aarjaneiro/planetmath huggingface dataset:
proof of Dehn’s theorem

proof of Dehn’s theorem

We define the Dehn’s

invariant

, which is a number given to any

polyhedron

which does not change under scissor-equivalence.

Choose an

additive function

f

:

ℝ

→

ℝ

such that

f

⁢

(

π

)

=

f

⁢

(

0

)

=

0

and define for any polyhedron

P

the number (Dehn’s invariant)

D

⁢

(

P

)

=

∑

e

∈

{

edges of

P

}

f

⁢

(

θ

e

)

⁢

ℓ

⁢

(

e

)

where

θ

e

is the angle between the two faces of

P

joining in

e

, and

ℓ

⁢

(

e

)

is the length of the edge

e

.

We want to prove that if we decompose

P

into smaller polyhedra

P

1

,

…

,

P

N

as in the definition of scissor-equivalence, we have

D

⁢

(

P

)

=

∑

k

=

1

N

D

⁢

(

P

k

)

(1)

which means that if

P

is scissor

equivalent

to

Q

then

D

⁢

(

P

)

=

D

⁢

(

Q

)

.

Let

P

1

,

…

,

P

N

be such a

decomposition

of

P

. Given any edge

e

of a piece

P

k

the following cases arise:

1.

e

is

contained

in the interior of

P

. Since an entire neighbourhood of

e

is contained in

P

the angles of the pieces which have

e

as an edge (or part of an edge) must have sum

2

⁢

π

. So in the right hand side of (

1

) the edge

e

gives a contribution of

f

⁢

(

2

⁢

π

)

⁢

ℓ

⁢

(

e

)

(recall
that

f

is

additive

).

2.

e

is contained in a facet of

P

. The same

argument

as before is valid, only we find that the total contribution is

f

⁢

(

π

)

⁢

ℓ

⁢

(

e

)

.

3.

e

is contained in an edge

e

′

of

P

. In this case the total contribution given by

e

to the right hand side of (

1

) is given by

f

⁢

(

θ

e

′

)

⁢

ℓ

⁢

(

e

)

.

Since we have choosen

f

so that

f

⁢

(

π

)

=

0

and hence also

f

⁢

(

2

⁢

π

)

=

0

(since

f

is additive) we conclude that the equivalence (

1

) is valid.

Now we are able to prove Dehn’s

Theorem

. Choose

T

to be a

regular tetrahedron

with edges of length

1

. Then

D

⁢

(

T

)

=

6

⁢

f

⁢

(

θ

)

where

θ

is the angle between two faces of

T

. We know that

θ

/

π

is irrational, hence there exists an additive function

f

such that

f

⁢

(

θ

)

=

1

while

f

⁢

(

π

/

2

)

=

0

(as there exist additive functions which are not linear).

So if

P

is any

parallelepiped

we find that

D

⁢

(

P

)

=

0

(since each angle between facets of

P

is

π

/

2

and

f

⁢

(

π

/

2

)

=

0

) while

D

⁢

(

T

)

=

6

. This means that

P

and

T

cannot be

scissor-equivalent

.

Title

proof of Dehn’s theorem

Canonical name

ProofOfDehnsTheorem

Date of creation

2013-03-22 16:18:07

Last modified on

2013-03-22 16:18:07

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Proof

Classification

msc 51M04

Defines

Dehn invariant

Defines

Dehn’s invariant

Generated on Fri Feb 9 22:03:59 2018 by

LaTeXML
proof of pivot theorem

proof of pivot theorem

Let

△

⁢

A

⁢

B

⁢

C

be a

triangle

, and let

D

,

E

, and

F

be points
on

B

⁢

C

,

C

⁢

A

, and

A

⁢

B

, respectively. The

circumcircles

of

△

⁢

A

⁢

E

⁢

F

and

△

⁢

B

⁢

F

⁢

D

intersect in

F

and in another
point, which we call

P

. Then

A

⁢

E

⁢

P

⁢

F

and

B

⁢

F

⁢

P

⁢

D

are

cyclic
quadrilaterals

, so

∠

⁢

A

+

∠

⁢

E

⁢

P

⁢

F

=

π

and

∠

⁢

B

+

∠

⁢

F

⁢

P

⁢

D

=

π

Combining this with

∠

⁢

A

+

∠

⁢

B

+

∠

⁢

C

=

π

and

∠

⁢

E

⁢

P

⁢

F

+

∠

⁢

F

⁢

P

⁢

D

+

∠

⁢

D

⁢

P

⁢

E

=

2

⁢

π

, we get

∠

⁢

C

+

∠

⁢

D

⁢

P

⁢

E

=

π

.

This implies that

C

⁢

D

⁢

P

⁢

E

is a cyclic quadrilateral as well, so that

P

lies on the circumcircle of

△

⁢

C

⁢

D

⁢

E

. Therefore, the
circumcircles of the triangles

A

⁢

E

⁢

F

,

B

⁢

F

⁢

D

, and

C

⁢

D

⁢

E

have a common
point,

P

.

Title

proof of pivot theorem

Canonical name

ProofOfPivotTheorem

Date of creation

2013-03-22 13:12:37

Last modified on

2013-03-22 13:12:37

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

4

Author

pbruin (1001)

Entry type

Proof

Classification

msc 51M04

Generated on Fri Feb 9 22:04:08 2018 by

LaTeXML
radian

radian

The

radian

is a of angle used in the higher mathematics.  The magnitude of an angle

α

is one radian, if the arc corresponding the angle

α

as a

central angle

of a circle is equally long as the radius of the circle.  Thus, a radian is equal to

180

π

, minutes and seconds approximately

57

o

⁢

17

′

⁢

44.80625

′′

.

In degrees, a circle has 360 degrees, while in radians a circle has

2

⁢

π

radians. In fact, many angles of

equilateral polygons

are equal to a multiple of

π

divided by some integer: for example, the

interior angle

of an

equilateral triangle

’s vertex is

π

3

, while the interior angle of an equilateral pentagon’s vertex is

3

⁢

π

5

.

Title

radian

Canonical name

Radian

Date of creation

2013-03-22 14:48:58

Last modified on

2013-03-22 14:48:58

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

7

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 51M04

Synonym

absolute unit of angle

Related topic

SolidAngle

Generated on Fri Feb 9 22:04:11 2018 by

LaTeXML
regular decagon inscribed in circle

regular decagon inscribed in circle

If a

line segment

has been divided into two parts such that the greater part is the

central proportional

of the whole

segment

and the smaller part, then one has performed the

golden section

(Latin

sectio aurea

) of the line segment.

Theorem

.

The side of the

regular

(

http://planetmath.org/RegularPolygon

)

decagon

(

http://planetmath.org/

Polygon

),

inscribed

in a circle, is equal to the greater part of the radius divided with the .

Proof.

A

regular polygon

can be inscribed in a circle (

http://planetmath.org/RegularPolygonAndCircles

). In the picture below, there is seen an

isosceles

central triangle

O

⁢

A

⁢

B

of a

regular decagon

with the

central angle

O

=

360

∘

:

10

=

36

∘

; the base angles are

(

180

∘

-

36

∘

)

:

2

=

72

∘

. One of the base angles is halved with the line

A

⁢

C

, when one gets a smaller

isosceles triangle

A

⁢

B

⁢

C

with equal angles as in the

triangle

O

⁢

A

⁢

B

. From these

similar triangles

we obtain the proportion equation

r

:

s

=

s

:

(

r

-

s

)

,

(1)

which shows that the side

s

of the regular decagon is the central proportional of the radius

r

of the circle and the

difference

r

-

s

.

Note.

(1) can be simplified to the

quadratic equation

(

http://planetmath.org/QuadraticFormula

)

s

2

+

r

⁢

s

-

r

2

=

0

which yields the

positive

solution

s

=

-

1

+

5

2

⁢

r

≈

0.618

⁢

r

.

Cf. also the golden ratio.

Title

regular decagon inscribed in circle

Canonical

name

RegularDecagonInscribedInCircle

Date of creation

2013-03-22 17:34:26

Last modified on

2013-03-22 17:34:26

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Synonym

regular decagon

Related topic

RegularPolygonAndCircles

Related topic

HomogeneousEquation

Related topic

Pentagon

Defines

golden section

Generated on Fri Feb 9 22:04:13 2018 by

LaTeXML
regular polygon and circles

regular polygon and circles

Theorem.

Every

regular polygon

has a circumscribed circle and an inscribed circle.

Proof.

Given a

regular

n

-gon, draw the

angle bisectors

of its

interior angles

. Since the interior angles of a regular

n

-gon are

congruent

, one gets

n

isosceles triangles

. (See determining from angles that a

triangle

is isosceles for more details.) Moreover, since the sides of the regular

n

-gon are congruent, these isosceles triangles have congruent bases. Thus, these triangles are congruent (ASA). Therefore, the sides

adjacent

to the

vertex angles

are congruent. Hence, the vertices of two adjacent triangles and thus of all triangles coincide. This common vertex point is equidistant from all vertices of the

polygon

and also from all sides of the polygon, whence it is simultaneously the center (

http://planetmath.org/Center8

) of the circumscribed circle and the inscribed circle.
∎

To illustrate what is going on in the proof, the procedure explained in the proof will be demonstrated for a

regular pentagon

and a

regular hexagon

. In the pictures below, the regular pentagon is on the left, and the regular hexagon is on the right.

In the first picture, the

n

angle bisectors are drawn in blue. Note how they all intersect at one point. This point is the center of the regular polygon.

In the second picture, the

n

angle bisectors are only drawn to the center. Note that the resulting picture for the regular hexagon is no different than the previous picture.

In the last picture, the inscribed circle is drawn in green, and the circumscribed circle is drawn in cyan.

References

1

K. Väisälä

:

Geometria

. Tenth edition.  Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1971).

Title

regular polygon and circles

Canonical name

RegularPolygonAndCircles

Date of creation

2013-03-22 17:17:06

Last modified on

2013-03-22 17:17:06

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51-00

Related topic

Circumscribe

Related topic

Inscribe

Related topic

CircleHasOneCenter

Related topic

RegularDecagonInscribedInCircle

Generated on Fri Feb 9 22:04:16 2018 by

LaTeXML
sine of angle of triangle

sine of angle of triangle

The

cosines law

allows to express the cosine of an angle of

triangle

through the sides:

cos

⁡

α

=

b

2

+

c

2

-

a

2

2

⁢

b

⁢

c

.

(1)

Substituting this to the “fundamental formula of

trigonometry

”,

sin

2

⁡

α

+

cos

2

⁡

α

=

1

,

we can calculate as follows:

sin

⁡

α

=

+

1

-

(

b

2

+

c

2

-

a

2

2

⁢

b

⁢

c

)

2

=

(

2

⁢

b

⁢

c

)

2

-

(

b

2

+

c

2

-

a

2

)

2

(

2

⁢

b

⁢

c

)

2

=

(

2

⁢

b

⁢

c

+

b

2

+

c

2

-

a

2

)

⁢

(

2

⁢

b

⁢

c

-

b

2

-

c

2

+

a

2

)

2

⁢

b

⁢

c

=

[

(

b

+

c

)

2

-

a

2

]

⁢

[

a

2

-

(

b

-

c

)

2

]

2

⁢

b

⁢

c

=

(

b

+

c

+

a

)

⁢

(

b

+

c

-

a

)

⁢

(

a

+

b

-

c

)

⁢

(

a

-

b

+

c

)

2

⁢

b

⁢

c

Thus we have the beautiful formula

sin

⁡

α

=

(

-

a

+

b

+

c

)

⁢

(

a

-

b

+

c

)

⁢

(

a

+

b

-

c

)

⁢

(

a

+

b

+

c

)

2

⁢

b

⁢

c

.

Substituting (1) similarly to the general formula for the sine of half-angle (

http://planetmath.org/GoniometricFormulae

)

sin

⁡

α

2

=

±

1

-

cos

⁡

α

2

,

one can obtain the formula

sin

⁡

α

2

=

(

a

-

b

+

c

)

⁢

(

a

+

b

-

c

)

4

⁢

b

⁢

c

.

Title

sine of angle of triangle

Canonical name

SineOfAngleOfTriangle

Date of creation

2013-03-22 18:27:16

Last modified on

2013-03-22 18:27:16

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51M04

Related topic

DifferenceOfSquares

Generated on Fri Feb 9 22:04:19 2018 by

LaTeXML
spherical trigonometry

spherical trigonometry

Cosine law

.

In the following we deduce the cosine law for a spherical

trihedron

.

Let

𝐞

𝟏

,

𝐞

𝟐

,

𝐞

𝟑

be the vertex

unitary

vectors as shown
in the figure.

Figure 1:

Spherical Trihedron used to deduce trigonometric relations

The cosine of the angle

α

formed by the plane
defined by

𝐞

𝟏

,

𝐞

𝟐

and the plane defined by

𝐞

𝟏

,

𝐞

𝟑

is:

cos

⁡

α

=

(

𝐞

𝟏

×

𝐞

𝟑

)

⋅

(

𝐞

𝟏

×

𝐞

𝟐

)

∥

𝐞

𝟏

×

𝐞

𝟑

∥

⁢

∥

𝐞

𝟏

×

𝐞

𝟐

∥

=

(

𝐞

𝟏

×

𝐞

𝟑

)

⋅

(

𝐞

𝟏

×

𝐞

𝟐

)

sin

⁡

b

⁢

sin

⁡

c

Now, using the cyclic property of the

triple vector product

and Lagrange’s formula (

http://planetmath.org/TripleCrossProduct

), we can write:

cos

⁡

α

=

𝐞

𝟏

⋅

(

𝐞

𝟑

×

(

𝐞

𝟏

×

𝐞

𝟐

)

)

sin

⁡

b

⁢

sin

⁡

c

=

𝐞

𝟏

⋅

(

(

𝐞

𝟑

⋅

𝐞

𝟐

)

⁢

𝐞

𝟏

-

(

𝐞

𝟑

⋅

𝐞

𝟏

)

⁢

𝐞

𝟐

)

sin

⁡

b

⁢

sin

⁡

c

=

cos

⁡

a

-

cos

⁡

b

⁢

cos

⁡

c

sin

⁡

b

⁢

sin

⁡

c

Hence:

cos

⁡

a

=

cos

⁡

b

⁢

cos

⁡

c

+

sin

⁡

b

⁢

sin

⁡

c

⁢

cos

⁡

α

Title

spherical trigonometry

Canonical name

SphericalTrigonometry

Date of creation

2013-03-22 17:08:13

Last modified on

2013-03-22 17:08:13

Owner

fernsanz (8869)

Last modified by

fernsanz (8869)

Numerical id

12

Author

fernsanz (8869)

Entry type

Topic

Classification

msc 51M04

Related topic

AreaOfASphericalTriangle

Generated on Fri Feb 9 22:04:23 2018 by

LaTeXML
supplementary angles

supplementary angles

Two angles are called

supplementary angles

of each other if the sum of their measures (

http://planetmath.org/AngleMeasure

) is equal to the

straight angle

π

, i.e. (

http://planetmath.org/Ie

)

180

∘

.

For example, when two lines intersect each other, they the plane into four disjoint domains (

http://planetmath.org/Domain2

) corresponding to four convex angles; then any of these angles has a supplementary angle on either side of it (see

linear pair

). However, two angles that are

supplementary

to each other do not need to have a common side — see e.g. (

http://planetmath.org/Eg

) an entry regarding opposing angles in a

cyclic quadrilateral

(

http://planetmath.org/OpposingAnglesInACyclicQuadrilateralAreSupplementary

).

Supplementary angles have always equal sines, but the cosines are opposite numbers:

sin

⁡

(

π

-

α

)

=

sin

⁡

α

,

cos

⁡

(

π

-

α

)

=

-

cos

⁡

α

These formulae may be proved by using the subtraction formulas of sine and cosine.

Title

supplementary angles

Canonical name

SupplementaryAngles

Date of creation

2013-03-22 17:34:59

Last modified on

2013-03-22 17:34:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Definition

Classification

msc 51M04

Classification

msc 51F20

Synonym

supplementary

Related topic

Supplement

Related topic

Angle

Related topic

ComplementaryAngles

Related topic

GoniometricFormulae

Generated on Fri Feb 9 22:04:25 2018 by

LaTeXML
theorems of Euclid

theorems of Euclid

This is a list of theorems

contained

in Euclid’s

The

Elements

.
At present, the list is incomplete, but as time goes on, more

theorems

will
systematically be added until it is

complete

. Note that both
theorems and constructions are combined together in the elements; for a listing
of the constructions, please see

http://planetmath.org/node/9531

their entry.

•

I 4

SAS theorem

•

I 5 (the

pons asinorum

) The

base angles

of an

isosceles triangle

are equal to each other. Moreover, the angles under
the bases are also equal.

•

I 6 If two angles of a

triangle

are equal, then that triangle is
isosceles. (

converse

of the preceding

proposition

)

•

I 7 Given a

line segment

, it is not possible to construct
upon it two distinct triangles which lie upon the same side of the segment
whose remaining sides are equal to each other pairwise.

•

I 8

SSS theorem

Title

theorems of Euclid

Canonical name

TheoremsOfEuclid

Date of creation

2013-03-22 17:13:23

Last modified on

2013-03-22 17:13:23

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Topic

Classification

msc 51M04

Generated on Fri Feb 9 22:04:27 2018 by

LaTeXML
triangle mid-segment theorem

triangle mid-segment theorem

Theorem.

The segment connecting the

midpoints

of any two sides of a

triangle

is

parallel

to the third side and is half as long.

Proof.

In the triangle

A

⁢

B

⁢

C

, let

A

′

be the midpoint of

A

⁢

C

and

B

′

the midpoint of

B

⁢

C

.  Using the side-vectors

A

⁢

C

→

and

C

⁢

B

→

as a basis (

http://planetmath.org/Basis

) of the plane, we calculate the mid-segment

A

′

⁢

B

′

as a vector:

A

′

⁢

B

′

→

=

A

′

⁢

C

→

+

C

⁢

B

′

→

=

1

2

⁢

A

⁢

C

→

+

1

2

⁢

C

⁢

B

→

=

1

2

⁢

(

A

⁢

C

→

+

C

⁢

B

→

)

=

1

2

⁢

A

⁢

B

→

The last expression indicates that the segment

A

′

⁢

B

′

is such as asserted.

Corollary (Varignon’s theorem).

If one connects the midpoints of the of a

quadrilateral

, one obtains a

parallelogram

.

Title

triangle mid-segment theorem

Canonical name

TriangleMidsegmentTheorem

Date of creation

2013-03-22 17:46:35

Last modified on

2013-03-22 17:46:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51M25

Synonym

mid-segment theorem

Related topic

MutualPositionsOfVectors

Related topic

ParallelogramTheorems

Related topic

MedianOfTrapezoid

Related topic

CommonPointOfTriangleMedians

Related topic

Grafix

Related topic

SimonStevin

Related topic

InterceptTheorem

Generated on Fri Feb 9 22:04:30 2018 by

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truncated cone

truncated cone

Think a general cone (not necessarily a

circular

one). If a plane intersects the

lateral surface

of the cone, but not the base of this cone, then the solid remaining from the cone between the intersecting plane and the plane of the base

A

is called a

truncated cone

. Also the name

frustum

(the Latin

frustum

= ‘piece, fragment’) is used, althouh this may mean the portion of any solid which lies between two

parallel planes

cutting the solid. Sometimes one sees the (wrong) variant name

frustrum

. As the plural form of

frustum

both

frustums

and

frusta

are used.

We restrict our to the frustum of cone with the cutting plane

parallel

to the plane of the base. The part of its surface contained in the intersecting plane is the other

base

A

′

of the frustum. Since the bases

A

and

A

′

are

homothetic

with respect to the apex of the whole cone, they are

similar

planar figures. The

height

h

of the frustum is the part of the height

H

of the whole cone between the both base planes. Denote

h

′

:=

H

-

h

.

The volume of the frustum is obtained as the

difference

of the volumes of two cones:

V

=

1

3

⁢

A

⁢

H

-

1

3

⁢

A

′

⁢

h

′

=

1

3

⁢

[

A

⁢

h

+

(

A

-

A

′

)

⁢

h

′

]

(1)

One needs the term

(

A

-

A

′

)

⁢

h

′

.  The ratio of the similar areas

A

and

A

′

equals the square of the line ratio:

A

A

′

=

(

h

+

h

′

h

′

)

2

Thus we have the

proportion equation

A

A

′

=

h

+

h

′

h

′

;

subtracting 1 from both sides, it may be written

A

-

A

′

A

′

=

h

h

′

,

or

A

-

A

′

A

⁢

A

′

+

A

′

=

h

h

′

,

whence

(

A

-

A

′

)

⁢

h

′

=

(

A

⁢

A

′

+

A

′

)

⁢

h

.

Considering this in (1) we arrive to the volume formula of the frustum:

V

=

A

+

A

⁢

A

′

+

A

′

3

⋅

h

(2)

The

quotient

in front of

h

is called the

Heronian mean

of the positive numbers

A

and

A

′

.

References

1

K. Väisälä

:

Geometria

.  Reprint of the tenth edition.  Werner Söderström Osakeyhtiö, Porvoo & Helsinki (1971).

Title

truncated cone

Canonical

name

TruncatedCone

Date of creation

2013-03-22 17:48:01

Last modified on

2013-03-22 17:48:01

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51M04

Classification

msc 01A20

Classification

msc 51M20

Synonym

frustum

Synonym

frusta

Synonym

frustrum

Related topic

Prismatoid

Related topic

KalleVaisala

Defines

height of frustum

Defines

bases of frustum

Defines

Heronian mean

Generated on Fri Feb 9 22:04:32 2018 by

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antipodal

antipodal

Definition

Suppose

x

and

y

are points on the

n

-sphere (

http://planetmath.org/Sphere

)

S

n

.
If

x

=

-

y

then

x

and

y

are called

antipodal points

. The

antipodal map

is the map

A

:

S

n

→

S

n

defined as

A

⁢

(

x

)

=

-

x

.

Properties

1.

The antipodal map

A

:

S

n

→

S

n

is
homotopic (

http://planetmath.org/HomotopyOfMaps

)
to the

identity map

if

n

is odd

[

1

]

.

2.

The degree (

http://planetmath.org/DegreeMapOfSpheres

) of the
antipodal map is

(

-

1

)

n

+

1

.

References

1

V. Guillemin, A. Pollack,

Differential topology

, Prentice-Hall Inc., 1974.

Title

antipodal

Canonical

name

Antipodal

Date of creation

2013-03-22 13:57:45

Last modified on

2013-03-22 13:57:45

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Definition

Classification

msc 51M05

Classification

msc 15-00

Related topic

DiametralPoints

Defines

antipodal points

Defines

antipodal map

Generated on Fri Feb 9 22:04:37 2018 by

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antipodal map on S^n is homotopic to the identity if and only if n is odd

antipodal map on

S

n

is homotopic to the identity if and only if

n

is odd

Lemma

.

If

X

:

S

n

→

S

n

is a

unit vector

field, then
there is a homotopy between the

antipodal map

on

S

n

and the

identity map

.

Proof.

Regard

S

n

as a

subspace

of

R

n

+

1

and define

H

:

S

n

×

[

0

,

1

]

→

R

n

+

1

by

H

⁢

(

v

,

t

)

=

(

cos

⁡

π

⁢

t

)

⁢

v

+

(

sin

⁡

π

⁢

t

)

⁢

X

⁢

(

v

)

. Since

X

is a unit
vector field,

X

⁢

(

v

)

⟂

v

for any

v

∈

S

n

. Hence

∥

H

⁢

(

v

,

t

)

∥

=

1

, so

H

is into

S

n

. Finally observe that

H

⁢

(

v

,

0

)

=

v

and

H

⁢

(

v

,

1

)

=

-

v

. Thus

H

is a homotopy between
the antipodal map and the identity map.
∎

Proposition

.

The antipodal map

A

:

S

n

→

S

n

is homotopic
to the identity if and only if

n

is odd.

Proof.

If

n

is even, then the antipodal map

A

is the

composition

of an odd of

reflections

. It
therefore has degree

-

1

. Since the degree of the identity
map is

+

1

, the two maps are not homotopic.

Now suppose

n

is odd, say

n

=

2

⁢

k

-

1

. Regard

S

n

has a
subspace of

ℝ

2

⁢

k

. So each point of

S

n

has

coordinates

(

x

1

,

…

,

x

2

⁢

k

)

with

∑

i

x

i

2

=

1

. Define
a map

X

:

ℝ

2

⁢

k

→

ℝ

2

⁢

k

by

X

⁢

(

x

1

,

x

2

,

…

,

x

2

⁢

k

-

1

,

x

2

⁢

k

)

=

(

-

x

2

,

x

1

,

…

,

-

x

2

⁢

k

,

x

2

⁢

k

-

1

)

,
pairwise swapping coordinates and negating the even coordinates.
By construction, for any

v

∈

S

n

, we have that

∥

X

⁢

(

v

)

∥

=

1

and

X

⁢

(

v

)

⟂

v

. Hence

X

is a unit vector field. Applying the
lemma, we conclude that the antipodal map is homotopic to the identity.
∎

References

1

Hatcher, A.

Algebraic topology

, Cambridge University Press, 2002.

2

Munkres, J.

Elements of

algebraic topology

, Addison-Wesley, 1984.

Title

antipodal map on

S

n

is homotopic to the identity if and only if

n

is odd

Canonical name

AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd

Date of creation

2013-03-22 15:47:33

Last modified on

2013-03-22 15:47:33

Owner

mps (409)

Last modified by

mps (409)

Numerical id

5

Author

mps (409)

Entry type

Derivation

Classification

msc 51M05

Classification

msc 15-00

Generated on Fri Feb 9 22:04:40 2018 by

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