HT AREA AND PURE COMPLEX CROFTON MEASURE FOR COMPLEX

L1

SPACE

LOUIS YANG LIU Abstract.

We nd a discrete Crofton measure on the space of ane lines

Gr1 (R2 ) in real L1 plane, and show that there is no pure complex Crofton measure for complex L1 space.

1. Discrete Crofton Measure for Real

2

Let (R , || · ||1 ) (R2 , F ), we know,

be the

1

L

-plane.

For any

L1

P1 := (x1 , y1 )

Plane

and

P2 := (x2 , y2 )

||P1 P2 ||1 = ||P1 Q||1 + ||QP2 ||1 , where

in

obviously, that

Q := (x2 , y1 ).

(1.1)

Furthermore, we have the following lemma.

P1 := (x1 , y1 ) and P2 := (x2 , y2 ) are distinct two arbitrary (R2 , || · ||1 ), then all rectiable monotonic curves joining P1 and P2 have 1 2 the same L - length, here a curve C(t) : [a, b] → R , C(t) = (x(t), y(t)) is called monotone if both x(t) and y(t) are monotonic functions.

Lemma 1.1. Suppose

points in

Proof. Since the length of a rectiable curve is the least upper bound of the length



of inscribed polygonal path and (1.1), then the lemma follows.

First, let us explore some properties of this space. In Figure 1.1 on page 2, the curve

C

is a rectiable monotonic curve , so it has the same

straight line segment

P 1 P2 .

L1

-length with the

However, one always has

#(l ∩ P1 P2 ) 6 #(l ∩ C) for all

l ∈ Gr1 (R2 )

but

l

coinciding

Additionally, if one rotates

√

0<

P1 P2

(1.2)

P1 P 2 . by angle

α,

denoted as

Rα (P1 P2 ),

√ 2 ||P1 P2 ||1 6 ||Rα (P1 P2 )||1 6 2||P1 P2 ||1 . 2

then (1.3)

Now, let us prove the following theorem. Theorem 1.2. The Crofton measure is supported on the set of lines which are

horizontal or vertical in

L1

plane.

Proof. Assume that there is a Crofton measure

ˆ

||P1 P2 ||1 = #(l ∩ P1 P2 ) 6 #(l ∩ C)

Date : November 4, 2008.

a.e.

length, in other words,

#(l ∩ P1 P2 )dµ(l), l∈Gr1 (R2 )

then

µ for the L1

l ∈ Gr1 (R2 ). 1

(1.4)

2

LOUIS YANG LIU

Figure 1.1. Paths Intersecting with Ane Lines

P1

Choose

to be the origin and

P2 = (0, 1)

on the

y -axis,

let

I := P1 P2 ,

and

dene

in which

θ

is the angle of

ϕ : Gr1 (R2 ) → S 1 × R ϕ(l) = (θ, r) l to x-axis and r is the euclidean

(1.5) distance of

l

to the

origin. Then by (1.4),

o n (ϕ∗ (µ))([0, π] × [0, 1]) = µ( l ∈ Gr1 (R2 ) : #(l ∩ I) > 0 ) = 1. Moreover, if one chooses

(0, r sin φ),

P1

to be the origin,

P2 = (r cos φ, r sin φ)

(1.6) and

n o ϕ( l ∈ Gr1 (R2 ) : #(l ∩ (P1 Q ∪ QP2 )) = 2 ) o n h π ) 6 θ 6 : h ∈ [0, r sin φ] = ( π2 − φ)(1 − r sin φ 2 for

φ ∈ (0, π2 )

(1.7)

and

n o ϕ( l ∈ Gr1 (R2 ) : #(l ∩ (P1 Q ∪ QP2 )) = 2 ) n o π−φ π = 6 θ 6 φ + h : h ∈ [0, r sin φ] 2 r sin φ for

Q =

see Figure 1.2 on page 3, then

(1.8)

φ ∈ ( π2 , π). On the other hand, since

P1 Q ∪ QP2

has the same length with

P1 P2 ,

then, by

(1.2) and (1.4), we have

n o µ( l ∈ Gr1 (R2 ) : #(l ∩ (P1 Q ∪ QP2 )) = 2 ) = 0.

(1.9)

HT AREA AND PURE COMPLEX CROFTON MEASURE FOR COMPLEX L1 SPACE

3

Figure 1.2. Lines with Two Intersection Points

Therefore, by (1.7) and (1.8),

  π h π (ϕ∗ (µ))( ( − φ)(1 − ) 6 θ 6 : h ∈ [0, r sin φ] ) = 0 2 r sin φ 2 for

φ ∈ (0, π2 )

and

  π π−φ (ϕ∗ (µ))( 6θ 6φ+ h : h ∈ [0, r sin φ] ) = 0 2 r sin φ for

(1.10)

φ ∈ ( π2 , π).

By taking

r=

1 φ and

φ→0

in (1.10), and

r=

1 π−φ and

(1.11)

φ→π

in

(1.11), one obtains that

(ϕ∗ (µ))((0,

π π ) ∪ ( , π)) × [0, 1]) = 0. 2 2

(1.12)

4

LOUIS YANG LIU

By rescaling

P1 P2 ,

we conclude that

π π ) ∪ ( , π)) × R) = 0, (1.13) 2 2 π thus ϕ∗ (µ) is supported on {0} × R and 2 × R, which implies that µ is supported 1 on the set of lines which are horizontal or vertical in L plane.  (ϕ∗ (µ))((0,

Let's dene a discrete measure supported on

( (δx + δy )(l) :=

1 0

if

l

coincides

x-axis

and

x-axis

or

y -axis

by

y -axis,

(1.14)

otherwise.

We have the following theorem

L1

Theorem 1.3. The Crofton measure for real

Proof. For any straight line segment

P 1 P1

in

1

L

plane is

δx + δy

.

plane, for which one can see Figure

1.1 on page 2,Since

Length(P1 P1 ) = |πy (P1 P1 )| + |πx (P1 P1 )|,

(1.15)

ˆ

then

#(l ∩ P1 P1 )d(δx + δy ),

Length(P1 P1 ) =

(1.16)

Gr1 (R2 )



and so the claim follows. 2. Crofton Measure on Suppose E is contained in ||(z, w)|| = |z| + |w|, and µ is 2 for HT , i.e.

a complex line

L

Gr2 (R4 ) in

C2

with complex

L1

norm

the Crofton measure, for which one can see [BF],

ˆ HT 2 (E) =

vol2 (πP (E))dµ.

(2.1)

Gr2 (R4 ) From the intersection map

π : Gr3 (R4 ) × Gr3 (R4 ) \ 4 → Gr2 (R4 ), as we know,

in

µ

is supported on

{C × {0}} ∪ {{0} × C} ∪ TAB ,

(2.2) where

TAB := {span(vA , vB ) : vA ∈ C × {0} , vB ∈ {0} × C} . (2.3) Since TAB ∼ = T2 , where T2 := SA × SB in which SA and SB are the unit circles C × {0} and {0} × C respectively, then in terms of the push-forward measure by

(2.2), one has that

´

TAB

vol(πP (E))dµ =

However,

´ (θA ,θB )∈T2

vol(πspan(θA ,θB ) (E))

vol2 (πspan(θA ,θB ) (E))dθA dθB .

doesn't vanish for most

(θA , θB ) ∈ T2 ,

(2.4) so the

integrals in (2.4)doesn't vanish. To see an example, we choose

  L = (z, w) ∈ C2 : z = w = (x, y, x, y) ∈ R4 E to be a real 2-disk in L. For vA = (x1 , x2 , 0, 0) ⊥ ⊥ vA := (−x2 , x1 , 0, 0) and vB := (0, 0, −x4 , x3 ), then

and

and

⊥ ⊥ R4 = span(vA , vB ) ⊕ span(vA , vB ).

(2.5)

vB = (0, 0, v3 , v4 ),

let

(2.6)

HT AREA AND PURE COMPLEX CROFTON MEASURE FOR COMPLEX L1 SPACE

5

Therefore

(x, y, x, y) = (

xx1 + yx2 xx3 + yx4 yx1 − xx2 ⊥ yx3 − xx4 ⊥ vA + 2 vB )+( 2 v + 2 v ) 2 2 2 x1 + x2 x3 + x4 x1 + x22 A x3 + x24 B

and so

xx1 + yx2 xx3 + yx4 vA + 2 vB x21 + x22 x3 + x24

πspan(vA ,vB ) ((x, y, x, y)) = for any

(x, y, x, y) ∈ L.

Thus

πspan(vA ,vB ) (L) = span(vA , vB )   x1 x2 det 6= 0. x3 x4

Therefore it is easy to see that

vol2 (πspan(θA ,θB ) (E)) 6= 0

(2.7)

(2.8)

as long as (2.9)

provided (2.9) that holds

span(vA , vB ) ∈ TAB .

for most

3. Crofton Measure on

Gr1 (C2 )

2 Lθ := span((cos θ, sin θ)√C ) be a complex line in C , √ spanned by (cos θ, sin θ) and ( −1 cos θ, −1 sin θ), then Let

and

U

be the rectangle



area(πspan(θA ,θB ) (U ))



cos θ cos θA sin θ cos θB cos θ sin θA sin θ sin θB = | sin θ cos θ sin(θB − θA )|, = |det

|

(3.1)

ˆ

therefore

area(πspan(θA ,θB ) (U ))dθA dθB = | sin θ cos θ|.

(3.2)

(θA ,θB )∈T2 √ On the other hand, for any complex line

Lφ,ψ := span((cos φ, e

−1ψ

C2 ,

sin φ)C )

area(π  Lφ,ψ (U ))  cos θ cos φ + sin θ sin φ cos ψ − sin θ sin φ sin ψ = |det | sin θ sin φ sin ψ cos θ cos φ + sin θ sin φ cos ψ = cos2 θ cos2 φ + sin2 θ sin2 φ + 2 sin θ cos θ sin φ cos φ cos ψ.

in

(3.3)

As we know, the problem on the existence of Crofton measure on complex lines for area becomes whether there exists some function

ˆ

2π

ˆ

such that

ˆ

2π

area(πLφ,ψ (U ))f (φ)dφdψ = 0

f (φ),

area(πspan(θA ,θB ) (U ))dθA dθB .

0 (θA ,θB

)∈T2 (3.4)

But from (3.3),

ˆ

2π

ˆ

ˆ

2π

0

0

2π

(cos2 θ cos2 φ + sin2 θ sin2 φ)f (φ)dφ,

area(πLφ,ψ (U ))f (φ)dφdψ = 0

(3.5) thus by (3.2) we need to have

ˆ

2π

(cos2 θ cos2 φ + sin2 θ sin2 φ)f (φ)dφ = | sin θ cos θ|. 0

(3.6)

6

LOUIS YANG LIU

Taking

θ=0

π 2 in (3.6), we have ˆ 2π 2

and

ˆ

which implies

0

´ 2π

However, let

2π

sin2 φf (φ)dφ = 0,

cos φf (φ)dφ =

(3.7)

0

f (φ)dφ = 0. θ ∈ (0, π2 ) and

0

ˆ

one has

take the derivative with respect to

θ

in (3.6), then

2π

(− sin 2θ cos 2φ)f (φ)dφ = cos 2θ, which implies that

´ 2π 0

(3.8)

0

f (φ) cos 2φdφ = − cot 2θ,

that is a contradiction.

So we obtain the following Theorem 3.1. There is no Crofton measure which is only supported on complex

lines for complex

L1

space.

Acknowledgement. Thanks to Dr. Joseph H. G. Fu for his illuminating questions, helpful advices and some corrections. References

[BF] A. Bernig, J. H. G. Fu, Convolution of convex valuations, Geometriae Dedicata, 2006, Springer