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.
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
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.
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
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).
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.