Example 2 (new version): the generators are. Example 4: the generators are

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1 First, let us quickly dismiss Example 3 (and Example 6): the second generator may be removed (being the square of the third or, respectively, the fourth one), and then the quotient is clearly the simplest case of a flat Klein bottle. This leaves us with seven examples: 1, 2, 4, 5, 7, 8 and 9. Second, in the three five-generator examples (7, 8, 9), one generator (the fifth in the first two of them, the fourth in the last) is redundant and should be removed, Specifically, in Example 7 the third generator followed by the fourth yields the fifth; in Example 8 the second generator followed by the fourth and then by the third yields the fifth; in Example 9 the third generator followed by the fifth yields the fourth. Third, Examples 8 and 9, now reduced to four generators, are essentially identical: Example 9 becomes Example 8 if one uses (x a/4, y) as the new coordinates. From now on we are just considering six examples: 1 2, 4 5, 7 8, each with only four generators. Next, in all six examples, one of the generators is a reflection about a point or a line, and a simple coordinate change turns that point or line into the origin or, respectively, one of the coordinate axes. Specifically, the new coordinates are (x, y b/4) in Examples 5 and 8; in the other four examples we leave the coordinates unchanged. Using the old symbols x, y for the new coordinates, we see that the new version of Example 8 (with the fifth generator removed, and with the entry y b/2 changed to y + b/2, which is achieved when the fourth generator is followed by the second) is the same as Example 7 (also with the fifth generator removed). We are thus left with five examples: 1, 2, 4, new version of 5, and 7. Let us further modify Examples 2, 5 and 7, switching, in all three, the names of x and y as well as those of a and b. Thus, we obtain: Example 1: the generators are ψ 4 : (x, y) (x + a/2, y). Example 2 (new version): the generators are Example 4: the generators are ψ 3 : (x, y) ( x, y) ψ 4 : (x, y) ( x, y + b/2). ψ 4 : (x, y) (x + a/2, y + b/2). 1

2 2 Example 5 (new version): the generators are ψ 3 : (x, y) ( x, y) ψ 4 : (x, y) ( x + a/2, y + b/2). Example 7 (new version): the generators are ψ 4 : (x, y) ( x + a/2, y + b/2). Remark 0.1. For each of the above examples the image H of G under the linear-part homomorphism is contained in the matrix group {diag(δ, ε) : δ, ε {1, 1}}, isomorphic to Z 2 Z 2. Examples 2 and 5 have H = {diag(1,1), diag(1, 1)}. On the other hand, H = {diag(1,1), diag( 1, 1)} in Example 4. Lemma 0.2. Every G-orbit intersects the rectangle Q given by Q = [0, a/2] [0, b/2] in Example 2, and Q = [0, a/4] [ b/2, b/2] in Examples 1, 4, 5, 7. Proof. If a group G of affine transformations of a real vector space contain the translations by all elements of a basis e 1,..., e n, then every orbit of G intersects the set {µ 1 e µ n e n : µ 1,..., µ n [ 1/2, 1/2]}. In fact, any x IR differs by an integer from some y [ 1/2, 1/2]. This fact, applied to n = 2 and e 1 = (a, 0), e 2 = (0, b), shows that all G-orbits intersect the rectangle R = [ a/2, a/2] [ b/2, b/2]. Let the phrase we use ψ to eliminate S now mean that a transformation ψ G sends a set S R onto a subset of R S (and so the stated property still holds when R is replaced by R S). First, in Examples 1, 4, 5, 7 (or, Example 2), ψ 3 (or, ψ 4 ) may be used to eliminate [ a/2, 0) [ b/2, b/2]. Similarly, in Example 1 (or, 2) the set (a/4, a/2) [ b/2, b/2] (or, (0, a/2] [ b/2, 0)) is eliminated with the aid of ψ 4 ψ 3 (or, respectively, ψ 3 ). Next, in Example 4, we eliminate (a/4, a/2) [0, b/2] (or, (a/4, a/2) [ b/2, 0]) using ψ 4 ψ 3 (or, respectively, ψ2 1 ψ 4 ψ 3 ). Finally, in both Examples 5 and 7, we use ψ2 1 ψ 4 (or, ψ 4 ) to eliminate (a/4, a/2) [0, b/2] or, respectively, (a/4, a/2) [ b/2, 0]. Lemma 0.3. In Examples 1, 2, 4, 5, 7, every ψ G is given by (0.1) ψ(x, y) = (δx + ka/2, εy + lb/2) with some δ, ε {1, 1} and k, l Z,

3 3 for all (x, y) IR 2. Furthermore, whenever ψ G is of the form (0.1), we have (0.2) i) in Example 1 : ( 1) k δε = ( 1) l = 1, ii) in Example 2 : ε = ( 1) k = 1, iii) in Example 4 : δε = ( 1) k+l = 1, iv) in Example 5 : ε = ( 1) k+l = 1, v) in Example 7 : ( 1) k δε = ( 1) k+l = 1. Proof. Affine transformations of type (0.1) form a group, and each of the mappings assigning to ψ with (0.1) the value δ, or ε, or ( 1) k, or ( 1) l, is a group homomorphisms into {1, 1}. (This is immediate since, expressing (0.1) as ψ (δ, ε, k, l), we have ψ ˆψ (δˆδ, εˆε, k + ˆk, l + εˆl) and ψ 1 (δ, ε, δk, εl) whenever ψ (δ, ε, k, l) and ˆψ (ˆδ, ˆε, ˆk, ˆl), while Id (1, 1, 0, 0).) Each of the terms equated to 1 in (0.2) thus represents a group homomorphism, and G is, in each case, contained in its kernel, since so is, clearly, the generator set for each G, specified above. We can now describe the quotient space IR 2 /G of the action of G. Proposition 0.4. In Example 2, two different points of Q = [0, a/2] [0, b/2] lie in one G-orbit if and only if these points are (x, 0) and (x, b/2) with some x [0, a/2]. In the other four examples, a two-element subset of Q = [0, a/4] [ b/2, b/2] is contained in a single orbit of G if and only if it equals {(x, b/2), (x, b/2)}, where x [0, a/4], or has one of the following additional forms. (a) Examples 1, 4 and 7 only: {(0, y), (0, y)} for some y (0, b/2]. (b) Example 4 only: {(a/4, y), (a/4, y + b/2)} for some y [0, b/4). (c) Example 4 only: {(a/4, y), (a/4, y b/2)} for some y ( b/4, 0]. (d) Examples 5 and 7 only: {(a/4, y), (a/4, y + b/2)} for some y [ b/2, 0]. Proof. The if part of our assertion follows if one uses: ψ 4 ψ 3 and ψ 2 in the first or, respectively, second paragraph of the lemma; ψ 3 for (a); ψ 4 ψ 3 and ψ2 1 ψ 4 ψ 3 in (b) (c); ψ 4 and ψ2 1 ψ 4 for (d) (e). Let Q now contain both (x, y) and (ˆx, ŷ) = ψ(x, y) (x, y). Note that (ˆx, ŷ) = (δx + ka/2, εy + lb/2) by (0.1). We will repeatedly invoke the following obvious fact: (0.3) p q c whenever p and q lie in a closed interval of length c, with equality only if p and q are the endpoints. In Example 2, Q = [0, a/2] [0, b/2]. Thus, 2x/a and k + 2δx/a lie in [0, 1], while k is even by (0.2.ii), and so (0.3) gives k = 0. Therefore ˆx = x (as δ = 1 unless x = 0. The resulting relation y ŷ = y + lb/2 (with ε = 1, cf. (0.2.ii)) means that l 0, while 2y/b and l + 2y/b lie in [0, 1]. Our claim about Example 2 is now obvious from (0.3).

4 4 In Examples 1, 4, 5, 7, our assumption that Q = [0, a/4] [ b/2, b/2] contains both (x, y) and (ˆx, ŷ) (x, y) has an immediate consequence: (0.4) ˆx = x, and (δ, k) = (1, 0) or (δ, k, x) = ( 1, 0, 0) or (δ, k, x) = ( 1, 1, a/4). (Namely, both 4x/a and 2k + 4δx/a lie in [0, 1], and so, due to (0.3), k = 0 if δ = 1, while the case δ = 1 allows just two possibilities: (x, k) = (0, 0) or (x, k) = (a/4, 1)).) The relation (x, y) (ˆx, ŷ) must thus be realized by the y components, that is, y and εy + lb/2 are two different numbers in [ b/2, b/2] or, equivalently, (0.5) 2y/b is different from l + 2εy/b and both lie in [ 1, 1], while 2 l 2, the last conclusion being obvious from (0.3). First, if l = 2, (0.3) yields the case mentioned in the second paragraph of the lemma. We may therefore assume from now on that l 1. Next, suppose that l = 0, and so ε = 1 by (0.5). In view of (0.2.iv), this excludes Example 5. In the remaining Examples 1, 4 and 7, (0.2) gives ( 1) k δ = 1 which, combined with (0.4) and (0.1), shows that ˆx = x = 0 and ŷ = y. Hence (a) follows. Finally, let l = 1, so that (0.2) excludes Example 1, and implies oddness of k in Examples 4, 5, 7. Thus, by (0.4), (δ, k, x) = ( 1, 1, a/4), while (0.2) provides a specific value of ε in each case. Using (0.4) and (0.1) again, we obtain (b), (c), (d) or (e). We refer to a set as a triangle if it has a fixed bijectrive identification with a Euclidean triangle. A nonempty set Σ will be called a triangulated compact surface if it is a union of finitely many such triangles, any two of which, if different and non-disjoint, intersect along a common edge, while any edge can be shared by no more than two of the triangles. With the vertices removed, Σ then becomes a flat C Riemannian manifold, the Riemannian metric and the manifold structure being extended across the edges in the obvious way. Any given Σ as above may or may not admit an embedding in a Euclidean space, isometric on each of its triangles. Even if it does, speaking of isometries involving Σ we will always mean its intrinsic metric rather than the Euclidean distance restricted to Σ. Remark 0.5. Let Σ be the triangulated compact surface obtained from four mutually isometric Euclidean isosceles triangles by gluing them along edges so as to obtain a tetrahedron. Clearly, Σ can be realized in a Euclidean 3-space (or, any inner-product space) if and only if the vertex angle θ of its triangles is less than π/2. (When θ π/2, the long-base edge joining two verices is not, or not the only, shortest path joining them, as another such path joins the two vertices to the midpoints of their opposite bases.) Remark 0.6. The translational identification of two parallel sides in a Euclidean rectangle obviously gives rise to a cylinder with a flat (product) Riemannian metric. Remark 0.7. If, in Remark 0.6, the two halves of one of remaining sides are additionally glued together with the aid of a translation, the resulting quotient space is a Möbius strip

5 with boundary, carrying a (smooth) flat Riemannian metric. In fact, up to an isometry, the cylinder of Remark 0.6 may be treated as the surface S r [0, q] C IR, for the circle S r = {z C : z = r} and some r, q (0, ), the additional gluing taking place between (z, 0) and ( z, 0). The quotient is clearly the same as that of S r [ q, q] under the free isometric action of Z 2 genertated by (z, t) ( z, t) (which yields the required flat metric), but also arises from the standard Möbius-strip identifications (r, t) ( r, t) in the subset S + r [ q, q], where S + r = {z S r : Re z 0}. 5