Theorema Egregium, Intrinsic Curvature

Transcription

1 Theorema Egregium, Intrinsic Curvature Cartan s second structural equation dω 12 = K θ 1 θ 2, (1) is, as we have seen, a powerful tool for computing curvature. But it also has far reaching theoretical consequences. An isometry Φ : M N is a bijection which preserves infinitesimal length in the sense that v, w = Φ (v), Φ (w) for all v, w T p (M). In fact, isometries also preserve angles and, by integrating, the lengths of curves as well. Theorema Egregium: The Gaussian curvature of a surface is invariant under isometries. In other words, it is completely independent of how the surface is imbedded in R 3. This is sometimes expressed by saying that two-dimensional beings living on a two-dimensional surface could determine the curvature without even being aware of the surrounding three-dimensional space. Later on, we will see how such measurements could be carried out in practice.

2 For the proof, we first note that the claim is local so it is really enough to prove it in a neighborhood of an arbitrary point (which we can still call M). Locally, we can always construct orthonormal frame fields by the Gram- Schmidt procedure, and we can then reformulate the first structural equations intrinsically on M. In fact, if we start from an orthonormal frame field F 1, F 2 on M, then we first extend to a frame field F 1, F 2, F 3 in a neighborhood of M which gives structural equations dθ 1 = ω 11 θ 1 + ω 12 θ 2 + ω 13 θ 3, dθ 2 = ω 21 θ 1 + ω 22 θ 2 + ω 23 θ 3. But if we now take the restriction back to M (i.e. take the pull-back by the inclusion map i : M R 3 ) and note that ω 11 = ω 22 = 0, ω 21 = ω 12 and i (dθ 3 ) = 0. we obtain dθ 1 = ω 12 θ 2, dθ 2 = ω 12 θ 1, where now everything is defined on M alone. Now let Φ : M M be an isometry. Since Φ is in particular a diffeomorphism, there is no

3 problem in defining the direct image ( pushforward ) F 1, F 2 = Φ (F ), Φ (F 2 ) of the frame field F 1, F 2. If we let θ 1, θ 2 and ω 12 denote the corresponding dual one-forms and connection form, then we have Lemma: θ 1 =Φ ( θ 1 ), θ 2 =Φ ( θ 2 ), ω 12 =Φ ( ω 12 ). Proof: To prove that F 1, F 2 is a frame field is trivial since Φ preserves scalar products; F i (p), F j (p) = Φ (F i )(p), Φ (F j )(p) = F i (Φ 1 (p)), F j (Φ 1 (p)) = δ ij. Similarly, to see that Φ ( θ i ) = θ i, we need only check that Φ ( θ i ) is the dual of F i ; Φ ( θ i )(F j ) = θ i (Φ (F j )) = θ i ( F j ) = δ ij. To prove the last relation, we observe that from the first structural equations on M, d θ 1 = ω 12 θ 2, d θ 2 = ω 12 θ 1, we obtain by taking pull-back (in view of what we have just proved), dθ 1 = Φ ( ω 12 ) θ 2, dθ 2 = Φ ( ω 12 ) θ 1.

4 These equations are just the first structural equations on M with ω 12 replaced by Φ ( ω 12 ). to prove that these forms are equal, it is therefore enough to prove that the first structural equations determine ω 12 uniquely. Since a oneform is determined at each point p by its values on the basis F 1 (p), F 2 (p), this follows from applying the equations to F 1 and F 2, since by the definition of the -product, dθ 1 (F 1, F 2 ) = ω 12 (F 1 )θ 2 (F 2 ) ω 12 (F 2 )θ 2 (F 1 ) = ω 12 (F 1 ) and similarly, dθ 2 (F 1, F 2 ) = ω 12 (F 2 ). Proof of the Theorem: Taking pull-back along Φ of the second structural equation on M d ω 12 = K θ 1 θ 2, gives in view of 1 (since Φ d = d Φ ), dω 12 = ( K Φ)θ 1 θ 2. On the other hand, we have the usual second structural equation (1) on M, dω 12 = Kθ 1 θ 2. Since this equation of forms determine the coefficient K uniquely, it follows by comparison

5 that K Φ = K, i.e. K(Φ(p)) = K(p) for every p M. Q.E.D. A consequence of Theorema Egregium is that you can not straighten out a surface with K 0 without changing area or distances. In particular, you can not make a plane map of the earth without deforming the picture. Let Γ M (M oriented) be a unit speed curve with parameter t. The tangent vector field T (t) is clearly intrinsically defined on M. We can define N simply by rotating T (t) through the angle π 2 in the positive direction in the tangent space T Γ(t) (M). The usual argument shows that T (t) is orthogonal to T (t), but there is no reason to suppose that it is parallel to N(t). But since we assume Γ M, what we need to describe is really the component of T (t) which is parallel to T Γ(t) (M): Definition: Let T π(t) be the orthogonal projection of T (t) onto T Γ(t) (M) The geodesic curvature of Γ is defined to be the (unique) function κ g (t) such that T π(t) = κ g (t) N(t).

6 If we similarly write N π (t) for the orthogonal projection of N (t) we have The Frenet formulas on M: { T π (t) = κ g (t) N(t), N π(t) = κ g (t) T (t). Proof: The first equation is just the definition of κ g. To prove the second, observe that since N π (t) lies in T Γ(t) (M) and is orthogonal to N(t) (since N(t) has unit length), there must be some function, say ν, such that N π(t) = ν(t) T (t). To prove that κ g = ν, observe that 0 = d dt T (t), N(t) = T (t), N(t) + T (t), N (t) = T π (t), N(t) + T (t), N π (t) = κ g(t) ν(t). The theory is very similar to the theory in R 2. But curvature does have consequences: it is no longer true that the integral of κ g around closed curves is equal to ±2π. Let U M be an open and simply connected and let Γ = U. have the induced orientation. Then Theorem Γ κ gdλ + U KdM = 2π. Next time we will study the consequences of this result. In particular, we will se how it can be used to compute curvature intrinsically.