8.5.3 The Hopf fibration

8.5.3 The Hopf fibration

We will first construct a

structure

of

H-space

on the circle

𝕊

1

, hence by

\autoref

lem:hopf-construction we will get a

fibration

over

𝕊

2

with fiber

𝕊

1

and total space

𝕊

1

*

𝕊

1

. We will then prove that this join is

equivalent

to

𝕊

3

.

Lemma 8.5.1

.

There is an H-space structure on the circle

S

1

.

Proof.

For the

base point

of the H-space structure we choose

𝖻𝖺𝗌𝖾

.
Now we need to define the

multiplication

operation

μ

:

𝕊

1

×

𝕊

1

→

𝕊

1

.
We will define the curried form

μ

~

:

𝕊

1

→

(

𝕊

1

→

𝕊

1

)

of

μ

by

recursion

on

𝕊

1

:

μ

~

(

𝖻𝖺𝗌𝖾

)

:

≡

𝗂𝖽

𝕊

1

,

and

μ

~

(

𝗅𝗈𝗈𝗉

)

:=

𝖿𝗎𝗇𝖾𝗑𝗍

(

h

)

.

where

h

:

∏

(

x

:

𝕊

1

)

(

x

=

x

)

is the

function

defined in

\autoref

thm:S1-autohtpy,
which has the property that

h

(

𝖻𝖺𝗌𝖾

)

:

≡

𝗅𝗈𝗈𝗉

.

Now we just have to prove that

μ

⁢

(

x

,

𝖻𝖺𝗌𝖾

)

=

μ

⁢

(

𝖻𝖺𝗌𝖾

,

x

)

=

x

for every

x

:

𝕊

1

.
By definition, if

x

:

𝕊

1

we have

μ

⁢

(

𝖻𝖺𝗌𝖾

,

x

)

=

μ

~

⁢

(

𝖻𝖺𝗌𝖾

)

⁢

(

x

)

=

𝗂𝖽

𝕊

1

⁢

(

x

)

=

x

. For the equality

μ

⁢

(

x

,

𝖻𝖺𝗌𝖾

)

=

x

we do it by

induction

on

x

:

𝕊

1

:

•

If

x

is

𝖻𝖺𝗌𝖾

then

μ

⁢

(

𝖻𝖺𝗌𝖾

,

𝖻𝖺𝗌𝖾

)

=

𝖻𝖺𝗌𝖾

by definition, so we
have

𝗋𝖾𝖿𝗅

𝖻𝖺𝗌𝖾

:

μ

⁢

(

𝖻𝖺𝗌𝖾

,

𝖻𝖺𝗌𝖾

)

=

𝖻𝖺𝗌𝖾

.

•

When

x

varies along

𝗅𝗈𝗈𝗉

, we need to prove that

𝗋𝖾𝖿𝗅

𝖻𝖺𝗌𝖾

⁢

\centerdot

⁢

𝖺𝗉

λ

⁢

x

.

x

⁢

(

𝗅𝗈𝗈𝗉

)

=

𝖺𝗉

λ

⁢

x

.

μ

⁢

(

x

,

𝖻𝖺𝗌𝖾

)

⁢

(

𝗅𝗈𝗈𝗉

)

⁢

\centerdot

⁢

𝗋𝖾𝖿𝗅

𝖻𝖺𝗌𝖾

.

The left-hand side is equal to

𝗅𝗈𝗈𝗉

, and for the right-hand side we have:

𝖺𝗉

λ

⁢

x

.

μ

⁢

(

x

,

𝖻𝖺𝗌𝖾

)

⁢

(

𝗅𝗈𝗈𝗉

)

⁢

\centerdot

⁢

𝗋𝖾𝖿𝗅

𝖻𝖺𝗌𝖾

=

𝖺𝗉

λ

⁢

x

.

(

μ

~

⁢

(

x

)

)

⁢

(

𝖻𝖺𝗌𝖾

)

⁢

(

𝗅𝗈𝗈𝗉

)

=

𝗁𝖺𝗉𝗉𝗅𝗒

⁢

(

𝖺𝗉

λ

⁢

x

.

(

μ

~

⁢

(

x

)

)

⁢

(

𝗅𝗈𝗈𝗉

)

,

𝖻𝖺𝗌𝖾

)

=

𝗁𝖺𝗉𝗉𝗅𝗒

⁢

(

𝖿𝗎𝗇𝖾𝗑𝗍

⁢

(

h

)

,

𝖻𝖺𝗌𝖾

)

=

h

⁢

(

𝖻𝖺𝗌𝖾

)

=

𝗅𝗈𝗈𝗉

.

∎

Now recall from

\autoref

sec:colimits that the

join

A

*

B

of types

A

and

B

is the pushout of the diagram

A

←

𝗉𝗋

1

A

×

B

→

𝗉𝗋

2

B

.

Lemma 8.5.2

.

The operation of join is

associative

: if

A

,

B

and

C

are three types
then we have an

equivalence

(

A

*

B

)

*

C

≃

A

*

(

B

*

C

)

.

Proof.

We define a map

f

:

(

A

*

B

)

*

C

→

A

*

(

B

*

C

)

by induction. We first need to define

f

∘

𝗂𝗇𝗅

:

A

*

B

→

A

*

(

B

*

C

)

which will be done by induction, then

f

∘

𝗂𝗇𝗋

:

C

→

A

*

(

B

*

C

)

, and then

𝖺𝗉

f

∘

𝗀𝗅𝗎𝖾

:

∏

(

t

:

(

A

*

B

)

×

C

)

f

⁢

(

𝗂𝗇𝗅

⁢

(

𝗉𝗋

1

⁢

(

t

)

)

)

=

f

⁢

(

𝗂𝗇𝗋

⁢

(

𝗉𝗋

2

⁢

(

t

)

)

)

which will be done by induction on the
first

component

of

t

:

(

f

∘

𝗂𝗇𝗅

)

(

𝗂𝗇𝗅

(

a

)

)

)

:

≡

𝗂𝗇𝗅

(

a

)

,

(

f

∘

𝗂𝗇𝗅

)

(

𝗂𝗇𝗋

(

b

)

)

)

:

≡

𝗂𝗇𝗋

(

𝗂𝗇𝗅

(

b

)

)

,

𝖺𝗉

f

∘

𝗂𝗇𝗅

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

a

,

b

)

)

:=

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗅

⁢

(

b

)

)

,

f

⁢

(

𝗂𝗇𝗋

⁢

(

c

)

)

:

≡

𝗂𝗇𝗋

(

𝗂𝗇𝗋

(

c

)

)

,

𝖺𝗉

f

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

𝗂𝗇𝗅

⁢

(

a

)

,

c

)

)

:=

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗋

⁢

(

c

)

)

,

𝖺𝗉

f

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

𝗂𝗇𝗋

⁢

(

b

)

,

c

)

)

:=

𝖺𝗉

𝗂𝗇𝗋

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

b

,

c

)

)

,

𝖺𝗉𝖽

λ

⁢

x

.

𝖺𝗉

f

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

x

,

c

)

)

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

a

,

b

)

)

:=

`

⁢

`

⁢

𝖺𝗉𝖽

λ

⁢

x

.

𝗀𝗅𝗎𝖾

⁢

(

a

,

x

)

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

b

,

c

)

)

′′

.

For the last

equation

, note that the right-hand side is of type

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

λ

⁢

x

.

𝗂𝗇𝗅

⁢

(

a

)

=

𝗂𝗇𝗋

⁢

(

x

)

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

b

,

c

)

,

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗅

⁢

(

b

)

)

)

=

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗋

⁢

(

c

)

)

whereas it is supposed to be of type

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

λ

⁢

x

.

f

⁢

(

𝗂𝗇𝗅

⁢

(

x

)

)

=

f

⁢

(

𝗂𝗇𝗋

⁢

(

c

)

)

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

a

,

b

)

,

𝖺𝗉

f

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

𝗂𝗇𝗅

⁢

(

a

)

,

c

)

)

)

=

𝖺𝗉

f

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

𝗂𝗇𝗋

⁢

(

b

)

,

c

)

)

.

But by the previous clauses in the definition, both of these types are equivalent to the following type:

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗋

⁢

(

c

)

)

=

𝗀𝗅𝗎𝖾

⁢

(

a

,

𝗂𝗇𝗅

⁢

(

b

)

)

⁢

\centerdot

⁢

𝖺𝗉

𝗂𝗇𝗋

⁢

(

𝗀𝗅𝗎𝖾

⁢

(

b

,

c

)

)

,

and so we can coerce by an equivalence to obtain the necessary

element

.
Similarly, we can define a map

g

:

A

*

(

B

*

C

)

→

(

A

*

B

)

*

C

, and checking that

f

and

g

are

inverse

to each other is a long and tedious but essentially
straightforward computation.
∎

A more conceptual proof sketch is as follows.

Proof.

Let us consider the following diagram where the maps are the obvious
projections:

\xymatrix

⁢

A

⁢

&

⁢

A

×

C

⁢

\ar

⁢

[

l

]

⁢

\ar

⁢

[

r

]

⁢

&

⁢

A

×

C

⁢

A

×

B

⁢

\ar

⁢

[

u

]

⁢

\ar

⁢

[

d

]

⁢

&

⁢

A

×

B

×

C

⁢

\ar

⁢

[

l

]

⁢

\ar

⁢

[

u

]

⁢

\ar

⁢

[

r

]

⁢

\ar

⁢

[

d

]

⁢

&

⁢

A

×

C

⁢

\ar

⁢

[

u

]

⁢

\ar

⁢

[

d

]

⁢

B

⁢

&

⁢

B

×

C

⁢

\ar

⁢

[

l

]

⁢

\ar

⁢

[

r

]

⁢

&

⁢

C

Taking the colimit of the columns gives the following
diagram, whose colimit is

(

A

*

B

)

*

C

:

\xymatrix

⁢

A

*

B

⁢

&

⁢

(

A

*

B

)

×

C

⁢

\ar

⁢

[

l

]

⁢

\ar

⁢

[

r

]

⁢

&

⁢

C

On the other hand, taking the colimit of the lines gives a diagram whose
colimit is

A

*

(

B

*

C

)

.

Hence using a Fubini-like

theorem

for colimits (that we havenÃ¢â¬â¢t proved) we
have an equivalence

(

A

*

B

)

*

C

≃

A

*

(

B

*

C

)

. The proof of this Fubini theorem

for colimits still requires the long and tedious computation, though.
∎

Lemma 8.5.3

.

For any type

A

, there is an equivalence

Σ

⁢

A

≃

2

*

A

.

Proof.

It is easy to define the two maps

back and forth

and to prove that they are
inverse to each other. The details are left as an exercise to the reader.
∎

We can now construct the

Hopf fibration

:

Theorem 8.5.4

.

There is a fibration over

S

2

of fiber

S

1

and total space

S

3

.

Proof.

We proved that

𝕊

1

has a structure of H-space (cf

\autoref

lem:hspace-S1)
hence by

\autoref

lem:hopf-construction there is a fibration over

𝕊

2

of
fiber

𝕊

1

and total space

𝕊

1

*

𝕊

1

. But by the two previous results
and

\autoref

thm:suspbool we have:

𝕊

1

*

𝕊

1

=

(

Σ

⁢

𝟐

)

*

𝕊

1

=

(

𝟐

*

𝟐

)

*

𝕊

1

=

𝟐

*

(

𝟐

*

𝕊

1

)

=

Σ

⁢

(

Σ

⁢

𝕊

1

)

=

𝕊

3

.

∎

Title

8.5.3 The Hopf fibration

\metatable

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