◦ f : Rn → Rk is differentiable at a point x0 and the differential is
given by
d(G ◦ f ) = dG ◦ dF
Reminder: dF : Tx0 Rn → Ty0 Rm and dG : Ty0 Rm → TG(y0 ) Rk .
∂F
Implicit function theorem Let F : Rm × Rn → Rn and F (x0 , y0 ) = z0 . If ∂y (x0 ,y0 ) =rankn, then
there exists y = y(x) such that y0 = y(x0 ) and F (x, y(x)) = z0
Exercise: Look up the inverse function theorem
Theorem Let γ : R → R2 be a smooth curve with γ ′ (t0 ) = (x′ (t0 ), y ′ (t0 )) ̸= 0. Then if x′ (t0 ) ̸= 0,
locally γ(t(x)) = (x, y(x)) near (x(t0 ), y(t0 )) also locally γ is given by an implicit equation F (x, y) = 0,
with dF |x0 ,y0 ̸= 0
Lecture 2 - 07-02-2024
Setting: Given a smooth map γ : (−ϵ, ϵ) → R2 . Then, assuming γ ′ (0) ̸= 0, there exists (a possibly
smaller neighbourhood (−δ, δ) of 0 ∈ R such that
• γ|(−δ,δ) is a regular curve, i.e. γ ′ (t) ̸= 0 for all t ∈ (−δ, δ) and it has no self-intersections (one says
γ is a simple curve).
• The locus X = {γ(t) = (x(t), y(t)) ∈ R2 | t ∈ (−δ, δ)} coincides with the level set of a function
F : u(0) ⊂ R2 → R : F (x, y) = 0 (i.e. X is a solution set to F = 0).
• X = {(x, f (x)) | x ∈ (−η, η)} (i.e. γ is a graph of a function).
Moreover, for suitable δ, η, u(0), all three conditions are equivalent which is a corollary of the implicit
function theorem.
Remark: A similar statement holds for smooth maps γ : (−ϵ, ϵ) → Rn with γ ′ (0) ̸= 0 with the second
condition replaced by ?? (exercise) (Hint: Implicit function theorem)
Def: A smooth curve γ : (a, b) → Rn is called regular if γ ′ (t) ̸= 0 for all t ∈ (a, b). If γ ′ (t0 ) = 0, then t0
is called a singular point of γ or γ(t0 ) is called a singular value.
Similarly, if n = 2 and the curve is defined as F (x, y) = 0 (implicit equation), then a point (x0 , y0 ) is
singular if dF |(x0 ,y0 ) = 0 and F (x0 , y0 ) = 0
Similarly, if F : R2 → R is a smooth map and dF (x0 , y0 ) = 0, then (x0 , y0 ) is a singular point and
F (x0 , y0 ) is a singular value.
We will mostly concern ourselves with curves with a finite amount of singular points.
Examples of curves:
β
In polar coordinates (r, ϕ), r > 0 constant defines a circle. More generally, r = 1+e cos(ϕ) , β > 0 defines
a conic section. e is called eccentricity. For e < 1 you get an ellipse, for e = 1 a parabola and for e > 1
a hyperbola.
Conic sections: sections of a cone together with a two-dimensional plane. Q(x, y) = ax2 + by 2 + cxy +
dx + ey + f = 0 defines conics algebraically as implicit equations.
2
,Remark: Conics are solutions to Keplers problem (apparently we are already supposed to know this??)
→
−̈ GM −
x =− 2 →
er
r
Generic Mathematica promotion
Examples of curves:
• Bernoulli’s lemniscate:
t3
t
γ= ,
1 + t4 1 + t4
– Is γ smooth? Yes, because it is defined as a rational function.
– Is gamma regular? To check: γ ′ (t) ̸= 0. Not going to do this in class
– Is it a simple curve? Depends, as parametric curve it is (bc it only meets at ∞) but in
Cartesian coordinates it does have a self-intersection. We could also reparameterize this. BE
CAREFUL WITH THIS ON EXAM
• Conic sections (see earlier)
• Cycloids
1
z = z(ϕ) = (meiϕ + eimϕ )
m+1
curve depends on m ∈ R\{−1}. They look cool as fuck. Curves tangent to a circle. Rotating a
sphere of radius r along one of radius R (either on the inside or the outside). Like with those weird
little stencils. We have m = Rr . There are hypocycloids and epicycloids.
• We can also roll a circle on a line. These are solutions to the Brachistochrone and Tautochrone
problems. Something about things rolling. Cool animations on Wikipedia. Proof is hard, but he
wants to talk about it in private with smart people.
• Many many more examples. Try to plot things with common sense first before resorting to software.
Good exercise.
WARNING: material will get heavier around week 3 for a little bit.
Envelopes Suppose we have a family of lines in affine space R2 parameterized by a parameter t. Call
this family Dt , t ∈ I ⊂ R. An envelope of Dt is a smooth curve γ(t) such that
1. γ(t) ∈ Dt
2. γ ′ (t)∥Dt
No promises about existence and uniqueness at this point, this needs restrictions on Dt .
How to construct envelopes? Say Dt is given by Dt = {u(t)x + v(t)y + w = 0}
Let γ(t) = (x(t), y(t)). Then γ(t) ∈ Dt implies that
1. u(t)x(t) + v(t)y(t) + w(t) = 0
2. u(t)x′ (t) + v(t)y ′ (t) = 0
We then get by differentiating shit (see lecture notes for details)
u(t) v(t) x(t) w(t)
=
u′ (t) v ′ (t) y(t) w′ (t)
If
u(t) v(t)
det ′ ̸= 0
u (t) v ′ (t)
3
, then −1
x(t) u(t) v(t) −w(t)
= ′
y(t) u (t) v ′ (t) −w′ (t)
is a solution to the envelope problem.
Funny drawing, see lecture notes.
Parabolas appear as the envelope of perpendicular bisectors.
Cycloids appear as envelopes of the family of lines M1 M2 , M1 = R · eiω1 t , M2 = R · eiω2 t . The
cycloid depends on the ratio ω2 /ω1 , you get different epi- and hypocycloids. The envelope is G(t) =
1 iω1 t
ω1 +ω2 e ω2 + eiω2 t ω1 .
Lecture 3 - 13-02-2024
Envelopes
Given R2 (affine 2-plane). We take a point F ∈ R2 and a line L not on that point. Take a point
M = (x, −1/4a) and draw a line from F to M . Then Dx consists of perpendicular bisectors of this line.
Claim: Dx is a parabola. Nikolay thinks his drawings are not of the best quality.
Now apparently we will compute these envelopes ourselves.
−−→
We have that the midpoint of F M is given by
1 t2
Dt = {−tx + y+ = 0}
2a 2
Solution to envelope problem is
−1
x(t) u(t) v(t) −w(t)
= ′
y(t) u (t) v ′ (t) −w′ (t)
1 t2
We have ut = −t, v(t) = 2a , w(t) = 2 so
1 −1
t2
x(t) −t 2a −2
=
y(t) −1 0 −t
Solving gives us t2
x(t) 0 −1 −2 t
= =
y(t) 2a −2at −t at2
or y = ax2 .
4
Lecture 1 - 06-02-2024
Course info
Differential geometry bit:
Ch7- Curves (envelopes and evolutes)
Ch8- Surfaces (first and second fundamental form and gaussian curvature)
Algebraic geometry bit:
Ch1- Affine geometry
Ch2,3,4- fast recollection/introduction to Euclidean geometry.
Ch5- Projective geometry
Ch6- Quadrics. Ignore this part.
If you want to do self-study: Coxeter- Introduction to Geometry. Mainly relevant as far as the geometry
of surfaces is concerned. It has a proper discussion of geodescics that Audin misses. Might get exercises
on homeworks.
Teaching team:
• Nikolay Martynchuck (lecturer, 4th floor, integrable systems)
• Tobias V. Henriksen (TA)
• Leendert Los (TA)
Exercises require mostly time. Hints will be distributed. Solutions will be posted at some point.
Affine geometry
Let Kn = {(x1 , . . . , xn ) |xi ∈ K} with K a given field.
Definition Let (E, K) be a vector space over a field K. Let E be a set and
f :E ×E →E
−−→
f (A, B) 7→ AB
such that
1. ∀ A ∈ R, f (A, ·) : E → E is a bijection;
2. ∀ A, B, C ∈ E, f (A, B) + f (B, C) = f (A, C)
Then E is called an affine space.
−−→
Vectors are not points. If you have a pair of points, then f (A, B) = AB connects these points.
Using the vector notation, we have that
−−→
1. b ∈ E 7→ AB, where A is fixed, is a bijection.
−−→ −−→ −→
2. AB + BC = AC
A bit more on affine spaces.
−−→ −−→
Theorem (paralellogram rule) If A, B, C, D are 4 points in an affine space such that AD = BC, then
−−→ −−→
AB = DC
→
−
Definition (Addition of points and vectors in an affine space) If A ∈ E, V ∈ E, then C =
−→
A+→ −
v ∈ E such that AC = →
−
v.
1
,Examples
• Rn as an affine space or Cn as an affine space. You can also have them as vector spaces.
Remark Affine spaces are vector spaces without the origin
Curves in Rn or Cn are given by parametric equations
γ − (x1 (t), . . . , xn (t)) : R → Rn
Chain rule: Let f : Rn → Rm which is differentiable at a point x0 ∈ Rn . Let G : Rm → Rk which is
differentiable at y0 = f (x0 ). Then G ◦ f : Rn → Rk is differentiable at a point x0 and the differential is
given by
d(G ◦ f ) = dG ◦ dF
Reminder: dF : Tx0 Rn → Ty0 Rm and dG : Ty0 Rm → TG(y0 ) Rk .
∂F
Implicit function theorem Let F : Rm × Rn → Rn and F (x0 , y0 ) = z0 . If ∂y (x0 ,y0 ) =rankn, then
there exists y = y(x) such that y0 = y(x0 ) and F (x, y(x)) = z0
Exercise: Look up the inverse function theorem
Theorem Let γ : R → R2 be a smooth curve with γ ′ (t0 ) = (x′ (t0 ), y ′ (t0 )) ̸= 0. Then if x′ (t0 ) ̸= 0,
locally γ(t(x)) = (x, y(x)) near (x(t0 ), y(t0 )) also locally γ is given by an implicit equation F (x, y) = 0,
with dF |x0 ,y0 ̸= 0
Lecture 2 - 07-02-2024
Setting: Given a smooth map γ : (−ϵ, ϵ) → R2 . Then, assuming γ ′ (0) ̸= 0, there exists (a possibly
smaller neighbourhood (−δ, δ) of 0 ∈ R such that
• γ|(−δ,δ) is a regular curve, i.e. γ ′ (t) ̸= 0 for all t ∈ (−δ, δ) and it has no self-intersections (one says
γ is a simple curve).
• The locus X = {γ(t) = (x(t), y(t)) ∈ R2 | t ∈ (−δ, δ)} coincides with the level set of a function
F : u(0) ⊂ R2 → R : F (x, y) = 0 (i.e. X is a solution set to F = 0).
• X = {(x, f (x)) | x ∈ (−η, η)} (i.e. γ is a graph of a function).
Moreover, for suitable δ, η, u(0), all three conditions are equivalent which is a corollary of the implicit
function theorem.
Remark: A similar statement holds for smooth maps γ : (−ϵ, ϵ) → Rn with γ ′ (0) ̸= 0 with the second
condition replaced by ?? (exercise) (Hint: Implicit function theorem)
Def: A smooth curve γ : (a, b) → Rn is called regular if γ ′ (t) ̸= 0 for all t ∈ (a, b). If γ ′ (t0 ) = 0, then t0
is called a singular point of γ or γ(t0 ) is called a singular value.
Similarly, if n = 2 and the curve is defined as F (x, y) = 0 (implicit equation), then a point (x0 , y0 ) is
singular if dF |(x0 ,y0 ) = 0 and F (x0 , y0 ) = 0
Similarly, if F : R2 → R is a smooth map and dF (x0 , y0 ) = 0, then (x0 , y0 ) is a singular point and
F (x0 , y0 ) is a singular value.
We will mostly concern ourselves with curves with a finite amount of singular points.
Examples of curves:
β
In polar coordinates (r, ϕ), r > 0 constant defines a circle. More generally, r = 1+e cos(ϕ) , β > 0 defines
a conic section. e is called eccentricity. For e < 1 you get an ellipse, for e = 1 a parabola and for e > 1
a hyperbola.
Conic sections: sections of a cone together with a two-dimensional plane. Q(x, y) = ax2 + by 2 + cxy +
dx + ey + f = 0 defines conics algebraically as implicit equations.
2
,Remark: Conics are solutions to Keplers problem (apparently we are already supposed to know this??)
→
−̈ GM −
x =− 2 →
er
r
Generic Mathematica promotion
Examples of curves:
• Bernoulli’s lemniscate:
t3
t
γ= ,
1 + t4 1 + t4
– Is γ smooth? Yes, because it is defined as a rational function.
– Is gamma regular? To check: γ ′ (t) ̸= 0. Not going to do this in class
– Is it a simple curve? Depends, as parametric curve it is (bc it only meets at ∞) but in
Cartesian coordinates it does have a self-intersection. We could also reparameterize this. BE
CAREFUL WITH THIS ON EXAM
• Conic sections (see earlier)
• Cycloids
1
z = z(ϕ) = (meiϕ + eimϕ )
m+1
curve depends on m ∈ R\{−1}. They look cool as fuck. Curves tangent to a circle. Rotating a
sphere of radius r along one of radius R (either on the inside or the outside). Like with those weird
little stencils. We have m = Rr . There are hypocycloids and epicycloids.
• We can also roll a circle on a line. These are solutions to the Brachistochrone and Tautochrone
problems. Something about things rolling. Cool animations on Wikipedia. Proof is hard, but he
wants to talk about it in private with smart people.
• Many many more examples. Try to plot things with common sense first before resorting to software.
Good exercise.
WARNING: material will get heavier around week 3 for a little bit.
Envelopes Suppose we have a family of lines in affine space R2 parameterized by a parameter t. Call
this family Dt , t ∈ I ⊂ R. An envelope of Dt is a smooth curve γ(t) such that
1. γ(t) ∈ Dt
2. γ ′ (t)∥Dt
No promises about existence and uniqueness at this point, this needs restrictions on Dt .
How to construct envelopes? Say Dt is given by Dt = {u(t)x + v(t)y + w = 0}
Let γ(t) = (x(t), y(t)). Then γ(t) ∈ Dt implies that
1. u(t)x(t) + v(t)y(t) + w(t) = 0
2. u(t)x′ (t) + v(t)y ′ (t) = 0
We then get by differentiating shit (see lecture notes for details)
u(t) v(t) x(t) w(t)
=
u′ (t) v ′ (t) y(t) w′ (t)
If
u(t) v(t)
det ′ ̸= 0
u (t) v ′ (t)
3
, then −1
x(t) u(t) v(t) −w(t)
= ′
y(t) u (t) v ′ (t) −w′ (t)
is a solution to the envelope problem.
Funny drawing, see lecture notes.
Parabolas appear as the envelope of perpendicular bisectors.
Cycloids appear as envelopes of the family of lines M1 M2 , M1 = R · eiω1 t , M2 = R · eiω2 t . The
cycloid depends on the ratio ω2 /ω1 , you get different epi- and hypocycloids. The envelope is G(t) =
1 iω1 t
ω1 +ω2 e ω2 + eiω2 t ω1 .
Lecture 3 - 13-02-2024
Envelopes
Given R2 (affine 2-plane). We take a point F ∈ R2 and a line L not on that point. Take a point
M = (x, −1/4a) and draw a line from F to M . Then Dx consists of perpendicular bisectors of this line.
Claim: Dx is a parabola. Nikolay thinks his drawings are not of the best quality.
Now apparently we will compute these envelopes ourselves.
−−→
We have that the midpoint of F M is given by
1 t2
Dt = {−tx + y+ = 0}
2a 2
Solution to envelope problem is
−1
x(t) u(t) v(t) −w(t)
= ′
y(t) u (t) v ′ (t) −w′ (t)
1 t2
We have ut = −t, v(t) = 2a , w(t) = 2 so
1 −1
t2
x(t) −t 2a −2
=
y(t) −1 0 −t
Solving gives us t2
x(t) 0 −1 −2 t
= =
y(t) 2a −2at −t at2
or y = ax2 .
4
