Hessian and inflexion points

Hessian and inflexion points

Theorem 1

.

Suppose that

C

is a curve in the real projective plane

R

⁢

P

2

given by a

homogeneous
equation

F

⁢

(

x

,

y

,

z

)

=

0

of
degree of homogeneity (

http://planetmath.org/HomogeneousFunction

)

n

.
If

F

has

continuous

first derivatives

in a neighborhood of
a point

P

and the

gradient

of

F

is non-zero at

P

and

P

is an

inflection point

of

C

, then

H

⁢

(

P

)

=

0

, where

H

is
the

Hessian

determinant:

H

=

|

∂

2

⁡

F

∂

⁡

x

2

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

z

∂

2

⁡

F

∂

⁡

y

⁢

∂

⁡

x

∂

2

⁡

F

∂

⁡

y

2

∂

2

⁡

F

∂

⁡

y

⁢

∂

⁡

z

∂

2

⁡

F

∂

⁡

z

⁢

∂

⁡

x

∂

2

⁡

F

∂

⁡

z

⁢

∂

⁡

y

∂

2

⁡

F

∂

⁡

z

2

|

Proof.

We may choose a system

x

,

y

,

z

of homogenous

coordinates

such that the point

P

lies at

(

0

,

0

,

1

)

and the equation
of the

tangent

to

C

at

P

is

y

=

0

. Using the

implicit function theorem

, we may conclude that there
exists an

interval

(

-

ϵ

,

ϵ

)

and a

function

f

:

(

-

ϵ

,

ϵ

)

→

ℝ

such that

F

⁢

(

t

,

f

⁢

(

t

)

,

1

)

=

0

when

-

ϵ

<

t

<

ϵ

. In
other words, the portion of curve near

P

may be described
in non-homogenous coordinates by

y

=

f

⁢

(

x

)

. By the way
the coordinates were chosen,

f

⁢

(

0

)

=

0

and

f

′

⁢

(

0

)

=

0

.
Because

P

is an inflection point, we also have

f

′′

⁢

(

0

)

=

0

.

Differentiating the equation

F

⁢

(

t

,

f

⁢

(

t

)

,

1

)

=

0

twice,
we obtain the following:

0

=

d

d

⁢

t

⁢

F

⁢

(

t

,

f

⁢

(

t

)

,

1

)

=

∂

⁡

F

∂

⁡

x

⁢

(

t

,

f

⁢

(

t

)

,

1

)

+

f

′

⁢

(

t

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

t

,

f

⁢

(

t

)

,

1

)

0

=

d

2

d

⁢

t

2

⁢

F

⁢

(

t

,

f

⁢

(

t

)

,

1

)

=

∂

2

⁡

F

∂

⁡

x

2

⁢

(

t

,

f

⁢

(

t

)

,

1

)

+

f

′

⁢

(

t

)

⁢

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

⁢

(

t

,

f

⁢

(

t

)

,

1

)

+

(

f

′

⁢

(

t

)

)

2

⁢

∂

2

⁡

F

∂

⁡

y

2

⁢

(

t

,

f

⁢

(

t

)

,

1

)

+

f

′′

⁢

(

t

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

t

,

f

⁢

(

t

)

,

1

)

We will now put

t

=

0

but, for reasons which will be
explained later, we do not yet want to make use of the
fact that

f

′′

⁢

(

0

)

=

0

:

∂

⁡

F

∂

⁡

x

⁢

(

0

,

0

,

1

)

=

0

∂

2

⁡

F

∂

⁡

x

2

⁢

(

0

,

0

,

1

)

=

-

f

′′

⁢

(

0

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

Since

F

is homogenous, Euler’s

formula

holds:

x

⁢

∂

⁡

F

∂

⁡

x

+

y

⁢

∂

⁡

F

∂

⁡

y

+

z

⁢

∂

⁡

F

∂

⁡

z

=

n

⁢

F

Taking

partial derivatives

, we obtain the following:

x

⁢

∂

2

⁡

F

∂

⁡

x

2

+

y

⁢

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

+

z

⁢

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

z

=

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

x

x

⁢

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

+

y

⁢

∂

2

⁡

F

∂

⁡

y

2

+

z

⁢

∂

2

⁡

F

∂

⁡

y

⁢

∂

⁡

z

=

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

y

x

⁢

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

z

+

y

⁢

∂

2

⁡

F

∂

⁡

y

⁢

∂

⁡

z

+

z

⁢

∂

2

⁡

F

∂

⁡

z

2

=

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

z

Evaluating at

(

0

,

0

,

1

)

and making use of the
equations deduced above, we obtain the following:

∂

⁡

F

∂

⁡

z

⁢

(

0

,

0

,

1

)

=

0

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

z

⁢

(

0

,

0

,

1

)

=

0

∂

2

⁡

F

∂

⁡

y

⁢

∂

⁡

z

⁢

(

0

,

0

,

1

)

=

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

∂

2

⁡

F

∂

⁡

z

2

⁢

(

0

,

0

,

1

)

=

0

Making use of these facts, we may now evaluate the determinant:

H

⁢

(

0

,

0

,

1

)

=

|

-

f

′′

⁢

(

0

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

⁢

(

0

,

0

,

1

)

0

∂

2

⁡

F

∂

⁡

x

⁢

∂

⁡

y

⁢

(

0

,

0

,

1

)

∂

2

⁡

F

∂

2

⁡

y

⁢

(

0

,

0

,

1

)

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

0

(

n

-

1

)

⁢

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

0

|

=

(

n

-

1

)

2

⁢

(

∂

⁡

F

∂

⁡

y

⁢

(

0

,

0

,

1

)

)

2

⁢

f

′′

⁢

(

0

)

Since

P

is an inflection point,

f

′′

⁢

(

0

)

=

0

, so we
have

H

⁢

(

0

,

0

,

1

)

=

0

.
∎

Actually, we proved slightly more than what was stated.
Because the gradient is assumed not to vanish at

P

,
but

∂

⁡

F

/

∂

⁡

x

=

0

and

∂

⁡

F

/

∂

⁡

z

=

0

by the way we set up our

coordinate
system

, we must have

∂

⁡

F

/

∂

⁡

y

≠

0

.
Thus, we see that, if

n

≠

1

, then

H

⁢

(

0

,

0

,

1

)

=

0

if and only if

f

′′

⁢

(

0

)

. However, note that this does
not mean that the Hessian

vanishes

if and only if

P

is
an inflection point since the

definition

of inflection
point not only requires that

f

′′

⁢

(

0

)

=

0

but that the
sign of

f

′′

⁢

(

t

)

change as

t

passes through

0

.

This result is used quite often in

algebraic geometry

,
where

F

is a homogenous

polynomial

. In such a

context

,
it is desirable to keep demonstrations purely

algebraic

and avoid introducing

analysis

where possible,
so a variant of this result is preferred. The

theorem

may be restated as follows:

Theorem 2

.

Suppose that

C

is a curve in the real projective plane

R

⁢

P

2

given by an equation

F

⁢

(

x

,

y

,

z

)

=

0

where

F

is a homogenous polynomial of degree

n

.
If

C

is

regular

at a point

P

and

P

is an inflection point
of

C

, then

H

⁢

(

P

)

=

0

, where

H

is the Hessian determinant.

To make our proof purely algebraic, we replace the use of
the implicit function theorem to obtain

f

with an

expansion

in a formal power series. As above, we choose our

x

,

y

,

z

coordinates so as to place

P

at

(

0

,

0

,

1

)

and make

C

tangent to the line

y

=

0

at

P

. Then, since

P

is a

regular
point

of

C

, we may parameterize

C

by a formal power series

f

⁢

(

t

)

=

∑

k

=

0

∞

c

k

⁢

t

k

such that

F

⁢

(

t

,

f

⁢

(

t

)

,

1

)

=

0

.
Then, if we
define

derivatives

algebraically (

http://planetmath.org/DerivativeOfPolynomial

),
we may proceed with the rest of the proof exactly as above.

Title

Hessian and inflexion points

Canonical

name

HessianAndInflexionPoints

Date of creation

2013-03-22 18:22:26

Last modified on

2013-03-22 18:22:26

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 53A04

Classification

msc 26A51

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