Table Of ContentGeneralized Timelike Mannheim Curves in
1 Minkowski space-time E4
1 1
0
2
M. AKYI˙GˇI˙T, S. ERSOY, I˙. O¨ZGU¨R, M. TOSUN
n
a
J
9 Department of Mathematics, Faculty of Arts and Sciences
1
Sakarya University, 54187 Sakarya/TURKEY
]
G Abstract
D WegiveadefinitionofgeneralizedtimelikeMannheimcurveinMinkowski
. space-time E14. The necessary and sufficient conditions for the general-
h izedtimelikeMannheimcurveobtain. Weshowsomecharacterizationsof
t
a generalized Mannheim curve.
m MathematicsSubjectClassification(2010): 53B30,53A35,53A04.
Keywords: Mannheim curve, Minkowski space-time.
[
1
v
1 Introduction
6
8
6
The geometry of curves has long captivated the interests of mathematicians,
3
from the ancient Greeks through to the era of Isaac Newton (1647-1727) and
.
1 the invention of the calculus. It is branch of geometry that deals with smooth
0
curves in the plane and in the space by methods of differential and integral
1
calculus. The theory of curves is the simpler and narrower in scope because
1
: a regular curve in a Euclidean space has no intrinsic geometry. One of the
v
mostimportanttoolsusedtoanalyzecurveistheFrenetframe,amovingframe
i
X thatprovidesacoordinatesystemateachpoint ofcurvethatis ”bestadopted”
r to the curve near that point. Every person of classical differential geometry
a
meets early in his course the subject of Bertrand curves, discovered in 1850 by
J. Bertrand. A Bertrand curve is a curve such that its principal normals are
the principal normals of a second curve. There are many works related with
Bertrand curves in the Euclidean space and Minkowski space, [1]-[3].
AnotherkindofassociatedcurveiscalledMannheimcurveandMannheimpart-
ner curve. The notion of Mannheim curves was discovered by A. Mannheim in
1878. These curves in Euclidean 3-space are characterized in terms of the cur-
vature and torsionas follows: A space curveis a Mannheim curve if andonly if
its curvature κ and torsion τ satisfy the relation
k =β k2+k2
1 1 2
(cid:0) (cid:1)
1
for some constant β. The articles concerning Mannheim curves are rather few.
In [4], a remarkable class of Mannheim curves is studied. General Mannheim
curves in the Euclidean 3-space are obtained in [5]-[7] Recently, Mannheim
curves are generalized and some characterizations and examples of generalized
Mannheim curves are given in Euclidean 4-space E4 by [8].
In this paper, we study the generalized spacelike Mannheim partner curves in
4 dimensionalMinkowskispace-time. We willgive the necessaryandsufficient
−
conditions for the generalized spacelike Mannheim partner curves.
2 Preliminaries
To meet the requirements in the next sections,the basic elements of the theory
of curves in Minkowski space-time E4 are briefly presented in this section. A
1
more complete elementary treatment can be found in [9].
Minkowskispace-timeE4 isanEuclideanspaceprovidedwiththe standardflat
1
metric given by
, = dx2+dx2+dx2+dx2
h i − 1 2 3 4
where (x , x , x , x ) is a rectangular coordinate system in E4.
1 2 3 4 1
Since , is anindefinite metric, recallthat a v E4 canhaveone ofthe three
causalhchiaracters;itcanbespacelikeif v,v >0∈orv1 =0,timelikeif v,v <0
andnull(ligthlike)if v,v =0andv=h0. Siimilarly,anarbitrarycurvehc=ic(t)
h i 6
in E4 can locally be spacelike, timelike or null (lightlike) if all of its velocity
1
vectorsc′(t)are,respectively,spacelike,timelikeornull. Thenormofv E4 is
givenby v = v,v . If c′(t) = c′(t),c′(t) =0forallt I,∈then1C
k k |h i| k k |h i|6 ∈
isaregularcurveinE4. Atimelike(spacelike)regularcurveC isparameterized
p 1 p
byarc-lengthparametertwhichisgivenbyc:I E4,thenthetangentvector
c′(t)alongC hasunit length,thatis, c(t),c(t)→=11, ( c(t),c(t) = 1)for
h i h i −
all t I.
∈
Hereafter, curves considered are timelike and regular C∞ curves in E4. Let
1
T (t)= c′(t) for all t I, then the vector field T (t) is timelike and it is called
∈
timelike unit tangent vector field on C.
The timelike curve C is called special timelike Frenet curve if there exist three
smoothfunctionsk , k , k onCandsmoothnon-nullframefield T, N, B , B
1 2 3 1 2
{ }
along the curve C. Also, the functions k , k and k are called the first, the
1 2 3
secondand the third curvature function on C, respectively. For the C∞ special
timelike Frenet curve C, the following Frenet formula is
T′ 0 k 0 0 T
1
N′ k 0 k 0 N
B ′ = 01 k 02 k B
1 2 3 1
B2′ 0 −0 −k3 0 B2
[9].
2
Here, due to characters of Frenet vectors of the timelike curve, T, N, B and
1
B are mutually orthogonalvector fields satisfying equations
2
T , T = 1 , N, N = B , B = B , B =1 .
1 1 2 2
h i − h i h i h i
For t I, the non-null frame field T, N, B , B and curvature functions
1 2
∈ { }
k , k and k are determined as follows
1 2 3
1st step T(t)=c′(t)
2nd step k (t)= T′(t) >0
1
k k
N(t)= 1 T′(t)
k1(t)
3rd step k (t)= N′(t) k (t)T (t) >0
2 1
k − k
B (t)= 1 (N′(t) k (t)T (t))
1 k2(t) − 1
4th step B (t)=ε 1 B ′(t)+k (t)N(t)
2 kB1′(t)+k2(t)N(t)k 1 2
where ε is determined by the fact that orthono(cid:0)rmal frame field T(cid:1)(t), N(t),
{
B (t), B (t) , is of positive orientation. The function k is determined by
1 2 3
}
k (t)= B ′(t) , B (t) =0.
3 1 2
6
So the function k3 never vanishes(cid:10). (cid:11)
Inorderto makesurethat the curve C is a specialtimelike Frenetcurve,above
steps must be checked, from 1st step to 4th step, for t I.
∈
Let T, N, B , B be the moving Frenet frame along a unit speed timelike
1 2
{ }
curveC inE4,consistingofthetangent,theprincipalnormal,thefirstbinormal
1
and the second binormal vector field, respectively. Since C is a timelike curve,
its Frenet frame contains only non-null vector fields.
3 Generalized timelike Mannheim curves in E4
1
Mannheim curves are generalized in by [8]. In this paper, we have investigated
generalizationof timelike Mannheim curves in Minkowski space E4.
1
Definition 3.1 AspecialtimelikecurveC inE4 isageneralizedtimelikeMannheim
1
curve if there exists a special timelike Frenet curve C∗ in E4 such that the first
1
normal line at each point of C is included in the plane generated by the second
normal line and the third normal line of C∗ at the corresponding point under
φ. Here φ is a bijection from C to C∗. The curve C∗ is called the generalized
timelike Mannheim mate curve of C.
By the definition, a generalized Mannheim mate curve C∗ is given by the map
c∗ :I∗ E4 such that
→ 1
c∗(t)=c(t)+β(t)N(t), t I. (3.1)
∈
Hereβ isasmoothfunctiononI.Generally,the parametertisn’tanarc-length
of C∗. Let t∗ be the arc-length of C∗ defined by
3
t
dc∗(t)
t∗ = dt.
dt
Z (cid:13) (cid:13)
0 (cid:13) (cid:13)
If a smooth function f :I I∗ is giv(cid:13)(cid:13)en by f(cid:13)(cid:13)(t)=t∗, then for t I, we have
→ ∀ ∈
dt∗ dc∗(t)
f′(t)= = = (1+β(t)k (t))2+(β′(t))2+(β(t)k (t))2 .
dt dt − 1 2
(cid:13) (cid:13) r(cid:12) (cid:12)
The representati(cid:13)(cid:13)on of tim(cid:13)(cid:13)elike(cid:12)(cid:12)curve C∗ with arc-length parameter t∗ is (cid:12)(cid:12)
(cid:13) (cid:13)
c∗ : I∗ E4
t∗ → c∗1(t∗).
→
For a bijection φ : C C∗ defined by φ(c(t)) = c∗(f(t)), the reparameteri-
→
zation of C∗ is
c∗(f(t))=c(t)+β(t)N(t)
where β is a smooth function on I. Thus, we have
dc∗(f(t)) dc∗(t∗)
= f′(t)=f′(t)T∗(f(t)), t I.
dt dt ∈
(cid:12)t∗=f(t)
(cid:12)
Theorem 3.1 If a special time(cid:12)(cid:12)like Frenet curve C in E14 is a generalized time-
like Mannheim curve, then the following relation between the first curvature
function k and the second curvature function k holds:
1 2
k (t)= β k2(t) k2(t) , t I (3.2)
1 − 1 − 2 ∈
where β is a constant number. (cid:0) (cid:1)
Proof Let C be a generalized timelike Mannheim curve and C∗ be the gener-
alized timelike Mannheim mate curve of C, as following diagram
c c∗
·· ··
f : I I∗
→
↓ ↓
φ : E4 E4
1 → 1
A smooth function h is defined by f(t) = dc∗(t) dt = t∗ and t∗ is the arc-
dt
length parameter of C∗. Also φ is a bijectio(cid:13)n whic(cid:13)h is defined by φ(c(t)) =
R (cid:13) (cid:13)
c∗(f(t)). Thus, the timelike curve C∗ is repa(cid:13)rametr(cid:13)izedas follows
c∗(f(t))=c(t)+β(t)N(t) (3.3)
where β : I R R is a smooth function. By differentiating both sides of
⊂ →
equation (3.3) with respect to t, we have
f′(t)T∗(f(t))=(1+β(t)k (t))T +β′(t)N(t)+β(t)k (t)B (t). (3.4)
1 2 1
4
On the other hand, since the first normalline at the each point of C is lying in
the plane generated by the second normal line and the third normal line of C∗
at the corresponding points under bijection φ, the vector field N(t) is given by
N(t)=g(t)B ∗(f(t))+h(t)B ∗(f(t))
1 2
wheregandharesomesmoothfunctionsonI R. Ifwetakeintoconsideration
⊂
T∗(f(t)), g(t)B ∗(f(t))+h(t)B ∗(f(t)) =0
1 2
h i
andthe equation(3.4),thenwe haveβ′(t)=0Sowerewritethe equation(3.4)
as
f′(t)T∗(f(t))=(1+βk (t))T (t)+βk (t)B (t), (3.5)
1 2 1
that is,
(1+βk (t)) βk (t)
T∗(f(t))= 1 T (t)+ 2 B (t)
f′(t) f′(t) 1
where
f′(t)= (1+βk (t))2+(βk (t))2 .
1 2
−
r
(cid:12) (cid:12)
Bytaking differentiationboth(cid:12)sides ofthe equations(3.5)w(cid:12) ithrespecttot I,
(cid:12) (cid:12) ∈
we get
′
f′(t)k∗(f(t))N∗(f(t))= 1+βk1(t) T(t)
1 f′(t)
+ (1+βk1(cid:16)(t))k1(t)−β(cid:17)(k2(t))2 N(t) (3.6)
f′(t)
′
+(cid:16)βk2(t) B (t)+ βk2(t)(cid:17)k3(t) B (t).
f′(t) 1 f′(t) 2
(cid:16) (cid:17) (cid:16) (cid:17)
Since
N∗(f(t)), g(t)B ∗(f(t))+h(t)B ∗(f(t)) =0.
1 2
h i
The coefficient of N(t) in equation (3.6) vanishes, that is,
(1+βk (t))k (t) β(k (t))2 =0.
1 1 2
−
Thus, this completes the proof.
Theorem 3.2 In E4, let C be a special timelike Frenet curve such that its
1
non-constant first and second curvature functions satisfy the equality k (s) =
1
β k2(t) k2(t) for all t I R. If the timelike curve C∗ given by
− 1 − 2 ∈ ⊂
(cid:0) (cid:1) c∗(t)=c(t)+βN(t)
is a special timelike Frenet curve, then C∗ is a generalized timelike Mannheim
mate curve of C.
5
Proof The arc-length parameter of C∗ is given by
t
dc∗(t)
t∗ = dt , t I.
dt ∈
Z (cid:13) (cid:13)
0 (cid:13) (cid:13)
(cid:13) (cid:13)
Under the assumption of (cid:13) (cid:13)
k (t)= β k2(t) k2(t) ,
1 − 1 − 2
we obtain f′(t)= 1+βk (t) , t (cid:0)I. (cid:1)
1
Differentiating the eq|uation c∗(|f(t)∈)= c(t)+βN(t) with respect to t the we
p
reach
f′(t)T∗(f(t))=(1+βk (t))T (t)+βk (t)B (t).
1 2 1
Thus, it is seen that
1+βk (t) βk (t)
T∗(f(t))= 1 T (t)+ 2 B (t) , t I. (3.7)
1
1+βk1(t) 1+βk1(t) ! ∈
| | | |
The differentiationpof the last equation wpith respect to t is
′
f′(t)k∗(f(t))N∗(f(t))= 1+βk (t) T(t)
1 | 1 |
+ (1+βk√1((cid:16)|t1)p+)kβ1k(1t)(−t)β|k22(t) N(cid:17) (t) (3.8)
(cid:18) ′ (cid:19)
+ βk2(t) B (t)+ βk2(t)k3(t) B (t).
√|1+βk1(t)| 1 √|1+βk1(t)| 2
(cid:18) (cid:19) (cid:18) (cid:19)
From our assumption, we have
k (t)+βk2(t) βk2(t)
1 1 − 2 =0.
1+βk (t)
1
| |
Thus, the coefficient of N(t)pin the equation (3.8) is zero. It is seen from
the equation (3.7), T∗(f(t)) is a linear combination of T (t) and B (t). Ad-
1
ditionally, from equation (3.8), N∗(f(t)) is given by linear combination of
T (t), B (t) and B (t). On the otherhand, C∗ is a special timelike Frenet
1 2
curve that the vector N(t) is given by linear combination of T∗(f(t)) and
N∗(f(t)).
Therefore, the first normal line C lies in the plane generated by the second
normal line and third normal line of C∗ at the corresponding points under a
bijection φ which is defined by φ(c(t))=c∗(f(t)).
This,completes the proof.
Remark 3.1 In 4-diemsional Minkowski space E4, a special timelike Frenet
1
curveC withcurvaturefunctionsk andk satisfyingk (t)= β k2(t) k2(t) ,
1 2 1 − 1 − 2
it is not clear that a smooth timelike curve C∗ given by (3.1) is a special Frenet
(cid:0) (cid:1)
curve. Thus, it is unknown whether the reverse of Theorem 3.1 is true or false.
6
Theorem 3.3 Let C be a special timelike curve in E4 with non-zero third cur-
1
vature function k . If there exists a timelike special Frenet curve C∗ in E4 such
3 1
that the first normal line of C is linearly dependent with the third normal line
of C∗ at the corresponding points c(t) and c∗(t), respectively, under a bijection
φ:C C∗, iff the curvatures k and k of C are constant functions.
1 2
→
Proof Let C be a timelike Frenet curve in E4 with the Frenet frame field
1
T, N,B , B andcurvaturefunctionsk , k andk . Also,weassumethatC∗
1 2 1 2 3
b{eatimelikesp}ecialFrenetcurveinE4withtheFrenetframefield T∗, N∗, B ∗,
B ∗ and curvature functions k∗, k∗1and k∗. Let the first norma{l line of C 1be
2 } 1 2 3
linearlydependentwiththe thirdnormallineofC∗ atthe correspondingpoints
C and C∗, respectively. Then the parameterization of C∗ is
c∗(f(t))=c(t)+β(t)N(t), t I. (3.9)
∈
If the arc-length parameter of C∗ is given t∗, then
t
t∗ = (1+β(t)k (t))2+(β′(t))+(β(t)k (t))2 dt (3.10)
1 2
−
Z0 r(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
and
f : I I∗
→
t f(t)=t∗.
→
Moreover, φ:C C∗ is a bijection given by φ(c(t))=c∗(f(t)).
→
Differentiating the equation (3.9) with respect to t and using Frenet formulas,
we get
f′(t)T∗(f(t))=(1+β(t)k (t))T(t)+β′(t)N(t)
1 (3.11)
+β(t)k (t)B (t).
2 1
Since B ∗(f(t))= N(t), then
2
∓
(1+β(t)k (t))T (t)+β′(t)N(t)
f′(t)T∗(f(t)), B ∗(f(t)) = 1 ,
h 2 i +β(t)k2(t)B1(t), N(t)
(cid:28) ∓ (cid:29)
that is,
0= β′(t).
∓
From last equation, it is easily seen that β is a constant. Hereafter, we can
denote β(t)=β, for all t I.
∈
From the equation (3.10), we have
f′(t)= (1+βk (t))2+(βk (t))2 >0.
1 2
−
r
(cid:12) (cid:12)
(cid:12) (cid:12)
Thus, we rewrite the equatio(cid:12)n (3.11) as follows; (cid:12)
1+βk (t) βk (t)
T∗(f(t))= 1 T (t)+ 2 B (t).
f′(t) f′(t) 1
(cid:18) (cid:19) (cid:18) (cid:19)
7
The differentiation of the last equation with respect to t is
′
f′(t)k∗(f(t))N∗(f(t))= 1+βk1(t) T (t)
1 f′(t)
+ (1+(cid:16)βk1(t))k1(t(cid:17))−βk22(t) N(t) (3.12)
f′(t)
′
+(cid:16)βk2(t) B (t)+ βk2(cid:17)(t)k3(t) B (t).
f′(t) 1 f′(t) 2
(cid:16) (cid:17) (cid:16) (cid:17)
Since f′(t)k∗(f(t))N∗(f(t)), B ∗(f(t)) = 0 and B ∗(f(t)) = N(t) for
h 1 2 i 2 ∓
all t I, we obtain
∈
k (t)+βk2(t) βk2(t)=0
1 1 − 2
is satisfied. Then
k (t)
β = 1 (3.13)
−k2(t) k2(t)
1 − 2
is a non-zero constant number. Thus, from the equation (3.12), we reach
′ ′
N∗(f(t))= 1 1+βk1(t) T (t)+ 1 βk2(t) B (t)
f′(t)K(t) f′(t) f′(t)K(t) f′(t) 1
+ 1 β(cid:16)k2(t)k3(t) (cid:17)B (t) (cid:16) (cid:17)
f′(t)K(t) f′(t) 2
(cid:16) (cid:17)
where K(t) = k∗(f(t)) for all t I. Differentiating the last equation with
1 ∈
respect to t, then we have
′ ′
f′(t)[k∗(f(t))T∗(f(t))+k∗(f(t))B (f(t))]= 1 1+βk1(t) T(t)
1 2 1 f′(t)K(t) f′(t)
′ (cid:18) (cid:16) ′ (cid:17)(cid:19)
+ k1(t) 1+βk1(t) k2(t) βk2(t) N(t)
f′(t)K(t) f′(t) − f′(t)K(t) f′(t)
(cid:18) (cid:16) (cid:17)′ ′ (cid:16) (cid:17)(cid:19)
+ 1 βk2(t) k3(t) βk2(t)k3(t) B (t)
(cid:18)f′(t)K(t)(cid:16) f′(t) (cid:17)(cid:19) − f′(t)K(t)(cid:16) f′(t) (cid:17)! 1
′ ′
+ 1 βk2(t)k3(t) + k3(t) βk2(t) B (t)
f′(t)K(t) f′(t) f′(t)K(t) f′(t) 2
(cid:18)(cid:16) (cid:16) (cid:17)(cid:17) (cid:16) (cid:17)(cid:19)
for all t I. Considering
∈
f′(t)(k∗(f(t))T∗(f(t))+k∗(f(t))B ∗(f(t))) , B ∗(f(t)) =0
h 1 2 1 2 i
and
B ∗(f(t))= N(t),
2
∓
then we get
1+βk (t) ′ βk (t) ′
k (t) 1 k (t) 2 =0.
1 f′(t) − 2 f′(t)
(cid:18) (cid:19) (cid:18) (cid:19)
Arranging the last equation, we find
8
β k (t)k ′(t) k (t)k ′(t) f′(t) k (t)+βk2(t) βk2(t) f′′(t)=0.
1 1 − 2 2 − 1 1 − 2
(3.14)
(cid:2) (cid:3) (cid:2) (cid:3)
Moreover,the differentiation of the equation (3.13) with respect to t is
k′ (t)+2β(k (t)k′ (t) k (t)k′ (t))=0.
1 1 1 − 2 2
From the above equation, it is seen that
k′ (t)
1 =β(k (t)k′ (t) k (t)k′ (t)). (3.15)
1 1 2 2
− 2 −
Substituting the equations(3.13)and(3.15)intothe equation(3.14),weobtain
k′ (t)
1 =0.
− 2
This means that the first curvature function is constant (that is, positive con-
stant). Additionally, from the equation (3.15) it is seen that the second curva-
ture function k is positive constant, too.
2
Conversely, suppose that C is a timelike Frenet curve E4 in with the Frenet
1
frame field T, N, B , B and curvature functions k , k and k . The first
1 2 1 2 3
{ }
curvature function k and the second curvature function k of C are of positive
1 2
constant. Thus, k1 is a positive constant number, say β.
k22−k12
The representation of timelike curve C∗ with arc-length parameter t is
c∗ : I E4
t → c1∗(t)=c(t)+β(t)N(t). (3.16)
→
Let t∗ denote the arc-length parameter of C∗, we have
f : I I∗
→
t t∗ =f(t)= 1+βk t.
1
→ | |
Then, we obtain f′(t)= 1+βk and p
1
| |
f′(t)T∗p(f(t))=T (t)+βN′(t)
=(1+βk )T (t)+βk B (t),
1 2 1
that is
βk
T∗(f(t))= 1+βk T (t)+ 2 B (t). (3.17)
1 1
| | 1+βk
1
p | |
By differentiating both sides of the above equaplity with respect to t we find
dT∗(t∗) βk
f′(t) = 1+βk T′(t)+ 2 B ′(t)
dt∗ (cid:12)(cid:12)t∗=f(t) p| 1| |1+βk1| 1
(cid:12) p
(cid:12)
9
= k1(1+βk1)−βk22 N(t)+ βk2k3(t) B (t)
√|1+βk1| √|1+βk1| 2
(cid:20) (cid:21) (cid:20) (cid:21)
= βk2k3(t) B (t).
√|1+βk1| 2
(cid:20) (cid:21)
Hence, since k doesn’t vanish, we get
3
dT∗(t∗) βk k (t)
k∗(f(t))= =ε 2 3 >0
where ε=sign(k13) denotes(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)thedsti∗gn o(cid:12)(cid:12)(cid:12)(cid:12)tf∗=fufn(tc)t(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)ion k31. T+hβakt1is, ε is −1 or +1.
We can put
1 dT∗(t∗)
N∗(t∗)= , t I.
k∗(t∗) dt∗ ∈
1
Then, we get
N∗(f(t))= B (t).
2
∓
Differentiating of the last equation with respect to t, we reach
dN∗(t∗) k
f′(t) = ε 3 B (t)
dt∗ (cid:12)t∗=f(t) − |1+βk1| 1
(cid:12)
and we have (cid:12) p
(cid:12)
dN∗(t∗) βk k (t)
f′(t) k∗(f(t))T∗(f(t))= ε 2 3 T (t) ε 1+βk B (t).
dt∗ (cid:12)(cid:12)t∗=f(t)− 1 − |1+βk1| − p| 1| 1
Since εk3(t) is(cid:12)(cid:12)positive for t I, we have p
∈
k∗(f(t))= dN∗(t∗) k∗(f(t))T∗(f(t))
2 dt∗ t∗=f(t)− 1
= (cid:13)(cid:13)(cid:13)(cid:13)−β2k122+(kβ(cid:12)(cid:12)(cid:12)3k(1t))2 +(1+βk1)(k3(t))2 (cid:13)(cid:13)(cid:13)(cid:13)
= q(k (t))2 =εk (t)>0.
3 3
Thus, we can put q
B ∗(f(t))= 1 dN∗(t∗) k∗(f(t))T∗(f(t))
1 k2∗(f(t)) dt∗ t∗=f(t)− 1
= −√|1β+k2βk1(cid:18)|T (t)−(cid:12)(cid:12)(cid:12) |1+βk1|B1(t) , t∈I(cid:19).
Differentiation of the above with respecpt to t, we get
dB ∗(t∗) k
f′(t) 1 = 2 N(t) k (t) 1+βk B (t).
dt∗ (cid:12)(cid:12)t∗=f(t) |1+βk1| − 3 p| 1| 2
(cid:12) p
(cid:12)
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