On the Level Set of a Function with Degenerate Minimum Point

For n ≥ 2, let M be an n-dimensional smooth closed manifold and f : M → R a smooth function. We set minf(M) = m and assume thatm is attained by unique point p ∈ M such that p is a nondegenerate critical point. Then the Morse lemma tells us that if a is slightly bigger thanm, f−1(a) is diffeomorphic to S. In this paper, we relax the condition on p from being nondegenerate to being an isolated critical point and obtain the same consequence. Some application to the topology of polygon spaces is also included.


Introduction and Statement of the Result
Throughout this paper,   denotes the standard topological sphere equipped with the standard differential structure.For  ≥ 2, let  be an -dimensional smooth closed manifold and  :  → R a smooth function.We set min() =  and assume that  is attained by unique point  ∈ .Then the following result is a consequence of the Morse lemma (see, e.g., [1]): If  is a nondegenerate critical point and there are no critical points in  −1 ((, ]) (where  < ), then there is a diffeomorphism  −1 () ≅  −1 .
The purpose of this paper is to study the question of whether the same result holds if we relax the condition on  from being nondegenerate to being an isolated critical point.We also give an application of our result to the topology of polygon spaces.Now our main result is the following.
Theorem A. For  ≥ 2, let  be an -dimensional smooth closed manifold and  :  → R a smooth function.One sets min() =  and assumes that  is attained by unique point  ∈ .If  is an isolated critical point and there are no critical points in  −1 ((, ]) (where  < ), then the following results hold: (i) If  ̸ = 5, there is a diffeomorphism  −1 () ≅  −1 .

Corollary B.
For  ≥ 2, let  be an -dimensional smooth closed manifold and  :  → R a smooth function.One sets max() =  and assumes that  is attained by unique point  ∈ .If  is an isolated critical point and there are no critical points in  −1 ([, )) (where  < ), then the following results hold: This paper is organized as follows.In Section 2 we prove Theorem A. In Section 3 we study an application of it.Theorem 4 is the main result in this section.Remark 7(ii) states the essential difference between the known map and ours.

Proof of Theorem A
We keep the notations of Theorem A. For  ∈ R, we set ( (Such  indeed exists because  ∈ ∘   .)Note that  ∉   \ and   \  is compact.Note also that  is compact.Hence there exist  0 and  1 ∈ R (where  <  0 <  1 ) such that We fix such  0 and  1 .
For a compact set  ⊂ ∘   , since max() <  holds, there exists  such that max() <  < .This implies that By assumption, there are no critical points in  −1 ((, ]).Hence there is a diffeomorphism We set  := ().Then  is an open set of Proof.Since  −1 () is a manifold, the assertion is clear for  = 2.We assume  ≥ 3. We claim that To prove (5), note that   is a manifold with boundary such that   =  −1 ().Lefschetz duality implies that   (  ,   ; Z) ≅  − (  ; Z) ∀.
Recall that the inclusion  : ∘   →   is a homotopy equivalence.Since ∘   ≅ R  by Lemma 1,   has the cohomology of a point.In the homology long exact sequence of the pair (  ,   ), we apply (6).Then we obtain (5).

An Application
Starting in [9][10][11], the topology of the configuration space of planar polygons has been considered by many authors.We refer to [12] for an excellent exposition.
The configuration space of spatial polygons has also been studied by many authors.(See, e.g., [12,14,15].)In this case, our object is defined by Recently, motivated by chemistry, Crippen [16], Goto and Komatsu [17], and O'Hara [18] studied certain subspaces of  ,1 .Namely, they studied the configuration space of equilateral polygons with restriction on the splay angle of each vertex.
First, we define the angle   to be (7/12) and arccos(−1/ 3) as  = 5 and  ≥ 6, respectively.Goto and Komatsu [17] chose the angle   with molecular model in mind.Then they studied the space defined by where ⟨, ⟩ denotes the standard inner product on R 3 .The closed chains in   are equilateral polygons in R 3 with  vertices such that the interior angles are all equal to   except for the two angles at the successive vertices O and ∑ −1 =1   .The main result in [17] states that when  = 5, 6, 7,   is a manifold homeomorphic to  −4 .Since they use Reeb's theorem, they state their result as a homeomorphism.But actually   and  −4 are diffeomorphic because the differential structure on   is unique for  ≤ 3.But what is more important is that it is not known whether   is a manifold for  ≥ 8.
Second, for all  ∈ R, we define a space by Here we understand  +1 to be  1 .The closed chains in   () are equilateral polygons in R 3 with  vertices such that the interior angles are all equal to .Crippen [16] studied the topological type of   () for various  when  = 3, 4, 5.
The result is that   () is either ⌀, one point, or two points depending on .Next, O'Hara [18] studied  6 () for various .The result is that  6 () is disjoint union of a certain number of  1 's and points.It is to be noted that since dim ,1 = 2 − 6, it is natural to expect that dim  () = 2−6− = −6.But the above results imply that the defining equations for   () do not intersect transversally when  is small.Note that the above results in [16][17][18] concentrate on the case for small .This is understandable because imposing some conditions on the interior angles causes difficulties in computations.Nevertheless, we would like to prove some assertion which holds for general .Modifying the definition of   , we define a space as follows: and ℓ such that the interior angles are all equal to /2 except for the two angles at the endpoints of the edge of length ℓ.
For the rest of this paper, we prove Theorem 4. For that purpose, we define Moreover, similarly to   in (10), we define a map   :   → R by Then (15) gives an identification  ,ℓ =  −1  (ℓ 2 ) and Theorem 3 tells us that max  (  ) is attained by   .In order to compute the Hessian matrix of   at   , we construct the commutative diagram shown in Figure 4.
First, we set Namely,   is the universal cover.Second, we construct   by induction on .We define  3 to be the unique map between one-point spaces.
Proof.We fix ( 1 , . . .,  −3 ) ∈  −4 .In order to prove the lemma by contradiction, assume that  = 0 were an accumulation point of   (  (ii) Although we have not used the diffeomorphism   in the above arguments, it is to be noted that the map   ∘   :  −3 → R is much more difficult than the known map   :  −2 → R in (10).

( 1 ) 1 . 2 International
LemmaThere is a diffeomorphism ∘   ≅ R  .Proof.By [2, Lemma 3], it will suffice to prove that, for any compact set  ⊂ ∘   , there exists an open set  of ∘   such that  ⊂  and  ≅ R  .Journal of Mathematics and Mathematical Sciences We fix an open set  which satisfies that  ∈  ⊂  ⊂ ∘   ,  ≅ R  .
B. If  satisfies the assumption of Corollary B, the function  := − satisfies the assumption of  in Theorem A. Hence Corollary B follows.