卡坦-狄翁定理

第二十六章 The Cartan–Dieudonn´e Theorem 卡坦-狄翁定理 In this chapter the structure of the orthogonal group is studied in more depth. In particular, we prove that every isometry in O(n) is the composition of at most n reflections about hyperplanes (for n ≥ 2, see Theorem 26.1). This important result is a special case of the “Cartan–Dieudonn´e theorem” (Cartan [33], Dieudonn´e [51]). We also prove that every rotation in SO(n) is the composition of at most n flips (for n ≥ 3). 本章对正交群的结构进行了较深入的研究。特别地，我们证明了O（n）中的每一个等值线都是超平面上至多n个反射的组合（n≥2，见定理26.1）。这个重要的结果是“卡坦-迪乌顿定理”的一个特例（卡坦[33]，迪乌顿[51]）。我们还证明了so（n）中的每一个旋转都是最多n个翻转（n≥3）的组合。 Affine isometries are defined, and their fixed points are investigated. First, we characterize the set of fixed points of an affine map. Then we show that the Cartan–Dieudonn´e theorem can be generalized to affine isometries: Every rigid motion in Is(n) is the composition of at most n affine reflections if it has a fixed point, or else of at most n + 2 affine reflections. We prove that every rigid motion in SE(n) is the composition of at most n affine flips (for n ≥ 3). 定义了仿射等距线，研究了它们的不动点。首先，我们描述了仿射映射的不动点集。然后证明了卡坦-迪乌顿定理可以推广到仿射等轴测：在is（n）中的每一个刚性运动都是至多n个具有固定点的仿射反射的组合，或者至多n+2个仿射反射的组合。我们证明了SE（n）中的每一个刚性运动都是至多n个仿射翻转（n≥3）的组合。 26.1 The Cartan–Dieudonn´e Theorem for Linear Isometries 26.1线性等轴测的卡坦-迪乌顿定理 The fact that the group O(n) of linear isometries is generated by the reflections is a special case of a theorem known as the Cartan–Dieudonn´e theorem. Elie Cartan proved a version of this theorem early in the twentieth century. A proof can be found in his book on spinors [33], which appeared in 1937 (Chapter I, Section 10, pages 10–12). Cartan’s version applies to nondegenerate quadratic forms over R or C. The theorem was generalized to quadratic forms over arbitrary fields by Dieudonn´e [51]. One should also consult Emil Artin’s book [6], which contains an in-depth study of the orthogonal group and another proof of the Cartan–Dieudonn´e theorem. 线性等距图的O（n）组是由反射产生的，这是一个称为卡坦-迪乌登定理的定理的特例。伊莱·卡坦在二十世纪初证明了这个定理的一个版本。他在1937年出版的关于旋转器的书[33]中找到了证据（第一章，第10节，第10-12页）。Cartan的版本适用于r或c上的非退化二次型。Dieudonn’e[51]将该定理推广到任意场上的二次型。我们还应该参考埃米尔·阿丁的书[6]，其中包括对正交群的深入研究和卡坦-迪乌登定理的另一个证明。 Theorem 26.1. Let E be a Euclidean space of dimension n ≥ 1. Every isometry f ∈ O(E) that is not the identity is the composition of at most n reflections. When n ≥ 2, the identity is the composition of any reflection with itself. 定理26.1。设e为尺寸n≥1的欧几里得空间。每一个非同一性的等距f∈o（e）至多是n个反射的组成。当n≥2时，同一性是任何反射本身的组成。 859 八百五十九 Proof. We proceed by induction on n. When n = 1, every isometry f ∈ O(E) is either the identity or −id, but −id is a reflection about H = {0}. When n ≥ 2, we have id = s ◦ s for every reflection s. Let us now consider the case where n ≥ 2 and f is not the identity. There are two subcases. 证据。我们对n进行归纳，当n=1时，每一个等距f∈o（e）要么是恒等式，要么是−id，但−id是关于h=0_的反映。当n≥2时，每个反射s的id=s s。现在让我们考虑n≥2和f不是同一性的情况。有两个子类。 Case 1. The map f admits 1 as an eigenvalue, i.e., there is some nonnull vector w such that f(w) = w. In this case, let H be the hyperplane orthogonal to w, so that E = H ⊕Rw. We claim that f(H) ⊆ H. Indeed, if 案例1。映射f承认1为特征值，即存在一些非零向量w，使得f（w）=w。在这种情况下，让h是与w正交的超平面，因此e=h rw。我们声称F（H）H。事实上，如果 v · w = 0 V·W=0 for any v ∈ H, since f is an isometry, we get 对于任何v∈h，因为f是一个等距，我们得到 f(v) · f(w) = v · w = 0, F（V）·F（W）=V·W=0， and since f(w) = w, we get f(v) · w = f(v) · f(w) = 0, 因为f（w）=w，我们得到f（v）·w=f（v）·f（w）=0， and thus f(v) ∈ H. Furthermore, since f is not the identity, f is not the identity of H. Since H has dimension n − 1, by the induction hypothesis applied to H, there are at most k ≤ n − 1 reflections s1,…,sk about some hyperplanes H1,…,Hk in H, such that the restriction of f to H is the composition sk ◦···◦ s1. Each si can be extended to a reflection in E as follows: If H = Hi ⊕ Li (where Li = Hi⊥, the orthogonal complement of Hi in H), L = Rw, and Fi = Hi ⊕ L, since H and L are orthogonal, Fi is indeed a hyperplane, E = Fi ⊕ Li = Hi ⊕ L ⊕ Li, and for every u = h + λw ∈ H ⊕ L = E, since 因此f（v）∈h。此外，由于f不是恒等式，f不是h的恒等式。由于h的维数为n-1，根据应用于h的诱导假设，在h中的某些超平面h1，…，h k最多有k≤n-1反射s1，…，sk，因此f对h的限制是组成sk·····s1。每个si可以扩展到e中的反射，如下所示：如果h=hi li（其中li=hi，h中hi的正交补码）、l=rw和fi=hi l，因为h和l是正交的，fi确实是一个超平面，e=fi li=hi l li，并且对于每个u=h+λw∈h l=e，因为 si(h) = pHi(h) − pLi(h), si（h）=phi（h）−pli（h）、 we can define si on E such that 我们可以在e上定义si，这样 si(h + λw) = pHi(h) + λw − pLi(h), si（h+λw）=phi（h）+λw−pli（h）， and since h ∈ H, w ∈ L, Fi = Hi ⊕ L, and H = Hi ⊕ Li, we have 既然h∈h，w∈l，fi=hi l，h=hi li，我们有 si(h + λw) = pFi(h + λw) − pLi(h + λw), si（h+λw）=pfi（h+λw）−pli（h+λw）、 which defines a reflection about Fi = Hi ⊕ L. Now, since f is the identity on L = Rw, it is immediately verified that f = sk ◦ ··· ◦ s1, with k ≤ n − 1. See Figure 26.1. 它定义了关于fi=hi l的反射。现在，由于f是l=rw上的标识，因此立即验证f=sk····s1，k≤n−1。见图26.1。 Case 2. The map f does not admit 1 as an eigenvalue, i.e., f(u) =6 u for all u = 06 . Pick any w = 06 in E, and let H be the hyperplane orthogonal to f(w) − w. Since f is an isometry, we have kf(w)k = kwk, and by Lemma 12.2, we know that s(w) = f(w), where s is the reflection about H, and we claim that s ◦ f leaves w invariant. Indeed, since s2 = id, we have 案例2。映射f不承认1为特征值，即f（u）=6 u（对于所有u=06）。选取e中的任意w=06，设h为与f（w）−w正交的超平面。由于f是一个等距线，我们得到kf（w）k=kwk，根据引理12.2，我们知道s（w）=f（w），其中s是关于h的反射，我们声称s f留下w不变量。事实上，因为s2=id，我们有 s(f(w)) = s(s(w)) = w. S（F（W））=S（S（W））=W。 See Figure 26.2. 见图26.2。

Figure 26.1: An illustration of how to extend the reflection si of Case 1 in Theorem 26.1 to E. The result of this extended reflection is the bold green vector. 图26.1：关于如何将定理26.1中情况1的反射si扩展到e的说明。扩展反射的结果是粗体绿色矢量。 Since s2 = id, we cannot have s◦f = id, since this would imply that f = s, where s is the identity on H, contradicting the fact that f is not the identity on any vector. Thus, we are back to Case 1. Thus, there are k ≤ n−1 hyperplane reflections such that s◦f = sk ◦···◦s1, from which we get f = s ◦ sk ◦ ··· ◦ s1, 因为s2=id，我们不能有s f=id，因为这意味着f=s，其中s是h上的恒等式，这与f不是任何向量上的恒等式的事实相矛盾。因此，我们回到案例1。因此，存在k≤n−1超平面反射，因此s f=sk·····s1，从中我们得到f=s sk·····s1， with at most k + 1 ≤ n reflections.
最多k+1≤n反射。 Remarks: 评论： (1) A slightly different proof can be given. Either f is the identity, or there is some nonnull vector u such that f(u) =6 u. In the second case, proceed as in the second part of the proof, to get back to the case where f admits 1 as an eigenvalue. 可以给出稍微不同的证明。要么f是恒等式，要么有一些非零向量u，这样f（u）=6u。在第二种情况下，继续进行证明的第二部分，回到f承认1为特征值的情况。 (2) Theorem 26.1 still holds if the inner product on E is replaced by a nondegenerate symmetric bilinear form ϕ, but the proof is a lot harder; see Section 28.9. 定理26.1仍然适用，如果e上的内积被非退化对称双线性形式_替换，但证明要困难得多；见第28.9节。 (3) The proof of Theorem 26.1 shows more than stated. If 1 is an eigenvalue of f, for any eigenvector w associated with 1 (i.e., f(w) = w, w = 0)6 , then f is the composition of k ≤ n − 1 reflections about hyperplanes Fi such that Fi = Hi ⊕ L, where L is the line Rw and the Hi are subspaces of dimension n − 2 all orthogonal to L (the Hi are hyperplanes in H). This situation is illustrated in Figure 26.3. 定理26.1的证明比所说的要多。如果1是f的特征值，对于与1（即f（w）=w，w=0）6相关的任何特征向量w，则f是关于超平面f i的k≤n-1反射的组成，因此fi=hi l，其中l是线rw，hi是尺寸n-2的子空间，均与l正交（hi是hy垂直高度）。这种情况如图26.3所示。 If 1 is not an eigenvalue of f, then f is the composition of k ≤ n reflections about hyperplanes H,F1,…,Fk−1, such that Fi = Hi ⊕ L, where L is a line intersecting H, and the Hi are subspaces of dimension n−2 all orthogonal to L (the Hi are hyperplanes in L⊥). This situation is illustrated in Figure 26.4. 如果1不是f的特征值，那么f是关于超平面h，f1，…，f k−1的k≤n反射的组成，这样fi=hi l，其中l是与h相交的线，hi是尺寸n−2的子空间，都与l正交（hi是l中的超平面）。这种情况如图26.4所示。

Figure 26.2: The construction of the hyperplane H for Case 2 of Theorem 26.1. 图26.2：定理26.1的情形2的超平面h的构造。

Figure 26.3: An isometry f as a composition of reflections, when 1 is an eigenvalue of f. 图26.3：当1是f的特征值时，作为反射组成的等距线f。 (4) It is natural to ask what is the minimal number of hyperplane reflections needed to obtain an isometry f. This has to do with the dimension of the eigenspace Ker(f − id) associated with the eigenvalue 1. We will prove later that every isometry is the composition of k hyperplane reflections, where 很自然地，我们会问，获得等距f所需的超平面反射的最小数目是多少。这与特征值1相关的特征空间ker（f-id）的维数有关。稍后我们将证明每个等距线都是k超平面反射的组成，其中 k = n − dim(Ker(f − id)), k=n−dim（ker（f−id））， and that this number is minimal (where n = dim(E)). 这个数字是最小的（其中n=dim（e））。 When n = 2, a reflection is a reflection about a line, and Theorem 26.1 shows that every isometry in O(2) is either a reflection about a line or a rotation, and that every rotation is the product of two reflections about some lines. In general, since det(s) = −1 for a reflection s, when n ≥ 3 is odd, every rotation is the product of an even number less than or equal 当n=2时，反射是关于一条直线的反射，定理26.1表明O（2）中的每个等距线要么是关于一条直线的反射，要么是关于一个旋转的反射，并且每个旋转都是关于一些直线的两个反射的乘积。一般来说，由于反射s的Det（s）=-1，当n≥3为奇数时，每个旋转都是小于或等于偶数的乘积。

Figure 26.4: An isometry f as a composition of reflections when 1 is not an eigenvalue of f. Note that the pink plane H is perpendicular to f(w) − w. 图26.4：当1不是f的特征值时，作为反射组成的等距f。注意，粉红色平面h垂直于f（w）−w。 to n − 1 of reflections, and when n is even, every improper orthogonal transformation is then − 1 of reflections. product of an odd number less than or equal to 到反射的n−1，当n为偶数时，每一个不正确的正交变换都是反射的−1。奇数小于或等于的积 In particular, for n = 3, every rotation is the product of two reflections about planes. When n is odd, we can say more about improper isometries. Indeed, when n is odd, every improper isometry admits the eigenvalue −1. This is because iff(u)kλ, then=Ekuis a Euclidean space ofk for every u ∈ E, if λ finite dimension and f : E → E is an isometry, because k is any eigenvalue of f and u is an eigenvector associated with 特别是，对于n=3，每次旋转都是平面上两个反射的产物。当n是奇数时，我们可以说更多关于不适当等距的内容。实际上，当n为奇数时，每一个不适当的等距测量都承认特征值−1。这是因为if f（u）kλ，那么=ekui是k的欧几里得空间，对于每个u∈e，如果λ有限维和f:e→e是一个等距，因为k是f的任何特征值，u是与 kf(u)k = kλuk = |λ|kuk = kuk, kf（u）k=kλuk=λkuk=kuk， which implies |λ| = 1, since u 6= 0. Thus, the real eigenvalues of an isometry are either 这意味着λ=1，因为u 6=0。因此，等距测量的实际特征值可以是 real root. As a consequence, the characteristic polynomial det( 真正的根。因此，特征多项式（ +1root, which is either +1 ordet(orf) −is the product of the eigenvalues, the real roots cannot all be +1, and thus1. However, it is well known that polynomials of odd degree always have some−1. Since f is an improper isometry, det(f − λid) off) =f −has some real1, and since−1 is anf, +1根，要么是+1，要么是（ORF）−是特征值的乘积，实际根不能都是+1和Thus1。然而，众所周知，奇数度的多项式总是有一些−1。因为f是一个不正确的等距测量，所以det（f−λid）off）=f−有一些real1，因为−1是anf， eigenvalue ofthere is some nonnull eigenvectorproof, we see that the hyperplanef. Going back to the proof of Theorem 26.1, sincew such that f(w) = −w. Using the second part of thew = −−21wis an eigenvalue ofis in fact orthogonalH,F1,…,Fk−1 这里的特征值是一些非空的特征向量，我们看到了超平面。回到定理26.1的证明，因为f（w）=−w。使用w=−21wis的第二部分，一个事实上正交的特征值，f1，…，fk−1 H orthogonal to f(w) − h与f（w）正交- to w, and thus f is the product of k ≤ n reflections about hyperplanesH, and the Hi are hyperplanes in such that Fi = Hi ⊕ L, where L is a line orthogonal to 到w，因此f是关于超平面的k≤n反射的产物，hi是超平面，其中fi=hi l，其中l是与 H = L⊥ orthogonal to L. However, k must be odd, and so k − 1 is even, and thus then is odd, an composition of the reflections about F1,…,Fk−1 is a rotation. Thus, when improper isometry is the composition of a reflection about a hyperplane H with a rotation consisting of reflections about hyperplanes F1,…,Fk−1 containing a line, L, orthogonal to H=L与L正交。但是，K必须是奇数，因此K−1是偶数，因此是奇数，关于f1，…，fk−1的反射的组成是旋转。因此，当不适当的等距测量是关于超平面H的反射的组成，其旋转包括关于超平面F1的反射，…，fk−1，其中包含一条直线，l，与 H. In particular, when n = 3, every improper orthogonal transformation is the product of a rotation with a reflection about a plane orthogonal to the axis of rotation. 特别是，当n=3时，每一个不适当的正交变换都是一个旋转与一个与旋转轴正交的平面上的反射的乘积。 Using Theorem 26.1, we can also give a rather simple proof of the classical fact that in a Euclidean space of odd dimension, every rotation leaves some nonnull vector invariant, and thus a line invariant. 利用定理26.1，我们还可以给出一个相当简单的证明，证明在奇数维的欧几里得空间中，每一个旋转都会留下一些非零向量不变量，因此是一个线不变量。 If λ is an eigenvalue of f, then the following lemma shows that the orthogonal complement Eλ(f)⊥ of the eigenspace associated with λ is closed under f. 如果λ是f的特征值，那么下面的引理表明与λ相关的特征空间的正交补码eλ（f）在f下闭合。 Proposition 26.2. Let E be a Euclidean space of finite dimension n, and let f : E → E be an isometry. For any subspace F of E, if f(F) = F, then f(F ⊥) ⊆ F ⊥ and E = F ⊕ F ⊥. 提案26.2.设e为有限维n的欧几里得空间，设f:e→e为等距线。对于e的任何子空间f，如果f（f）=f，则f（f）f和e=f f。 Proof. We just have to prove that if w ∈ E is orthogonal to every u ∈ F, then f(w) is also orthogonal to every u ∈ F. However, since f(F) = F, for every v ∈ F, there is some u ∈ F such that f(u) = v, and we have 证据。我们只需要证明，如果w∈e与每一个u∈f正交，那么f（w）也与每一个u∈f正交。然而，由于f（f）=f，对于每一个v∈f，有一些u∈f，这样f（u）=v，我们得到 f(w) · v = f(w) · f(u) = w · u, f（w）·v=f（w）·f（u）=w·u， since f is an isometry. Since we assumed that w ∈ E is orthogonal to every u ∈ F, we have 因为f是等距线。既然我们假设w∈e与每一个u∈f是正交的，我们有 w · u = 0, w·u=0， and thus f(w) · v = 0, 因此f（w）·v=0， and this for every v ∈ F. Thus, f(F ⊥) ⊆ F ⊥. The fact that E = F ⊕ F ⊥ follows from Lemma 11.11.
对于每一个v∈f，因此，f（f）f。e=f_f这一事实源自引理11.11。 Lemma 26.2 is the starting point of the proof that every orthogonal matrix can be diagonalized over the field of complex numbers. Indeed, if λ is any eigenvalue of f, then f(Eλ(f)) = Eλ(f), where Eλ(f) is the eigenspace associated with λ, and thus the orthogonal Eλ(f)⊥ is closed under f, and E = Eλ(f) ⊕ Eλ(f)⊥. The problem over R is that there may not be any real eigenvalues. However, when n is odd, the following lemma shows that every rotation admits 1 as an eigenvalue (and similarly, when n is even, every improper orthogonal transformation admits 1 as an eigenvalue). 引理26.2是证明每个正交矩阵都可以在复数域上对角化的起点。实际上，如果λ是f的特征值，那么f（eλ（f））=eλ（f），其中eλ（f）是与λ相关的特征空间，因此正交eλ（f）在f下闭合，e=eλ（f）eλ（f）。R上的问题是可能没有任何实际的特征值。然而，当n是奇数时，下面的引理表明每个旋转都承认1为特征值（同样，当n是偶数时，每个不适当的正交变换都承认1为特征值）。 Proposition 26.3. Let E be a Euclidean space. 提案26.3.设e为欧几里得空间。 (1) If E has odd dimension n = 2m + 1, then every rotation f admits 1 as an eigenvalue and the eigenspace F of all eigenvectors left invariant under f has an odd dimension 2p + 1. Furthermore, there is an orthonormal basis of E, in which f is represented by a matrix of the form 如果e的奇数维数n=2 m+1，那么每个旋转f都承认1为特征值，而f下所有特征向量的特征空间f都是奇数维数2p+1。此外，还有一个e的正交基，其中f由形式的矩阵表示。 , ， where R2(m−p) is a rotation matrix that does not have 1 as an eigenvalue. 其中，r2（m-p）是一个旋转矩阵，没有1作为特征值。 (2) If E has even dimension n = 2m, then every improper orthogonal transformation f admits 1 as an eigenvalue and the eigenspace F of all eigenvectors left invariant under f has an odd dimension 2p + 1. Furthermore, there is an orthonormal basis of E, in which f is represented by a matrix of the form 如果e的维数为偶数n=2 m，则每一个不适当的正交变换f都承认1为特征值，而f下所有特征向量的特征空间f都是奇数维2p+1。此外，还有一个e的正交基，其中f由形式的矩阵表示。 , ， where S2(m−p)−1 is an improper orthogonal matrix that does not have 1 as an eigenvalue. 其中s2（m−p）−1是不适当的正交矩阵，没有1作为特征值。 Proof. We prove only (1), the proof of (2) being similar. Since f is a rotation and n = 2m+1 is odd, by Theorem 26.1, f is the composition of an even number less than or equal to 2m of reflections. From Lemma 23.15, recall the Grassmann relation 证据。我们只证明（1），证明（2）相似。由于f是一个旋转，n=2m+1是奇数，根据定理26.1，f是小于或等于2m反射的偶数的组合。从引理23.15，回忆格拉斯曼关系 dim(M) + dim(N) = dim(M + N) + dim(M ∩ N), 尺寸（m）+尺寸（n）=dim（m+n）+尺寸（m n） where M and N are subspaces of E. Now, if M and N are hyperplanes, their dimension is n − 1, and thus dim(M ∩ N) ≥ n − 2. Thus, if we intersect k ≤ n hyperplanes, we see that the dimension of their intersection is at least n − k. Since each of the reflections is the identity on the hyperplane defining it, and since there are at most 2m = n − 1 reflections, their composition is the identity on a subspace of dimension at least 1. This proves that 1 is an eigenvalue of f. Let F be the eigenspace associated with 1, and assume that its dimension is q. Let G = F ⊥ be the orthogonal of F. By Lemma 26.2, G is stable under f, and E = F ⊕ G. Using Lemma 11.10, we can find an orthonormal basis of E consisting of an orthonormal basis for G and orthonormal basis for F. In this basis, the matrix of f is of the form 其中m和n是e的子空间。现在，如果m和n是超平面，那么它们的尺寸是n-1，因此dim（m n）≥n-2。因此，如果我们与k≤n超平面相交，我们可以看到它们相交的尺寸至少为n−k。由于每个反射都是定义它的超平面上的同一性，并且由于反射最多为2 m=n−1，因此它们的组成是Dimen子空间上的同一性。SION至少1.这证明了1是f的特征值。让f是与1相关的特征空间，并假设其维数为q。让g=f是f的正交。由引理26.2，g在f下是稳定的，e=f_g。利用引理11.10，我们可以找到由正交b组成的e的正交基。对于g和f的正交基，asis。在这个基中，f的矩阵是形式的。 . . Thus, det(f) = det(R), and R must be a rotation, since f is a rotation and det(f) = 1. Now, if f left some vector u = 06 in G invariant, this vector would be an eigenvector for 1, and we would have u ∈ F, the eigenspace associated with 1, which contradicts E = F ⊕ G. Thus, by the first part of the proof, the dimension of G must be even, since otherwise, the restriction of f to G would admit 1 as an eigenvalue. Consequently, q must be odd, and R does not admit 1 as an eigenvalue. Letting q = 2p + 1, the lemma is established. 因此，det（f）=det（r），r必须是一个旋转，因为f是一个旋转，det（f）=1。现在，如果f在g不变量中留下一个向量u=06，这个向量将是1的一个特征向量，我们将得到u∈f，这个与1相关的特征空间，它与e=f g相矛盾，因此，根据证明的第一部分，g的维数必须是偶数，否则，f对g的限制是偶数。将1作为特征值。因此，q必须是奇数，r不接受1作为特征值。设q=2p+1，建立引理。 An example showing that Lemma 26.3 fails for n even is the following rotation matrix (when n = 2): 一个例子表明，引理26.3对于n是失败的，甚至是如下的旋转矩阵（当n=2时）： . . The above matrix does not have real eigenvalues for θ =6 kπ. 上面的矩阵没有θ=6 kπ的实特征值。 It is easily shown that for n = 2, with respect to any chosen orthonormal basis (e1, e2), every rotation is represented by a matrix of form 可以很容易地证明，对于n=2，对于任何选定的正交基（e1，e2），每个旋转都由一个形式的矩阵表示。

where θ ∈ [0,2π[, and that every improper orthogonal transformation is represented by a matrix of the form 式中θ∈[0,2π[，并且每一个不适当的正交变换都由一个形式的矩阵表示。 . . In the first case, we call θ ∈ [0,2π[ the measure of the angle of rotation of R w.r.t. the orthonormal basis (e1, e2). In the second case, we have a reflection about a line, and it is easy to determine what this line is. It is also easy to see that S is the composition of a reflection about the x-axis with a rotation (of matrix R). 在第一种情况下，我们称之为θ∈[0,2π[r w.r.t旋转角度的测量.正交基（e1，e2）.在第二种情况下，我们对一条线进行了反射，很容易确定这条线是什么。也很容易看出，S是围绕X轴旋转（矩阵R）的反射的组合。  We refrained from calling θ “the angle of rotation,” because there are some subtleties involved in defining rigorously the notion of angle of two vectors (or two lines). For example, note that with respect to the “opposite basis” (e2, e1), the measure θ must be changed to 2π − θ (or −θ if we consider the quotient set R/2π of the real numbers modulo 我们避免称θ为“旋转角度”，因为严格定义两个向量（或两条线）的角度概念涉及到一些微妙之处。例如，请注意，对于“相反基”（e2，e1），如果我们考虑实数模的商集r/2π，则测量θ必须更改为2π−θ（或−θ）。 2π). 2π）。 It is easily shown that the group SO(2) of rotations in the plane is abelian. First, recall that every plane rotation is the product of two reflections (about lines), and that every isometry in O(2) is either a reflection or a rotation. To alleviate the notation, we will omit the composition operator ◦, and write rs instead of r ◦ s. Now, if r is a rotation and s is a reflection, rs being in O(2) must be a reflection (since det(rs) = det(r)det(s) = −1), and thus (rs)2 = id, since a reflection is an involution, which implies that 很容易证明平面上的转动群是阿贝尔的。首先，回想一下，每个平面的旋转都是两个反射（关于直线）的乘积，而O（2）中的每个等距线都是反射或旋转。为了减少符号，我们将省略组合运算符，并写r s而不是r_s。现在，如果r是一个旋转，s是一个反射，r在o（2）中必须是一个反射（因为det（rs）=det（r）det（s）=1），因此（rs）2=id，因为反射是一个对合，这意味着那个 srs = r−1. SRS=R−1。 Then, given two rotations r1 and r2, writing r1 as r1 = s2s1 for two reflections s1,s2, we have 然后，给定两个转动R1和R2，将R1写为两个反射的R1=S2S1，s1，s2，我们得到 r1r2r1−1 = s2s1r2(s2s1)−1 = s2s1r2s−1 1s−2 1 = s2s1r2s1s2 = s2r2−1s2 = r2, r1r2r1−1=s2s1 r2（s2s1）−1=s2s1r2−1 1s−2 1=s2s1r2s1s2=s2r2−1s2=r2， since srs = r−1 for all reflections s and rotations r, and thus r1r2 = r2r1. 因为所有反射和旋转的s r s=r−1，因此r1r2=r2r1。 We can also perform the following calculation, using some elementary trigonometry: 我们还可以使用一些初等三角法执行以下计算： . . The above also shows that the inverse of a rotation matrix 上面还显示了旋转矩阵的逆矩阵

is obtained by changing θ to −θ (or 2π − θ). Incidentally, note that in writing a rotation r as the product of two reflections r = s2s1, the first reflection s1 can be chosen arbitrarily, since, and rs1 is a reflection. 通过将θ更改为−θ（或2π−θ）获得。顺便说一句，请注意，在写一个旋转r作为两个反射r=s2s1的乘积时，第一个反射s1可以任意选择，因为，rs1是一个反射。 For n = 3, the only two choices for p are p = 1, which corresponds to the identity, or p = 0, in which case f is a rotation leaving a line invariant. This line D is called the axis of 对于n=3，p的唯一两个选择是p=1，它对应于恒等式，或者p=0，在这种情况下，f是一个保留直线不变量的旋转。这条线称为

Figure 26.5: 3D rotation as the composition of two reflections. 图26.5：两个反射组成的三维旋转。 rotation. The rotation R behaves like a two-dimensional rotation around the axis of rotation. Thus, the rotation R is the composition of two reflections about planes containing the axis of rotation D and forming an angle θ/2. This is illustrated in Figure 26.5. 旋转。旋转R的行为类似于围绕旋转轴的二维旋转。因此，旋转r是包含旋转d轴并形成角度θ/2的平面上的两个反射的组成。如图26.5所示。 The measure of the angle of rotation θ can be determined through its cosine via the formula cosθ = u · R(u), 旋转角θ的测量可以通过其余弦公式cosθ=u·r（u）来确定。 where u is any unit vector orthogonal to the direction of the axis of rotation. However, this does not determine θ ∈ [0,2π[ uniquely, since both θ and 2π − θ are possible candidates. What is missing is an orientation of the plane (through the origin) orthogonal to the axis of rotation. 其中u是与旋转轴方向垂直的任何单位向量。然而，由于θ和2π−θ都是可能的候选者，因此这不能确定θ∈[0,2π[唯一的]。缺少的是与旋转轴垂直的平面方向（通过原点）。 In the orthonormal basis of the lemma, a rotation is represented by a matrix of the form 在引理的正交基中，旋转由形式的矩阵表示。 . . Remark: For an arbitrary rotation matrix A, since a11 + a22 + a33 (the trace of A) is the sum of the eigenvalues of A, and since these eigenvalues are cosθ +isinθ, cosθ −isinθ, and 1, for some θ ∈ [0,2π[, we can compute cosθ from 注：对于任意旋转矩阵a，由于a11+a22+a33（a的迹线）是a的特征值之和，并且由于这些特征值是cosθ+isinθ，cosθ−isinθ，和1，对于某些θ∈[0,2π[，我们可以从 1 + 2cosθ = a11 + a22 + a33. 1+2cosθ=A11+A22+A33。 It is also possible to determine the axis of rotation (see the problems). 也可以确定旋转轴（参见问题）。 An improper transformation is either a reflection about a plane or the product of three reflections, or equivalently the product of a reflection about a plane with a rotation, and we noted in the discussion following Theorem 26.1 that the axis of rotation is orthogonal to the plane of the reflection. Thus, an improper transformation is represented by a matrix of the form 不适当的变换是关于一个平面的反射或三个反射的乘积，或者是关于一个平面的反射与一个旋转的乘积，我们在定理26.1的讨论中注意到，旋转轴与反射面的平面正交。因此，不适当的变换用形式矩阵表示。 . . When n ≥ 3, the group of rotations SO(n) is not only generated by hyperplane reflections, but also by flips (about subspaces of dimension n − 2). We will also see, in Section 26.2, that every proper affine rigid motion can be expressed as the composition of at most n flips, which is perhaps even more surprising! The proof of these results uses the following key lemma. 当n≥3时，旋转组so（n）不仅由超平面反射产生，也由翻转产生（关于尺寸n-2的子空间）。我们还将在第26.2节中看到，每一个适当的仿射刚性运动都可以表示为最多n个翻转的组合，这可能更令人惊讶！这些结果的证明使用以下关键引理。 Proposition 26.4. Given any Euclidean space E of dimension n ≥ 3, for any two reflections h1 and h2 about some hyperplanes H1 and H2, there exist two flips f1 and f2 such that h2 ◦ h1 = f2 ◦ f1. 提案26.4.对于尺寸n≥3的任何欧几里得空间e，对于某些超平面h1和h2的任意两个反射h1和h2，存在两个翻转f1和f2，使得h2 h1=f2 f1。 Proof. If h1 = h2, it is obvious that 证据。如果h1=h2，很明显 h1 ◦ h2 = h1 ◦ h1 = id = f1 ◦ f1 h1 h2=h1 h1=id=f1 f1 for any flip f1. If h1 =6 h2, then H1 ∩ H2 = F, where dim(F) = n − 2 (by the Grassmann relation). We can pick an orthonormal basis (e1,…,en) of E such that (e1,…,en−2) is an orthonormal basis of F. We can also extend (e1,…,en−2) to an orthonormal basis 对于任何翻转F1。如果h1=6 h2，则h1 h2=f，其中dim（f）=n−2（根据格拉斯曼关系）。我们可以选择e的正交基（e1，…，en），这样（e1，…，en-2）就是f的正交基。我们也可以将（e1，…，en-2）扩展到正交基。 (e1,…,en−2,u1,v1) of E, where (e1,…,en−2,u1) is an orthonormal basis of H1, in which case e的（e1，…，en-2，u1，v1），其中（e1，…，en-2，u1）是h1的正态基，在这种情况下 en−1 = cosθ1 u1 + sinθ1 v1, en = sinθ1 u1 − cosθ1 v1, en−1=cosθ1 u1+sinθ1 v1，en=sinθ1 u1−cosθ1 v1， for some θ1 ∈ [0,2π]. See Figure 26.6 对于某些θ1∈[0,2π]。见图26.6 Since h1 is the identity on H1 and v1 is orthogonal to H1, it follows that h1(u1) = u1, h1(v1) = −v1, and we get 因为h1是h1上的单位，v1与h1是正交的，所以h1（u1）=u1，h1（v1）=-v1，我们得到 , ， After some simple calculations, we get 经过一些简单的计算，我们得到 h1(en−1) = cos2θ1 en−1 + sin2θ1 en, h1(en) = sin2θ1 en−1 − cos2θ1 en. h1（en-1）=cos2θ1 en-1+sin2θ1 en，h1（en）=sin2θ1 en-1−cos2θ1 en。

Figure 26.6: An illustration of the hyperplanes H1, H2, their intersection F, and the two orthonormal basis utilized in the proof of Proposition 26.4. 图26.6：证明26.4的超平面h1、h2及其交点f和两个正交基的图解。 As a consequence, the matrix A1 of h1 over the basis (e1,…,en) is of the form 因此，基（e1，…，en）上的h1矩阵a1的形式为 . . Similarly, the matrix A2 of h2 over the basis (e1,…,en) is of the form 同样，基（e1，…，en）上的h2矩阵a2的形式为 . . Observe that both A1 and A2 have the eigenvalues −1 and +1 with multiplicity n − 1. The trick is to observe that if we change the last entry in In−2 from +1 to −1 (which is possible since n ≥ 3), we have the following product A2A1: 观察A1和A2都具有多重性n-1的特征值−1和+1。诀窍是观察到，如果我们将−2中的最后一个条目从+1更改为−1（这是可能的，因为n≥3），我们得到以下产品A2A1： . . Now, the two matrices above are clearly orthogonal, and they have the eigenvalues −1,−1, and +1 with multiplicity n−2, which implies that the corresponding isometries leave invariant a subspace of dimension n − 2 and act as −id on its orthogonal complement (which has dimension 2). This means that the above two matrices represent two flips f1 and f2 such that h2 ◦ h1 = f2 ◦ f1. See Figure 26.7.
现在，上面的两个矩阵显然是正交的，它们具有多重性n−2的特征值−1、−1和+1，这意味着相应的等轴测保持不变的维度n−2的子空间，并在其正交补码（具有维度2）上充当−id。这意味着上述两个矩阵表示两个翻转f1和f2，使得h2 h1=f2 f1。见图26.7。

Figure 26.7: The conversion of the hyperplane reflection h1 into the flip or 180◦ rotation around the green axis in the e2e3-plane. The green axis corresponds to the restriction of the eigenspace associated with eigenvalue 1. 图26.7：将超平面反射h1转换为翻转或围绕e2e3平面中绿色轴180°旋转。绿轴对应于与特征值1相关的特征空间的限制。 Using Lemma 26.4 and the Cartan–Dieudonn´e theorem, we obtain the following characterization of rotations when n ≥ 3. 利用引理26.4和卡坦-迪乌顿定理，我们得到了当n≥3时旋转的以下特征。 Theorem 26.5. Let E be a Euclidean space of dimension n ≥ 3. Every rotation f ∈ SO(E) is the composition of an even number of flips f = f2k ◦···◦ f1, where 2k ≤ n. Furthermore, if u = 06 is invariant under f (i.e., u ∈ Ker(f − id)), we can pick the last flip f2k such that , where F2k is the subspace of dimension n − 2 determining f2k. 定理26.5。设e为尺寸n≥3的欧几里得空间。每个旋转f∈so（e）是偶数个翻转f=f2k········f1的组合，其中2k≤n。此外，如果u=06在f（即u∈ker（f−id））下不变，我们可以选取最后一个翻转f2k，其中f2k是确定f2k的维度n−2的子空间。 Proof. By Theorem 26.1, the rotation f can be expressed as an even number of hyperplane reflections f = s2k◦s2k−1◦···◦s2◦s1, with 2k ≤ n. By Lemma 26.4, every composition of two reflections s2i ◦ s2i−1 can be replaced by the composition of two flips f2i ◦ f2i−1 (1 ≤ i ≤ k), which yields f = f2k ◦ ··· ◦ f1, where 2k ≤ n. 证据。根据定理26.1，旋转f可以表示为偶数个超平面反射f=s2 k s2k−1····s2 s1，其中2k≤n。根据引理26.4，两个反射的每一个组成s2 i s2i−1可以被两个翻转的组成f2i f2i−1（1≤i≤k）所代替，其中h产生f=f2k····f1，其中2k≤n。 Assume that f(u) = u, with u = 06 . We have already made the remark that in the case where 1 is an eigenvalue of f, the proof of Theorem 26.1 shows that the reflections si can be chosen so that si(u) = u. In particular, if each reflection si is a reflection about the hyperplane Hi, we have u ∈ H2k−1 ∩ H2k. Letting F = H2k−1 ∩ H2k, pick an orthonormal basis (e1,…,en−3,en−2) of F, where 假设f（u）=u，其中u=06。我们已经指出，在1是f的特征值的情况下，定理26.1的证明表明可以选择反射si，以便si（u）=u。特别是，如果每个反射si是关于超平面hi的反射，我们有u∈h2k h2k。让f=h2k−1h2k，选取f的正交基（e1，…，en-3，en-2），其中 . .

26.2. AFFINE ISOMETRIES (RIGID MOTIONS) 26.2。仿射等距线（刚性运动） The proof of Lemma 26.4 yields two flips f2k−1 and f2k such that 引理26.4的证明产生了两个翻转f2k−1和f2k，这样 f2k(en−2) = −en−2 and s2k ◦ s2k−1 = f2k ◦ f2k−1, f2k（en-2）=-en-2和s2k s2k-1=f2k f2k-1， since the (n − 2)th diagonal entry in both matrices is −1, which means that , where F2k is the subspace of dimension n − 2 determining f2k. Since u = kuken−2, we also have.
因为这两个矩阵中的（n-2）第（n-2）个对角线项都是−1，这意味着，其中f2k是确定f2k的维度n-2的子空间。由于u=kuken−2，我们也有。 Remarks: 评论： (1) It is easy to prove that if f is a rotation in SO(3) and if D is its axis and θ is its angle of rotation, then f is the composition of two flips about lines D1 and D2 orthogonal to D and making an angle θ/2. 可以很容易地证明，如果f是在so（3）中的旋转，如果d是它的轴，θ是它的旋转角，那么f是关于与d正交的d1和d2线的两个翻转的组合，并形成一个角θ/2。 (2) It is natural to ask what is the minimal number of flips needed to obtain a rotation f (when n ≥ 3). As for arbitrary isometries, we will prove later that every rotation is the composition of k flips, where 自然会问，获得旋转f所需的最小翻转次数是多少（当n≥3时）。对于任意等距图，我们稍后将证明每个旋转都是k翻转的组合，其中 k = n − dim(Ker(f − id)), k=n−dim（ker（f−id））， and that this number is minimal (where n = dim(E)). 这个数字是最小的（其中n=dim（e））。 We now turn to affine isometries. 现在我们来看看仿射等距图。 26.2 Affine Isometries (Rigid Motions) 26.2仿射等距图（刚性运动） In the remaining sections we study affine isometries. First, we characterize the set of fixed points of an affine map. Using this characterization, we prove that every affine isometry f can be written uniquely as 在剩下的部分，我们研究仿射等距线。首先，我们描述了仿射映射的不动点集。利用这个特征，我们证明了每一个仿射同构f都可以唯一地写成 f = t ◦ g, with t ◦ g = g ◦ t, f=t_g，其中t_g=g_t， where g is an isometry having a fixed point, and t is a translation by a vector τ such that →−f (τ) = τ, and with some additional nice properties (see Theorem 26.10). This is a generalization of a classical result of Chasles about (proper) rigid motions in R3 (screw motions). We prove a generalization of the Cartan–Dieudonn´e theorem for the affine isometries: Every isometry in Is(n) can be written as the composition of at most n affine reflections if it has a fixed point, or else as the composition of at most n+2 affine reflections. We also prove that every rigid motion in SE(n) is the composition of at most n affine flips (for n ≥ 3). This is somewhat surprising, in view of the previous theorem. 其中g是一个具有固定点的等距测量，t是一个矢量τ的平移，使得→−f（τ）=τ，并具有一些附加的优良性质（见定理26.10）。这是关于r3（螺旋运动）中（适当的）刚性运动的裂缝经典结果的推广。我们证明了仿射等距线的卡坦-迪乌顿定理的推广：在is（n）中的每个等距线如果有固定点，可以写成至多n个仿射反射的合成，或者写成至多n+2个仿射反射的合成。我们还证明了SE（n）中的每一个刚性运动都是至多n个仿射翻转（n≥3）的组合。根据前面的定理，这有点令人惊讶。 Definition 26.1. Given any two nontrivial Euclidean affine spaces E and F of the same finite dimension n, a function f : E → F is an affine isometry (or rigid map) if it is an affine map and 定义26.1.给定任意两个非平凡欧几里得仿射空间e和f的相同有限维n，函数f:e→f是仿射等值线（或刚性映射），如果它是仿射映射，并且 , ， for all a,b ∈ E. When E = F, an affine isometry f : E → E is also called a rigid motion. 对于所有a，b∈e，当e=f时，仿射等距f:e→e也被称为刚性运动。 Thus, an affine isometry is an affine map that preserves the distance. This is a rather strong requirement. In fact, we will show that for any function f : E → F, the assumption that 因此，仿射等值线是保持距离的仿射图。这是一个相当强烈的要求。事实上，我们将证明对于任何函数f:e→f，假设 , ， for all a,b ∈ E, forces f to be an affine map. 对于所有a，b∈e，强制f是仿射映射。 Remark: Sometimes, an affine isometry is defined as a bijective affine isometry. When E and F are of finite dimension, the definitions are equivalent. 注：有时，仿射等值线被定义为双射仿射等值线。当e和f为有限维时，定义是等价的。 The following simple lemma is left as an exercise. 下面的简单引理是作为练习留下的。 Proposition 26.6. Given any two nontrivial Euclidean affine spaces E and F of the same finite dimension , an affine map f : E → F is an affine isometry iff its associated linear map f : E → F is an isometry. An affine isometry is a bijection. 提案26.6.给定任意两个相同有限维的非平凡欧几里德仿射空间e和f，仿射映射f:e→f是仿射等值线，而其相关线性映射f:e→f是等值线。仿射等值线是双射。 Let us now consider affine isometries→− f : E → E. If →−f is a rotation, we call f a proper (or direct) affine isometry, and if f is an improper linear isometry, we call f an improper (or skew) affine isometry. It is easily shown that the set of affine isometries f : E → E forms a group, and those for which →−f is a rotation is a subgroup. The group of affine isometries, or rigid motions, is a subgroup of the affine group GA(E), denoted by Is(E) (or Is(n) when E = En). In Snapper and Troyer [157] the group of rigid motions is denoted by Mo(E). Since we denote the group of affine bijections as GA(E), perhaps we should denote the group of affine isometries by IA(E) (or EA(E)!). The subgroup of Is(E) consisting of the direct rigid motions is also a subgroup of SA(E), and it is denoted by SE(E) (or SE(n), when E = En). The translations are the affine isometries f for which = id, the identity map on →−E. The following lemma is the counterpart of Lemma 11.12 for isometries between Euclidean vector spaces. 现在让我们考虑仿射等轴测→−f:e→e。如果→−f是一个旋转，我们称之为适当（或直接）仿射等轴测，如果f是一个不适当的线性等轴测，我们称之为不适当（或歪斜）仿射等轴测。可以很容易地看出，仿射等距图f:e→e构成一个群，其中→−f是一个旋转的群是一个子群。仿射轴测或刚性运动组是仿射组ga（e）的一个子组，表示为is（e）（e=e n时为（n）。在Snapper和Troyer[157]中，刚性运动组用mo（e）表示。既然我们把仿射双射群表示为ga（e），也许我们应该用ia（e）（或ea（e）！）来表示仿射等轴测群。.由直接刚性运动组成的IS（E）子组也是SA（E）子组，当E=e n时，用SE（E）（或SE（N）表示。翻译是仿射等轴测图f，其中=id，在→−e上的标识映射。下面的引理是引理11.12的对应项，用于欧几里得向量空间之间的等轴测图。 Proposition 26.7. Given any two nontrivial Euclidean affine spaces E and F of the same finite dimension n, for every function f : E → F, the following properties are equivalent: 提案26.7。给定任意两个非平凡欧几里得仿射空间e和f，对于每个函数f:e→f，下列性质是等价的： (1) f is an affine map and kf−−−−−(a)f(→b)k = k→−abk, for all a,b ∈ E. （1）f是仿射图，kf−−−−（a）f（→b）k=k→−abk，对于所有a，b∈e。 , for all a,b ∈ E. ，对于所有a，b∈e。 Proof. Obviously, (1) implies (2). In order to prove that (2) implies (1), we proceed as follows. First, we pick some arbitrary point Ω ∈ E. We define the map g: →−E → →−F such that for all u ∈ E. Since 证据。显然，（1）意味着（2）。为了证明（2）意味着（1），我们继续如下。首先，我们选取一些任意的点Ω∈E，定义了图G：→−E→→−F，这样对于所有的u∈E。

26.3. FIXED POINTS OF AFFINE MAPS 26.3。仿射映射的不动点 for all u ∈ →−E, f will be affine if we can show thatg is a linear isometry. g is linear, and f will be an affine isometry if we can show that 对于所有u∈→−e，如果我们能证明g是一个线性等距，f将是仿射的。G是线性的，如果我们能证明，F是仿射等距线。 Observe that 注意

Then, the hypothesis 那么，假设 k−−−−−f(a)f(→b)k = k→−abk K−−−F（A）F（→B）K=K→−ABK for all a,b ∈ E, implies that 对于所有a，b∈e，意味着 . . Thus, g preserves the distance. Also, by definition, we have 因此，G保持距离。而且，根据定义，我们 g(0) = 0. G（0）=0。 Thus, we can apply Lemma 11.12, which shows that g is indeed a linear isometry, and thus f is an affine isometry.
因此，我们可以应用引理11.12，它表明g确实是一个线性等距，因此f是一个仿射等距。 In order to understand the structure of affine isometries, it is important to investigate the fixed points of an affine map. 为了了解仿射等距图的结构，研究仿射图的不动点十分重要。 26.3 Fixed Points of Affine Maps 26.3仿射图的不动点 Recall that denotes the eigenspace of the linear map →−f associated with the scalar 表示线性映射的特征空间→−f与标量相关 Ker →− →− . Given some origin Ω ∈ E, since∈ →−E such that →−f (u) = u. Clearly, 1, that is, the subspace consisting of all vectors u KER––––––。给定一个原点，Ω∈e，因为∈→−e使得→−f（u）=u。很明显，1，也就是说，由所有向量u组成的子空间 f(a) = f(Ω + −Ω→a) = f(Ω) + →−f (−Ω→a), F（A）=F（欧姆+欧姆→A）=F（欧姆）+→F（−欧姆→A）， we have), and thus 我们有），因此 . . From the above, we get 从上面我们可以看到

Using this, we show the following lemma, which holds for arbitrary affine spaces of finite dimension and for arbitrary affine maps. 利用这个，我们给出了以下引理，它适用于有限维的任意仿射空间和任意仿射映射。 Proposition 26.8. Let E be any affine space of finite dimension. For every affine map f : E → E, let Fix(f) = {a ∈ E | f(a) = a} be the set of fixed points of f. The following properties hold: 提案26.8。设e为有限维的任意仿射空间。对于每一个仿射映射f:e→e，让fix（f）=a∈e f（a）=a为f的不动点集。以下属性成立： (1) If f has some fixed point a, so that Fix(f) 6= ∅, then Fix(f) is an affine subspace of E such that 如果f有不动点a，那么fix（f）6=∅，那么fix（f）是e的仿射子空间，这样 Fix(, 修复，（ where is the eigenspace of the linear map →−f for the eigenvalue 1. 其中，线性映射的特征空间→−f表示特征值1。 (2) The affine map f has a unique fixed point iff = Ker . 仿射映射f有一个唯一的固定点iff=ker。 Proof. (1) Since the identity 证据。（1）自身份

holds for all Ω,b ∈ E, if f(a) = a, then −−−af(→a) = 0, and thus, letting Ω = a, for any b ∈ E we have and so 对于所有的Ω，b∈e，如果f（a）=a，那么−−af（→a）=0，因此，对于任何b∈e，让Ω=a，我们有，因此 iff 敌我识别 −−−af(→b) − →−ab = 0 −−af（→b）−→−ab=0 iff 敌我识别 →−f (→−ab) →−ab = 0 →−F（→−AB）→−AB=0 iff 敌我识别 , ， which proves that 这证明了 Fix(. 修理（…） (2) Again, fix some origin Ω. Some a satisfies f(a) = a iff （2）再次，固定一些原点Ω。一些a满足f（a）=iff −−−Ωf(→a) − −Ω→a = 0 −−ΩF（→A）−−Ω→A=0 iff 敌我识别 −−−−Ωf(Ω) +→ →−f (−Ω→a) −Ω→a = 0, −−−欧F（欧）+→→−F（−欧→A）−欧→A=0， which can be rewritten as 可以重写为 We have = Ker id is injective, and since →−E has finite 我们有=KER ID是内射的，因为→−E是有限的 dimension, f − id is also surjective, and thus, there is indeed some a ∈ E such that 维度，f−id也是主观的，因此，确实有一些a∈e这样 , ，

and it is unique, since →−f − id is injective. Conversely, if f has a unique fixed point, say a, from 它是独一无二的，因为→−F−ID是注射的。相反，如果f有一个唯一的固定点，比如a，from , ， we have (Ω) = Ω, and since a is the unique fixed point of f, we must have a = Ω, which shows that →−f − id is injective. 我们有（Ω）=Ω，由于a是f的唯一固定点，我们必须有a=Ω，这表明→−f−id是内射的。 Remark: The fact that E has finite dimension is used only to prove (2), and (1) holds in general. 注：E有有限维的事实仅用于证明（2），（1）一般成立。 If an affine isometry f leaves some point fixed, we can take such a point Ω as the origin, and then f(Ω) = Ω and we can view f as a rotation or an improper orthogonal transformation, depending on the nature of →−f . Note that it is quite possible that Fix(f) = ∅. For example, nontrivial translations have no fixed points. A more interesting example is provided by the composition of a plane reflection about a line composed with a a nontrivial translation parallel to this line. 如果一个仿射等值线f离开某个固定点，我们可以取这样一个点Ω作为原点，然后f（Ω）=Ω，我们可以将f视为旋转或不适当的正交变换，这取决于→−f的性质。注意，很有可能固定（f）=∅。例如，非平凡的翻译没有固定点。一个更有趣的例子是由一条平行于这条线的非平凡平移组成的线的平面反射组成。 Otherwise, we will see in Theorem 26.10 that every affine isometry is the (commutative) composition of a translation with an affine isometry that always has a fixed point. 否则，我们将在定理26.10中看到，每个仿射等值线都是一个具有固定点的仿射等值线的翻译的（交换）组合。 26.4 Affine Isometries and Fixed Points 26.4仿射等距线和固定点 Let E be an affine space. Given any two affine subspaces F,G, if F and G are orthogonal complements in E, which means that →−F and →−G are orthogonal subspaces of →−E such that →−E = →−F ⊕ →−G, for any point Ω ∈ F, we define q: E → →−G such that 设e为仿射空间。给定任意两个仿射子空间f，g，如果f和g是e中的正交互补，这意味着→−f和→−g是→−e的正交子空间，因此→−e=→−f→−g，对于任何点Ω∈f，我们定义q:e→→−g以便 . . Note that q(a) is independent of the choice of Ω ∈ F, since we have 注意，q（a）独立于Ω∈f的选择，因为我们有 , ， and for any Ω1 ∈ F, we have 对于任何Ω1∈f，我们有 , ， and since , this shows that 从那以后，这表明 . . Then the map g: E → E such that g(a) = a − 2q(a), or equivalently 然后图g:e→e，这样g（a）=a−2q（a），或相等 , ， does not depend on the choice of Ωin F, we note that g is identified with the symmetry with respect to∈ F. If we identify E to →−E by choosing any origin Ω→−F and parallel to →−G. 不依赖于f中的Ω的选择，我们注意到g与∈f对称。如果我们通过选择任何原点Ω→−f并与→−g平行来确定e到→−e。 Thus, the map g is an affine isometry, and it is called the affine orthogonal symmetry about F. Since 因此，图G是一个仿射等值线，它被称为关于f的仿射正交对称。

the (linear) symmetry about the subspacefor all Ω ∈ F and for all a ∈ E, we note that the linear map→−F (the direction of F), and parallel to→−g associated with→−G (gtheis direction of G). 关于子空间的（线性）对称，对于所有的Ω∈f和所有的a∈e，我们注意到线性映射→−f（f的方向），并平行于→−g与→−g（g的gtheis方向）相关联。 Remark: The map p: E → F such that p(a) = a − q(a), or equivalently 备注：图P:E→F，使得P（A）=A−Q（A），或等效 , ， is also independent of Ω ∈ F, and it is called the affine orthogonal projection onto F. 也独立于Ω∈F，称为F上的仿射正交投影。 The following amusing lemma shows the extra power afforded by affine orthogonal symmetries: Translations are subsumed! Given two parallel affine subspaces F1 and F2 in E, letting →−F be the common direction of F1 and F2 and →−G = →−F ⊥ be its orthogonal comple- 下面有趣的引理显示了仿射正交对称所提供的额外能力：翻译被包含在内！给定e中的两个平行仿射子空间f1和f2，让→−f为f1和f2的公共方向，→−g=→−f为其正交配位。- ment, for anybetween23.16). We define theF1 anda ∈FF1is independent of the choice of, the affine subspacedistance between F1aand+→−GFintersects2 asa →−abkF. It is easily seen that the distanceF1, and that it is the minimum of2 in a single point b (see Lemma k 在23.16之间）。我们定义了f1和∈ff1独立于f1aa和+之间的仿射子空间距离→−gf与s2 asa→−abkf的选择。很容易看出距离f1，在一个点b中是2的最小值（参见引理k 2 in 2英寸 k−xy→k for all x ∈ F1 and all y ∈ F2. 所有x∈f1和所有y∈f2的k−xy→k。 Proposition 26.9.defined by the vector Given any affine space2→−ab, where →−ab is any vector perpendicular to the common directionE, if f : E → FE2, thenand gg: E◦ f→is a translationE are affine→−F orthogonal symmetries about parallel affine subspaces F1 and 命题26.9.由给定任意仿射空间2→−ab的矢量定义，其中→−ab是垂直于公共方向的任意矢量，如果f:e→fe2，那么和gg:e f→是一个平移，即仿射→−f关于平行仿射子空间f1和 of F1 and F2 such that k→−abk is the distance betweenτ is obtained as the composition of two affineF1 and F2, with a ∈ F1 and b ∈ F2. 其中k→−abk是τ之间的距离，由两个仿射f1和f2组成，其中a∈f1和b∈f2。 Conversely, every translation by a vector 相反，每一个矢量的翻译 orthogonal symmetries about parallel affine subspaces F1 and F2 whose common direction is orthogonal to→−abkτ/2=. →−ab, for some a ∈ F1 and some b ∈ F2 such that the distance between F1 and F2 is k 平行仿射子空间f1和f2的正交对称性，其公共方向与→−abkτ/2=正交。→−a b，对于一些a∈f1和一些b∈f2，使得f1和f2之间的距离为k Proof. We observed earlier that the linear maps →−f and →−g associated with f and g are the linear reflections about the directions of F1 and F2. However, F1 and F2 have the same 证据。我们之前观察到，与f和g相关的线性映射→−f和→−g是关于f1和f2方向的线性反射。然而，f1和f2是一样的。 direction, and soevery reflection is an involution, we have→−f = →−g . Since −−g ◦→f = →−−−gg◦→◦f→−f=and sinceid, proving that→−f ◦ →−g g=◦→−ffaboutis a translation. If◦ →−f =F2id, because, and it is 方向，所以反射是对合的，我们有→−f=→−g。由于−−g→f=→−−−gg→f→−f=和sinceid，证明→−f→−g=→−ffabout是一个翻译。如果→−f=f2id，因为，它是 we pick a ∈ F1, then g ◦ f(a) = g(a), the affine reflection of 我们选取a∈f1，然后g f（a）=g（a），仿射 distance betweeneasily checked thatF1gand◦ f is the translation by the vectorF2. The second part of the lemma is left as an easy exercise.τ = ag(a) whose norm is twice the 检查之间的距离，其中f1gand f是矢量2的平移。引理的第二部分留作一个简单的练习。τ=Ag（a），其范数是

We conclude our quick study of affine isometries by proving a result that plays a major role in characterizing the affine isometries. This result may be viewed as a generalization of Chasles’s theorem about the direct rigid motions in E3. 通过对仿射等距图的研究，证明了仿射等距图在表征仿射等距图中起着重要作用。这个结果可以被看作是关于E3中直接刚性运动的Chales定理的推广。 Theorem 26.10. Let E be a Euclidean affine space of finite dimension n. For every affine 定理26.10。对于每个仿射，设e为有限维n的欧几里得仿射空间。 isometry f : →−Ef (→τ) =E, there is a unique affine isometryτ (i.e., τ ∈ Ker), such that the setgE: Eof direction→ EFix(and a unique translationg) = {a ∈ E | g(a) = t = tτ, with a} of fixed points of g is a nonempty affine subspace of 等距f：→−ef（→τ）=e，存在一个唯一的仿射等距τ（即τ∈ker），使得集合ge:e of方向→efix（和一个唯一的平移g）=a∈e g（a）=t=tτ，其中g的固定点a是一个非空的仿射子空间 →−G = Ker , →−G=KER， and such that 这样的话 f = t ◦ g and t ◦ g = g ◦ t. f=tg和t_g=g_t。 Furthermore, we have the following additional properties: 此外，我们还有以下附加属性： (a) f = g and τ = 0 iff f has some fixed point, i.e., iff Fix(f) 6= ∅. f=g，τ=0，iff f有固定点，即iff fix（f）6=∅。 (b) If f has no fixed points, i.e., Fix(f) = ∅, then dim Ker . 如果F没有固定点，即固定（F）=∅，则调暗KER。 Proof. The proof rests on the following two key facts: 证据。证据基于以下两个关键事实： (1) If we can find some x ∈ E such that belongs to Ker , we get the existence of g and τ. （1）如果我们能找到一些x∈e，那么我们就得到了g和τ的存在性。 , and the spaces Ker and 以及空格和 Im are orthogonal. This implies the uniqueness of g and τ. 我是正交的。这意味着g和τ的唯一性。 First, we prove that for every isometry h: →−E → →−E, Ker(h − id) and Im(h − id) are orthogonal and that 首先，我们证明对于每一个等距测量，h：→−e→−e，ker（h−id）和im（h−id）是正交的，并且 →−E = Ker(h − id) ⊕ Im(h − id). →−E=KER（H−ID）IM（H−ID）。 Recall that dim = dim(Kerϕ) + dim(Imϕ), 回想一下dim=dim（ker）+dim（im_） for any linear map ; see Theorem 5.11. To show that we have a direct sum, we prove orthogonality. Let u ∈ Ker(h − id), so that , and compute 对于任何线性映射，请参见定理5.11。为了证明我们有一个直接和，我们证明了正交性。让u∈ker（h-id），这样，然后计算 u · (h(v) − v) = u · h(v) − u · v = h(u) · h(v) − u · v = 0, U·（H（V）−V）=U·H（V）−U·V=H（U）·H（V）−U·V=0， since h(u) = u and h is an isometry. 因为h（u）=u和h是等距测量。 Next, assume that there is some x ∈ E such that belongs to the space Ker . If we define g: E → E such that 接下来，假设有一些x∈e，它属于空间ker。如果我们将g:e→e定义为 g = t(−τ) ◦ f, g=t（−τ）f， we have g(x) = f(x) − τ = x, 我们有g（x）=f（x）−τ=x， sinceis equivalent to x = f(x) − τ. As a composition of affine isometries, g is an affine isometry, x is a fixed point of g, and since τ ∈ Ker , we have sinceis等于x=f（x）−τ。作为仿射等距图的一个组成部分，G是仿射等距图，X是G的不动点，由于τ∈ker，我们得到 →−f (τ) = τ, →−f（τ）=τ， and since 从那以后 g(b) = f(b) − τ g（b）=f（b）−τ is an affine subspace offor all b ∈ E, we have →−gE=with direction Ker→−f . Since g has some fixed point= Kerx, by Lemma 26.8, Fix(→− . We also haveg) 是所有b∈e的仿射子空间，我们有→−ge=和方向ker→−f。由于g有一些固定点=kerx，根据引理26.8，fix（→−。我们也有g） f(b) = g(b) + τ for all b ∈ E, and thus f（b）=g（b）+τ表示所有b∈e，因此 (g ◦ tτ)(b) = g(b + τ) = g(b) + →−g (τ) = g(b) + →−f (τ) = g(b) + τ = f(b), （g_tτ）（b）=g（b+τ）=g（b）+→−g（τ）=g（b）+→−f（τ）=g（b）+τ=f（b）、 and 和 (tτ ◦ g)(b) = g(b) + τ = f(b), （tτg）（b）=g（b）+τ=f（b）， which proves that t ◦ g = g ◦ t. 证明t g=g t。 To prove the existence of x as above, pick any arbitrary point a ∈ E. Since 为了证明上述x的存在，选取任意点a∈e。 →−E = Ker , →E=KER， there is a unique vector τ ∈ Ker and some v ∈ →−E such that 有一个唯一的向量τ∈ker和一些v∈→−e，这样

For any x ∈ E, since we also have 对于任何x∈e，因为我们也有 , ， we get 我们得到 , ， which can be rewritten as 可以重写为 . . If we let −ax→ = −v, that is, x = a − v, we get 如果我们让−a x→−v，也就是x=a−v，我们得到

with τ ∈ Ker . 与τ∈ker。 Finally, we show that τ is unique. Assume two decompositions (g1,τ1) and (g2,τ2). Since 最后，我们证明τ是唯一的。假设两个分解（g1，τ1）和（g2，τ2）。自从 , we have Ker( id) = Ker . Since g1 has some fixed point b, we get ，我们有ker（id）=ker。因为g1有固定点b，我们得到 f(b) = g1(b) + τ1 = b + τ1, f（b）=g1（b）+τ1=b+τ1， that is, bf−−−(→b) = τ1, and τ1 ∈ . Similarly, for some fixed point c of g2, we get cf(c) = τ2 and cf(c) ∈ Ker . Then we have 也就是说，bf−−（→b）=τ1，τ1∈。同样地，对于g2的某个不动点c，我们得到了cf（c）=τ2和cf（c）∈ker。然后我们有了 , ， which shows that 这表明 τ2 − τ1 ∈ Ker , τ2−τ1∈ker， and thus that τ2 = τ1, since we have shown that 因此，τ2=τ1，因为我们已经证明 →−E = Ker . →−E=KER。 The fact that (a) holds is a consequence of the uniqueness of g and τ, since f and 0 clearly satisfy the required conditions. That (b) holds follows from Lemma 26.8 (2), since the affine map f has a unique fixed point iff = Ker . （a）成立的事实是g和τ的唯一性的结果，因为f和0显然满足要求的条件。这（b）符合引理26.8（2），因为仿射映射f有一个唯一的固定点iff=ker。 The determination of x is illustrated in Figure 26.8. X的测定如图26.8所示。

Figure 26.8: Affine rigid motion as f = t ◦ g, where g has some fixed point x. 图26.8：仿射刚性运动，f=t_g，其中g有一些固定点x。 Remarks: 评论： (1) Note that Keriff Fix(g) consists of a single element, which is the unique fixed point of f. However, even if f is not a translation, f may not have any fixed points. For example, this happens when E is the affine Euclidean plane and f is the composition of a reflection about a line composed with a nontrivial translation parallel to this line. 请注意，keriff fix（g）由单个元素组成，这是f的唯一固定点。但是，即使f不是翻译，f也可能没有任何固定点。例如，当e是仿射欧几里得平面，f是关于一条线的反射的合成，该线由平行于该线的非平凡平移组成。 (2) The fact that E has finite dimension is used only to prove (b). 事实上，E的有限维仅用于证明（b）。 (3) It is easily checked that Fix(g) consists of the set of points x such that is minimal. 可以很容易地检查fix（g）是否由点X组成，从而使其最小。 In the affine Euclidean plane it is easy to see that the affine isometries (besides the identity) are classified as follows. An affine isometry f that has a fixed point is a rotation if it is a direct isometry; otherwise, it is an affine reflection about a line. If f has no fixed point, then it is either a nontrivial translation or the composition of an affine reflection about a line with a nontrivial translation parallel to this line. 在仿射欧几里得平面上，很容易看出仿射等距线（除了同一性）被分类如下。一个具有固定点的仿射等距f是一个旋转，如果它是一个直接等距；否则，它是关于一条直线的仿射反射。如果f没有不动点，那么它要么是一个非平凡平移，要么是关于一条线的仿射反射的组合，其中一条线的非平凡平移与此线平行。 In an affine space of dimension 3 it is easy to see that the affine isometries (besides the identity) are classified as follows. There are three kinds of affine isometries that have a fixed point. A proper affine isometry with a fixed point is a rotation around a line D (its set of fixed points), as illustrated in Figure 26.9. 在维数3的仿射空间中，很容易看出仿射等距线（除了恒等式）被分类如下。有三种具有固定点的仿射等距线。具有固定点的适当仿射等值线是围绕线D（其固定点集）旋转的，如图26.9所示。

Figure 26.9: 3D proper affine rigid motion with line D of fixed points (rotation). 图26.9：固定点D线（旋转）的三维适当仿射刚性运动。 An improper affine isometry with a fixed point is either an affine reflection about a plane H (the set of fixed points) or the composition of a rotation followed by an affine reflection about a plane H orthogonal to the axis of rotation D, as illustrated in Figures 26.10 and 26.11. In the second case, there is a single fixed point O = D ∩ H. 具有固定点的不适当仿射等值线是关于平面H（固定点集）的仿射反射，或者是关于与旋转轴D正交的平面H的旋转后的仿射反射的组合，如图26.10和26.11所示。在第二种情况下，有一个固定点o=d h。 There are three types of affine isometries with no fixed point. The first kind is a nontrivial translation. The second kind is the composition of a rotation followed by a nontrivial translation parallel to the axis of rotation D. Such an affine rigid motion is proper, and is called a screw motion. A screw motion is illustrated in Figure 26.12. 仿射等角图有三种类型，没有固定点。第一种是非平凡的翻译。第二类是旋转的组合，随后是平行于旋转轴d的非平凡平移。这种仿射刚性运动是适当的，称为螺旋运动。图26.12说明了螺钉的运动。

26.5. THE CARTAN–DIEUDONNE THEOREM FOR AFFINE ISOMETRIES´ 26.5。仿射等距线的卡坦-迪乌顿定理

Figure 26.10: 3D improper affine rigid motion with a plane H of fixed points (reflection). 图26.10：固定点平面H（反射）的三维不适当仿射刚性运动。

Figure 26.11: 3D improper affine rigid motion with a unique fixed point. 图26.11：具有唯一固定点的三维不适当仿射刚性运动。 The third kind is the composition of an affine reflection about a plane followed by a nontrivial translation by a vector parallel to the direction of the plane of the reflection, as illustrated in Figure 26.13. 第三类是关于平面的仿射反射的合成，随后是平行于反射平面方向的向量的非平凡平移，如图26.13所示。 This last transformation is an improper affine isometry. 最后一个转换是一个不正确的仿射同构。 26.5 The Cartan–Dieudonn´e Theorem for Affine Isometries 26.5仿射等轴测的卡坦-迪乌顿定理 The Cartan–Dieudonn´e theorem also holds for affine isometries, with a small twist due to translations. The reader is referred to Berger [11], Snapper and Troyer [157], or Tisseron [170] for a detailed treatment of the Cartan–Dieudonn´e theorem and its variants. 卡坦-迪乌登定理也适用于仿射等距线，由于翻译，它有一个小的扭曲。读者可以参考Berger[11]、Snapper和Troyer[157]或Tisseron[170]来详细处理Cartan–Dieudonn’e定理及其变体。 Theorem 26.11. Let E be an affine Euclidean space of dimension n ≥ 1. Every affine isometry f ∈ Is(E) that has a fixed point and is not the identity is the composition of at most n affine reflections. Every affine isometry f ∈ Is(E) that has no fixed point is the 定理26.11。设e为尺寸n≥1的仿射欧几里德空间。每一个仿射同构f∈都是（e）有一个固定的点，而不是同一性，至多是n个仿射反射的组成。每一个没有固定点的仿射同构f∈是（e）是

(i.) （一） (ii.) （二） Figure 26.12: 3D proper affine rigid motion with no fixed point (screw motion). The second illustration demonstrates that a screw motion produces a helix path along the surface of a cylinder. 图26.12：没有固定点的三维适当仿射刚性运动（螺旋运动）。第二幅图显示了螺杆运动沿圆柱表面产生螺旋路径。

Figure 26.13: 3D improper affine rigid motion with no fixed points. 图26.13：没有固定点的三维不适当仿射刚性运动。 26.5. THE CARTAN–DIEUDONNE THEOREM FOR AFFINE ISOMETRIES´ 26.5。仿射等距线的卡坦-迪乌顿定理 composition of at most n + 2 affine reflections. When n ≥ 2, the identity is the composition of any reflection with itself. 最多由n+2个仿射反射组成。当n≥2时，同一性是任何反射本身的组成。 Proof. First, we use Theorem 26.10. If f has a fixed point Ω, we choose Ω as an origin and work in the vector space EΩ. Since f behaves as a linear isometry, the result follows from Theorem 26.1. More specifically, we can write hyperplane reflections →−si . We define the affine reflections si such that 证据。首先，我们使用定理26.10。如果F有一个固定点Ω，我们选择Ω作为原点，并在向量空间EΩ中工作。由于f表现为一个线性等距测量，其结果来自定理26.1。更具体地说，我们可以写超平面反射→−si。我们将仿射反射定义为 si(a) = Ω + →−si (Ω−→a) Si（A）=Ω+→Si（Ω-→A） for all a ∈ E, and we note that f = sk ◦ ··· ◦ s1, since 对于所有a∈e，我们注意到f=sk····s1，因为

for all a ∈ E. If f has no fixed point, then for some affine isometry g that has a fixed point Ω and some translation t = tτ, with f (τ) = τ. By the argument just given, we can write g = sk ◦ ··· ◦ s1 for some affine reflections (at most n). However, by Lemma 26.9, the translation t = tτ can be achieved by two affine reflections about parallel hyperplanes, and thus f = sk+2 ◦ ··· ◦ s1, for some affine reflections (at most n + 2). 对于所有a∈e，如果f没有不动点，那么对于某些具有不动点Ω的仿射等距G，某些平移t=tτ，其中f（τ）=τ。根据刚才给出的论点，我们可以写出g=sk·····s1来表示一些仿射反射（最多n）。然而，在引理26.9中，对于某些仿射反射（最多n+2），平移t=tτ可以通过两个关于平行超平面的仿射反射来实现，因此f=sk+2····s1。 When n ≥ 3, we can also characterize the affine isometries in SE(n) in terms of affine flips. Remarkably, not only we can do without translations, but we can even bound the number of affine flips by n. 当n≥3时，我们也可以用仿射翻转来描述SE（n）中的仿射等距图。值得注意的是，我们不仅可以不翻译，而且还可以将仿射翻转的数量限制为n。 Theorem 26.12. Let E be a Euclidean affine space of dimension n ≥ 3. Every affine rigid motion f ∈ SE(E) is the composition of an even number of affine flips f = f2k ◦ ··· ◦ f1, where 2k ≤ n. 定理26.12。设e为维数n≥3的欧氏仿射空间。每一个仿射刚性运动f∈se（e）是偶数个仿射翻转f=f2k···f1的组合，其中2k≤n。 Proof. As in the proof of Theorem 26.11, we distinguish between the two cases where f has some fixed point or not. If f has a fixed point Ω, we apply Theorem 26.5. More specifically, we can write →−f = f−→2k ◦ ··· ◦ →−f1 for some flips. We define the affine flips fi such that 证据。在定理26.11的证明中，我们区分了F是否有固定点的两种情况。如果f有一个固定点Ω，我们应用定理26.5。更具体地说，我们可以为一些翻转写→−f=f−→2K···→−f1。我们定义仿射翻转fi，以便

for all a ∈ E, and we note that f = f2k ◦ ··· ◦ f1, since 对于所有a∈e，我们注意到f=f2k···f1，因为

for all a ∈ E. 对于所有a∈e。 If f does not have a fixed point, as in the proof of Theorem 26.11, we get 如果f没有固定点，如定理26.11的证明，我们得到 f = tτ ◦ f2k ◦ ··· ◦ f1, f=tτf2k···f1， for some affine flips fi. We need to get rid of the translation. However,, and by the second part of Theorem 26.5, we can assume that , where F2k is the direction of the affine subspace defining the affine flip f2k. Finally, appealing to Lemma 26.9, since 对于一些仿射翻转fi。我们得把翻译掉。然而，根据定理26.5的第二部分，我们可以假设，其中f2k是定义仿射翻转f2k的仿射子空间的方向。最后，吸引引理26.9，因为 , the translation tτ can be expressed as the compositionof two affine flips andabout the two parallel subspaces Ω + and Ω +, whose distance is kτk/2. However, since and f2k are both the identity on Ω + F2k, we must have , and thus ，平移tτ可以表示为两个仿射翻转和两个平行子空间Ω+和Ω+的组合，其距离为kτk/2。但是，既然和f2k都是Ω+f2k上的标识，我们必须有，因此 f = tτ ◦ f2k ◦ f2k−1 ◦ ··· ◦ f1 f=tτf2k f2k−1···f1 = f20k ◦ f20k−1 ◦ f2k ◦ f2k−1 ◦ ··· ◦ f1 = f20k ◦ f2k−1 ◦ ··· ◦ f1, =f20k f20k−1 f2k f2k−1···f1=f20k f2k−1··f1， since and = id, since f2k is an affine symmetry.
因为和=id，因为f2k是仿射对称。 Remark: It is easy to prove that if f is a screw motion in SE(3), D its axis, θ is its angle of rotation, and τ the translation along the direction of D, then f is the composition of two affine flips about lines D1 and D2 orthogonal to D, at a distance kτk/2 and making an angle θ/2. 注：可以很容易地证明，如果f是在se（3）中的螺旋运动，d是它的轴，θ是它的旋转角，τ是沿着d的方向的平移，那么f是关于d1和d2直线的两个仿射翻转的组合，在kτk/2的距离上，作一个θ/2角。