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Extra info for A-stable block implicit one-step methods
For any contravariant two-tensor ω on M (such as Ric or Hess(f )) we define the contravariant one-tensor div(ω) as follows: For any vector field X we set div(ω)(X) = ∇∗ ω(X) = grs ∇r (ω)(X, ∂s ). 9. dR = 2div(Ric) = 2∇∗ Ric. For a proof see Proposition 6 of Chapter 2 on page 40 of . We shall also need a formula relating the connection Laplacian on contravariant one-tensors with the Ricci curvature. Recall that for a smooth function f , we defined the symmetric two-tensor ∇2 f by ∇2 f (X, Y ) = ∇X ∇Y (f ) − ∇∇X (Y ) (f ) = Hess(f )(X, Y ), and then defined the Laplacian △f = tr∇2 f = gij (∇2 f )ij .
Calabi, 1958) Let f (x) = d(p, x) be the distance function from p. If (M, g) has Ric ≥ 0, then n−1 △f ≤ f in the sense of distributions. [Compare , p. 284 Lemma 42]. 29. The statement that △f ≤ n−1 f in the sense of distributions (or equivalently in the weak sense) means that for any non-negative test function φ, that is to say for any compactly supported C ∞ -function φ, we have n−1 f △φdvol ≤ φdvol. f M M Since the triangle inequality implies that |f (x) − f (y)| ≤ d(x, y), it follows that f is Lipschitz, and hence that the restriction of ∇f to any compact subset of M is an L2 one-form.
The only way this can happen is if there are surgery caps that prevent extending the limit back to −∞. This means that the base points in our sequence are all within a fixed distance and time (after the rescaling) of a surgery region. But in this case results from the nature of the standard solution show that if we have taken δ > 0 sufficiently small, then the base points have canonical neighborhoods modeled on the canonical neighborhoods in the standard solution, again contradicting our assumption that none of the base points has a canonical neighborhood.