An improvement of transversality results and its applications
aa r X i v : . [ m a t h . O C ] F e b AN IMPROVEMENT OF TRANSVERSALITY RESULTSAND ITS APPLICATIONS
SHUNSUKE ICHIKI
Abstract.
In this paper, we establish a transversality theorem from the view-point of Hausdorff measures, which is an improvement of significant transver-sality results from the viewpoint of Lebesgue measures such as the basictransversality result due to Ren´e Thom and its strengthening which was givenby John Mather in 1973. By using the improved transversality theorem, weobtain two transversality theorems on generic linear perturbations and theirapplications. Furthermore, we also give an application to multiobjective opti-mization from the viewpoint of differential topology and singularity theory. Introduction
In this paper, unless otherwise stated, all manifolds are without boundary andassumed to have countable bases.First, we give the definition of transversality.
Definition 1.
Let X and Y be C r manifolds, and Z be a C r submanifold of Y ( r ≥ f : X → Y be a C mapping.(1) We say that f : X → Y is transverse to Z at x if f ( x ) Z or in the caseof f ( x ) ∈ Z , the following holds: df x ( T x X ) + T f ( x ) Z = T f ( x ) Y. (2) We say that f : X → Y is transverse to Z if for any x ∈ X , the mapping f is transverse to Z at x .Let X , A and Y be C r manifolds ( r ≥
1) and U be an open set of X × A . Inwhat follows, by π : U → X and π : U → A , we denote the natural projectionsdefined by π ( x, a ) = x,π ( x, a ) = a. Let F : U → Y be a C mapping. For any element a ∈ π ( U ), let F a : π ( U ∩ ( X × { a } )) → Y be the mapping defined by F a ( x ) = F ( x, a ). Here, note that π ( U ∩ ( X × { a } )) isopen in X . SetΣ( F, Z ) = { a ∈ π ( U ) | F a is not transverse to Z } . Mathematics Subject Classification.
Key words and phrases. transversality, Hausdorff measure, generic linear perturbation, topol-ogy of Pareto sets.
Some significant transversality results have been given so far (Proposition 1and Theorems 1 and 2). The following basic transversality result due to Ren´e Thomlies at the heart of most application of transversality.
Proposition 1 ([4]) . Let X , A and Y be C ∞ manifolds, Z be a C ∞ submanifoldof Y and F : X × A → Y be a C ∞ mapping. If F is transverse to Z , then Σ( F, Z ) has Lebesgue measure zero in A . In [12], an improvement of Proposition 1 is given by John Mather (for the result,see Theorem 1). In order to state the result, we give the following definition.
Definition 2.
Let X and Y be C r manifolds, and Z be a C r submanifold of Y ( r ≥ f : X → Y be a C mapping. For any x ∈ X , set δ ( f, x, Z ) = (cid:26) f ( x ) Z, dim Y − dim( df x ( T x X ) + T f ( x ) Z ) if f ( x ) ∈ Z,δ ( f, Z ) = sup { δ ( f, x, Z ) | x ∈ X } . In the case that all manifolds and mappings are of class C ∞ , Definition 2 is thedefinition of [12, p. 230]. As in [1], δ ( f, x, Z ) measures the extent to which f failsto be transverse to Z at x . Since δ ( f, Z ) = 0 if and only if f is transverse to Z , thefollowing result by Mather is a natural strengthening of Proposition 1. Theorem 1 ([12]) . Let X , A and Y be C ∞ manifolds, Z be a C ∞ submanifold of Y and F : X × A → Y be a C ∞ mapping. If for any ( x, a ) ∈ X × A , δ ( F a , x, Z ) = 0 or δ ( F, ( x, a ) , Z ) < δ ( F a , x, Z ) , then Σ( F, Z ) has Lebesgue measure zero in A . Theorem 1 is an important transversality result for investigating global proper-ties of differentiable mappings. For example, this result is an essential tool for theproofs of [12, Theorem 1] and [1, Theorem 2.2]. However, it is difficult to apply tomappings with elements ( x, a ) ∈ X × A satisfying δ ( F, ( x, a ) , Z ) = δ ( F a , x, Z ) > Definition 3.
Let X , A and Y be C r manifolds, and Z be a C r submanifold of Y ( r ≥ F : U → Y be a C mapping, where U is an open set of X × A .Then, we define W ( F, Z ) = { ( x, a ) ∈ U | δ ( F a , x, Z ) = δ ( F, ( x, a ) , Z ) > } ,δ ∗ ( F, Z ) = dim X + dim Z − dim Y + δ ( F, Z )= dim X − codim Z + δ ( F, Z ) , where codim Z = dim Y − dim Z . Theorem 2 ([9]) . Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C r mapping, where U is an open set of X × A . If r > max { δ ∗ ( F, Z ) , } , then the following ( α ) and ( β ) are equivalent. ( α ) The set π ( W ( F, Z )) has Lebesgue measure zero in π ( U ) . ( β ) The set Σ( F, Z ) has Lebesgue measure zero in π ( U ) . If a given C ∞ mapping F : X × A → Y satisfies the assumption of Theorem 1,then W ( F, Z ) = ∅ . Thus, Theorem 2 implies Mather’s Theorem 1.Although Proposition 1 and Theorems 1 and 2 are useful for investigating prop-erties of generic mappings, it is difficult to estimate the Hausdorff dimension of the N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 3 bad set Σ(
F, Z ) by these results from the viewpoint of Lebesgue measures. Hence,in this paper, we establish a transversality theorem from the viewpoint of Haus-dorff measures (see Theorem 3), which is an improvement of Theorem 2 (and henceProposition 1 and Theorem 1). By Theorem 3, we can also estimate the Hausdorffdimension of Σ(
F, Z ). In what follows, we denote the image of a given mapping f by Im f . Theorem 3.
Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C r mapping, where U is an open set of X × A and r is a positiveinteger. Then, the following hold: (1) Suppose δ ∗ ( F, Z ) ≥ . Then, for any real number s satisfying s ≥ dim A − δ ∗ ( F, Z ) + 1 r , (1.1) the following ( α ) and ( β ) are equivalent. ( α ) The set π ( W ( F, Z )) has s -dimensional Hausdorff measure zero in π ( U ) . ( β ) The set Σ( F, Z ) has s -dimensional Hausdorff measure zero in π ( U ) . (2) Suppose δ ∗ ( F, Z ) < . Then, the following hold: (2a) We have W ( F, Z ) = ∅ . (2b) For any non-negative real number s satisfying s > dim A + δ ∗ ( F, Z ) , theset Σ( F, Z ) has s -dimensional Hausdorff measure zero in π ( U ) . (2c) For any a ∈ π ( U ) − Σ( F, Z ) , we have Im F a ∩ Z = ∅ . Finally, we give some important remarks on Theorem 3.
Remark 1. (1) If r < δ ∗ ( F, Z )+1, both ( α ) and ( β ) of Theorem 3 (1) triviallyhold since s ≥ dim A − δ ∗ ( F, Z ) + 1 r > dim A. Namely, in Theorem 3 (1), note that the case r ≥ δ ∗ ( F, Z ) + 1 is essential.(2) We will show that Theorem 3 is an improvement of Theorem 2 (and henceProposition 1 and Theorem 1).Let F : U → Y be a C r mapping satisfying the assumption r > max { δ ∗ ( F, Z ) , } of Theorem 2.First, we consider the case δ ∗ ( F, Z ) ≥
0. Since r ≥ δ ∗ ( F, Z ) + 1, we canset s = dim A in (1.1). Since a subset of π ( U ) has (dim A )-dimensionalHausdorff measure zero in π ( U ) if and only if the subset has Lebesguemeasure zero in π ( U ), the assertion ( α ) (resp., ( β )) of Theorem 3 (1) isthe same as ( α ) (resp., ( β )) of Theorem 2. Thus, in the case δ ∗ ( F, Z ) ≥ δ ∗ ( F, Z ) <
0. Since W ( F, Z ) = ∅ by Theo-rem 3 (2a), we obtain ( α ) of Theorem 2. By Theorem 3 (2b), Σ( F, Z ) has(dim A )-dimensional Hausdorff measure zero in π ( U ). Thus, Σ( F, Z ) alsohas Lebesgue measure zero in π ( U ). Namely, we obtain ( β ) of Theorem 2.(3) There exists a C r mapping such that ( α ) ⇒ ( β ) of Theorem 3 (1) does nothold if a given non-negative real number s does not satisfy (1.1). Namely,in general, we cannot improve (1.1) (see Example 1 in Section 2).(4) In Theorem 3 (1), if all manifolds and mappings are of class C ∞ , then forany real number s such that s > dim A −
1, there exists a positive integer
SHUNSUKE ICHIKI r satisfying (1.1). Thus, in the C ∞ case, we can replace (1.1) by s > dim A − . Moreover, in the C ∞ case, there exists an example such that ( α ) ⇒ ( β ) ofTheorem 3 (1) does not hold if we replace s > dim A − s ≥ dim A − α ) ⇒ ( β ) of Theorem 3 (1), ( β ) ⇒ ( α ) always holds forany positive integer r and any non-negative real number s which do notnecessarily satisfy (1.1) (see Lemma 6 in Section 4).(6) There exists an example such that Theorem 3 (2) does not hold if we replace s > dim A + δ ∗ ( F, Z ) by s ≥ dim A + δ ∗ ( F, Z ) (see Example 3 in Section 2).(7) In the case F ( U ) ∩ Z = ∅ , there exists an example such that dim A + δ ∗ ( F, Z ) < s is non-negative.On the other hand, note that in the case F ( U ) ∩ Z = ∅ , we always havedim A + δ ∗ ( F, Z ) ≥ Some examples on Theorem 3 and Remark 1
First, as mentioned in Remark 1 (3), (4) and (6), we give Examples 1 to 3,respectively.
Example 1 (An example of Theorem 3 (1)) . As in [2, Example 4.2], there exists a C r mapping η = ( η , . . . , η n ) : R n → R n such that the Hausdorff dimension of theset consisting of all critical values of η is n − r , where r is a positive integer(for the definition of critical values, see Section 3).Let F : R n × R n → R n be the mapping defined by F ( x, a ) = ( a − η ( x ) , . . . , a n − η n ( x )) , with Z = { (0 , . . . , } ( ⊂ R n ), where a = ( a , . . . , a n ). Since δ ( F, Z ) = 0, we have δ ∗ ( F, Z ) = 0. Thus, by Theorem 3 (1), for any real number s satisfying s ≥ n − r , (2.1)the following ( α ) and ( β ) are equivalent.( α ) The set π ( W ( F, Z )) has s -dimensional Hausdorff measure zero in R n .( β ) The set Σ( F, Z ) has s -dimensional Hausdorff measure zero in R n .Since π ( W ( F, Z )) = ∅ , we have ( α ) for any s ≥
0. Hence, we obtain ( β ) for any s satisfying (2.1). N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 5
On the other hand, ( β ) does not hold in the case 0 ≤ s < n − r as follows.Since a ∈ Σ( F, Z ) if and only if there exists x ∈ R n such that F a ( x ) = (0 , . . . ,
0) (i.e., a = η ( x )) and rank dη x < n , it follows that Σ( F, Z ) is equal to the set consisting ofall critical values of η . Therefore, the Hausdorff dimension of Σ( F, Z ) is n − r .Thus, in the case 0 ≤ s < n − r , ( β ) does not hold. Namely, we cannot improvethe assumption (2.1). Example 2 (An example of Theorem 3 (1)) . Let F : R n × R → R be the mappingdefined by F ( x, a ) = (cid:0) , a − a (cid:1) , where a = ( a , a ). Set Z = { (0 , } ( ⊂ R ).Since JF ( x,a ) = (cid:18) · · · · · · a − a (cid:19) , we have δ ( F, Z ) = 2 and δ ∗ ( F, Z ) = n ( ≥ F is of class C ∞ , by Theo-rem 3 (1) (and Remark 1 (4)), for any real number s satisfying s > dim R − , the following ( α ) and ( β ) are equivalent.( α ) The set π ( W ( F, Z )) has s -dimensional Hausdorff measure zero in R .( β ) The set Σ( F, Z ) has s -dimensional Hausdorff measure zero in R .Since δ ( F, ( x, a ) , Z ) = a − a = 0 , a − a = 0 and a = 0 , a = a = 0 ,δ ( F a , x, Z ) = (cid:26) a − a = 0 , a − a = 0 , we obtain W ( F, Z ) = R n × { (0 , } . Since π ( W ( F, Z )) = { (0 , } ( ⊂ R ), wehave ( α ) for any s >
1. Hence, we also have ( β ) for any s > F, Z ) = { ( a , a ) ∈ R | a − a = 0 } , ( β ) does not hold in the case s = 1 although ( α ) holds even in the case s ≥ α ) ⇒ ( β ) does not hold if we replace s > s ≥ Example 3 (An example of Theorem 3 (2)) . Let F : R × R ℓ → R ℓ ( ℓ ≥
2) be themapping defined by F ( x, a ) = ( x + a , . . . , x + a ℓ ) , where a = ( a , . . . , a ℓ ) ∈ R ℓ . Set Z = { (0 , . . . , } ( ⊂ R ℓ ).Since δ ( F, Z ) = 0, we have δ ∗ ( F, Z ) = 1 − ℓ ( < W ( F, Z ) = ∅ .(2b) For any non-negative real number s satisfying s > dim R ℓ + δ ∗ ( F, Z ) = 1 , the set Σ( F, Z ) has s -dimensional Hausdorff measure zero in R ℓ . SHUNSUKE ICHIKI (2c) For any a ∈ R ℓ − Σ( F, Z ), we have Im F a ∩ Z = ∅ .On the other hand, sinceΣ( F, Z ) = { ( a , . . . , a ℓ ) ∈ R ℓ | a = · · · = a ℓ } , the set Σ( F, Z ) does not have 1-dimensional Hausdorff measure zero in R ℓ . Thus,we cannot replace s > s ≥ Example 4.
Let F : R × R → R n ( n ≥
3) be the mapping defined by F ( x, a ) =(0 , . . . , Z = { (1 , , . . . , } ( ⊂ R n ). Since δ ∗ ( F, Z ) = 1 − n , we havedim R + δ ∗ ( F, Z ) = 2 − n ( < A + δ ∗ ( F, Z ) <
0. 3.
Preliminaries for the proof of Theorem 3
Let s be an arbitrary non-negative real number. Then, the s -dimensional Haus-dorff outer measure on R n is defined as follows. Let B be a subset of R n . The0-dimensional Hausdorff outer measure of B is the number of points in B . For s >
0, the s -dimensional Hausdorff outer measure of B is defined bylim δ → H sδ ( B ) , where for each 0 < δ ≤ ∞ , H sδ ( B ) = inf ∞ X j =1 (diam C j ) s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B ⊂ ∞ [ j =1 C j , diam C j ≤ δ . Here, for a subset C of R n , we writediam C = sup { k x − y k | x, y ∈ C } , where k z k denotes the Euclidean norm of z ∈ R n . Note that the infimum in H sδ ( B )is over all coverings of B by subsets C , C , . . . of R n satisfying diam C j ≤ δ for allpositive integers j .Let s be an arbitrary non-negative real number. Let N be a C r manifold ( r ≥ n , and { ( U λ , ϕ λ ) } λ ∈ Λ be a coordinate neighborhood system of N .Then, a subset Σ of N has s -dimensional Hausdorff measure zero in N if for any λ ∈ Λ, the set ϕ λ (Σ ∩ U λ ) has s -dimensional Hausdorff (outer) measure zero in R n .Note that this definition does not depend on the choice of a coordinate neighborhoodsystem of N . Moreover, for a subset Σ of N , setHD N (Σ) = inf { s ∈ [0 , ∞ ) | Σ has s -dimensional Hausdorff measure zero in N } , which is called the Hausdorff dimension of
Σ.Let X and Y be C r manifolds ( r ≥ f : X → Y be a C mapping. Apoint x ∈ X is called a critical point of f if rank df x < dim Y . A point x ∈ X iscalled a regular point of f if it is not a critical point. We say that a point y ∈ Y is a critical value if it is the image of a critical point. A point y ∈ Y is called a regular value if it is not a critical value.The assertions (1) and (2) of the following result follow from [3, 3.4.3 (p. 316)]and [14, Theorem 1 (p. 253)], respectively. Theorem 4 ([3, 14]) . Let X and Y be C r manifolds, and f : X → Y be a C r mapping, where r is a positive integer. N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 7 (1)
Suppose dim X ≥ dim Y . Then, the set of critical values of f has s -dimensionalHausdorff measure zero, where s = dim Y − X − dim Y + 1 r . (2) Suppose dim
X < dim Y . Then, for any real number s > dim X , the set ofcritical values of f has s -dimensional Hausdorff measure zero. Lemma 1 ([9]) . Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C mapping, where U is an open set of X × A ( r ≥ . Then,for any ( x, a ) ∈ U , we have δ ( F a , x, Z ) ≥ δ ( F, ( x, a ) , Z ) . The following lemma has been shown as a part of the proof of [9, Theorem 2].For the sake of readers’ convenience, in this paper, we explicitly give the part as alemma with the proof.
Lemma 2 ([9]) . Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C mapping, where U is an open set of X × A ( r ≥ . Then,it follows that π ( W ( F, Z )) ∪ π ( f W ( F, Z )) = Σ(
F, Z ) , where f W ( F, Z ) = { ( x, a ) ∈ U | δ ( F a , x, Z ) > δ ( F, ( x, a ) , Z ) } .Proof of Lemma 2. It is clearly seen that π ( W ( F, Z )) ∪ π ( f W ( F, Z )) ⊂ Σ( F, Z ).Next, let a ∈ Σ( F, Z ). Then, there exists x ∈ π ( U ) satisfying δ ( F a , x, Z ) >
0. ByLemma 1, we have ( x, a ) ∈ W ( F, Z ) ∪ f W ( F, Z ). Thus, we obtain a ∈ π ( W ( F, Z )) ∪ π ( f W ( F, Z )). (cid:3)
Lemma 3.
Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C mapping, where U is an open set of X × A ( r ≥ . If F ( U ) ∩ Z = ∅ , then we have dim A + δ ∗ ( F, Z ) ≥ .Proof. Since there exists ( x, a ) ∈ U satisfying F ( x, a ) ∈ Z , we have δ ( F, Z ) ≥ dim Y − dim(Im dF ( x,a ) + T F ( x,a ) Z ) ≥ dim Y − (dim Im dF ( x,a ) + dim T F ( x,a ) Z ) ≥ dim Y − (dim U + dim Z )= dim Y − (dim X + dim A + dim Z ) . Thus, it follows thatdim A + δ ∗ ( F, Z ) = dim A + dim X + dim Z − dim Y + δ ( F, Z ) ≥ . (cid:3) In the following, for two sets V , V , a mapping f : V → V , and a subset V of V , the restriction of the mapping f to V is denoted by f | V : V → V . Lemma 4 ([9]) . Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C mapping, where U is an open set of X × A ( r ≥ . Forany integer ρ satisfying ≤ ρ ≤ δ ( F, Z ) , set f W ρ = { ( x, a ) ∈ U | δ ( F a , x, Z ) > δ ( F, ( x, a ) , Z ) = ρ } . SHUNSUKE ICHIKI
Then, for any ( x , a ) ∈ f W ρ , there exist an open neighborhood e U of ( x , a ) and a C r submanifold e Z of Y satisfying the following: (a) dim e Z = dim Z + ρ . (b) F ( e U ) ∩ Z ⊂ e Z . (c) The mapping F | e U : e U → Y is transverse to e Z . (d) For any ( x, a ) ∈ e U , it follows that δ ( F a , x, Z ) − δ ( F a , x, e Z ) ≤ ρ . Proof of Theorem 3
In Section 4.1, we give an essential result for the proof of Theorem 3 (see Propo-sition 2). In Section 4.2 (resp., Section 4.3), we show Theorem 3 (1) (resp., Theo-rem 3 (2)).4.1.
An essential result for the proof of Theorem 3.Proposition 2.
Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C r mapping, where U is an open set of X × A and r is apositive integer. Set f W ( F, Z ) = { ( x, a ) ∈ U | δ ( F a , x, Z ) > δ ( F, ( x, a ) , Z ) } . Then, the following hold: (1)
Suppose δ ∗ ( F, Z ) ≥ . Then, for any real number s satisfying s ≥ dim A − δ ∗ ( F, Z ) + 1 r , the set π ( f W ( F, Z )) has s -dimensional Hausdorff measure zero in π ( U ) . (2) Suppose δ ∗ ( F, Z ) < . Then, for any non-negative real number s satisfying s > dim A + δ ∗ ( F, Z ) , the set π ( f W ( F, Z )) has s -dimensional Hausdorff measure zero in π ( U ) .Proof of Proposition 2. The proof will be separated into STEP 1, STEP 2.1 andSTEP 2.2. In STEP 1, we give the former common part of the proofs of (1) and(2). In STEP 2.1 and STEP 2.2, we give the latter parts of the proofs of (1) and(2), respectively.STEP 1. In this step, we give the former common part of the proofs of Propo-sition 2 (1) and (2).If F ( U ) ∩ Z = ∅ , then π ( f W ( F, Z )) (= ∅ ) has 0-dimensional Hausdorff measurezero in π ( U ). Thus, in this case, both (1) and (2) hold.Now, we will consider the case F ( U ) ∩ Z = ∅ . By Lemma 3, we have dim A + δ ∗ ( F, Z ) ≥
0. Hence, in the proof of not only (1) but also (2), we can assume s > ρ satisfying 0 ≤ ρ ≤ δ ( F, Z ), set f W ρ = { ( x, a ) ∈ U | δ ( F a , x, Z ) > δ ( F, ( x, a ) , Z ) = ρ } . Then, we have f W ( F, Z ) = [ ≤ ρ ≤ δ ( F,Z ) f W ρ . For any integer ρ satisfying 0 ≤ ρ ≤ δ ( F, Z ), by Lemma 4, there exist countablymany open neighborhoods e U ρ, , e U ρ, , . . . such that f W ρ ⊂ S ∞ i =1 e U ρ,i ( f W ρ ∩ e U ρ,i = ∅ N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 9 for any i ) and countably many C r submanifolds e Z ρ, , e Z ρ, , . . . satisfying for anypositive integer i ,(a) dim e Z ρ,i = dim Z + ρ .(b) F ( e U ρ,i ) ∩ Z ⊂ e Z ρ,i .(c) The mapping F | e U ρ,i : e U ρ,i → Y is transverse to e Z ρ,i .(d) For any ( x, a ) ∈ e U ρ,i , it follows that δ ( F a , x, Z ) − δ ( F a , x, e Z ρ,i ) ≤ ρ .For any integer ρ satisfying 0 ≤ ρ ≤ δ ( F, Z ) and any positive integer i , set B ρ,i = ( F | e U ρ,i ) − ( e Z ρ,i ) . Now, we prepare the following lemma.
Lemma 5.
In the above situation, for any integer ρ satisfying ≤ ρ ≤ δ ( F, Z ) andany positive integer i , we have the following: (1) For any ( x, a ) ∈ f W ρ ∩ e U ρ,i , it follows that δ ( F a , x, e Z ρ,i ) > . (2) We have f W ρ ∩ e U ρ,i ⊂ B ρ,i , and hence B ρ,i = ∅ . (3) The set B ρ,i is a C r submanifold of e U ρ,i satisfying dim B ρ,i = dim X + dim A + dim e Z ρ,i − dim Y. (4) We have dim B ρ,i ≤ dim B δ ( F,Z ) ,i . (5) It follows that δ ∗ ( F, Z ) = dim X + dim Z + δ ( F, Z ) − dim Y = dim X + dim e Z δ ( F,Z ) ,i − dim Y = dim B δ ( F,Z ) ,i − dim A ≥ dim B ρ,i − dim A. Proof of Lemma 5.
Let ( x, a ) ∈ f W ρ ∩ e U ρ,i be any element. Since δ ( F a , x, Z ) > ρ ,we have δ ( F a , x, e Z ρ,i ) > x, a ) ∈ f W ρ ∩ e U ρ,i be an arbitrary element. Since F a ( x ) ∈ e Z ρ,i by (1), weobtain ( x, a ) ∈ B ρ,i . Since f W ρ ∩ e U ρ,i = ∅ , we also have B ρ,i = ∅ . Hence, (2) holds.The assertion (3) (resp., (4)) can be obtained by (c) (resp., (a) and (3)). In (5),the second equality, the third equality and the last inequality can be obtained by(a), (3) and (4), respectively. (cid:3) By Lemma 5 (5), note that for any positive integer i , we have δ ∗ ( F, Z ) ≥ ⇐⇒ dim B δ ( F,Z ) ,i ≥ dim A. (4.1)Since f W ρ = S ∞ i =1 ( f W ρ ∩ e U ρ,i ), we have f W ( F, Z ) = [ ≤ ρ ≤ δ ( F,Z ) ∞ [ i =1 ( f W ρ ∩ e U ρ,i ) ! . Hence, in order to show that π ( f W ( F, Z )) has s -dimensional Hausdorff measure zeroin π ( U ), it is sufficient to show that π ( f W ρ ∩ e U ρ,i ) has s -dimensional Hausdorffmeasure zero in π ( e U ρ,i ) for any ρ and i .Now, for the proof of Proposition 2, let ρ and i be arbitrary integers satisfying0 ≤ ρ ≤ δ ( F, Z ) and i ≥
1, respectively.
We consider the case dim B ρ,i = 0. Since B ρ,i is countable, we see that f W ρ ∩ e U ρ,i (and hence π ( f W ρ ∩ e U ρ,i )) is also countable by Lemma 5 (2). Since s >
0, the set π ( f W ρ ∩ e U ρ,i ) has s -dimensional Hausdorff measure zero in π ( e U ρ,i ).Therefore, in what follows, it is sufficient to consider the case dim B ρ,i > ρ,i be the set of the critical values of π | B ρ,i : B ρ,i → π ( e U ρ,i ). Let a ∈ π ( f W ρ ∩ e U ρ,i ) be an arbitrary element. Then, there exists x ∈ π ( e U ρ,i ) such that( x, a ) ∈ f W ρ ∩ e U ρ,i . Since ( F | e U ρ,i ) a is not transverse to e Z ρ,i by Lemma 5 (1), it iseasy to see that a ∈ Σ ρ,i . Namely, we obtain π ( f W ρ ∩ e U ρ,i ) ⊂ Σ ρ,i . Thus, in order to show that π ( f W ρ ∩ e U ρ,i ) has s -dimensional Hausdorff mea-sure zero in π ( e U ρ,i ), it is sufficient to show that Σ ρ,i has s -dimensional Hausdorffmeasure zero in π ( e U ρ,i ).STEP 2.1. In this step, we give the latter part of the proof of Proposition 2 (1).For the proof, it is sufficient to show that Σ ρ,i has s -dimensional Hausdorff measurezero in π ( e U ρ,i ) in the case δ ∗ ( F, Z ) ≥
0, where s is a real number satisfying s ≥ dim A − δ ∗ ( F, Z ) + 1 r .
Here, by Lemma 5 (4) and (4.1), note that in the case δ ∗ ( F, Z ) ≥
0, it is necessaryto consider the two cases dim B ρ,i ≥ dim A and dim B ρ,i < dim A .We consider the first case dim B ρ,i ≥ dim A . Since δ ∗ ( F, Z ) ≥ dim B ρ,i − dim A by Lemma 5 (5), we have s ≥ dim A − B ρ,i − dim A + 1 r . Hence, by applying Theorem 4 (1) to π | B ρ,i , it follows that Σ ρ,i has s -dimensionalHausdorff measure zero in π ( e U ρ,i ).In the second case dim B ρ,i < dim A , since s > dim A − ≥ dim B ρ,i , by applying Theorem 4 (2) to π | B ρ,i , the set Σ ρ,i has s -dimensional Hausdorffmeasure zero in π ( e U ρ,i ).STEP 2.2. In this step, we give the latter part of the proof of Proposition 2 (2).For the proof, it is sufficient to show that Σ ρ,i has s -dimensional Hausdorff measurezero in π ( e U ρ,i ) in the case δ ∗ ( F, Z ) <
0, where s is a real number satisfying s > dim A + δ ∗ ( F, Z ) ( ≥ . In the case δ ∗ ( F, Z ) <
0, by the last inequality of Lemma 5 (5), it is sufficient toconsider only the case dim B ρ,i < dim A . By the same inequality, we havedim B ρ,i ≤ dim A + δ ∗ ( F, Z ) ( < s ) . Thus, by applying Theorem 4 (2) to the mapping π | B ρ,i , it follows that Σ ρ,i has s -dimensional Hausdorff measure zero in π ( e U ρ,i ). (cid:3) N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 11
Proof of Theorem 3 (1).
By combining Lemma 2 and Proposition 2 (1),we obtain ( α ) ⇒ ( β ).In order to prove ( β ) ⇒ ( α ), it is sufficient to prepare the following. Lemma 6.
Let X , A and Y be C r manifolds, Z be a C r submanifold of Y and F : U → Y be a C mapping, where U is an open set of X × A ( r ≥ . Let s be anynon-negative real number. If Σ( F, Z ) has s -dimensional Hausdorff measure zero in π ( U ) , then π ( W ( F, Z )) has s -dimensional Hausdorff measure zero in π ( U ) .Proof of Lemma 6. Since π ( W ( F, Z )) ⊂ Σ( F, Z ) by Lemma 2, this lemma clearlyholds. (cid:3)
Proof of Theorem 3 (2).
First, we show W ( F, Z ) = ∅ . Suppose W ( F, Z ) = ∅ . Then, there exists ( x, a ) ∈ U such that δ ( F a , x, Z ) = δ ( F, ( x, a ) , Z ) > . Thus, we have δ ∗ ( F, Z ) = dim X + dim Z − dim Y + δ ( F, Z ) ≥ dim X + dim Z − dim Y + δ ( F a , x, Z )= dim X + dim Z − dim(Im( dF a ) x + T F a ( x ) Z ) ≥ dim X + dim Z − (dim Im( dF a ) x + dim T F a ( x ) Z ) ≥ δ ∗ ( F, Z ) <
0. Hence, we obtain W ( F, Z ) = ∅ .Thus, we have Σ( F, Z ) = π ( f W ( F, Z )) by Lemma 2, where f W ( F, Z ) is definedin Lemma 2. By Proposition 2 (2), we obtain (2b).Now, suppose that there exists a ∈ π ( U ) − Σ( F, Z ) such that Im F a ∩ Z = ∅ .Then, there exists x ∈ π ( U ) satisfyingIm( dF a ) x + T F a ( x ) Z = T F a ( x ) Y. Since dim Im( dF a ) x + dim T F a ( x ) Z − dim T F a ( x ) Y ≥ , we have dim X + dim Z − dim Y ≥ . This contradicts δ ∗ ( F, Z ) <
0. Hence, we have (2c). ✷ Two transversality theorems on generic linear perturbations
Let L ( R m , R ℓ ) be the space consisting of all linear mappings of R m into R ℓ . Inwhat follows, we will regard L ( R m , R ℓ ) as the Euclidean space ( R m ) ℓ in the obviousway.In [7], for a C ∞ immersion (resp., a C ∞ injection) f : X → V and an arbitrary C ∞ mapping g : V → R ℓ , a transversality theorem on the 1-jet extension (resp.,on crossings) of ( g + π ) ◦ f : X → R ℓ ( π ∈ L ( R m , R ℓ ) − Σ) is given, where X isa C ∞ manifold, V is an open subset of R m and Σ is a subset of L ( R m , R ℓ ) withLebesgue measure zero (for details, see Theorem 1 (resp., Theorem 2) in [7]). In[8], the two transversality theorems on generic linear perturbations described aboveare improved so that these work even in the case where manifolds and mappings are not necessarily of class C ∞ . However, these are transversality theorems fromthe viewpoint of Lebesgue measures (for details, see Theorems 1 and 2 in [8]).On the other hand, in this section, as applications of Theorem 3, we give twotransversality theorems on generic linear perturbations from the viewpoint of Haus-dorff measures (see Theorems 5 and 6), which are further improvements of thetransversality theorems from the viewpoint of Lebesgue measures described above.For the statement of Theorem 5, we prepare some definitions. Let X be a C r manifold ( r ≥
2) of dimension n and J ( X, R ℓ ) be the space of 1-jets of mappings of X into R ℓ . Then, note that J ( X, R ℓ ) is a C r − manifold. For a given C r mapping f : X → R ℓ , the 1-jet extension j f : X → J ( X, R ℓ ) is defined by q j f ( q ).Then, notice that j f is of class C r − . For details on J ( X, R ℓ ) and j f , see forexample, [4].Now, let { ( V λ , ϕ λ ) } λ ∈ Λ be a coordinate neighborhood system of X . Let Π : J ( X, R ℓ ) → X × R ℓ be the natural projection defined by Π( j f ( q )) = ( q, f ( q )).Let Φ λ : Π − ( V λ × R ℓ ) → ϕ λ ( V λ ) × R ℓ × J ( n, ℓ ) be the homeomorphism definedby Φ λ (cid:0) j f ( q ) (cid:1) = (cid:0) ϕ λ ( q ) , f ( q ) , j ( ψ λ ◦ f ◦ ϕ − λ ◦ e ϕ λ )(0) (cid:1) , where J ( n, ℓ ) = { j f (0) | f : ( R n , → ( R ℓ , } and e ϕ λ : R n → R n (resp., ψ λ : R ℓ → R ℓ ) is the translation given by e ϕ λ (0) = ϕ λ ( q ) (resp., ψ λ ( f ( q )) = 0). Then, { (Π − ( V λ × R ℓ ) , Φ λ ) } λ ∈ Λ is a coordinate neighborhood system of J ( X, R ℓ ).Set S k = { j f (0) ∈ J ( n, ℓ ) | corank Jf (0) = k } , where corank Jf (0) = min { n, ℓ } − rank Jf (0) and k = 1 , , . . . , min { n, ℓ } . Set S k ( X, R ℓ ) = [ λ ∈ Λ Φ − λ (cid:0) ϕ λ ( V λ ) × R ℓ × S k (cid:1) . Then, the set S k ( X, R ℓ ) is a submanifold of J ( X, R ℓ ) satisfyingcodim S k ( X, R ℓ ) = dim J ( X, R ℓ ) − dim S k ( X, R ℓ )= ( n − v + k )( ℓ − v + k ) , where v = min { n, ℓ } . (For details on S k and S k ( X, R ℓ ), see [4, pp. 60–61]).The following is the first transversality theorem on generic linear perturbationsfrom the viewpoint of Hausdorff measures. Theorem 5.
Let f : X → V be a C r immersion and g : V → R ℓ be a C r mapping,where r is an integer satisfying r ≥ , X is a C r manifold and V is an open subsetof R m . Let k be an integer satisfying ≤ k ≤ min { dim X, ℓ } . Set Σ k = { π ∈ L ( R m , R ℓ ) | j (( g + π ) ◦ f ) is not transverse to S k ( X, R ℓ ) } . Then, the following hold: (1)
Suppose dim X − codim S k ( X, R ℓ ) ≥ . Then, for any real number s satisfying s ≥ mℓ − X − codim S k ( X, R ℓ ) + 1 r − , (5.1) the set Σ k has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . (2) Suppose dim X − codim S k ( X, R ℓ ) < . Then, we have the following: N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 13 (2a)
For any real number s satisfying s > mℓ + dim X − codim S k ( X, R ℓ ) , the set Σ k has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . (2b) For any π ∈ L ( R m , R ℓ ) − Σ k , we have j (( g + π ) ◦ f )( X ) ∩ S k ( X, R ℓ ) = ∅ . Remark 2. (1) In Theorem 5 (1), if all manifolds and mappings are of class C ∞ , then wecan replace (5.1) by s > mℓ − f is an immersion, we have n ≤ m , where n = dim X .Thus, in Theorem 5 (2), since mℓ + dim X − codim S k ( X, R ℓ ) ≥ mℓ + n − nℓ = ( m − n ) ℓ + n ≥ n, it is not necessary to assume that s is non-negative.For the statement of Theorem 6, we prepare some definitions. Let X be a C r manifold ( r ≥ X ( d ) = { ( q , . . . , q d ) ∈ X d | q i = q j ( i = j ) } . Note that X ( d ) is an open submanifold of X d . For any mapping f : X → R ℓ , let f ( d ) : X ( d ) → ( R ℓ ) d be the mapping given by f ( d ) ( q , . . . , q d ) = ( f ( q ) , . . . , f ( q d )) . Set ∆ d = { ( y, . . . , y ) ∈ ( R ℓ ) d | y ∈ R ℓ } . Then, ∆ d is a submanifold of ( R ℓ ) d satis-fying codim ∆ d = dim ( R ℓ ) d − dim ∆ d = ℓ ( d − . As in [7], for any injection f : X → R m , set d f = max ( d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∀ ( q , . . . , q d ) ∈ X ( d ) , dim d X i =2 R −−−−−−−→ f ( q ) f ( q i ) = d − ) . Since the mapping f is an injection, we have 2 ≤ d f . Since f ( q ) , . . . , f ( q d f ) arepoints of R m , it follows that d f ≤ m + 1. Hence, we get2 ≤ d f ≤ m + 1 . The following is the second transversality theorem on generic linear perturbationsfrom the viewpoint of Hausdorff measures.
Theorem 6.
Let f : X → V be a C r injection and g : V → R ℓ be a C r mapping,where r is a positive integer, X is a C r manifold and V is an open subset of R m .Set Σ d = { π ∈ L ( R m , R ℓ ) | (( g + π ) ◦ f ) ( d ) is not transverse to ∆ d } , where d is an integer satisfying ≤ d ≤ d f . Then, the following hold: (1) Suppose dim X ( d ) − codim ∆ d ≥ . Then, for any real number s satisfying s ≥ mℓ − X ( d ) − codim ∆ d + 1 r , (5.2) the set Σ d has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . (2) Suppose dim X ( d ) − codim ∆ d < . Then, the following hold: (2a) For any real number s satisfying s > mℓ + dim X ( d ) − codim ∆ d , the set Σ d has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . (2b) For any π ∈ L ( R m , R ℓ ) − Σ d , we have (( g + π ) ◦ f ) ( d ) ( X ( d ) ) ∩ ∆ d = ∅ . Remark 3. (1) In Theorem 6 (1), if all manifolds and mappings are of class C ∞ , then wecan replace (5.2) by s > mℓ − n = dim X . In Theorem 6 (2), since mℓ + dim X ( d ) − codim ∆ d = mℓ + nd − ℓ ( d − ≥ nd, it is not necessary to assume that s is non-negative.(3) As in [7], there is a case of d f = 3 as follows. If n + 1 ≤ m , X = S n and f : S n → R m is the inclusion f ( x ) = ( x, , . . . , d f = 3.Indeed, suppose that there exists a point ( q , q , q ) ∈ ( S n ) (3) satisfyingdim P i =2 R −−−−−−−→ f ( q ) f ( q i ) = 1. Since the number of the intersections of f ( S n )and a straight line of R m is at most two, this contradicts the assumption.Thus, we have d f ≥
3. On the other hand, since S × { } ⊂ f ( S n ), we get d f <
4, where 0 = (0 , . . . , | {z } ( m − -tuple. Therefore, it follows that d f = 3. Remark 4.
In Theorems 5 and 6, there is an advantage that the domain of g : V → R ℓ is not R m but an arbitrary open subset V of R m . Suppose V = R . Let g : R → R be the function defined by g ( x ) = | x | . Since g is not differentiable at x = 0, we cannot apply Theorems 5 and 6 to g : R → R . On the other hand, if V = R − { } , then we can apply these theorems to g | V .6. Proof of Theorem 5
Set n = dim X . For a positive integer e n , we denote the e n × e n unit matrix by E e n .Let Γ : X × L ( R m , R ℓ ) → J ( X, R ℓ ) be the C r − mapping defined byΓ( q, π ) = j (( g + π ) ◦ f )( q ) . First, we will show that δ (Γ , S k ( X, R ℓ )) = 0. Let ( e q, e π ) ∈ X × L ( R m , R ℓ ) bean arbitrary element satisfying Γ( e q, e π ) ∈ S k ( X, R ℓ ). Then, in order to show that δ (Γ , S k ( X, R ℓ )) = 0, it is sufficient to show thatdim (cid:0) Im d Γ ( e q, e π ) + T Γ( e q, e π ) S k ( X, R ℓ ) (cid:1) = n + ℓ + nℓ. (6.1)As in Section 5, let { ( V λ , ϕ λ ) } λ ∈ Λ (resp., { (Π − ( V λ × R ℓ ) , Φ λ ) } λ ∈ Λ ) be a co-ordinate neighborhood system of X (resp., J ( X, R ℓ )). There exists a coordi-nate neighborhood (cid:0) V e λ × L ( R m , R ℓ ) , ϕ e λ × id (cid:1) containing ( e q, e π ) of X × L ( R m , R ℓ ),where id is the identity mapping of L ( R m , R ℓ ) into L ( R m , R ℓ ), and ϕ e λ × id : V e λ × L ( R m , R ℓ ) → ϕ e λ ( V e λ ) × L ( R m , R ℓ ) is given by (cid:0) ϕ e λ × id (cid:1) ( q, π ) = (cid:0) ϕ e λ ( q ) , id( π ) (cid:1) .Then, (cid:0) Π − ( V e λ × R ℓ ) , Φ e λ (cid:1) is a coordinate neighborhood containing the point Γ( e q, e π )of J ( X, R ℓ ). Let x = ( x , . . . , x n ) ∈ R n be a local coordinate on ϕ e λ ( V e λ ). N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 15
Here, let ( a ij ) ≤ i ≤ ℓ, ≤ j ≤ m be a representing matrix of a linear mapping π ∈L ( R m , R ℓ ). Then, ( g + π ) ◦ f : X → R ℓ is given as follows:( g + π ) ◦ f = (cid:18) g ◦ f + m X j =1 a j f j , . . . , g ℓ ◦ f + m X j =1 a ℓj f j (cid:19) , (6.2)where f = ( f , . . . , f m ), g = ( g , . . . , g ℓ ) and ( a , . . . , a m , . . . , a ℓ , . . . , a ℓm ) ∈ ( R m ) ℓ .Hence, the mapping Γ is locally given by the following: Φ e λ ◦ Γ ◦ ( ϕ e λ × id) − ( x, π )= Φ e λ ◦ j (( g + π ) ◦ f ) ◦ ϕ − e λ ( x )= (cid:16) x, ( g + π ) ◦ f ◦ ϕ − e λ ( x ) ,∂ (( g + π ) ◦ f ◦ ϕ − e λ ) ∂x ( x ) , . . . , ∂ (( g + π ) ◦ f ◦ ϕ − e λ ) ∂x n ( x ) , · · · · · · · · · ,∂ (( g + π ) ℓ ◦ f ◦ ϕ − e λ ) ∂x ( x ) , . . . , ∂ (( g + π ) ℓ ◦ f ◦ ϕ − e λ ) ∂x n ( x ) ! = (cid:16) x, ( g + π ) ◦ f ◦ ϕ − e λ ( x ) ,∂g ◦ e f∂x ( x ) + m X j =1 a j ∂ e f j ∂x ( x ) , . . . , ∂g ◦ e f∂x n ( x ) + m X j =1 a j ∂ e f j ∂x n ( x ) , · · · · · · · · · ,∂g ℓ ◦ e f∂x ( x ) + m X j =1 a ℓj ∂ e f j ∂x ( x ) , . . . , ∂g ℓ ◦ e f∂x n ( x ) + m X j =1 a ℓj ∂ e f j ∂x n ( x ) ! , where g + π = (( g + π ) , . . . , ( g + π ) ℓ ) and e f = ( e f , . . . , e f m ) = ( f ◦ ϕ − e λ , . . . , f m ◦ ϕ − e λ ) = f ◦ ϕ − e λ . The Jacobian matrix of Γ at ( e q, e π ) is the following: J Γ ( e q, e π ) = E n · · · · · · ∗ · · · · · · ∗ t ( Jf e q ) ∗ t ( Jf e q ) . . . t ( Jf e q ) ( x,π )=( ϕ e λ ( e q ) , e π ) , where Jf e q is the Jacobian matrix of f at e q . Notice that t ( Jf e q ) is the transpose of Jf e q and that there are ℓ copies of t ( Jf e q ) in the above description of J Γ ( e q, e π ) . Since S k ( X, R ℓ ) is a subfiber-bundle of J ( X, R ℓ ) with the fiber S k , in order to show(6.1), it is sufficient to prove that the matrix M given below has rank n + ℓ + nℓ : M = E n + ℓ ∗ · · · · · · ∗ t ( Jf e q ) t ( Jf e q ) . . . t ( Jf e q ) ( x,π )=( ϕ e λ ( e q ) , e π ) . Notice that there are ℓ copies of t ( Jf e q ) in the above description of M . Note thatfor any i (1 ≤ i ≤ mℓ ), the ( n + ℓ + i )-th column vector of M coincides with the( n + i )-th column vector of J Γ ( e q, e π ) . Since f is an immersion ( n ≤ m ), the rank of M is equal to n + ℓ + nℓ . Therefore, we get (6.1).Since δ (Γ , S k ( X, R ℓ )) = 0, we have δ ∗ (Γ , S k ( X, R ℓ )) = dim X − codim S k ( X, R ℓ ) . Note that Σ k = Σ(Γ , S k ( X, R ℓ )) . Next, we will show Theorem 5 (1). Since dim X − codim S k ( X, R ℓ ) ≥
0, weobtain δ ∗ (Γ , S k ( X, R ℓ )) ≥
0. Notice that Γ is of class C r − ( r ≥
2) and we have s ≥ mℓ − X − codim S k ( X, R ℓ ) + 1 r − mℓ − δ ∗ (Γ , S k ( X, R ℓ )) + 1 r − . Since δ (Γ , S k ( X, R ℓ )) = 0, we obtain W (Γ , S k ( X, R ℓ )) = ∅ . Therefore, the set π ( W (Γ , S k ( X, R ℓ ))) has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ). Thus,by Theorem 3 (1), the set Σ(Γ , S k ( X, R ℓ )) has s -dimensional Hausdorff measurezero in L ( R m , R ℓ ). Since Σ k = Σ(Γ , S k ( X, R ℓ )), we have Theorem 5 (1).Finally, we will show Theorem 5 (2). Since dim X − codim S k ( X, R ℓ ) <
0, weobtain δ ∗ (Γ , S k ( X, R ℓ )) <
0. Since Γ is of class C r − ( r ≥
2) and we have s > mℓ + dim X − codim S k ( X, R ℓ ) = mℓ + δ ∗ (Γ , S k ( X, R ℓ )) , the set Σ(Γ , S k ( X, R ℓ )) has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) byTheorem 3 (2b). By Theorem 3 (2c), for any π ∈ L ( R m , R ℓ ) − Σ(Γ , S k ( X, R ℓ )), wehave Im Γ π ∩ S k ( X, R ℓ ) = ∅ . Since Σ k = Σ(Γ , S k ( X, R ℓ )), we obtain Theorem 5 (2). ✷ Proof of Theorem 6
Set n = dim X . For a positive integer e n , we denote the e n × e n unit matrix by E e n .Let Γ : X ( d ) × L ( R m , R ℓ ) → ( R ℓ ) d be the C r mapping given byΓ( q, π ) = ((( g + π ) ◦ f )( q ) , . . . , (( g + π ) ◦ f )( q d )) , where q = ( q , . . . , q d ).First, we will show that δ (Γ , ∆ d ) = 0. Let ( e q, e π ) ∈ X ( d ) × L ( R m , R ℓ ) be anarbitrary element satisfying Γ( e q, e π ) ∈ ∆ d . Then, in order to show that δ (Γ , ∆ d ) = 0,it is sufficient to show thatdim (cid:0) Im d Γ ( e q, e π ) + T Γ( e q, e π ) ∆ d (cid:1) = dℓ. (7.1)Let { ( V λ , ϕ λ ) } λ ∈ Λ be a coordinate neighborhood system of X . There exists acoordinate neighborhood ( V e λ ×· · ·× V e λ d ×L ( R m , R ℓ ) , ϕ e λ ×· · ·× ϕ e λ d × id) containing( e q, e π ) of X ( d ) × L ( R m , R ℓ ), where id : L ( R m , R ℓ ) → L ( R m , R ℓ ) is the identitymapping, and ϕ e λ × · · · × ϕ e λ d × id : V e λ × · · · × V e λ d × L ( R m , R ℓ ) → ϕ e λ ( V e λ ) ×· · · × ϕ e λ d ( V e λ d ) × L ( R m , R ℓ ) is defined by ( ϕ e λ × · · · × ϕ e λ d × id)( q , . . . , q d , π ) =( ϕ e λ ( q ) , . . . , ϕ e λ d ( q d ) , id( π )). Let x = ( x , . . . , x d ) ∈ ( R n ) d be a local coordinate on ϕ e λ ( V e λ ) × · · · × ϕ e λ d ( V e λ d ). N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 17
Let ( a ij ) ≤ i ≤ ℓ, ≤ j ≤ m be a representing matrix of a linear mapping π ∈ L ( R m , R ℓ ).As in the proof of Theorem 5, the mapping ( g + π ) ◦ f : X → R ℓ is given by thesame expression as (6.2). Hence, Γ is locally given by the following:Γ ◦ (cid:16) ϕ e λ × · · · × ϕ e λ d × id (cid:17) − ( x , . . . , x d , π )= (cid:16) ( g + π ) ◦ f ◦ ϕ − e λ ( x ) , ( g + π ) ◦ f ◦ ϕ − e λ ( x ) , . . . , ( g + π ) ◦ f ◦ ϕ − e λ d ( x d ) (cid:17) = g ◦ e f ( x ) + m X j =1 a j e f j ( x ) , . . . , g ℓ ◦ e f ( x ) + m X j =1 a ℓj e f j ( x ) ,g ◦ e f ( x ) + m X j =1 a j e f j ( x ) , . . . , g ℓ ◦ e f ( x ) + m X j =1 a ℓj e f j ( x ) , · · · · · · · · · ,g ◦ e f ( x d ) + m X j =1 a j e f j ( x d ) , . . . , g ℓ ◦ e f ( x d ) + m X j =1 a ℓj e f j ( x d ) , where e f ( x i ) = ( e f ( x i ) , . . . , e f m ( x i )) = ( f ◦ ϕ − e λ i ( x i ) , . . . , f m ◦ ϕ − e λ i ( x i )) (1 ≤ i ≤ d ).For simplicity, set e x = ( ϕ e λ × · · · × ϕ e λ d )( e q ).The Jacobian matrix of Γ at ( e q, e π ) is the following: J Γ ( e q, e π ) = ∗ B ( x ) ∗ B ( x )... ... ∗ B ( x d ) ( x,π )=( e x, e π ) , where B ( x i ) = b ( x i ) b ( x i ) . . . b ( x i ) ℓ rowsand b ( x i ) = ( e f ( x i ) , . . . , e f m ( x i )). By the construction of T Γ( e q, e π ) ∆ d , in order toprove (7.1), it is sufficient to prove that the rank of the following matrix M isequal to dℓ : M = E ℓ B ( x ) E ℓ B ( x )... ... E ℓ B ( x d ) x = e x . There exists a dℓ × dℓ regular matrix Q satisfying Q M = E ℓ B ( x )0 B ( x ) − B ( x )... ...0 B ( x d ) − B ( x ) x = e x . There exists an ( ℓ + mℓ ) × ( ℓ + mℓ ) regular matrix Q satisfying Q M Q = E ℓ B ( x ) − B ( x )... ...0 B ( x d ) − B ( x ) x = e x = E ℓ ℓ rows −−−−−−−→ e f ( x ) e f ( x ) −−−−−−−→ e f ( x ) e f ( x ) . . . −−−−−−−→ e f ( x ) e f ( x )... ... ... ... ... ℓ rows −−−−−−−→ e f ( x ) e f ( x d ) −−−−−−−→ e f ( x ) e f ( x d ) . . . −−−−−−−→ e f ( x ) e f ( x d ) , where −−−−−−−→ e f ( x ) e f ( x i ) = ( e f ( x i ) − e f ( x ) , . . . , e f m ( x i ) − e f m ( x )) (2 ≤ i ≤ d ) and x = e x .From d − ≤ d f − d f , we havedim d X i =2 R −−−−−−−→ e f ( x ) e f ( x i ) = d − , where x = e x . Hence, by the construction of Q M Q and d − ≤ m , the rank of Q M Q is equal to dℓ . Therefore, the rank of M must be equal to dℓ . Hence, weget (7.1).Since δ (Γ , ∆ d ) = 0, we have δ ∗ (Γ , ∆ d ) = dim X ( d ) − codim ∆ d . Note that Σ d = Σ(Γ , ∆ d ) . Next, we will show Theorem 6 (1). Since dim X ( d ) − codim ∆ d ≥
0, we obtain δ ∗ (Γ , ∆ d ) ≥
0. Notice that Γ is of class C r ( r ≥
1) and we have s ≥ mℓ − X ( d ) − codim ∆ d + 1 r = mℓ − δ ∗ (Γ , ∆ d ) + 1 r . Since δ (Γ , ∆ d ) = 0, we get W (Γ , ∆ d ) = ∅ . Hence, the set π ( W (Γ , ∆ d )) has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ). Thus, by Theorem 3 (1), theset Σ(Γ , ∆ d ) has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ). Since Σ d =Σ(Γ , ∆ d ), we have Theorem 6 (1). N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 19
Finally, we will show Theorem 6 (2). Since dim X ( d ) − codim ∆ d <
0, we obtain δ ∗ (Γ , ∆ d ) <
0. Since Γ is of class C r ( r ≥
1) and we have s > mℓ + dim X ( d ) − codim ∆ d = mℓ + δ ∗ (Γ , ∆ d ) , the set Σ(Γ , ∆ d ) has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) by The-orem 3 (2b). By Theorem 3 (2c), for any π ∈ L ( R m , R ℓ ) − Σ(Γ , ∆ d ), we haveIm Γ π ∩ ∆ d = ∅ . Since Σ d = Σ(Γ , ∆ d ), we have Theorem 6 (2). ✷ Applications of Theorems 5 and 6
In Section 8.1 (resp., Section 8.2), applications of Theorem 5 (resp., Theorem 6)are stated and proved. In Section 8.2, applications obtained by combining Theo-rems 5 and 6 are also given. These are also improvements of some results from theviewpoint of Lebesgue measures in [7, 8].8.1.
Applications of Theorem 5. A C r function f : X → R ( r ≥
2) is called a
Morse function if all of the critical points of f are nondegenerate, where X is a C r manifold. For details on Morse functions, see for example, [4, p. 63]. In the case ℓ = 1, we have the following. Corollary 1.
Let f be a C r immersion of a C r manifold X into an open subset V of R m and g : V → R be a C r function, where r is an integer satisfying r ≥ . Set Σ = { π ∈ L ( R m , R ) | ( g + π ) ◦ f : X → R is not a Morse function } . Then, for any real number s satisfying s ≥ m − r − , (8.1) the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ) . Remark 5.
In Corollary 1, if all manifolds and mappings are of class C ∞ , thenwe can replace (8.1) by s > m − Proof of Corollary 1.
It is clearly seen that Σ is the set consisting of all elements π ∈ L ( R m , R ) satisfying that j (( g + π ) ◦ f ) is not transverse to S ( X, R ). Sincedim X − codim S ( X, R ) = 0, by Theorem 5 (1), for any real number s satisfying(8.1), the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ). (cid:3) Example 5 (An example of Corollary 1) . Set X = V = R and f ( x ) = x inCorollary 1. Let g : R → R be a C r function, where r is a positive integer satisfying r ≥
2. As in Corollary 1, setΣ = { π ∈ L ( R , R ) | g + π : R → R is not a Morse function } . By Corollary 1, for any real number s satisfying s ≥ r − , the set Σ has s -dimensional Hausdorff measure zero in L ( R , R ). We regard the Cantor set K of R as a subset of L ( R , R ).In the case r ≥
3, since Σ has -dimensional Hausdorff measure zero in L ( R , R ),we obtain HD L ( R , R ) (Σ) < HD L ( R , R ) ( K ) = log 2log 3 = 0 . · · · . On the other hand, in the case r = 2, there exists an example such that Σ = K as follows. Let ξ : R → R be a C function such that the set consisting of all criticalvalues of ξ is the Cantor set. Note that the existence of the mapping ξ can be easily shown from [13, Proposition 2 (p. 1485)] . Set g ( x ) = − R x ξ ( x ) dx . Note that g isof class C . Then, π ∈ Σ if and only if there exists x ∈ R such that a = ξ ( x ) and dξdx ( x ) = 0, where π ( x ) = ax ( a ∈ R ). Thus, we obtain Σ = K .Finally, we explain an advantage of using a result from the viewpoint of Hausdorffmeasures with this example. Since any subset of L ( R , R ) whose Hausdorff dimensionis less than 1 has Lebesgue measure zero in L ( R , R ), we cannot investigate whetherthe bad set Σ is equal to K or not by transversality results from the viewpoint ofLebesgue measures, such as Proposition 1 and Theorems 1 and 2. On the otherhand, in the case r ≥ K by Corollary 1 which is a result from the viewpoint of Hausdorff measures.Let f : X → R n − ( n ≥
2) be a C ∞ mapping, where X is a C ∞ manifold ofdimension n . A singular point q ∈ X of f is called a singular point of Whitneyumbrella if there exist two germs of C ∞ diffeomorphisms H : ( R n − , f ( q )) → ( R n − ,
0) and h : ( X, q ) → ( R n ,
0) such that H ◦ f ◦ h − ( x , x , . . . , x n ) =( x , x x , . . . , x x n , x , . . . , x n ), where ( x , x , . . . , x n ) is a local coordinate around h ( q ) = 0 ∈ R n . In the case ℓ = 2 dim X − X ≥ Corollary 2.
Let f be a C ∞ immersion of an n -dimensional C ∞ manifold X ( n ≥ into an open subset V of R m and g : V → R n − be a C ∞ mapping.Let Σ be the set consisting of all elements π ∈ L ( R m , R n − ) not satisfying thatany singular point of ( g + π ) ◦ f : X → R n − is a singular point of Whitneyumbrella. Then, for any real number s satisfying s > m (2 n − − , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R n − ) .Proof of Corollary 2. Let k be an integer satisfying 1 ≤ k ≤ n . As in Theorem 5,setΣ k = { π ∈ L ( R m , R n − ) | j (( g + π ) ◦ f ) is not transverse to S k ( X, R n − ) } . Note that dim X − codim S k ( X, R n − ) = n − k ( n − k ) . First, we consider the case k = 1. Since dim X − codim S ( X, R n − ) = 0and s > m (2 n − −
1, the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R n − ) by Theorem 5 (1) and Remark 2 (1).Next, we consider the case 2 ≤ k ≤ n . In this case, sincedim X − codim S k ( X, R n − ) < ,s > m (2 n − − ≥ m (2 n −
1) + dim X − codim S k ( X, R n − ) , by Theorem 5 (2), we have the following:(a) The set Σ k has s -dimensional Hausdorff measure zero in L ( R m , R n − ).(b) For any π ∈ L ( R m , R n − ) − Σ k , we have j (( g + π ) ◦ f )( X ) ∩ S k ( X, R n − ) = ∅ .Note that a point q ∈ X is a singular point of Whitney umbrella of ( g + π ) ◦ f if and only if j (( g + π ) ◦ f )( q ) ∈ S ( X, R n − ) and j (( g + π ) ◦ f ) is transverse to S ( X, R n − ) at q (for this fact, see for example [4, p. 179]). By this fact and (b), Proposition 2 of [13] is as follows: If e K is a compact subset of the closed interval [0,1], then e K has Lebesgue measure zero if and only if the set consisting of all critical values of e ξ is equal to e K for some C function e ξ : R → R .Since the Cantor set K is a compact subset of [0 ,
1] with Lebesgue measure zero, we canguarantee the existence of the above function ξ : R → R . N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 21 we can easily obtain Σ = S nk =1 Σ k . Since Σ , . . . , Σ n have s -dimensional Hausdorffmeasure zero in L ( R m , R n − ), the set Σ also has s -dimensional Hausdorff measurezero in L ( R m , R n − ). (cid:3) In the case ℓ ≥ X , we have the following. Corollary 3.
Let f be a C r immersion of an n -dimensional C r manifold X intoan open subset V of R m and g : V → R ℓ be a C r mapping, where ℓ ≥ n and r ≥ .Set Σ = { π ∈ L ( R m , R ℓ ) | ( g + π ) ◦ f : X → R ℓ is not an immersion } . Then, for any real number s satisfying s > mℓ + (2 n − ℓ − , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) .Proof of Corollary 3. Let k be an integer satisfying 1 ≤ k ≤ n . As in Theorem 5,set Σ k = { π ∈ L ( R m , R ℓ ) | j (( g + π ) ◦ f ) is not transverse to S k ( X, R ℓ ) } . Since ℓ ≥ n , we havedim X − codim S k ( X, R ℓ ) ≤ dim X − codim S ( X, R ℓ ) = 2 n − ℓ − < . Hence, since s > mℓ + (2 n − ℓ − ≥ mℓ + (dim X − codim S k ( X, R ℓ )) , by Theorem 5 (2), we have the following:(a) The set Σ k has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).(b) For any π ∈ L ( R m , R ℓ ) − Σ k , we have j (( g + π ) ◦ f )( X ) ∩ S k ( X, R ℓ ) = ∅ .By (b), we can easily obtain Σ = S nk =1 Σ k . By (a), the set Σ has s -dimensionalHausdorff measure zero in L ( R m , R ℓ ). (cid:3) Example 6 (An example of Corollary 3) . Set X = V = R and f ( x ) = x inCorollary 3. Let g : R → R ℓ ( ℓ ≥
2) be the C ∞ mapping defined by g ( x ) =( x , . . . , x ). As in Corollary 3, setΣ = { π ∈ L ( R , R ℓ ) | g + π : R → R ℓ is not an immersion } . Then, for any real number s satisfying s >
1, the set Σ has s -dimensional Hausdorffmeasure zero in L ( R , R ℓ ) by Corollary 3. Indeed, by the following direct calculation,we obtain Σ = B , where B = { π = ( π , . . . , π ℓ ) ∈ L ( R , R ℓ ) | π = · · · = π ℓ } . Since B does not have 1-dimensional Hausdorff measure zero in L ( R , R ℓ ), we cannotimprove the assumption s > B . First, we show Σ ⊂ B . Let π = ( π , . . . , π ℓ ) ∈ Σ be anarbitrary element. Set π i ( x ) = a i x ( a i ∈ R ) for i = 1 , . . . , ℓ . Then, there exists e x ∈ R such that 2 e x + a i = 0 for all i = 1 , . . . , ℓ . Since a = · · · = a ℓ , we have π ∈ B .Next, we show B ⊂ Σ. Let π ∈ B be an arbitrary element. Then, we can express π i ( x ) = ax ( a ∈ R ) for all i = 1 , . . . , ℓ . Set e x = − a . Since d ( g + π ) e x = 0, we obtain π ∈ Σ.Finally, we explain an advantage of using a result from the viewpoint of Haus-dorff measures with this example. Since any subset of L ( R , R ℓ ) whose Hausdorff dimension is less than ℓ has Lebesgue measure zero in L ( R , R ℓ ), we cannot esti-mate the Hausdorff dimension of the bad set Σ by transversality results from theviewpoint of Lebesgue measures. On the other hand, by Corollary 3 which is aresult from the view point of Hausdorff measures, we can estimate the Hausdorffdimension of the bad set, such as HD L ( R , R ℓ ) Σ ≤ Applications of Theorem 6.Definition 4.
Let f : X → R ℓ be a C mapping, where X is a C r manifold( r ≥ f is called a mapping with normal crossings if for any integer d satisfying d ≥
2, the mapping f ( d ) : X ( d ) → ( R ℓ ) d is transverse to ∆ d .In the following, for a set S , we denote the number of its elements (or its cardi-nality) by | S | .In the case dim X > ℓ , we have the following.
Corollary 4.
Let f be a C r injection of an n -dimensional C r manifold X intoan open subset V of R m and g : V → R ℓ be a C r mapping, where r is a positiveinteger. Suppose n > ℓ . Let Σ be the set consisting of all elements π ∈ L ( R m , R ℓ ) not satisfying that (( g + π ) ◦ f ) ( d ) : X ( d ) → ( R ℓ ) d is transverse to ∆ d for any integer d satisfying ≤ d ≤ d f . Then, for any real number s satisfying s ≥ mℓ − d f ( n − ℓ ) + ℓ + 1 r , (8.2) the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . Moreover, if amapping ( g + π ) ◦ f : X → R ℓ ( π ∈ L ( R m , R ℓ ) − Σ) satisfies that | (( g + π ) ◦ f ) − ( y ) | ≤ d f for any y ∈ R ℓ , then ( g + π ) ◦ f is a C r mapping with normal crossings. Remark 6.
In Corollary 4, if all manifolds and mappings are of class C ∞ , thenwe can replace (8.2) by s > mℓ − Proof of Corollary 4.
Let d be an integer satisfying 2 ≤ d ≤ d f . As in Theorem 6,set Σ d = { π ∈ L ( R m , R ℓ ) | (( g + π ) ◦ f ) ( d ) is not transverse to ∆ d } . Since n > ℓ , we obtaindim X ( d ) − codim ∆ d = nd − ℓ ( d −
1) = d ( n − ℓ ) + ℓ ( ≥ ,d f ( n − ℓ ) + ℓ ≥ d ( n − ℓ ) + ℓ = dim X ( d ) − codim ∆ d . Thus, we also have s ≥ mℓ − d f ( n − ℓ ) + ℓ + 1 r ≥ mℓ − X ( d ) − codim ∆ d + 1 r . Hence, by Theorem 6 (1), Σ d has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).Since Σ = S d f d =2 Σ d , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).Now, suppose that a mapping ( g + π ) ◦ f : X → R ℓ ( π ∈ L ( R m , R ℓ ) − Σ) satisfiesthat | (( g + π ) ◦ f ) − ( y ) | ≤ d f for any y ∈ R ℓ . Then, since (( g + π ) ◦ f ) ( d ) ( X ( d ) ) ∩ ∆ d = ∅ for any integer d satisfying d > d f , the mapping ( g + π ) ◦ f is a C r mapping withnormal crossings. (cid:3) In the case dim X ≤ ℓ ≤ X , we have the following. N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 23
Corollary 5.
Let f be a C r injection of an n -dimensional C r manifold X into anopen subset V of R m and g : V → R ℓ be a C r mapping, where r is a positive integer.Suppose n ≤ ℓ ≤ n . Let Σ be the set consisting of all elements π ∈ L ( R m , R ℓ ) notsatisfying that (( g + π ) ◦ f ) ( d ) : X ( d ) → ( R ℓ ) d is transverse to ∆ d for any integer d satisfying ≤ d ≤ d f . Then, for any real number s satisfying s ≥ mℓ − n − ℓ + 1 r , (8.3) the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . Moreover, if amapping ( g + π ) ◦ f : X → R ℓ ( π ∈ L ( R m , R ℓ ) − Σ) satisfies that | (( g + π ) ◦ f ) − ( y ) | ≤ d f for any y ∈ R ℓ , then ( g + π ) ◦ f is a C r mapping with normal crossings. Remark 7.
In Corollary 5, if all manifolds and mappings are of class C ∞ , thenwe can replace (8.3) by s > mℓ − Proof of Corollary 5.
Let d be an integer satisfying 2 ≤ d ≤ d f . As in Theorem 6,set Σ d = { π ∈ L ( R m , R ℓ ) | (( g + π ) ◦ f ) ( d ) is not transverse to ∆ d } . Since n ≤ ℓ ≤ n , we obtaindim X ( d ) − codim ∆ d = nd − ℓ ( d −
1) = d ( n − ℓ ) + ℓ ≤ n − ℓ ) + ℓ = 2 n − ℓ. First, we consider the case dim X ( d ) − codim ∆ d ≥
0. Since s ≥ mℓ − n − ℓ + 1 r ≥ mℓ − X ( d ) − codim ∆ d + 1 r , by Theorem 6 (1), Σ d has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).Secondly, we consider the case dim X ( d ) − codim ∆ d <
0. Since s ≥ mℓ − n − ℓ + 1 r > mℓ + dim X ( d ) − codim ∆ d , by Theorem 6 (2), Σ d has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).Since Σ = S d f d =2 Σ d , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).The latter assertion can be shown by the same argument as in the proof ofCorollary 4. (cid:3) In the case 2 dim
X < ℓ , we have the following.
Corollary 6.
Let f be a C r injection of an n -dimensional C r manifold X intoan open subset V of R m and g : V → R ℓ be a C r mapping ( r ≥ . Suppose n < ℓ . Let Σ be the set consisting of all elements π ∈ L ( R m , R ℓ ) not satisfyingthat ( g + π ) ◦ f : X → R ℓ is injective. Then, for any real number s satisfying s > mℓ + 2 n − ℓ , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) .Proof of Corollary 6. As in Theorem 6, setΣ = { π ∈ L ( R m , R ℓ ) | (( g + π ) ◦ f ) (2) is not transverse to ∆ } . Since 2 n < ℓ , we obtaindim X (2) − codim ∆ = 2 n − ℓ < . Hence, since s > mℓ + 2 n − ℓ = mℓ + dim X (2) − codim ∆ , by Theorem 6 (2), we have the following:(a) The set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ).(b) For any π ∈ L ( R m , R ℓ ) − Σ , we have (( g + π ) ◦ f ) (2) ( X (2) ) ∩ ∆ = ∅ .By (b), we obtain Σ = Σ . Since Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) by (a), this corollary holds. (cid:3) By combining Corollaries 3 and 6, we have the following.
Corollary 7.
Let f be a C r injective immersion of an n -dimensional C r manifold X into an open subset V of R m and g : V → R ℓ be a C r mapping ( r ≥ . Suppose n < ℓ . Let Σ be the set consisting of all elements π ∈ L ( R m , R ℓ ) not satisfyingthat ( g + π ) ◦ f : X → R ℓ is an injective immersion. Then, for any real number s satisfying s > mℓ + 2 n − ℓ , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . By Corollary 7, we obtain the following.
Corollary 8.
Let X be a compact C r manifold ( r ≥ of dimension n . Let f be a C r embedding of X into an open subset V of R m and g : V → R ℓ be a C r mapping.Suppose n < ℓ . Let Σ be the set consisting of all elements π ∈ L ( R m , R ℓ ) notsatisfying that ( g + π ) ◦ f : X → R ℓ is an embedding. Then, for any real number s satisfying s > mℓ + 2 n − ℓ , the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . An application to multiobjective optimization
The purpose of this section is to give an application of Theorem 5 (and henceTheorem 3) to multiobjective optimization (see Theorem 9) from the viewpoint ofdifferential topology and singularity theory. For a positive integer ℓ , set L = { , . . . , ℓ } . We consider the problem of optimizing several functions simultaneously. Moreprecisely, let f : X → R ℓ be a mapping, where X is a given arbitrary set. A point x ∈ X is called a Pareto solution of f if there does not exist another point y ∈ X such that f i ( y ) ≤ f i ( x ) for all i ∈ L and f j ( y ) < f j ( x ) for at least one index j ∈ L .We denote the set consisting of all Pareto solutions of f by X ∗ ( f ), which is calledthe Pareto set of f . The set f ( X ∗ ( f )) is called the Pareto front of f . The problemof determining X ∗ ( f ) is called the problem of minimizing f .Let f = ( f , . . . , f ℓ ) : X → R ℓ be a mapping, where X is a given arbitrary set.For a non-empty subset I = { i , . . . , i k } of L such that i < · · · < i k , set f I = ( f i , . . . , f i k ) . The problem of determining X ∗ ( f I ) is called a subproblem of the problem of mini-mizing f . Set ∆ ℓ − = ( ( w , . . . , w ℓ ) ∈ R ℓ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ℓ X i =1 w i = 1 , w i ≥ ) . We also denote a face of ∆ ℓ − for a non-empty subset I of L by∆ I = { ( w , . . . , w ℓ ) ∈ ∆ ℓ − | w i = 0 ( i I ) } . N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 25
In this section, for a C r manifold N (possibly with corners) and a subset V of R ℓ , a mapping g : N → V is called a C r mapping (resp., a C r diffeomorphism ) if g : N → R ℓ is of class C r (resp., g : N → R ℓ is a C r immersion and g : N → V is ahomeomorphism), where r is a positive integer or r = ∞ . Here, C mappings and C diffeomorphisms are continuous mappings and homeomorphisms, respectively. Definition 5 ([5, 6]) . Let f = ( f , . . . , f ℓ ) : X → R ℓ be a mapping, where X isa subset of R m . Let r be an integer satisfying r ≥ r = ∞ . The problem ofminimizing f is C r simplicial if there exists a C r mapping Φ : ∆ ℓ − → X ∗ ( f ) suchthat both the mappings Φ | ∆ I : ∆ I → X ∗ ( f I ) and f | X ∗ ( f I ) : X ∗ ( f I ) → f ( X ∗ ( f I ))are C r diffeomorphisms for any non-empty subset I of L . The problem of minimiz-ing f is C r weakly simplicial if there exists a C r mapping φ : ∆ ℓ − → X ∗ ( f ) suchthat φ (∆ I ) = X ∗ ( f I ) for any non-empty subset I of L .As described in [5], simpliciality is an important property, which can be seenin several practical problems ranging from facility location studied half a centuryago [11] to sparse modeling actively developed today [5]. If a problem is simplicial,then we can efficiently compute a parametric-surface approximation of the entirePareto set with few sample points [10].A subset X of R m is convex if tx + (1 − t ) y ∈ X for all x, y ∈ X and all t ∈ [0 , X be a convex set in R m . A function f : X → R is strongly convex if thereexists α > f ( tx + (1 − t ) y ) ≤ tf ( x ) + (1 − t ) f ( y ) − αt (1 − t ) k x − y k for all x, y ∈ X and all t ∈ [0 , k z k is the Euclidean norm of z ∈ R m .The constant α is called a convexity parameter of the function f . A mapping f = ( f , . . . , f ℓ ) : X → R ℓ is strongly convex if f i is strongly convex for any i ∈ L . Theorem 7 ([5, 6]) . Let f : R m → R ℓ be a strongly convex C r mapping, where r is a positive integer or r = ∞ . Then, the problem of minimizing f is C r − weaklysimplicial. Moreover, this problem is C r − simplicial if the rank of the differential df x is equal to ℓ − for any x ∈ X ∗ ( f ) . For the assertion on simpliciality (resp., weak simpliciality) of Theorem 7 in thecase r ≥
2, see [5, Theorem 1.1] (resp., [6, Theorem 5]). For the case r = 1 ofTheorem 7, see [6, Theorem 2].Moreover, in [5], we have the following result. Here, note that strong convexityis preserved under linear perturbations (see Lemma 8 in Section 10). Theorem 8 ([5]) . Let f : R m → R ℓ ( m ≥ ℓ ) be a strongly convex C r mapping,where r is an integer satisfying r ≥ or r = ∞ . Set Σ = { π ∈ L ( R m , R ℓ ) | The problem of minimizing f + π is not C r − simplicial } . If m − ℓ + 4 > , then Σ has Lebesgue measure zero in L ( R m , R ℓ ) . In [5], the problem of minimizing f : X → R ℓ is said to be C r weakly simplicial if there existsa C r mapping φ : ∆ ℓ − → f ( X ∗ ( f )) satisfying φ (∆ I ) = f ( X ∗ ( f I )) for any non-empty subset I of L . On the other hand, a surjective mapping of ∆ ℓ − into X ∗ ( f ) is important to describe X ∗ ( f ).Hence, the definition of weak simpliciality in [6] is updated from that in [5]. Thus, in this paper,we adopt the definition in [6]. As an improvement of Theorem 8, we give the following application of Theo-rem 5 (and hence Theorem 3) to multiobjective optimization from the viewpoint ofdifferential topology and singularity theory.
Theorem 9.
Let f : R m → R ℓ ( m ≥ ℓ ) be a strongly convex C r mapping, where r is an integer satisfying r ≥ or r = ∞ . Set Σ = { π ∈ L ( R m , R ℓ ) | The problem of minimizing f + π is not C r − simplicial } . If m − ℓ + 4 > , then for any non-negative real number s satisfying s > mℓ − ( m − ℓ + 4) , (9.1) the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . Remark 8.
There is an example of Theorem 9 such that we cannot improve theinequality (9.1) (see Example 7).In order to show that a given mapping in Example 7 is strongly convex, weprepare Lemma 7, which is a well-known result (for the proof, for example, see [6]).Let X be a convex subset of R m . A function f : X → R is said to be convex if f ( tx + (1 − t ) y ) ≤ tf ( x ) + (1 − t ) f ( y )for all x, y ∈ X and all t ∈ [0 , Lemma 7.
Let X be a convex subset of R m . Then, a function f : X → R is stronglyconvex with a convexity parameter α > if and only if the function g : X → R defined by g ( x ) = f ( x ) − α k x k is convex. Example 7 (An example of Theorem 9) . Let f = ( f , f ) : R → R be themapping defined by f i ( x , x ) = x + x for i = 1 ,
2. Since g ( x ) = f i ( x ) − k x k = 0is convex, f is strongly convex by Lemma 7, where x = ( x , x ). As in Theorem 9,setΣ = { π ∈ L ( R , R ) | The problem of minimizing f + π is not C ∞ simplicial } . Then, for any real number s satisfying s >
2, the set Σ has s -dimensional Hausdorffmeasure zero in L ( R , R ) by Theorem 9.Indeed, by the following direct calculation, we obtain Σ = B , where B = { π = ( π , π ) ∈ L ( R , R ) | π = π } . Since B does not have 2-dimensional Hausdorff measure zero in L ( R , R ), wecannot improve the assumption s > B . First, in order to show that Σ ⊂ B , we will show that L ( R , R ) − B ⊂ L ( R , R ) − Σ. Let π ∈ L ( R , R ) − B be an arbitrary element.Let H : R → R be the diffeomorphism defined by H ( X , X ) = ( X − X , X ).As f = f , we obtain H ◦ ( f + π ) = ( π − π , f + π ). Since π − π is a linearfunction satisfying π − π = 0, it follows that rank d ( H ◦ ( f + π )) x ≥ x ∈ R . As H is a diffeomorphism, we have that rank d ( f + π ) x ≥ x ∈ R .By Theorem 7, the problem of minimizing f + π is C ∞ simplicial. Namely, weobtain π ∈ L ( R , R ) − Σ. N IMPROVEMENT OF TRANSVERSALITY RESULTS AND ITS APPLICATIONS 27
Next, we will show that B ⊂ Σ. Let π = ( π , π ) ∈ B be an arbitrary element.Set π ( x , x ) = π ( x , x ) = a x + a x , where a , a ∈ R . Since( f i + π i )( x , x ) = x + x + a x + a x = (cid:16) x + a (cid:17) + (cid:16) x + a (cid:17) − a + a i = 1 ,
2, we obtain X ∗ ( f + π ) = (cid:8) ( − a , − a ) (cid:9) ( ⊂ R ). Hence, the problem ofminimizing f + π is not C simplicial (and hence, not C ∞ simplicial). Namely, weobtain π ∈ Σ. 10.
Proof of Theorem 9
Since Theorem 9 clearly holds by combining the following two results (Lemmas 8and 9) and Theorem 7, it is sufficient to prove Lemma 9.
Lemma 8 ([5]) . Let f : R m → R ℓ be a strongly convex mapping. Then, for any π ∈ L ( R m , R ℓ ) , the mapping f + π : R m → R ℓ is also strongly convex. Lemma 9.
Let f : R m → R ℓ ( m ≥ ℓ ) be a C r mapping ( r ≥ . Set Σ = { π ∈ L ( R m , R ℓ ) | There exists x ∈ R m such that rank d ( f + π ) x ≤ ℓ − } . If m − ℓ + 4 > , then for any non-negative real number s satisfying s > mℓ − ( m − ℓ + 4) , (10.1) the set Σ has s -dimensional Hausdorff measure zero in L ( R m , R ℓ ) . On Lemma 9, we give the following remark.
Remark 9. (1) In the case ℓ = 1, note that Σ = ∅ and mℓ − ( m − ℓ +4) = − ∅ ) has 0-dimensional Hausdorff measurezero in L ( R m , R ), Lemma 9 clearly holds.(2) In the case ℓ ≥
2, since m ≥ ℓ , we have codim S ( R m , R ℓ ) = 2( m − ℓ + 2).Thus, the inequality (10.1) implies that s > mℓ − ( m − ℓ + 4) = mℓ + m − m − ℓ + 2) = mℓ + m − codim S ( R m , R ℓ ) . Proof of Lemma 9.
By Remark 9 (1), it is sufficient to consider the case ℓ ≥
2. Asin Remark 9 (2), we havecodim S ( R m , R ℓ ) = 2( m − ℓ + 2) . Since m − ℓ + 4 >
0, we also have codim S ( R m , R ℓ ) > m .Let k be an integer satisfying 2 ≤ k ≤ ℓ . As in Theorem 5, setΣ k = { π ∈ L ( R m , R ℓ ) | j ( f + π ) is not transverse to S k ( R m , R ℓ ) } . It follows that m − codim S k ( R m , R ℓ ) ≤ m − codim S ( R m , R ℓ ) < . By Remark 9 (2), note that a given real number s satisfies that s > mℓ + m − codim S k ( R m , R ℓ ) . Since r ≥
2, by Theorem 5 (2), we have the following:(a) The set Σ k has s-dimensional Hausdorff measure zero in L ( R m , R ℓ ).(b) For any π ∈ L ( R m , R ℓ ) − Σ k , we have j ( f + π )( R m ) ∩ S k ( R m , R ℓ ) = ∅ . By (b), it is clearly seen that Σ = S ℓk =2 Σ k . By (a), the set Σ has s -dimensionalHausdorff measure zero in L ( R m , R ℓ ). (cid:3) Acknowledgements
The author was supported by JSPS KAKENHI Grant Number JP19J00650.
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Department of Mathematical and Computing Science, School of Computing, TokyoInstitute of Technology, Tokyo 152-8552, Japan
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