Solid angles and Seifert hypersurfaces

Given a smooth closed oriented manifold M of dimension n embedded in Rn+2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^{n+2}$$\end{document}, we study properties of the ‘solid angle’ function Φ:Rn+2\M→S1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPhi :{\mathbb {R}}^{n+2}{{\setminus }} M\rightarrow S^1$$\end{document}. It turns out that a non-critical level set of Φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPhi$$\end{document} is an explicit Seifert hypersurface for M. This gives an explicit analytic construction of a Seifert surface in higher dimensions.


Introduction
It has been known since Seifert [13] that every oriented link L = ∐ S 1 ⊂ ℝ 3 possesses a Seifert surface, that is, a compact oriented surface ⊂ ℝ 3 such that = L . Seifert gave an explicit algorithm for finding a Seifert surface from a link diagram.
In 1969 Erle [7] proved that any embedding M n ⊂ ℝ n+2 of codimension two of a closed oriented connected manifold M has a trivial normal bundle and admits a Seifert hypersurface n+1 ⊂ ℝ n+2 with = M ⊂ ℝ n+2 . The proof of the existence of the latter fact is not constructive; it relies on the Pontryagin-Thom construction applied to any smooth map f ∶ cl.(ℝ n+2 �M × D 2 ) → S 1 representing the generator Sadly, Andrew Ranicki passed away during the final stage of the preparation of the manuscript. with = f −1 (t) for a regular value t ∈ S 1 . An easy adjustment has to be made, because f is defined outside the tubular neighborhood of M; we refer to [7] for details.
In this paper, we use intuitions from physics to construct a concrete smooth map ∶ ℝ n+2 ⧵M → ℝ∕ℤ = S 1 . Namely, suppose M ⊂ ℝ 3 is a loop with constant electric current. The scalar magnetic potential ̃ of M at a point x ∉ M is the solid angle subtended by M, that is, the signed area of a spherical surface bounded by the image of M under the radial projection, as seen from x (Fig. 1); see [10,Chapter III] or [9,Section 8.3]. As the complement ℝ 3 ⧵M is not simply connected, the potential ̃ is defined only modulo a constant, which we normalize to be 1. The potential induces a well-defined function ∶ ℝ 3 ⧵M → ℝ∕ℤ . This physical interpretation suggests that there exists an open neighborhood N of M such that | N⧵M is a locally trivial fibration. In particular, a level set −1 (t) should be a (possibly disconnected) Seifert surface for M. In [4, Chapter VII] the second author proved that this is indeed the case, although the proof is rather involved. Even for a circle, the exact formula for is complicated; it was given by Maxwell in [10,Chapter XIV] in terms of power series and also by Paxton in [11]. The formulae for for the circle show that the analytic behavior of near M is quite intricate, although we can show that is a locally trivial fibration in U⧵M for some small neighborhood U of M; see Sect. 5.3. The construction can be generalized to higher dimensions, even though the physical interpretation seems to be a little less clear. For any closed oriented submanifold M n ⊂ ℝ n+2 , by the result of Erle [7] there exists a Seifert hypersurface. For any such hypersurface and a point x ∉ , we define ̃ (x) to be the high-dimensional solid angle of M, that is, the signed area of the image of the radial projection of to the (n + 1)-sphere of radius 1 and center x. The value of ̃ (x) depends on the choice of the hypersurface , but it turns out that under a suitable normalization, (x) ∶=̃ (x) mod 1 is independent of the choice of the Seifert hypersurface. Moreover, there is a formula for (x) in terms of integrals of some concrete differential forms over M, so the existence of is needed only to show that is well defined.
As long as t ≠ 0 ∈ ℝ∕ℤ , the preimage −1 (t) is a bounded hypersurface in ℝ n+2 ⧵M . If, additionally, t is a non-critical value, −1 (t) is smooth. To prove that −1 (t) is actually a Seifert hypersurface for M, we need to study the local behavior of near M. It turns out 1 ∈ cl. ℝ n+2 �M × D 2 , S 1 = H 1 ℝ n+2 �M = H n (M) = ℤ M O Fig. 1 Scalar magnetic potential as a solid angle. The scalar magnetic potential is calculated at point O that the closure of −1 (t) is smooth except possibly at the boundary. We obtain the following result, which we can state as follows.
• On the set of points {x ∈ ℝ n+2 ⧵M ∶ (0, 0, … , 0, 1) ∉ Sec x (M)} , the map is given by where Sec x is the secant map Sec x (y) = y − x ‖y − x‖ and is an explicit function depending on the dimension n described in (2.18). • Let t ≠ 0 be a non-critical value of . Then −1 (t) is a smooth (open) hypersurface whose closure is −1 (t) ∪ M . The closure of −1 (t) is a possibly disconnected Seifert hypersurface for M (in the sense of Definition 2.1), which is a topological submanifold of ℝ n+2 , smooth up to boundary. • For t ≠ 0 , the preimage −1 (t) has finite (n + 1)-dimensional volume.
To the best of our knowledge, up until now, there has been no known high-dimensional analogue of Seifert's algorithm for constructing Seifert hypersurfaces for general links. Our method of constructing a Seifert hypersurface is by a mixture of differential geometry and analysis, so is not strictly speaking algorithmic. Nevertheless, it is the first explicit construction for general links.
The fact that the hypersurface −1 (t) is a level set of the scalar magnetic potential function leads to the following question. We have, however, not been able to answer it so far. Question 1 Does the physical interpretation of −1 (t) imply some specific geometric or topological properties, like being a local minimizer for some energy function?
The structure of the paper is the following. Section 1 defines rigorously the solid angle map . Then a formula for in terms of an integral of an n-form over M is given. In Sect. 3 we prove that for t ≠ 0 the inverse images of −1 (t) ⊂ ℝ n+1 are bounded. It is also proved that extends to a smooth map S n+2 ⧵M → ℝ∕ℤ . In Sect. 4 we calculate explicitly for a linear subspace. The resulting simple formula is used later in the proof of the local behavior of for general M. In Sect. 5 we derive the Maxwell-Paxton formula for if M is a circle. These explicit calculations allow us to study the local behavior of in detail and give insight for the general case. In Sects. 6 and 7 we study the local behavior of for general M. This is the most technical part of the paper. We prove Theorem 1.1 in Sect. 8.

Definition of the map C
onsider a point x ∈ ℝ n+2 and define the map Sec x ∶ ℝ n+2 ⧵{x} → S n+1 given by

Definition 2.3 The solid angle of viewed from x is defined as
The map ̃ (x) is a signed area of a spherical surface spanned by Sec x (M) , that is, the radial projection of M from the point x.
We have the following fact.

Lemma 2.5
The value ̃ (x) mod 1 does not depend on the choice of . In particular, ̃ induces a well-defined function Proof Take another hypersurface ′ . Let X and X ′ be abstract models of and ′ , that is, X and X ′ are smooth compact (n + 2)-dimensional manifolds and ∶ X → , � ∶ X � → � are embeddings. Define to be the closed (n + 2)-dimensional manifold obtained by gluing X with X ′ along −1 (M) and � −1 (M) . Let ∶ → ℝ n+2 be the map equal to on X and to ′ on X ′ .
By functoriality of the integral, we have Therefore, The integral on the right-hand side is the evaluation of an (n + 1)-form * Sec * x 1 n+1 n+1 on the fundamental cycle of . The form 1 n+1 n+1 represents an integral cohomology class in H n+1 (S n+1 ) ; hence, * Sec * x 1 n+1 n+1 represents an integral cohomology class on . Therefore, is an integer. This means that ◻ Remark 2.6 One should not confuse the solid angle with the cone angle studied extensively by many authors, like [2,3,5]. To begin with, the cone angle is unsigned and takes values in ℝ ⩾0 , whereas the solid angle is an element in ℝ∕ℤ . This indicates that there exist fundamental differences between the two notions.
From the definition of , we recover its first important property.

Proposition 2.7
The map is smooth away from the complement of M ⊂ ℝ n+2 .
Proof Take a point y ∉ M . There exists a smooth compact surface such that = M and y ∉ . Then, a small neighborhood U of y is disjoint from . Thus, the map Sec x depends smoothly on the parameter x. This means that Sec * x n+1 depends smoothly on x. Integrating over a finite measure hypersurface preserves smooth dependence of a parameter. It follows that ̃ is smooth in U. ◻

˚ via integrals over M
The fact that the definition of (x) involves a choice of a Seifert hypersurface is quite embarrassing. In fact, it might be hard to find estimates for because we have little control over . We want to define via integrals over M itself. The key tool will be the Stokes' formula. We use the fact that while the volume form n+1 on S n+1 itself is not exact, its restriction ′ to the punctured sphere S n+1 ⧵{z} is.

3
We need the following result.

Remark 2.9
The result is non-trivial in the sense that one can construct a Seifert surface even for an unknot in ℝ 3 such that the restriction Sec x | is onto. However, notice that since Proof Let H be the half-line {x + tz, t > 0} . In other words H = Sec −1 x (z) . Choose any Seifert hypersurface . We might assume that H is transverse to . The set of intersection points of H and is bounded and discrete, hence finite. Let {w 1 , … , w m } = H ∩ and assume these points are ordered in such a way that on H the point w 1 appears first (with the smallest value of t), then w 2 , and so on (Fig. 2).
Choose the last point w m of this intersection and a small disk D ⊂ with center w m . We can make D small enough so that for any w � ∈ D the intersection is empty. Set now Consider a sphere S = S(x, r) , where r is large. Set S � = S⧵(S ∩ T) . Increasing r if necessary, we may and shall assume that S ′ is disjoint from . The new Seifert hypersurface is defined as With this construction, we have H ∩ � = {w 1 , … , w m−1 } . Repeating this construction finitely many times, we obtain a Seifert hypersurface disjoint from H. ◻ Let z be an n-form on S n+1 ⧵{z} such that d z = n+1 , and suppose is a Seifert hypersurface for M such that z ∉ Sec x ( ) . By Stokes' formula Therefore, we obtain the following formula for : The necessity of making the map modulo 1 comes now from different choices of the point z ∈ S n+1 ⧵M.
We shall need an explicit formula for z . For simplicity, we consider the case when z = (0, 0, … , 1) ∈ S n+1 ⊂ ℝ n+2 and define ∶= z ; the general case can be obtained by rotating the coordinate system. We start with the following proposition.  To obtain a formula for , it remains to solve (2.12). Rewriting it, we have The integrating factor of this ordinary differential equation is (1 − u 2 n+2 ) n+1 2 , so the general solution of (2.12) can be written as The requirement that the solution be smooth at u n+2 = −1 translates into the following formula The integral in (2.14) can be explicitly calculated. If n is odd, the result is a polynomial. If n is even, successive integration by parts eventually reduces the integral to ∫ √ 1 − s 2 ds . For small values of n, the function is as follows.

Definition 2.15
From now on, we shall assume that = (u n+2 ) n , where is as in (2.14).
We see that is smooth for u n+2 ∈ [−1, 1) and has a pole at u n+2 = 1 . We shall work mostly in regions, where u n+2 is bounded away from 1, so that and its derivatives will be bounded.

The pullback of the form
We shall gather some formulae for evaluating the pullback Sec * x . This will allow us to estimate the derivative of .
First, notice that where d y means that we take the exterior derivative with respect to the y variable (we treat x as a constant). Consider the expression To calculate the pullback, we replace du i by d y Notice that if in the wedge product the term (y i − x i )d y ‖y − x‖ −1 from (2.16) appears twice or more, this term will be zero. Therefore, the pullback takes the form where (i, j) is equal to j − 1 if j < i and j − 2 if j > i . Using the above expression together with (2.14) we can calculate the pullback of the form We calculate Notice that we can change the order of the sums in the last term of the above expression to be ∑ n+1 j=1 ∑ i≠j . Since i + (i, j) = j + (j, i) ± 1 , the last two sums cancel out. Hence, we obtain In particular, using Definition 2.15 we get a proof of the first part of Theorem 1.1.
If n = 1 we obtain the following explicit formula, used in [4].
It is worth mentioning the formula for n = 1 and a general z (not necessarily (0, 0, 1)), which was given in [4,Theorem 5.3.7].
We conclude by remarking that if (2.18) then analogous arguments as those that led to formula (2.17) imply that

Estimates for derivatives of Sec * x
The following results are direct consequences of the pullback formula for , (2.18). We record them for future use in Sects. 3 and 6 . Recall from Sect. 2.2 that was defined as a form on S n+1 ⧵(0, … , 0, 1) . The form z for general z ∈ S n+1 is obtained by rotation of the coordinate system.

Lemma 2.21 For any
Proof If m = 0 , the proof is a direct consequence of (2.18). The general case follows by an easy induction. ◻ As a consequence of Lemma 2.21, we prove the following fact.

Lemma 2.23 For any D < 1 and for any integer
Proof Apply a linear orthogonal map of ℝ n+2 that takes z to (0, 0, … , 0, 1) . Let x ′ and y ′ be the images of x and y, respectively, under this map. We have ‖y � − x � ‖ = ‖y − x‖ and the condition ⟨ y−x ‖y−x‖ , z⟩ < D becomes ‖y−x‖ < D . We shall use (2.22). As D < 1 , on the interval [−1, D] the function and its derivatives up to m-th inclusive are bounded above by some constant C D,m depending on D and m. The constant C D n,m can be chosen as In other words, all fibers of except −1 (0) are bounded.
Proof Choose a Seifert hypersurface for M. We may assume that it is contained in a ball B(0, r) for some r > 0 . As is compact and smooth, there exists a constant C such that if an (n + 1)-form n+1 on ℝ n+2 has all the coefficients bounded from above by T, then Now take R ≫ 0 and suppose x ∉ B(0, R + r) . Then the distance of x to any point y ∈ is at least R. Then Sec * x n+1 has all the coefficients bounded by R −n−1 , see (2.20), and therefore,

Sketch of proof Smoothness of
at infinity is equivalent to the smoothness of w ↦ ( w ‖w‖ 2 ) at w = 0 . The proof of Theorem 3.1 generalizes to show that for any m > 0 there exists C m with a property that �D ( Here is a multi-index. Now by the di Bruno's formula for higher derivatives of the composite function, we infer that �D ( w ‖w‖ 2 )� ⩽ C‖w‖ n+2−� � . (The worst case occurs when is differentiated only once, while w ‖w‖ 2 is differentiated | | times.) Hence, the limit at w → 0 of all derivatives of We can also strengthen the argument of Theorem 3.1 to obtain more detailed information about the behavior of at a large scale.

Theorem 3.3 Suppose is a Seifert hypersurface and r is such that
where C depends solely on and not on R and r.
This implies that where and Write also Now suppose ‖x‖ > R and ‖y‖ < r . Then − 1 − 3 has all the coefficients bounded by rR n+2 (R−r) n+2 ‖x‖ −n−2 . Likewise, notice that where we used Schwarz' inequality in the last estimate. Therefore, − 2 − 3 has all the coefficients bounded by ‖y‖ ‖y−x‖ n+2 , and by assumptions on ‖x‖ and ‖y‖ we have that We conclude that As (x) = ∫ 3 , we obtain the statement. ◻ The statement of Theorem 3.3, in theory, can be used to obtain information about C from the behavior of at infinity. The left-hand side of (3.4) is equal to and does not depend on . Therefore, if we know and its derivatives, we can find a lower bound for C , which roughly tells, how complicated Sec * x n+1 Sec * x n+1 might be. Unfortunately, we do not know of any examples where this can be used effectively.

Calculations
The analysis of near M will rely on replacing M by its tangent space and estimate the error. Therefore, we shall now define M for an n-dimensional surface. Define The first technical problem arises because H is not compact; therefore, the map does not even have to be defined. In what follows, we shall define H for H as above. We need to choose an analogue of the 'Seifert hypersurface', and our choice will be a half-hyperplane. As H is not compact, we cannot apply the argument of Lemma 2.5 to conclude that H does not depend on the choice of a half-hyperplane. And indeed, H will depend on this choice. In fact, H will be well defined up to an overall constant. In particular, the derivatives of H are well defined. This dependence is considered as a feature. Calculations for x j will be important in Sect. 6.
is (up to a sign) the area of the image Sec x ( ) . This image can be calculated explicitly.
Choose a point y = (y 1 , … , y n+2 ) ∈ S n+1 . The half-line from x ∉ through x + y is given by t ↦ x + ty , t ⩾ 0 ; see Fig. 3. By definition, y ∈ Sec x ( ) if and only if this halfline intersects , that is, for some t 0 > 0 we have Note that if x 2 = 0 , then the half-line through x and any point in will meet H, which results in an n-dimensional image Sec x ( ) in S n+1 . Suppose x 2 ≠ 0 . The condition hitting the Seifert hypersurface t 0 > 0 together with (4.1) implies that the signs of x 2 and y 2 must be opposite. Plugging t 0 from the first equation of (4.1) into the second one, we obtain The calculation of boils down to the study of the set of x 1 , x 2 satisfying (4.2). Write x 1 = r cos 2 and x 2 = r sin 2 . Multiply (4.2) by (which is negative) to obtain the inequality There are four cases depending on in which quadrant of the plane contains (x 1 , x 2 ) ; see Fig. 4. We next calculate the area of the image Sec x ( ) of each of those four cases. To do so, we first deal with the calculations and then discuss the choice of the sign. For the moment, we choose a sign for the area as ∈ {−1, +1} ; refer to Sect. 4.2 for the discussion of the sign convention.
Notice that the area of the two-dimensional circular sector in Fig. 4 is (up to normalization) equal to the (n + 1)-dimensional area of the image Sec x ( ) . This is because the defining equations are homogeneous, and other variables y 3 , … , y n+2 do not enter in the definition of the region.
Case 1 x 1 ⩾ 0 and x 2 > 0 . The region Sec x ( ) is given by y 2 < 0 (because the sign of y 2 is opposite to the sign of x 2 ), y 1 ⩽ y 2 ∕ tan 2 and tan 2 ∈ (0, ∞) . The area of the sector corresponding to Case 1 is equal to , and hence, (x) = , where is a sign. Case 2 x 1 ⩽ 0 and x 2 > 0 . The region Sec x ( ) is given by y 2 < 0 , y 1 ⩽ y 2 ∕ tan 2 , where tan 2 ∈ (−∞, 0) . The area of the sector is equal to and so (x) = . Case 3 x 1 ⩽ 0 and x 2 < 0 . Then y 2 > 0 and tan 2 ∈ (0, ∞) . The area of the sector is − , but now the hypersurface is seen from the other side; hence, the signed area is ( − ) . After normalizing and taking modulo 1, we obtain that (x) = . Case 4 x 1 ⩾ 0 , x 2 < 0 . Then y 2 > 0 and tan 2 ∈ (−∞, 0) . As in Case 3, we deduce that the area is − and we obtain (x) = .
Putting all the cases together, we see that (x) = .
Suppose we take another 'Seifert surface' for H, denoted ′ , given by u 1 = 0 , u 2 ⩽ 0 . Let ′ be the map defined relatively to ′ . To calculate ′ , we could repeat the above procedure, yet we present a quicker argument. A counterclockwise rotation A in the (u 1 , u 2 )-plane by angle 2 fixes H and takes to ′ . In particular, � (x) = (Ax) . Hence, We notice that ′ ≠ , but on the other hand � − is a constant. This approach shows that if we take a linear hypersurface (a half-space) for the 'Seifert surface' of , then it is well defined up to a constant, and so the derivatives are well defined.

The sign convention
Given that is defined as an integral of a differential form, changing the orientation of H induces a reversal of the sign of . We use the example of a linear surface to show how the sign is computed.
Choose the orientation of H in such a way that is a positive basis of TH.
Stokes' theorem is applicable if is oriented by the rule "normal outwards first", see [14,Chapter 5], so that is an oriented basis of T .
The way of seeing the sign is by calculating ∫ Sec * x n+1 . By (2.20) we know that Given the orientation of , we have Notice that on the left-hand side we have an integral of a differential form, whereas on the right-hand side the integral is with respect to the (n + 1)-dimensional Lebesgue measure on a subset of ℝ n+1 . The function ∫ 1 ‖y−x‖ 2 dy 3 … dy n+2 is positive; therefore, ̃ is positive for x 2 > 0 , negative for x 2 < 0 and 0 for x 2 = 0, x 1 > 0 (notice that (4.3) is not defined if x 2 = 0 and x 1 ⩽ 0 : if this holds, the point (x 1 , x 2 , … , x n+2 ) lies on and the integral diverges). Therefore, This is possible only if the choice of sign is = +1.

˚ for a circle
In Sect. 5 the manifold M is the round unit circle, that is . We now use the formula for via the integrals of the pullback of , see (2.18), to give an explicit formula for . The output is given in terms of elliptic integrals. Detailed calculations can be found, e.g., in [4]; therefore, we omit some tedious computations. We focus on the analysis of the behavior of near the circle.

Elliptic integrals
For the reader's convenience, we give a quick review of elliptic integrals and their properties. We shall use these definitions in future calculations. This section is based on [1].

The integral
is called an elliptic integral of the first kind. If = ∕2 , it is called a complete elliptic integral of the first kind, denoted by (k) ∶= ( ∕2, k).

The integral
is called an elliptic integral of the second kind. If = ∕2 , it is called a complete elliptic integral of the second kind, denoted by (k) ∶= ( ∕2, k).

The integral
is called an elliptic integral of the third kind. If = ∕2 , it is called a complete elliptic integral of the third kind, denoted by ( 2 , k) ∶= ( ∕2, 2 , k). 4. Heuman's Lambda function 0 ( , k) can be defined by the formula Although (k) blows up at k = 1 , we know how fast it goes to infinity as k approaches 1 from below.

In particular,
The differentials of (k) and (
We have some special cases where we can compute the integral explicitly. If x 3 = 0 , we use the identity and deal with improper integrals; there are two situations: • r < 1 : we have

Remark 5.11
The inverse image −1 (0) contains (and actually it is equal) to the set . This shows that the assumption that t ≠ 0 in Theorem 3.1 is necessary.
We now express (r, x 3 ) in terms of elliptic integrals. We use the following simplification, which follows by explicit computations. We may write the formula in terms of Heuman's Lambda function 0 using the formula relating and 0 ; see [1, page 228] or [11]. After straightforward but tedious calculations, we obtain the following explicit formula.

Proposition 5.13 (see [4, Proposition 6.3.1]) Let
Another approach in computing the solid angle for an unknot was given by F. Paxton; see [11]. He showed that the solid angle subtended at a point P with height L from the unknot and with distance r 0 from the axis of the unknot is equal to where R max = √ (1 + r 0 ) 2 + L 2 , = arctan L |1−r 0 | and k is given by (5.14). It can be shown that the Paxton formula agrees with the result of Proposition 5.13.
Finally, we remark that the computation of the solid angle of the unknot was already studied by Maxwell. He gave the formulae in terms of infinite series; see [10, Chapter XIV].

Behavior of ˚ near U
We shall now investigate the behavior of and its partial derivatives near U. Let us write where > 0 is small and ∈ [0, 1] . We have the following result.

Proposition 5.15 (see [4, Proposition 6.4.2]) The limit as → 0 + is given by
Sketch of proof Use r = 1 + cos 2 , x 3 = sin 2 and apply Proposition 5.13 together with a fact that ◻ Remark 5.16 The sign of the limit is − and not + . It is not hard to see that the orientation convention for the circle, that is, such that t ↦ (cos t, sin t, 0) is an oriented parameterization of U is opposite to the convention adopted in Sect. 4.2.
Next we compute the derivatives of near U. It is clear that the map for the circle is invariant with respect to the rotational symmetry around the z-axis. Hence, if is the longitudinal coordinate near U, then = 0 . The two coordinates we have to deal with are the meridional and radial coordinates and . The first result is the following.

Proposition 5.17 (see [4, Proposition 6.4.3]) We have
We observe that as → 0 + , we have k → 1 − by (5.14). The numerator (k) − (k) blows up, so the right-hand side of the formula in Proposition 5.17 is divergent as → 0 . For future use, we remark that by (5.4) and Proposition 5.17 we have for some constant C lin , which can be explicitly calculated.

Remark 5.19
This extension of through { = 0} will be generalized in the Continuous Extension Lemma 7.10.
We now estimate the derivative of with respect to .

Corollary 5.21
The derivatives x 2 , x 3 ) , j = 1, 2, 3 have at most a logarithmic pole at points (x 1 , x 2 , x 3 ) close to U. More precisely, there exists a constant C circ such that We remark that from (2.18), we get much weaker estimates on the derivative. We do not know whether these weaker estimates can be improved for general manifolds M.
To conclude, we show level sets of the function (r, x 3 ) ↦ (r, x 3 ) for the circle in Fig. 6. Notice that in the figure the half-lines stemming from point (1, 0) (and not parallel to the x 3 = 0 line) intersect infinitely many level sets near the point (1,0). This suggests that the radial derivative (1 + cos 2 , sin 2 ) is unbounded as → 0 + . We proved this fact rigorously in Proposition 5.17.

Derivatives of ˚ near M
We begin by recalling a well-known fact in differential geometry. Proposition 6.1 Let X ⊂ ℝ n+2 be a k-dimensional, smooth, compact submanifold with smooth boundary. Then, there exists a constant C X such that for every x ∈ ℝ n+2 and for any r > 0 we have lim →0 + (1 + cos 2 , 0, sin 2 ) = − (1) = −1.
vol k (X ∩ B(x, r)) ⩽ C X r k . Moreover, increasing C X if necessary, we may assume that if is a k-form on ℝ n+2 whose coefficients are bounded by T, then | ∫ X∩B(x,r) | ⩽ C X Tr k .
The result is well known to the experts; therefore, we present only a sketchy proof.
Sketch of proof Let k be the volume of unit k-dimensional ball. By smoothness of X, we infer that lim r→0 vol k (X ∩ B(x, r)) r k is 0, 1 2 k or k depending on whether x ∉ X , x ∈ X or x ∈ X⧵ X . Using Vitali's covering theorem, see, e.g., [8,Section 1.5], one shows that there exists r 0 independent of x such that vol k (X ∩ B(x, r)) r k ⩽ 2 k for all r < r 0 . We take C X to be the maximum of 2 k and vol k (X)∕r k 0 . The second part is standard and left to the reader. ◻

The Separation Lemma
The form z used in Sect. 2.2 has a pole at z ∈ S n+1 . In the applications for given x ∈ ℝ n+2 ⧵M , we choose a point z such that z ∉ Sec x (M) . Such a point exists; see Remark 2.9. However, in order to obtain a meaningful bound for Sec * x z , we need to know that z is separated from Sec x (M) , in the sense that there exists a constant D such that ⟨y, z⟩ ⩽ D for any y ∈ Sec x (M) . In this section, we show that the constant D < 1 can be chosen independently of x.

Remark 6.3
In general, it is impossible for a given point x ∈ M to find an element z ∈ S n+1 and a neighborhood U ⊂ ℝ n+2 of x, such that for every x � ∈ U we have z ∉ Sec x � (M) . In fact, the opposite holds. For any z ∈ S n+1 the sequence x n = x − z n has the property that z ∈ Sec x n (M) and x n → x . This is the main reason why the proof of an apparently obvious lemma is not trivial.
Put differently, the subtlety of the proof of Lemma 6.2 lies in the fact that the image Sec x (M) can be defined for x ∈ M as a closure of Sec x (M⧵{x}) , but we cannot argue that Proof Suppose the contrary, that is, for any n there exists a point y n ∈ M such that ‖x − y n ‖ < 1 n and Sec x (y n ) ∉ U . In particular, y n → x . As M is a smooth submanifold of ℝ n+2 , the tangent space T x M is the linear space of limits of secant lines through x. This means that if y n → x and y n ∈ M , then, up to passing to a subsequence, Sec x (y n ) converges to a point in S x . But then, starting with some n 0 > 0 , we must have Sec x (y n ) ∈ U for all n > n 0 . Contradiction. ◻ Choose now a neighborhood U of S x and r from Lemma 6.4. As S x is invariant with respect to the symmetry y ↦ −y , we may and shall assume that U also is. We assume that U is small neighborhood of S x , but in fact we shall only need that U is not dense in S n+1 . We shall need the following technical result. B(x, r)).

Proof of Lemma 6.5
We argue by contradiction. Assume the statement of the lemma does not hold. That is, there is a sequence x n converging to x such that both v and −v belong to Sec x n (M ∩ B(x, r)) . This means that for any n the line l x n ∶= {x n + tv, t ∈ ℝ} intersects M ∩ B(x, r) in at least one point for t > 0 and at least one point for t < 0 . For each n, choose a point y + In particular, Sec x n (y + n ) = v and Sec x n (y − n ) = −v. By taking subsequences of {y + n } and {y − n } we can assume that y + n → y + and y − n → y − for some y + , y − ∈ M ∩ B(x, r) ; compare Fig. 7.
If y + ≠ x , then the line l x = {x + tv} passes through y + , but this means that v ∈ Sec x (M ∩ B(x, r)) , but the assumption was that v ∉ U , so we obtain a contradiction. So y + = x . Analogously we prove that y − = x.
Finally, suppose y + = y − = x . The line l x = {x + tv} is the limit of secant lines passing through y + n and y − n ; therefore, l x is tangent to M at x. But then v ∈ S x ⊂ U . Contradiction. ◻ We extend the argument of Lemma 6.5 in the following way.  B(x, r)).
Proof The proof is a modification of the proof of Lemma 6.5. We leave the details for the reader. The next step in the proof of the Separation Lemma 6.2 is the following.

Moreover, there are two subsets
x and v � ∈ V x , then Sec x � (M) does not contain ±v � .
Proof Let U and r > 0 be as in the statement of Lemma 6.4. The set Sec x (M⧵B(x, r)) is the image of the compact manifold M⧵B(x, r) of dimension n under a smooth map; hence, its interior is empty. Therefore, there exists a point v ∈ S n+1 such that neither v nor −v is in the image Sec x (M⧵B(x, r)) and also neither v nor −v is in U. By the continuity of Sec x , there exist a small ball B gl ⊂ B(x, r) with center x and a small ball V gl ⊂ S n+1 containing v such that if v � ∈ V gl and x � ∈ B gl , then neither v ′ nor −v � belongs to Sec x � (M⧵B(x, r)) . Let V loc and B loc be from Lemma 6.6. Define V x = V loc ∩ V gl and B x = B loc ∩ B gl . Then for any We define B ± x as intersections of B loc± with B x , where B loc± are as in Lemma 6.7. ◻ We resume the proof of the Separation Lemma 6.2. Define As V x is an open set containing v x , we have x < 1 . This means that if x � ∈ B x and y ∈ Sec x (M) , then either ⟨y, v x ⟩ ⩽ x or ⟨y, −v x ⟩ ⩽ x . Cover now M by open sets B x for x ∈ M . As M is compact, there exists a finite set x 1 , … , x n such that M ⊂ B x 1 ∪ ⋯ ∪ B x n . The compactness of M implies also that there exists 0 > 0 such that the set M + B(0, 0 ) , that is, the set of points at distance less than (M ∩ B(x, r))}.
Then B ± i cover N 0 and for any i, if x � ∈ B ± i and y ∈ Sec x � (M) , then ⟨y, ±v i ⟩ ⩽ D as desired. ◻ For points x that are at distance greater than 0 from M, the statement of the Separation Lemma 6.2 holds as well. Let is the image of an n-dimensional compact manifold under a smooth map, so it is a closed nowhere dense subset of S n+1 . Thus, there exist a point z x ∈ S n+1 and a neighborhood of U x of z x such that U x ∩ Sec x (M) = � . Shrinking U x if necessary, we may guarantee that there exists a neighborhood W x ⊂ ℝ n+2 of x such that if x � ∈ W x and y ∈ Sec x � (M) , then y ∉ U x . We define again The sets W x cover P, and we take a finite subcover W x 1 , … , W x M . We define D mid as the maximum of x 1 , … , x M . The constant D mid works for points that are at distance between at least 0 from M, but stay inside B(0, R).
It is enough to take D = max(D close , D mid , D far ) . ◻ From now on, we assume that D < 1 is fixed.

The Drilled Ball Lemma
We begin to bound the value of . To this end, we shall differentiate the coefficients of Sec * x z . The point z will always be chosen in such a way that ⟨ Sec x � (y), z⟩ < D for all y ∈ M and for all x ′ sufficiently close to x.
The next result estimates the contribution to where = n − (n + 1) and C drill = C M C D n,1 is independent of , and x.
Proof By Lemma 2.23 and the Separation Lemma 6.2, the derivative of the pullback Sec * x is an n-form whose coefficients are bounded from above by Sec * x is integrated over M . We use Proposition 6.1 twice: first to conclude that the volume of M is bounded from above by C M n and second to conclude that the integral is bounded by C D n,1 C M n −(n+1) . ◻ The next result shows that if M is locally parameterized by some , then if we take a first-order approximation, the contribution to the derivative of (x) from the local piece does not change much. We need to set up some assumptions.
Choose > 0 and ∈ ( 1 2 , 1) . For a fixed point x at distance from M, we set M = M ∩ B(x, ) . We assume that , are such that M can parameterized by where B ′ is some bounded open subset in ℝ n and (0) is the point on M that is nearest to x. We also assume that B ′ is a star-shaped, that is, if w ∈ B � , then tw is also in B ′ for t ∈ [0, 1] . For simplicity of the formulae, we may transform B ′ in such a way that Choose > 0 in such a way that B ′ is a subset of an n-dimensional ball B(0, ) and B ′ is not a subset of B(0, ∕2) . Let 1 be the first-order approximation of , that is Lemma 6.14 We have ⩽ 4 .
Proof of Lemma 6.14 Suppose w ∈ B(0, ) , w ≠ 0 . Write w for the vector in ℝ n+2 given by 1 ‖w‖ (w, 0, 0) . Define a function w ∶ ℝ → ℝ by the formula w (t) = ⟨ (t w ‖w‖ ),w⟩ . By the definition, we have w (0) = 0 . From (6.11) we calculate that d dt w (0) = 1 . Furthermore, as the second derivative of is bounded by C 2 we have that | d 2 dt 2 w (t)| < C 2 . It follows that by the assumptions, we have . This shows that t 0 w ‖w‖ cannot belong to B ′ . As w was an arbitrary point in B ′ , this implies that no element in B ′ can have norm 2 . As B ′ is connected, this implies that B ′ must be contained in B(0, 2 ) . By the definition of , we immediately recover that ⩽ 4 . ◻ We resume the proof of Lemma 6.12. Choose w ∈ B(0, ) . Write y 0 = (w) , y 1 = 1 (w) . By the Taylor formula, we have We have ‖w‖ < 4 and 2 > 1 . Using the assumption that < (32C 2 ) −1∕(2 −1) we infer that C 2 (4 ) 2 ⩽ 1 2 so that ‖ 1 (w) − (w)‖ ⩽ 1 2 . Therefore, as dist (x, M ) = , we infer that for each point y ′ in the interval connecting y 0 and y 1 we have ‖x − y � ‖ ⩾ 1 2 ; see Fig. 8. Write now for i = 1, … , n + 1: (6.15) ‖y 0 − y 1 ‖ = ‖ 1 (w) − (w)‖ ⩽ C 2 ‖w‖ 2 . By the mean value theorem, for any i, j there exists a point y ′ in the interval connecting y 0 and y 1 such that where v is the directional derivative in the direction of the vector where the factor −n−1 comes from the estimate of F ij (x, (w)) and the constant C depends on previous constants, that is, C depends on C 1 , C 2 .
We use now (6.21) together with (6.16) and the definitions of G ij , H ij . After straightforward calculations, we obtain for some constant C: The last expression is bounded by C app ( (n+3) −(n+2) + (n+2) −(n+1) ) , where C app is a new constant. As < 1 and ≪ 1 , the term (n+3) −(n+2) is dominating. ◻ In the following result, we show that the constants in the Approximation Lemma 6.12 can be made universal, that is, depending only on M and and not on x and .

Proposition 6.22
For any ∈ ( 1 2 , 1) there exist constants C and 1 > 0 such that for any Here V is a map defined relatively to the plane V that is tangent to M at a point y such that dist (x, M) = ‖x − y‖.
Proof Cover M by a finite number of subsets U i such that each of these subsets can be parameterized by a map i ∶ V i → U i , where V i is a bounded subset of ℝ n . By the compactness of M, there exists 1 > 0 such that if U ⊂ M has diameter less than 1 , then U is contained in one of the U i . Shrinking 1 if necessary, we may and shall assume that if dist (x, M) < 1 , then there is a unique point y ∈ M such that dist (x, M) = ‖x − y‖.
Set C 1 and C 2 to be the upper bound on the first and the second derivatives of all of the i . The derivative D i (w) is injective for all w ∈ V i . We assume that C 0 > 0 is such that ‖D i (w)v‖ ⩾ C 0 ‖v‖ for all v ∈ ℝ n , i = 1, … , n and w ∈ V i . Choose a point x at distance > 0 to M such that 2 < 1 and < 1 . Let M = B(x, ) ∩ M . As this set has diameter less than 1 , we infer that M ⊂ U i for some i. Let B = −1 i (M ) ⊂ V i . Let y ∈ M be the unique point realizing dist (x, M) = ‖x − y‖ . We translate the set B in such a way that i (0) = y . Next, we rotate the coordinate system in ℝ n+2 in such a way that the image of D (0) has a block structure A ⊕ 0 0 0 0 for some invertible matrix A. We know that A −1 is a matrix with coefficients bounded by a universal constant depending on c 1 and C 1 . Define now B 0 = A −1 (B) and x = i •A . Then has first and second derivatives bounded by a constant depending on C 1 , C 2 and c 1 . Denote these constants by Such constant exists because D (w) is injective and we use the mean value theorem. Moreover, C 0 (x) is bounded below by a constant depending on C 1 , C 2 and C 0 . It remains to ensure that the following two conditions are satisfied. First, the set B = −1 x (M ) has to be star-shaped, second the inequality (6.13) is satisfied. The second condition is obviously guaranteed by taking 1 sufficiently small. We claim that the first condition can also be guaranteed by taking small 1 . To see this, we first notice that if 1 is sufficiently small, then M is connected for all < 1 . Next, we take a closer look at the definition of B ⊂ V i . Namely, we can think of B as the set of points w ∈ V i satisfying the inequality R(w) ⩽ , where in this case. We use repeatedly the Drilled Ball Lemma 6.10 for = k+1 and = k , k = 0, … , k 0 − 1 . We are allowed to do that because for k < k 0 − 1 we have and n k 0 − (n + 1) k 0 −1 ⩾ − . Summing the inequality from the Drilled Ball Lemma 6.10 for k from 0 to k 0 − 1 , we arrive at Recall that k 0 = 0 . Equation (6.26) does not cover the part of M outside of B(x, 1). However, on M⧵B(x, 1) , the form is easily seen to have coefficients bounded above by a constant independent of and x; hence, for some constant C ext depending on M but not on x and . It remains to show We cannot use the Drilled Ball Lemma 6.10 directly, because V is unbounded. However, we shall use similar ideas as in the proof of the Drilled Ball Lemma 6.10. The form is an n-form whose coefficients on V are bounded by C D � n,1 ‖y − x‖ −(n+1) , where D ′ is such that −1 x (V) ⊂ {u n+2 < D � } . Its restriction to V is equal to some function F x (y) times the volume form on V, where �F x (y)� ⩽ C V C D � n,1 ‖y − x‖ −(n+1) (it is easy to see that as V is a half-plane, C V exists). Therefore, we need to bound The method is standard. Introduce radial coordinates on V centered at y 0 , and notice that ‖y − x‖ ⩽ 2‖y 0 − y‖ as long as y ∈ V⧵B(x, 0 ) . Perform first the integral (6.29) over radial coordinates obtaining the integral over the radius only, that is where n−1 is the volume of a unit sphere of dimension n − 1 . This proves (6.28) with C flat = 2 n−1 n−1 C V C D � n,1 . Combining (6.25), (6.26),(6.27) and (6.28), we obtain the desired statement. ◻

Local triviality of ˚ near M
Let now X = (0, 0 ] × S 1 × M . Let ∶ X → N⧵M be a parameterization given by The composition • ∶ X → S 1 will still be denoted by . We are going to show that this map is a locally trivial fibration whose fibers have bounded (n + 1)-dimensional volume.
Again by (7.2) | | > 1 2 > 0 , so by the implicit function theorem we infer that t is in fact a smooth map. Then t is a smooth parameterization of the fiber of . It remains to show that is locally trivial.
To this end, we choose a point t ∈ S 1 and let U ⊂ S 1 be a neighborhood of t. Define the map ̃ ∶ (0, 0 ] × U × M → −1 (U) by the formula Clearly, ̃ is a bijection. As t depends smoothly on the parameter t, we infer that ̃ is a smooth map and the map −1 (U) → (0, 0 ] × U × M given by (r, , x) ↦ (r, (r, , x), x) is its inverse. Therefore, ̃ is a local trivialization. As a consequence of Fibration Lemma 7.3, we show that ∶ ℝ n+2 ⧵M → S 1 does not have too many critical points. This is a consequence of Sard's theorem and the control of near M provided by Lemma 7.3. Proof Extend to a map from S n+2 ⧵M → S 1 as in Corollary 3.2. We split the S n+2 ⧵M as a union of S n+2 ⧵N and N⧵M , where, recall, N is the set of points at distance less than or equal to 0 . By Sard's theorem, the map restricted to S n+2 ⧵N has a set of critical points which is closed boundary and of measure zero. On the other hand, on N⧵M the map has no critical points at all, because by the Fibration Lemma 7.3 the map restricted to N⧵M ≅ X is a locally trivial fibration. ◻ ∶ (r, , x) ↦ x + v 1 r cos + v 2 r sin .
Proof By (7.8) t r is bounded by 2C r − , so we the same argument as in the proof of Lemma 7.10 shows that t (r, x) converges as r → 0 uniformly with respect to x. Therefore, t is well defined. The composition M • t ∶ (0, 0 ] × M → (0, 0 ] × M is an identity. Hence, M • t ∶ [0, 0 ] × M → [0, 0 ] × M is also an identity. In particular, t is injective. ◻ We next prove the surjectivity of t . Before we state the proof, we indicate a possible problem in Fig. 9.   Fig. 9 The picture indicates the necessity of proving the surjectivity of t ; the map t is not onto. In Theorem 7.15 we show that the situation as on the picture cannot happen