1. Klein geometries 1.1. affine space, Euclidean space; define Euclidean motions as dis- tance preserving set maps. Proposition. (1) The Euclidean motions are exactly the maps of the form f(x) = Ax + b for A ∈ O(n) and b ∈ Rn. (2) Any Euclidean motion f is a diffeomorphism and writing f(x) = Ax + b as in (1), we have Df(x) = A for all x ∈ Rn. In particular, f is an isometry for the standard Riemannian metric on Rn. Proof. (1) Evidently, x 7→ Ax + b is a Euclidean motion. Conversely, suppose that f : Rn → Rn is a Euclidean motion and define A : Rn → Rn by A(x) := f(x) − f(0). Putting b := f(0), we then have f(x) = A(x) + b, so we have to show that A is an orthogonal linear map. By definition, we obtain kA(x) − A(y)k = kf(x) − f(y)k = kx − yk, and since A(0) = 0, this also shows that kA(x)k = kxk. The polar- 1 2 2 2 isation formula hx, yi = 2 (kxk + kyk − kx − yk ) then shows that hA(x),A(y)i = hx, yi for all x, y ∈ Rn. In particular, for an orthonor- n mal basis {e1, . , en} of R also {A(e1),...,A(en)} is an orthonormal n P basis. Expanding x ∈ R in the basis {ei}, we obtain x = ihx, eiiei. On the other hand, expanding A(x) in the orthonormal basis {A(ei)}, we obtain P P A(x) = ihA(x),A(ei)iA(ei) = ihx, eiiA(ei). P P Together with the above, this shows that A( xiei) = xiA(ei), so A is linear, and since we have already observed that A is compatible with the inner product, this shows that A ∈ O(n). (2) is easy. ∼ n n Euclidean motions form a group Euc(n) = R × O(n), can view R as homogeneous space Euc(n)/O(n). 1.2. Affine motions, and their relation to affine lines, affine space An as Aff(n)/GL(n). Realization of An as affine hyperplane in Rn+1 and resulting realization of Aff(n) as a subgroup of GL(n + 1). Analogous realization of Euc(n). Structure of euc(n) and Lie bracket, euc(n) ∼= Rn ⊕ o(n) as a representation of O(n) ⊂ Euc(n). Completion of Rn to RP n, viewing RP n as GL(n + 1, R)/P and projective geometry. 1.3. Recall properties of Maurer–Cartan form. Let ω ∈ Ω1(G, g) be the Maurer–Cartan form on a Lie group G. For a closed subgroup H ⊂ G consider the homogeneous space G/H and let p : G → G/H be the canonical projection. For each g ∈ G the map ω(g): TgG → g induces a linear isomorphism ϕg : TgH (G/H) → g/h which is characterized by ϕg(Tgp·ξ) = ω(ξ) + h. For h ∈ H, we have −1 ϕgh = Ad(h ) ◦ ϕg. 1 2 Let us specialize to G = Euc(n) and H = O(n). We have seen that euc(n) = Rn ⊕ o(n) and this decomposition is invariant under Ad(h) for each h ∈ H. According to this splitting, we can write ω = θ ⊕ γ for θ ∈ Ω1(G, Rn) and γ ∈ Ω1(G, o(n)). Since the decomposition g = Rn ⊕ o(n) is H–invariant, the two components θ and γ are in- dividually H–equivariant. Explicitly, this means for h ∈ H = O(n) that (rh)∗θ = h−1 ◦ θ and ((rh)∗γ)(ξ) = hγ(ξ)h−1. In particular, we n obtain an identification of g/h with R . Viewing ϕg as a linear isomor- n phism TgH (G/H) → R , it is characterized by ϕg(Tgp·ξ) = θ(ξ) and −1 we get ϕgh = h ◦ ϕg. We can pull back the standard inner product on Rn to the tangent space TgH (G/H) via ϕg. Since each h ∈ H is orthogonal, the result is independent of the choice of g ∈ gH. For a local smooth section σ of p and a vector field ξ ∈ X(G/H) we obtain T p ◦ T σ ◦ ξ = ξ and hence ∗ ϕσ(x)(ξ(x)) = θ(σ(x))(Txσ·ξ(x)) = σ θ(ξ)(x). Hence the inner product of ξ, η ∈ X(G/H) can be locally written as hσ∗θ(ξ), σ∗θ(η)i, which evidently is a smooth function. Hence we have obtained a Riemannian metric on G/H. For each g ∈ G, we have the map `g : G/H → G/H which is characterized by `g ◦ p = p ◦ λg. Left invariance of the Maurer–Cartan ∗ 0 form implies that (λg) θ = θ for all g ∈ G. Now for g, g ∈ G and ξ ∈ Tg0 G we obtain T `g ·T p·ξ = T p·T λg ·ξ and using this, we compute ∗ 0 ϕgg0 (T `g ·T p·ξ) = ϕgg0 (T p·T λg ·ξ) = ((λg) θ(g ))(ξ) 0 = θ(g )(ξ) = ϕg0 (Tp ·ξ). This shows that the map T `g : Tg0H (G/H) → Tgg0H (G/H) is orthogo- nal, i.e. that G acts on G/H by isometries. In particular, our Riemann- ian metric is invariant under translations, and hence coincides with the standard Riemannian metric on Rn. Proposition. Let U, V ⊂ Rn be connected open subsets and let f : U → V be an isometry for the Riemannian metrics induced from Rn. Then f is the restriction of a uniquely determined Euclidean motion of Rn. Proof. Put U˜ := p−1(U) ⊂ G. We first claim that there is a unique smooth function F : U˜ → G such that • F (gh) = F (g)h for all g ∈ G and h ∈ H • p ◦ F = f ◦ p • F ∗θ = θ Take a point x ∈ G/H. For g ∈ G with x = gH we have the map ϕg : n Tx(G/H) → R , which by construction is an orthogonal isomorphism. −1 Since ϕgh = h ◦ ϕg, the set of all these maps is exactly the set of all orthogonal isomorphisms between the two spaces. Now suppose that −1 −1 n x ∈ U and fix g ∈ p (x). Then ϕg ◦ (Txf) : Tf(x)(G/H) → R 3 is an orthogonal isomorphisms, so there is a unique element F (g) ∈ −1 −1 p (f(x)) such that ϕg ◦ (Txf) = ϕF (g). Equivalently, we can write the defining property of F (g) as ϕg = ϕF (g) ◦ Txf. In this way, we obtain a well defined set map F : U˜ → G, which by construction satisfies p ◦ F = f ◦ p. For h ∈ H we then get −1 −1 ϕF (gh) ◦ Txf = ϕgh = h ◦ ϕg = h ◦ ϕF (g) ◦ Txf = ϕF (g)h ◦ Txf, which shows that F (gh) = F (g)h. To see that F is smooth, consider local smooth sections σ andσ ˆ of p : G → G/H which are defined locally around x respectively f(x). Then there is an open neighborhood W of x in G/H such that both σ andσ ˆ ◦ f are defined on W . Then both σ∗θ and (ˆσ ◦ f)∗θ are smooth Rn valued one forms on W , whose values are orthogonal. Hence we can define a smooth map ψ : W → O(n) by ψ(y)(v) := ((ˆσ ◦ f)∗θ)(y)((σ∗θ(y))−1(v)). This implies that for ξ ∈ Ty(G/H) we obtain ∗ −1 ∗ ∗ ϕσ(y) = σ θ(y) = (ψ(y)) ◦ (f (ˆσ θ))(y) −1 ∗ −1 = (ψ(y)) ◦ (ˆσ θ)(f(y)) ◦ Tyf = (ψ(y)) ◦ ϕσˆ(f(y)) ◦ Tyf = ϕσˆ(f(y))ψ(y) ◦ Tyf. Hence we conclude that F (σ(y)) =σ ˆ(f(y))ψ(y) and thus F (σ(y)h) = σˆ(f(y))ψ(y)h for all y ∈ W . Since the map W × H → p−1(W ) defined by (y, h) 7→ σ(y)h is a diffeomorphism, this shows that F is smooth. For g ∈ G and ξ ∈ TgG we now compute θ(F (g))(TF ·ξ) = ϕF (g)(T p·TF ·ξ) = ϕF (g)(T f ·T p·ξ) = ϕg(T p·ξ) = θ(g)(ξ), which shows that F ∗θ = θ and completes the proof of the claim. Now we claim that we even have F ∗ω = ω for the Maurer–Cartan form ω. Since the inclusion i : U˜ → G also satisfies i∗ω = ω and U is connected, there is an element g0 ∈ G such that F = λg0 ◦ i which ∗ implies that f = `g0 |U and completes the proof. Since F θ = θ, we get F ∗ω − ω = F ∗γ − γ, and this has values in h ⊂ g. Moreover, consider A ∈ h and the corresponding left invariant vector field LA ∈ X(G). d By definition, exp(tA) ∈ H and LA(g) = dt |t=0g exp(tA). But then F (g exp(tA)) = F (g) exp(tA) and differentiating at t = 0 show that ∗ TgF ·LA(g) = LA(F (g)). This implies that (F ω − ω)(g) vanishes on the subspace {LA(g): A ∈ h}. Consequently, there is a linear map m Φg : R → h such that ∗ (F γ − γ)(g)(ξ) = Φg(θ(ξ)) 4 for all ξ ∈ TgG. Now that last ingredient to use is that ω satisfies the Maurer–Cartan equation. Vanishing of the Rn–component reads as 0 = dθ(ξ, η) + [γ(ξ), θ(η)] + [θ(ξ), γ(η)] = dθ(ξ, η) + γ(ξ)(θ(η)) − γ(η)(θ(ξ)). Of course, also F ∗ω satisfies the Maurer–Cartan equation, and since F ∗θ = θ we get an analogous equation with γ replaced by F ∗(γ). Subtracting these two, we see that 0 = (F ∗γ − γ)(ξ)(θ(η)) − (F ∗γ − γ)(η)(θ(ξ)) = Φg(θ(ξ))(θ(η)) − Φg(θ(η))(θ(ξ)).