sectional curvature determines Riemann curvature tensor

sectional curvature determines Riemann curvature tensor

Theorem 1

.

The

sectional curvature

operator

Π

↦

K

⁢

(

Π

)

completely determines the

Riemann curvature tensor

.

In fact, a more general result is true. Recall the Riemann

(

1

,

3

)

-curvature tensor

R

:

T

⁢

M

⊗

T

⁢

M

⊗

T

⁢

M

→

T

⁢

M

satisfies

(

x

,

y

,

z

,

t

)

+

(

y

,

z

,

x

,

t

)

+

(

z

,

x

,

y

,

t

)

=

0

First Bianchi identity

(1)

(

x

,

y

,

z

,

t

)

+

(

y

,

x

,

z

,

t

)

=

0

(2)

(

x

,

y

,

z

,

t

)

-

(

z

,

t

,

x

,

y

)

=

0

,

(3)

where

(

x

,

y

,

z

,

t

)

:=

g

⁢

(

R

⁢

(

x

,

y

,

z

)

,

t

)

, and the sectional curvature is defined by

K

(

Π

=

span

{

x

,

y

}

)

=

g

⁢

(

R

⁢

(

x

,

y

,

x

)

,

y

)

g

⁢

(

x

,

x

)

⁢

g

⁢

(

y

,

y

)

-

g

⁢

(

x

,

y

)

2

.

(4)

Thus Theorem

1

is implied by

Theorem 2

.

Let

V

be a real

inner product space

, with

inner product

⟨

-

,

-

⟩

. Let

R

and

R

′

be

linear maps

V

⊗

3

→

V

. Suppose

R

and

R

′

satisfies

•

Equations (

1

),(

2

),(

3

), and

•

K

⁢

(

σ

)

=

K

′

⁢

(

σ

)

for all

2

-planes

σ

, where

K

,

K

′

are defined by (

4

) using

⟨

-

,

-

⟩

in of

g

⁢

(

-

,

-

)

.

Then

R

=

R

′

.

Write

(

x

,

y

,

z

,

t

)

:=

⟨

R

⁢

(

x

,

y

,

z

)

,

t

⟩

(

x

,

y

,

z

,

t

)

′

:=

⟨

R

′

⁢

(

x

,

y

,

z

)

,

t

⟩

.

Proof of Theorem

2

.

We need to prove, for all

x

,

y

,

z

,

t

∈

V

,

(

x

,

y

,

z

,

t

)

=

(

x

,

y

,

z

,

t

)

′

.

From

K

=

K

′

, we get

(

x

,

y

,

x

,

y

)

=

(

x

,

y

,

x

,

y

)

′

for all

x

,

y

∈

V

. The first step is to use

polarization identity

to change this

quadratic form

(in

x

) into its associated

symmetric bilinear form

. Expand

(

x

+

z

,

y

,

x

+

z

,

y

)

=

(

x

+

z

,

y

,

x

+

z

,

y

)

′

and use (

3

), we get

(

x

,

y

,

x

,

y

)

+

2

⁢

(

x

,

y

,

z

,

y

)

+

(

z

,

y

,

z

,

y

)

=

(

x

,

y

,

x

,

y

)

′

+

2

⁢

(

x

,

y

,

z

,

y

)

′

+

(

z

,

y

,

z

,

y

)

′

.

So

(

x

,

y

,

z

,

y

)

=

(

x

,

y

,

z

,

y

)

′

for all

x

,

y

,

z

∈

V

.

Unfortunately, the form

(

x

,

y

,

z

,

t

)

is not

symmetric

in

y

and

t

, so we need to work harder. Expand

(

x

,

y

+

t

,

z

,

y

+

t

)

=

(

x

,

y

+

t

,

z

,

y

+

t

)

′

, we get

(

x

,

y

,

z

,

t

)

+

(

x

,

t

,

z

,

y

)

=

(

x

,

y

,

z

,

t

)

′

+

(

x

,

t

,

z

,

y

)

′

.

Now use (

2

) and (

3

), we get

(

x

,

y

,

z

,

t

)

-

(

x

,

y

,

z

,

t

)

′

=

(

x

,

t

,

z

,

y

)

′

-

(

x

,

t

,

z

,

y

)

=

(

z

,

y

,

x

,

t

)

′

-

(

z

,

y

,

x

,

t

)

=

(

y

,

z

,

x

,

t

)

-

(

y

,

z

,

x

,

t

)

′

.

So

(

x

,

y

,

z

,

t

)

-

(

x

,

y

,

z

,

t

)

′

is

invariant

under

cyclic permutation

of

x

,

y

,

z

. But the cyclic sum is zero by (

1

). So

(

x

,

y

,

z

,

t

)

=

(

x

,

y

,

z

,

t

)

′

∀

x

,

y

,

z

,

t

∈

V

as desired.
∎

Title

sectional curvature determines Riemann curvature tensor

Canonical name

SectionalCurvatureDeterminesRiemannCurvatureTensor

Date of creation

2013-03-22 15:55:09

Last modified on

2013-03-22 15:55:09

Owner

kerwinhui (11200)

Last modified by

kerwinhui (11200)

Numerical id

10

Author

kerwinhui (11200)

Entry type

Theorem

Classification

msc 53B21

Classification

msc 53B20

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