Vey’s Isochoric Morse Lemma

Now look at the following commutative diagram

π(C{z,u,v}<[1], [∂u₁H], . . . , [∂e v_lH]e >) ^⊂ >R^Ω,par

The hooked arrows are inclusions, and the upper map to the right is the isomorphismC^µ∼=C^µwhereµ=µ(H)=µ(f). By hypothesis (1) in the state-ment of the versality theorem, the moduleC{z}<[1], [∂u1Fe|u=0], . . . , [∂ukFe|u=0] is equal toR^Ω_f, hence the last line is an equality, so is the first. This completes the proof.

2.5. Vey’s Isochoric Morse Lemma

Letf : (Cⁿ, 0)→(C, 0) be a holomorphic function germ which is critical and Morse at the origin. The classical Morse lemma asserts that there is a coor-dinate transformationΦ∈Aut(Cⁿ, 0) such that

f ◦Φ(x₁, . . . ,x_n)=x²₁+. . .+x_n².

If one stipulates that the coordinate change has to be volume-preserving one has the "lemme de Morse isochore" by J. Vey ([Vey77]):

Theorem 2.11(Volume-Preserving Morse Lemma).

Let f: (Cⁿ, 0)→(C, 0)be a critical Morse singularity. Then there exists a volume-preserving automorphismΦ∈Aut(Cⁿ, 0)and a holomorphic germΨ: (C, 0)→ (C, 0)such that

f ◦Φ=Ψ◦Q_f.

Here Q_f denotes the quadratic part of f at the origin. It is possible to demand

Ψ(t)=t+O(t²)and with this requirementΨis uniquely determined by f . The rest of this section is devoted to a proof of this theorem. The existence proof will make use of the isochoric unfolding theorem by Garay when the Milnor number is one ([Gar07]). For the uniqueness ofΨwe discuss the lin-ear problem and integrals over cycles in the Milnor fibre off.

We start by looking at the linear problem. That is, we consider the group action

SL_nC×Sym_nC→Sym_nC, (S,A)7→S^tAS. (2.11) Here Sym_nCis the vector space of complex, symmetric (n×n)-matrices. The following easy proposition classifies the orbits of the group action.

Proposition 2.12. Two symmetric matrices A,B∈C^n×nare equivalent (under the congruence action ofSL_nC) if and only if either

• they have both the same rank k<n, or

• they have both the full rank k=n anddet(A)=det(B).

Proof. The neccessity of the rank and determinant conditions is clear. So let AandBhave the same rankk. SinceAis symmetric of rankk, by Sylvester there is an invertible matrixSe∈GLnCsuch thatSe^tAeSis of the form

det(S) and with the definitione S:=1

If we writeBin the same manner, we are left to check when the resulting matrices

are congruent under SL_nC.

2.5. Vey’s Isochoric Morse Lemma by a well-known theorem for computing the determinant of a block-matrix withF=0 we get multiples of the identity matices of the corresponding dimension and such that (*) holds. This shows the first part of the proposition.

However, when the rankk of A, is equal ton, then there is noHwhich can force the determinant ofS to be equal to one and so in this case we really need det(E)=1 (in factS=E). Using the above notation, it suffices to show that we can always find suchS ∈SL_nCwithS^{t 1}

λ²S= _µ¹2I. By the very definition ofλandµwe have det(A)=_λ¹2n and det(B)=_µ¹2n. Now since

det(A)=det(B), we get (λ²/µ²)ⁿ=1. Thus the desired result follows from the next lemma withc:=^λ_µ²2.

Lemma 2.13. There always exists a solution S∈SLnCof the equation S^tS= c I for any nth root of unity c.

Proof. Choose annth rootλofc. Sincecⁿ=1 we haveλⁿ=1 orλⁿ = −1.

In the first case putS=λI whereI is then×nidentity matrix; in the sec-ond case putS=λdiag(−1, 1, . . . , 1). Then in both casesS^tS=λ²I=c I and det(S)=1.

This completes the proof of proposition 2.12.

Given two holomorphic germsf^A(x)=x^tAx+O(kxk²) andf^B(x)=x^tB x+ O(kxk²) for someA,B∈C^n×nand assume that there exist holomorphic germs Φ: (Cⁿ, 0)→(Cⁿ, 0) andΨ: (C, 0)→(C, 0) such that

(Cⁿ, 0) ^Φ> (Cⁿ, 0)

(C, 0)

f^B

∨

Ψ> (C, 0)

f^A

∨

commutes. Then we havef^A(Φ(x))=Ψ(f^B(x)) and by comparing the quadratic parts we can deduce, usingS:=DΦ(0), that

S^tAS=Ψ⁰(0)B. (2.12)

This makes clear the role ofΨ⁰(0) in Vey’s theorem: If det(A)=det(B)6=0 (as is the case in the statement of theorem 2.11), we haveΨ⁰(0)ⁿ=1. By lemma 2.13 we cannot get rid of this nonuniqueness ofΨ⁰(0), so we have to put it as a requirement.

Lemma 2.14. Let f : (Cⁿ, 0)→Cbe critical and Morse at the origin. Then for any s-parametric unfolding G of f there is a commutative diagram

2.5. Vey’s Isochoric Morse Lemma

(Cⁿ×C^s, 0) ^Φ > (Cⁿ, 0)

(C×C^s,G(0))

G

∨

Ψ > (C,f(0))

f

∨

withΦaΩ-preserving and holomorphic germ andΨa holomorphic germ.

Proof. We can considerf itself as an unfolding off (i.e.k=0 in the versality theorem). Sincef is Morse and critical, its Milnor number isµ=1 so that R^Ω_f ∼=C{t}¹by the proposition 2.8. Therefore we can apply theA^Ω-versality theorem,(2)⇒(3), toF:=f which shows the assertion.

The above lemma makes it now possible to prove the existence ofΨin Vey’s lemma. Putf_t:=(1−t)Q_f +t f,t∈C. Then for any valuet_i∈[0, 1], by abuse of notation, the germf_s|s≈tiis an unfolding off_ti. To be more precise, byf_s|s≈tiwe denote the first map in the unfoldingF_ioff_ti:

F_i: (Cⁿ, 0)×(C,t_i)→(C, 0)×(C,t_i), (x,s)7→(f_s(x),s).

The Hessian matrix offtat the origin inCⁿ(!) is

Hess0(ft)=(1−t) Hess0(Q_f)+tHess0(f)=Hess0(f).

Hence each germft_i,ti ∈[0, 1], has the origin as a critical nondegenerate point so that we can apply lemma 2.14 for everyt_i∈[0, 1] to get the following commutative diagram

(Cⁿ, 0)×(C,ti) ^Φi > (Cⁿ, 0)

(C, 0)×(C,t_i)

F_i

∨

Ψi > (C, 0),

f_ti

∨

where theΦi is relative isochore. The domains of definition of the maps involved, give, whent_i varies over [0, 1], an open cover of the compact set [0, 1], so that we may assume that we have a sequence 0=t₀<t₁<. . .<t_N= 1 of points in [0, 1] and a commutative diagram

(Cⁿ, 0) ^Φ1(·,t0) > (Cⁿ, 0) . . . ^{ΦN−1(·,tN−2)} > (Cⁿ, 0) ^{ΦN(·,tN−1)} > (Cⁿ, 0) dia-gram, as on the right, withΦ holomor-phic and isochore and Ψ

SinceΦis isochore, it is a local biholomorphism by the inverse function theorem. It remains to show thatΨ(t)=t+O(t²). Equation (2.12), applied to our situation (since det(Hess₀(f))=det(Hess₀(Q_f))), obviously gives us Ψ⁰(0)ⁿ =1. IfΨ⁰(0)=1, the proof is completed. If this does not hold we apply the previous considerations in linear algebra toc:= _Ψ¹0(0)to obtain a commutative diagram

whereSis a volume-preserving matrix. To this diagram we compose the pre-vious one to get an overallΨwhich is finally normed (Ψ⁰(0)=1).

Let us prove thatΨis quite unique. Lett7→δf(t)∈H_n−1(X(t);C) be a lo-cally constant section of the homological Milnor fibration of f. This gives us a cycle in the Milnor fibreX(t) off over eacht≈0,t6=0, and well de-termined by f up to sign. Letλbe any holomorphicn−1-form which is primitive for the standard volume-formΩ.

The association

I_f: (C, 0)→C,I_f(t) := Z

δf(t)λ|X(t) (2.13)

2.5. Vey’s Isochoric Morse Lemma defines a holomorphic multivalued function with monodromy (−1)ⁿ. For a givenf,I_f is only defined up to a global sign, which corresponds to the inital choice of a vanishing cycle.

Proposition 2.15. Let Q(x)=x^tAx be any nondegenerate quadratic form (in n complex variables). Then

Note that the there are two choices ofp

det(A) and the monodromy ofp cycle ofy→y^ty and recall e.g. from [Lam81], p. 37 a concrete coordinate description using the real sphereSⁿ⁻¹. We get

Z

Proposition 2.16. Let f: (Cⁿ, 0)→(C, 0)be a critical Morse singularity with quadratic part Q. Assume that we know f◦Φ=Ψ◦Q for someΦ∈Aut(Cⁿ, 0).

Letλbe a holomorphic primitive of the standard volume formΩonCⁿ. Then Z

δf(t)λ= π^n/2 Γ(ⁿ₂+1)

(Ψ⁻¹(t))^n/2

pdet(Q) (2.14)

as multivalued germs in(C^∗, 0).

Proof. It is essential to note thatΦ∗mapsδQ(t) toδf(Ψ(t)). This follows since Hn−1(MilQ(t);Z)∼==Zand the fact thatδQ(t) is a generator of this group and so must be mapped byΦ∗to a generator ofHn−1(Mil_f(Ψ(t));Z)∼= Z^µ=Z. From this we obtain

Z

δf(Ψ(t))λ= Z

(Φ)∗δQ(t)λ

= Z

δQ(t)

(Φ)^∗λ

SinceΦpreservesdλand exact forms are irrelevant for integration we get together with proposition 2.15

Z

δf(Ψ(t))λ= π^n/2 Γ(ⁿ₂+1)

t^n/2 pdet(Q) as asserted.

The last proposition gives us a geometric interpretation of Vey’sΨ-function.

It measures the volume in each fibre.

Corollary 2.17. Ψf is unique up to the multiplication with an nth root of unity.

Im Dokument The Geometry of the Milnor Number (Seite 41-49)