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Chapter 3
Euclidean Geometry
We recall some familiar features of plane geometry. First of all, two trian- gles are congruent if there is a rigid motion of the plane that carries one tri- angle exactly onto the other. Corresponding angles of congruent triangles are equal, corresponding sides have the same length, the areas enclosed are equal, and so on. Indeed, any geometric property of a given triangle is automati- cally shared by every congruent triangle. Conversely, there are a number of simple ways in which one can decide whether two given triangles are con- gruent—for example, if for each the same three numbers occur as lengths of sides. In this chapter we shall investigate the rigid motions (isometries) of Euclid- ean space, and see how these remarks about triangles can be extended to other geometric objects.
3.1 Isometries of R^3
An isometry, or rigid motion, of Euclidean space is a mapping that preserves the Euclidean distance d between points (Definition 1.2, Chapter 2).
1.1 Definition An isometry of R^3 is a mapping F : R^3 Æ R^3 such that
for all points p , q in R^3.
1.2 Example (1) Translations. Fix a point a in R^3 and let T be the
mapping that adds a to every point of R^3. Thus T ( p ) = p + a for all
d F ( ( ) p , F ( ) q ) = d ( p q , )
3.1 Isometries of R^3
points p. T is called translation by a. It is easy to see that T is an isometry, since
(2) Rotation around a coordinate axis****. A rotation of the xy plane through an angle J carries the point ( p 1, p 2) to the point ( q 1, q 2) with coordinates (Fig. 3.1)
Thus a rotation C of three-dimensional Euclidean space R^3 around the z axis, through an angle J, has the formula
Evidently, the mapping C is a linear transformation. A straightforward com- putation shows that C preserves Euclidean distance, so it is an isometry.
Recall that if F and G are mappings of R^3 , the composite function GF is a mapping of R^3 obtained by applying first F , then G.
1.3 Lemma If F and G are isometries of R^3 , then the composite mapping
GF is also an isometry of R^3.
Proof. Since G is an isometry, the distance from G ( F ( p )) to G ( F ( q )) is d ( F ( p ), F ( q )). But since F is an isometry, this distance equals d ( p , q ). Thus
GF preserves distance; hence it is an isometry. ◆
C ( ) = p C p ( (^) 1 , p (^) 2 , p (^) 3 ) = ( p (^) 1 cos J - p (^) 2 sin J, p (^) 1 sin J + p (^) 2 cos J, p 3 ).
q (^) 2 = p 1 (^) sin J + p 2 cos J.
q (^) 1 = p 1 (^) cos J - p 2 sinJ,
= p - q = d ( p q , ).
= ( p + a ) - ( q + a )
d T ( ( ) p , T ( ) q ) = d ( p + a , q + a )
FIG. 3.
3.1 Isometries of R^3
Our goal now is Theorem 1.7, which asserts that every isometry can be expressed as an orthogonal transformation followed by a translation. The main part of the proof is the following converse of Lemma 1.5.
1.6 Lemma If F is an isometry of R^3 such that F ( 0 ) = 0 , then F is an
orthogonal transformation.
Proof. First we show that F preserves dot products; then we show that F is a linear transformation. Note that by definition of Euclidean distance, the norm || p || of a point p is just the Euclidean distance d ( 0 , p ) from the origin to p. By hypothesis, F preserves Euclidean distance, and F ( 0 ) = 0 ; hence
Thus F preserves norms. Now by a standard trick (“polarization”), we shall deduce that it also preserves dot products. Since F is an isometry,
for any pair of points. Hence
By the definition of norm, this implies
Hence
The norm terms here cancel, since F preserves norms, and we find
as required. It remains to prove that F is linear. Let u 1 , u 2 , u 3 be the unit points (1, 0, 0), (0, 1, 0), (0, 0, 1), respectively. Then we have the identity
p = ( p 1 (^) , p (^) 2 , p 3 ) = (^) Â pi u i.
F ( ) p • F ( ) = q p • q ,
F ( ) p - F ( ) p F ( ) + q F ( ) q = p - p q + q 2 2 2 2 2 • 2 •.
( F ( ) - p F ( ) q ) • ( F ( ) - p F ( ) q ) = ( p - q ) • ( p - q ).
F ( ) - p F ( ) = q p - q.
d F ( ( ) p , F ( ) q ) = d ( p q , )
F ( ) = p d ( 0 , F ( ) p ) = d F ( ( ) 0 , F ( ) p ) = d ( 0 p , ) = p.
= d ( p q , ) for all p q ,.
d C ( ( ) p , C ( ) q ) = C ( ) - p C ( ) = q C ( p - q ) = p - q
104 3. Euclidean Geometry
Also, the points u 1, u 2, u 3 are orthonormal; that is, u i • u j = d ij. We know that F preserves dot products, so F ( u 1), F ( u 2), F ( u 3 ) must also be orthonormal. Thus orthonormal expansion gives
But
so
Using this identity, it is a simple matter to check the linearity condition
We now give a concrete description of an arbitrary isometry.
1.7 Theorem If F is an isometry of R^3 , then there exist a unique trans-
lation T and a unique orthogonal transformation C such that
Proof. Let T be translation by F ( 0 ). Then Lemma 1.4 shows that T -^1 is translation by - F ( 0 ). But T -^1 F is an isometry, by Lemma 1.3, and furthermore,
Thus by Lemma 1.6, T -^1 F is an orthogonal transformation, say T -^1 F = C. Applying T on the left, we get F = TC. To prove the required uniqueness, we suppose that F can also be expressed as , where is a translation and an orthogonal transformation. We must prove = T and = C. Now TC = ; hence C = T -^1. Since C and are linear transformations, they of course send the origin to itself. It follows that ( T -^1 )( 0 ) = 0. But since T -^1 is a translation, we conclude that T -^1 = I ; hence = T. Then the equation TC = becomes TC = T.
Applying T -^1 gives C = .. ◆
Thus every isometry of R^3 can be uniquely described as an orthogonal trans- formation followed by a translation. When F = TC as in Theorem 1.7, we call
C
T T TC C
T T
C
T C TC TC
TC T C
( T -^1 F )( ) = 0 T -^1 ( F ( ) 0 ) = F ( ) - 0 F ( ) = 0 0.
F = TC.
F a ( p + b q ) = aF ( ) + p bF ( ) q.
F ( ) = p (^) Â p Fi ( u (^) i ).
F ( ) p • F ( u (^) i ) = p • u i = pi ,
F ( ) = p (^) Â F ( ) p • F ( u (^) i ) F ( u (^) i ).
106 3. Euclidean Geometry
3. Show that an isometry F = T (^) a C has an inverse mapping F -^1 , which is also an isometry. Find the translation and orthogonal parts of F -^1. 4. If
show that C is orthogonal; then compute C ( p ) and C ( q ), and check that C ( p ) • C ( q ) = p • q.
5. Let F = T (^) aC , where a = (1, 3, - 1) and
If p = (2, - 2, 8), find the coordinates of the point q for which (a) q = F ( p ). (b) q = F -^1 ( p ). (c) q = ( CTa ) ( p ).
6. In each case decide whether F is an isometry of R^3. If so, find its trans- lation and orthogonal parts. (a) F ( p ) = - p. (b) F ( p ) = ( p • a ) a , where || a || = 1. (c) F ( p ) = ( p 3 - 1, p 2 - 2, p 1 - 3). (d) F ( p ) = ( p 1, p 2, 1).
A group G is a set furnished with an operation that assigns to each pair g 1 , g 2 of elements of G an element g 1 g 2, subject to these rules: (1) associative law: ( g 1 g 2) g 3 = g 1 ( g 2 g 3), (2) there is a unique identity element e such that eg = ge = g for all g in G , and (3) inverses: For each g in G there is an element g -^1 in G such that gg -^1 = g -^1 g = e. Groups occur naturally in many parts of geometry, and we shall mention a few in subsequent exercises. Basic properties of groups may be found in a variety of elementary textbooks.
7. Prove that the set E(3) of all isometries of R^3 forms a group—with com- position of functions as the operation. E(3) is called the Euclidean group of order 3.
A subset H of a group G is a subgroup of G provided (1) if g 1 and g 2 are in H , then so is g 1 g 2, (2) is g is in H , so is g -^1 , and hence (3) the identity element e of G is in H. A subgroup H of G is automatically a group.
C =
Ê -
Ë
Á
Á
Á
C =
Ê
Ë
Á
Á
= ( - ) = ( )
Ï
Ì
Ó
and
p q
3.2 The Tangent Map of an Isometry 107
8. Prove that the set T (3) of all translations of R^3 and the set O (3) of all orthogonal transformations of R^3 are each subgroups of the Euclidean group E(3). O (3) is called the orthogonal group of order 3. Which isometries of R^3 are in both these subgroups?
It is easy to check that the results of this section, though stated for R^3 , remain valid for Euclidean spaces R n^ of any dimension.
9. (a) Give an explicit description of an arbitrary 2 ¥ 2 orthogonal matrix C. ( Hint: Use an angle and a sign.) (b) Give a formula for an arbitrary isometry F of R = R^1.
3.2 The Tangent Map of an Isometry
In Chapter 1 we showed that an arbitrary mapping F : R^3 Æ R^3 has a tangent map F * that carries each tangent vector v at p to a tangent vector F *( v ) at F ( p ). If F is an isometry, its tangent map is remarkably simple. (Since the dis- tinction between tangent vector and point is crucial here, we temporarily restore the point of application to the notation.)
2.1 Theorem Let F be an isometry of R^3 with orthogonal part C.
Then
for all tangent vectors v p to R^3. Verbally: To get F *( v p ), first shift the tangent vector v p to the canonically corresponding point v of R^3 , then apply the orthogonal part C of F , and finally shift this point C ( v ) to the canonically corresponding tangent vector at F ( p ) (Fig. 3.2). Thus all tangent vectors at all points p of R^3 are “rotated” in exactly the same way by F * —only the new point of application F ( p ) depends on p.
Proof. Write F = TC as in Theorem 1.7. Let T be translation by a , so F ( p ) = a + C ( p ). If v p is a tangent vector to R^3 , then by Definition 7.4 of Chapter 1, F *( v p ) is the initial velocity of the curve t Æ F ( p + t v ). But using the linearity of C , we obtain
= F ( ) + p tC ( ) v.
F ( p + t v ) = TC ( p + t v ) = T C ( ( ) + p tC ( ) v ) = a + C ( ) + p tC ( ) v
F (^) *( v (^) p ) = C ( ) v F (^) ( p )
3.2 The Tangent Map of an Isometry 109
Assertion (3) of Lemma 1.4 shows how two points uniquely determine a translation. We now show that two frames uniquely determine an isometry.
2.3 Theorem Given any two frames on R^3 , say e 1, e 2 , e 3 at the point p
and f 1 , f 2 , f 3 at the point q , there exists a unique isometry F of R^3 such that F *( e i ) = f i for 1 i 3.
Proof. First we show that there is such an isometry. Let eˆ 1, eˆ 2, eˆ 3, and fˆ 1 , fˆ 2, fˆ 3 be the points of R^3 canonically corresponding to the vectors in the two frames. Let C be the unique linear transformation of R^3 such that C ( eˆ i ) = fˆ i for 1 i 3. It is easy to check that C is orthogonal. Then let T be a translation by the point q - C ( p ). Now we assert that the isometry F = TC carries the e frame to the f frame. First note that
Then using Theorem 2.1 we get
for 1 i 3. To prove uniqueness, we observe that by Theorem 2.1 this choice of C is the only possibility for the orthogonal part of the required isometry. The translation part is then completely determined also, since it must carry C ( p )
to q. Thus the isometry F = TC is uniquely determined. ◆
To compute the isometry in the theorem, recall that the attitude matrix A of the e frame has the Euclidean coordinates of e i as its i th row: a (^) i 1 , a (^) i 2 , a (^) i 3. The attitude matrix B of the f frame is similar. We claim that C in the theorem (or strictly speaking, its matrix) is tBA. To verify this it suffices to check that tBA ( e i )^ =^ f i , since this uniquely characterizes^ C. For^ i^ =^ 1 we find, using the column-vector conventions,
that is, tBA ( e 1) = f 1. The cases i = 2, 3 are similar; hence C = tBA. As noted above, T is then necessarily translated by q - C ( p ).
Ê
Ë
Á
Á
Ê
Ë
Á
Á
Ê
Ë
Á
Á
Ê
Ë
Á
Á
tB
b b b b b b b b b
b b b
11 21 31 12 22 32 13 23 33
11 12 13
t (^) BA t
a a a
B
a a a a a a a a a
a a a
11 12 13
11 12 13 21 22 23 31 32 33
11 12 13
Ê
Ë
Á
Á
Ê
Ë
Á
Á
Ê
Ë
Á
Á
F (^) *( e (^) i ) = C (ˆ e (^) i ) F (^) ( p ) = ( f ˆ i^ (^) ) F (^) ( p )= ( f ˆ (^) i (^) ) q (^) = f i
F ( ) = p T C ( ( ) p ) = q - C ( ) + p C ( ) = p q.
110 3. Euclidean Geometry
Exercises
1. If T is a translation, show that for every tangent vector v the vector T ( v ) is parallel to v (same Euclidean coordinates). 2. Prove the general formulas ( GF )* = G * F * and ( F -^1 )* = ( F *)-^1 in the special case where F and G are isometries of R^3. 3. Given the frame
at p = (0, 1, 0) and the frame
at q = (3, - 1, 1), find a and C such that the isometry F = T (^) aC carries the e frame to the f frame.
4. (a) Prove that an isometry F = TC carries the plane through p orthog- onal to q π 0 to the plane through F ( p ) orthogonal to C ( q ). (b) If P is the plane through (1/2, - 1, 0) orthogonal to (0, 1, 0) find an isometry F = TC such that F ( P ) is the plane through (1, - 2, 1) orthogonal to (1, 0, - 1). 5. ( Computer .) (a) Verify that both sets of vectors in Exercise 3 form frames by showing that A tA = I for their attitude matrices. (b) Find the matrix C that carries each e (^) i to f (^) i , and check this for i = 1, 2, 3.
3.3 Orientation
We now come to one of the most interesting and elusive ideas in geometry. Intuitively, it is orientation that distinguishes between a right-handed glove and a left-handed glove in ordinary space. To handle this concept mathe- matically, we replace gloves by frames and separate all the frames on R^3 into two classes as follows. Recall that associated with each frame e 1, e 2 , e 3 at a point of R^3 is its attitude matrix A. According to the exercises for Section 1 of Chapter 2,
e (^) 1 • e (^) 2 ¥ e 3 = det A = ± 1.
f (^) 1 = (1 0 1 , , ) 2 , f (^) 2 = (0 1 0 , , ) , f 3 = (1 0 , ,- 1 ) 2
e (^) 1 = (2 2 1 , , ) 3 , e (^) 2 = (- 2 1 2, , ) 3 , e 3 = ( 1 , - 2 2, ) 3
112 3. Euclidean Geometry
We know that the tangent map of an isometry carries frames to frames. The following result tells what happens to their orientations.
3.2 Lemma If e 1, e 2, e 3 is a frame at some point of R^3 and F is an
isometry, then
Proof. If , then by the coordinate form of Theorem 2.1 we have
where C = ( c (^) ij ) is the orthogonal part of F. Thus the attitude matrix of the frame F *( e 1 ), F *( e 2), F *( e 3) is the matrix
But the triple scalar product of a frame is the determinant of its attitude matrix, and by definition, sgn F = det C. Consequently,
This lemma shows that if sgn F = +1, then F * carries positively oriented frames to positively oriented frames and carries negatively oriented frames to negatively oriented frames. On the other hand, if sgn F = -1, positive goes to negative and negative to positive.
3.3 Definition An isometry F of R^3 is said to be
where C is the orthogonal part of F.
orientation - reversing ifsgn F = det C = - 1 ,
orientation preserving - if sgn F = det C = + 1 ,
= ( sgn F ) e (^) 1 ◊ e (^) 2 ¥ e 3.
= det C ◊ det tA = det C ◊det A
F (^) * ( e (^) 1 ) • F (^) *( e (^) 2 ) ¥ F (^) * ( e 3 ) = det( C A t )
c aik jk c a C A k
ik
t kj k
Ê Â Â t Ë
Ê
Ë
¯ =^.
F (^) j c a Uik jk i
- (^) i k ( e ) = (^) Â , ,
e (^) j = (^) Â a Ujk k
F (^) *( e (^) 1 ) • F (^) *( e (^) 2 ) ¥ F (^) * ( e (^) 3 ) = (sgn F ) e (^) 1 • e (^) 2 ¥ e 3.
3.3 Orientation 113
3.4 Example (1) Translations. All translations are orientation-preserv-
ing. Geometrically this is clear, and in fact the orthogonal part of a transla- tion T is just the identity mapping I , so sgn T = det I = +1. (2) Rotations****. Consider the orthogonal transformation C given in Example 1.2, which rotates R^3 through angle q around the z axis. Its matrix is
Hence sgn C = det C = +1, so C is orientation-preserving (see Exercise 4).
(3) Reflections****. One can (literally) see reversal of orientation by using a mirror. Suppose the yz plane of R^3 is the mirror. If one looks toward that plane, the point p = ( p 1, p 2, p 3 ) appears to be located at the point
(Fig. 3.4). The mapping R so defined is called reflection in the yz plane. Evidently it is an orthogonal transformation, with matrix
Thus R is an orientation-reversing isometry, as confirmed by the experimen- tal fact that the mirror image of a right hand is a left hand. Both dot and cross product were originally defined in terms of Euclidean coordinates. We have seen that the dot product is given by the same formula,
v • w = ( (^) Â v (^) i e (^) i ) • ( (^) Â w (^) i e i ) =Â v wi i ,
Ê^ -
Ë
Á
Á
R ( ) = p (- p (^) 1 , p (^) 2 , p 3 )
cos sin sin cos.
q q q q
Ê^ -
Ë
Á
Á
FIG. 3.
3.3 Orientation 115
where
But U 1, U 2 , U 3 is positively oriented, so by Lemma 3.2, e = sgn F. ◆
Exercises
1. Prove
Deduce that sgn F = sgn ( F -^1 ).
2. If H 0 is an orientation-reversing isometry of R^3 , show that every orientation-reversing isometry has a unique expression H 0 F , where F is orientation-preserving. 3. Let v = (3, 1, - 1) and w = (-3, - 3, 1) be tangent vectors at some point. If C is the orthogonal transformation given in Exercise 4 of Section 1, check the formula 4. A rotation is an orthogonal transformation C such that det C = +1. Prove that C does, in fact, rotate R^3 around an axis. Explicitly, given a rotation C , show that there exists a number J and points e 1 , e 2 , e 3 with e i • e j = d ij such that (Fig. 3.5)
C ( e (^) 3 ) = e 3.
C ( e (^) 2 ) = - sin J e (^) 1 +cosJ e 2 ,
C ( e (^) 1 ) = cos J e (^) 1 +sinJ e 2 ,
C (^) *( v ¥ w ) = (sgn C C ) (^) * ( ) ¥ v C * ( w ).
sgn ( FG ) = sgn F ◊ sgn G = sgn ( GF ).
e = e (^) 1 • e (^) 2 ¥ e (^) 3 = F (^) *( U 1 (^) ( ) p ) • F * ( U 2 (^) ( ) p ) ¥ F * ( U 3 ( ) p ).
FIG. 3.
116 3. Euclidean Geometry
( Hint: The fact that the dimension of R^3 is odd means that C has an eigen- value +1, so there is a point p π 0 such that C ( p ) = p .)
5. Let a be a point of R^3 such that || a || = 1. Prove that the formula
defines an orthogonal transformation. Describe its general effect on R^3.
6. Prove (a) The set O +(3) of all rotations of R^3 is a subgroup of the orthogonal group O (3) (see Ex. 8 of Sec. 3.1). (b) The set E +(3) of all orientation-preserving isometries of R^3 is a sub- group of the Euclidean group E(3).
3.4 Euclidean Geometry
In the discussion at the beginning of this chapter, we recalled a fundamental feature of plane geometry: If there is an isometry carrying one triangle onto another, then the two (congruent) triangles have exactly the same geometric properties. A close examination of this statement will show that it does not admit a proof—it is, in fact, just the definition of “geometric property of a triangle.” More generally, Euclidean geometry can be defined as the totality of concepts that are preserved by isometries of Euclidean space. For example, Corollary 2.2 shows that the notion of dot product on tangent vectors belongs to Euclidean geometry. Similarly, Theorem 3.6 shows that the cross product is preserved by isometries (except possibly for sign). This famous definition of Euclidean geometry is somewhat generous, however. In practice, the label “Euclidean geometry” is usually attached only to those concepts that are preserved by isometries, but not by arbitrary map- pings, or even the more restrictive class of mappings (diffeomorphisms) that possess inverse mappings. An example should make this distinction clearer. If a = (a 1 , a2, a3) is a curve in R^3 , then the various derivatives
look pretty much alike. Now, Theorem 7.8 of Chapter 1 asserts that velocity is preserved by arbitrary mappings F : R^3 Æ R^3 , that is, if b = F (a), then b¢ = F *(a¢). But it is easy to see that acceleration is not preserved by arbitrary map- pings. For example, if a( t ) = ( t , 0, 0) and F = ( x^2 , y , z ), then a≤ = 0; hence F *(a≤) = 0. But b = F (a) has the formula b( t ) = ( t^2 , 0, 0), so b≤ = 2 U 1. Thus
Ê
Ë
Ê
Ë
a
a a a a
d a a a dt
d dt
d dt
d dt
d dt
d dt
1 2 3 2 1 2
2 2 2
2 3 , , , , , 2 ,...
C ( ) = p a ¥ p + ( p • a ) a
118 3. Euclidean Geometry
Thus by the coordinate version of Theorem 2.1, we get
On the other hand,
But each c (^) ij is constant, being by definition an entry in the matrix of the orthogonal part of the isometry F. Hence
Thus the vector fields F *( Y ¢) and ¢ are the same. ◆
We claimed earlier that isometries preserve acceleration: If = F (a), where F is an isometry, then ≤ = F *(a≤). This is an immediate consequence of the preceding result, for if we set Y = a¢, then by Theorem 7.8 of Chapter 1, = ¢; hence
Now we show that the Frenet apparatus of a curve is preserved by isome- tries. This is certainly to be expected on intuitive grounds, since a rigid motion ought to carry one curve into another that turns and twists in exactly the same way. And this is what happens when the isometry is orientation- preserving.
4.2 Theorem Let b be a unit-speed curve in R^3 with positive cur-
vature, and let = F (b) be the image curve of b under an isometry F of R^3. Then
where sgn F = ±1 is the sign of the isometry F.
Proof. Note that is also a unit-speed curve, since
b ¢ = F *( b ¢) = b¢ = 1.
b
B = ( sgn F F ) (^) *( B ),
t = (sgn F )t , N = F * ( N ),
k = k, T = F (^) *( T ),
b
a ¢¢ = Y ¢ = F (^) *( Y ¢) = F *( a ¢¢).
Y a
a
a
Y
Y ¢ = (^) Â ( ) =Â
d dt
c y U c
dy dt ij j i ij jUi.
Y = F (^) *( Y ) = (^) Â c y Uij j i.
F Y c
dy dt
3.4 Euclidean Geometry 119
Thus the definitions in Section 3 of Chapter 2 apply to both b and , so
Since F * preserves both acceleration and norms, it follows from the definition of curvature that
To get the full Frenet frame, we now use the hypothesis k > 0 (which implies > 0 , since = k). By definition, N = b≤/k; hence using preced- ing facts, we find
It remains only to prove the interesting cases B and t. Since the defini- tion B = T ¥ N involves a cross product, we use Theorem 3.6 to get
The definition of torsion is essentially t = - B ¢ • N = B • N ¢. Thus, using the results above for B and N , we get
The presence of sgn F in the formula for the torsion of F (b) shows that the torsion of a curve gives a more subtle description of the curve than has been apparent so far. The sign of t measures the orientation of the twisting of the curve. If F is orientation-reversing, the formula = -t proves that the twisting of the image of curve F (b) is exactly opposite to that of b itself. A simple example will illustrate this reversal.
4.3 Example Let b be the unit-speed helix
gotten from Example 3.3 of Chapter 2 by setting a = b = 1; hence c =. We know from the general formulas for helices that k = t = 1/2. Now let R be reflection in the xy plane, so R is the isometry R ( x , y , z ) = ( x , y , - z ). Thus the image curve = R (b) is the mirror image
b s
s c
s c
s c
( ) = -
Ê
Ë
cos , sin ,
b
b s
s c
s c
s c
( ) =
Ê
Ë
cos , sin , ,
t
t = B • N ¢ = (sgn F F ) (^) *( B ) • F (^) *( N ¢) = ( sgn F B ) • N ¢ = ( sgn F )t.
B = T ¥ N = F (^) *( T ) ¥ F (^) *( N ) = (sgn F F ) (^) *( T ¥ N ) = ( sgn F F ) (^) *( B ).
N
F
= F F N
( ¢¢)
Ê ¢¢
Ë
= ( )
b k
b k
b k
k k
k = b ¢¢ = F *( b ¢¢) = b ¢¢ =k.
T = b ¢ = F (^) *( b¢) = F (^) *( T ).
b
