射影几何

毕业论文 漫谈射影几何的几种子几何及其关系

英文翻译

2.1 ORTHOGONAL TRANSFORMATIONS

IN TWO DIMENSIONS

If T??(?2), then T is completely determined by its action on the basis

vectorse1?(1,0)ande2?(0,1).IfTe1?(?,?),then?2??2?1and

Te2??(??,?),since T preserves length and orthogonality. Choose

?,0???2?, such that cos??? and sin???.

If Te2?(??,?), then T is represented by the matrix

???????cos??sin? A??????,

???sin?co?s??? and it is clear that T is a counterclock wise rotation of the plane about the origin

through the angle? (see Figure 2.1). Observe that

det T??2??2?cos2??sin2??1.

If Te2?(?,??), then T is represented by the matrix

?????cos?B???????????sin?In this case observe that

sin???. ?cos??det T??cos2??sin2???1,

and that

??2??2B???02??10????, 22??????01?0

so that T2?1. It is easy to verify (Exercise 2.1) that the vector

x1?(cos?2,sin?2) is an eigenvector having eigenvalue 1 for T, so that the

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毕业论文 漫谈射影几何的几种子几何及其关系

line l???x1:???? is left pointwise fixed by T. Similarly, the vectorx2?(?sin?2,cos?2) is an eigevector with eigenvalue -1,and

x2?x1 [see Figure 2.2(a)]. With respect to the basis {x1,x2} the transformation T is represented by the matrix

?10?C???.

0?1??If x??1x1??2x2, then Tx??1x1??2x2, and T sends xto its mirror image with respect to the line l [see Figure 2.2(b)]. The transformation T is called

the reflectionthrough l or the reflectionalongx2. Observe that

Tx?x?2(x,x2)x2

For all x??2

We hsve shown that every orthogonal transformation of ?2 is either a rotation or a reflection.

2.2 FINITE GROUPS IN TWO DIMENSIONS

Suppose that dim V=2 and that ? is a finite subgroup of?(V). The set of all rotations in ? constitutes a subgroup? of ?. As was shown in Section 2.1, each T?? is a counterclockwise rotation of V though an angle

???(T) with 0???2?. If ??1, choose R?? with R?1, for which ?(R) is minimal. If T??, choose an integer m such that

m?(R)??(T)?(m?1)?(R).

Then 0??(T)?m?(R)??(R). But

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毕业论文 漫谈射影几何的几种子几何及其关系

?(T)?m?(R)??(R?mT),

since R?mT is a counterclockwise rotation through angle ?(T) followed by m clockwise rotations, each through angle ?(R). Since ?(R) was chosen to be minimal, we must have ?(R?mT)=0; R?mT=1 or T?Rm. In other words,

??R is a cyclic group. It also follows that ?(R)?2?n, wheren??. If ???, we have shown that ? is a cyclic group of order n , in which case ? will be denoted by ?2(the subscript calls attention to the fact that dim V=2).

Suppose next that ???. And choose a reflection S??. Since

det(SRk)?detS??1 for all integers k, the coset S? contains n??, distinct reflections. If T?? is a reflection, then

det(ST)?(detS)(detT)?(?1)(?1)?1,

so ST??; hence T?S?, since S?1?S. thus ? is a subgroup of index 2 in ?, and if ??R, as above, then

??R,S?{1,R,,Rn?1,S,SR,,SRn?1},

and ??2n. since RS is a reflection, we have (RS)2?1, or

RS?RS?1?RSn?1, completely determining the multiplication in ?. the

group ? is called the dihedral group of order 2n, and it will be denoted by.

?n2 We have proved.

Theorem 2.2.1

If dim V=2 and ? is afinite subgroup of ?(V), then ? is either a

ncyclic group ?n2 or a dihedral group ?2, n=1,2,3……

If we set T=RS in the dihedral group ?n2?S,R, then T is a reflection, since det T??1. Since TS?RS2?R, it is clear that orthonormal basis {x1,x2} of eigenvectors of S discussed in Section 2.1 coincides with the usual basis {e1,e2} in ?2, then we may assume that S and R are represented by the matrices

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毕业论文 漫谈射影几何的几种子几何及其关系

?10??cos2?n?sin2?n? and B?A???? ??sin2?ncos2?n??0?1?respectively. Thus T is represented by the matrix

?cos2?nsin2?n?C?BA???

?sin2?n?cos2?n?so T is areflection through a line l inclined at an angle of ?n to the positive x-axis. Let us use these ideas to give a geometrical interpretation of the group ?n2.

Denote by F the open wedege-shaped region in the first quadrant bounded by the x-axis and the line l. The x-axis is a reflecting line for the transformation S. and l is areflecting line for the transformation T. The 2n congruent regions in the plane obtained by rotating the region F through successive multiples of ?n can be labeled with the elements of ?n2 as follows. For each U??n, designate by U the region U(F) obtained by 2applying U to all points of the region F.

The procedure is illustrated in Figure 2.3 for the n=4. If two plane mirrors are set facing one another along the reflecting lines for S and T, with their common edge perpendicular to the plane at the origin, then the other lines may be seen in the mirrors as edges of F. This illustrates the principle of the kaleidoscope and shows a connection between the kaleidoscope and the dihedral groups.

Observe that the region F is open, that no point of F is mapped to any other point of F by any nonidentity element U of ?n, and that the union of the 22?closures (UF)?,U??n, is all of . 2

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