Mixed hypersurface singularities Mutsuo Oka Dept. of Math. Tokyo - PowerPoint PPT Presentation

Mixed hypersurface singularities Mutsuo Oka Dept. of Math. Tokyo University of Science FVJ2018 Nha Trang Septenber18-19 M. Oka Mixed hypersurface singularities

Lecture I. Mixed function This lecture is partially based on [19]. We consider the situation of real algebraic variety of codimension 2: V = { ( x , y ) ∈ R 2 n | g ( x , y ) = h ( x , y ) = 0 } where z j = x j + iy j and g , h ∈ R [ x , y ]. Under the canonical identification C n ∼ = R 2 n , z corresponds to x + i y . We study when the link is fibered over the circle. First in the complex Euclidean space, V can be described as V = { z ∈ C n | f ( z , ¯ z ) = 0 } , z ) := g ( z + ¯ z , z − ¯ z ) + ih ( z + ¯ z , z − ¯ z f ( z , ¯ ) 2 2 i 2 2 i We call f a mixed polynomial . f is a complex-valued real analytic function. M. Oka Mixed hypersurface singularities

Weighted homogeneous polynomials We first study elementary basic cases. A mixed polynomial z ) = ∑ z µ is called polar weighted homogeneous if ν,µ c ν,µ z ν ¯ f ( z , ¯ there exist integers p 1 , . . . , p n and a non-zero integer m p such that gcd( p 1 , . . . , p n ) = 1 , ∑ n j =1 p j ( ν j − µ j ) = m p , if c ν,µ ̸ = 0 . f ( z , ¯ z ) is called radially weighted homogeneous if there exist positive integers q 1 , . . . , q n and a non-zero integer m r such that n ∑ gcd( q 1 , . . . , q n ) = 1 , q j ( ν j + µ j ) = m r , if c ν,µ ̸ = 0 . j =1 We say f ( z , ¯ z ) is a mixed weighted homogeneous if f is radially weighted homogeneous of type, say ( q 1 , . . . , q n ; m r ), and f is also polar weighted homogeneous of type, say ( p 1 , . . . , p n ; m p )([10]). We define vectors of rational numbers ( u 1 , . . . , u n ) and ( v 1 , . . . , v n ) by u i = q i / m r , v i = p i / m p and we call them the normalized radial (respectively polar) weights . M. Oka Mixed hypersurface singularities Example 1

Using a polar coordinate ( r , η ) of C ∗ where r > 0 and η ∈ S 1 with S 1 = { η ∈ C | | η | = 1 } , we define a mixed R + × S 1 -action on C n by re i η = ( r , η ) ∈ R + × S 1 . ( r , η ) ◦ z := ( r q 1 η p 1 z 1 , . . . , r q n η p n z n ) , Assume that f ( z , ¯ z ) is mixed weighted homogeneous polynomial. Then f satisfies the functional equality z )) = r m r η m p f ( z , ¯ f (( r , η ) ◦ ( z , ¯ z ) . (1) This notion was introduced by Ruas-Seade-Verjovsky [23] implicitly and then by Cisneros-Molina [8]. It is easy to see that a mixed weighted homogeneous polynomial defines a global fibration f : C n − f − 1 (0) → C ∗ . M. Oka Mixed hypersurface singularities

For example, put U θ = { ρ e i η | θ − π < η < θ + π, ρ ̸ = 0 } ≡ R + × ( θ − π, θ + π ). The triviality over U θ is given as follows. Ψ( ρ e i ( θ + ξ ) , z ) = ( ρ 1 / m r , exp( i ξ/ m p )) ◦ z Ψ : U θ × f − 1 ( e i θ ) → f − 1 ( U θ ) , M. Oka Mixed hypersurface singularities

Theorem 2 \ K r → S 1 is a locally trivial fibration for any φ = f / | f | : S 2 n − 1 r r > 0 and it is equivalent to the restriction of the global fibration f : f − 1 ( S 1 ) → S 1 . Proof First, the triviality of φ is given similarly as follows. For example, over U = ( − π, π ), ψ : φ − 1 (1) × U → φ − 1 ( U ) , ψ ( z , θ ) = (1 , exp( i θ/ m p )) ◦ z . Define a fiber preserving diffeomorphism Φ( z ) = (1 / | f ( z ) | 1 / m r , 0) ◦ z Φ : S 2 n − 1 \ K r → f − 1 ( S 1 ) r M. Oka Mixed hypersurface singularities

This gives the commutative diagram: Φ S 2 n − 1 f − 1 ( S 1 ) \ K r − →   r   � φ � f S 1 S 1 = M. Oka Mixed hypersurface singularities

Mixed singular point for a generic mixed function Let f ( z , ¯ z ) be a mixed polynomial and we consider a hypersurface V = { z ∈ C n ; f ( z , ¯ z ) = 0 } . Put z j = x j + iy j . Then f ( z , ¯ z ) is a real analytic function of 2 n variables ( x , y ) with x = ( x 1 , . . . , x n ) and y = ( y 1 , . . . , y n ). Put f ( z , ¯ z ) = g ( x , y ) + i h ( x , y ) where g , h are real analytic functions. Recall that ( ) ( ) ∂ z j = 1 ∂ ∂ x j − i ∂ ∂ ∂ z j = 1 ∂ x j + i ∂ ∂ , 2 ∂ y j ∂ ¯ 2 ∂ y j We use the notations: ∂ z j = ∂ g z j = ∂ g ∂ f ∂ z j + i ∂ h ∂ f z j + i ∂ h ∂ z j , ∂ ¯ ∂ ¯ ∂ ¯ z j M. Oka Mixed hypersurface singularities

Put d R g ( x , y ) = ( ∂ g ∂ x 1 , . . . , ∂ g ∂ x n , ∂ g ∂ y 1 , . . . , ∂ g ∂ y n ) ∈ R 2 n d R h ( x , y ) = ( ∂ h ∂ x 1 , . . . , ∂ h ∂ x n , ∂ h ∂ y 1 , . . . , ∂ h ∂ y n ) ∈ R 2 n For a complex valued mixed polynomial, we use the notation: ¯ z ) = ( ∂ f ∂ z 1 , . . . , ∂ f z ) = ( ∂ f z 1 , . . . , ∂ f ∂ z n ) ∈ C n , z n ) ∈ C n ∂ f ( z , ¯ ∂ f ( z , ¯ ∂ ¯ ∂ ¯ We say that a point z ∈ V is a mixed-singular point of V if and only if df z : T z C n → T f ( z ) C is not surjective or equivalently the two vectors dg ( x , y ) , dh ( x , y ) are linearly dependent over R . Assume that p ∈ V is a mixed regular point. Then V is a real codimension two subvariety at p . M. Oka Mixed hypersurface singularities

Proposition 0.1 The following two conditions are equivalent. 1. z ∈ V is a mixed singular point. 2. dg , dh are linearly dependent over R . 3. There exists a complex number α, | α | = 1 such that z ) = α ¯ ∂ f ( z , ¯ ∂ f ( z , ¯ z ). M. Oka Mixed hypersurface singularities

First assume that d R g , d R h are linearly dependent at z . Suppose for example that dg ( x , y ) ̸ = 0 and write dh ( x , y ) = t dg ( x , y ) for some t ∈ R . This implies that ∂ x j = (1 + ti ) ∂ g ∂ f ∂ y j = (1 + ti ) ∂ g ∂ f ∂ x j , ∂ y j , thus ( ) ( ) ∂ f ∂ x j − i ∂ g ∂ g ∂ f ∂ x j + i ∂ g ∂ g ∂ z j = (1 + ti ) , z j = (1 + ti ) . ∂ y j ∂ ¯ ∂ y j Thus ( ) ∂ x 1 − i ∂ g ∂ g ∂ y 1 , . . . , ∂ g ∂ x n − i ∂ g ∂ f ( z , ¯ z ) = (1 + ti ) = 2(1 + ti ) ∂ g ( z , ¯ z ) ∂ y n ( ) ¯ ∂ x 1 + i ∂ g ∂ g ∂ y 1 , . . . , ∂ g ∂ x n + i ∂ g = 2(1 + ti )¯ ∂ f ( z , ¯ z ) = (1 + ti ) ∂ g ( z , ¯ z ) ∂ y n M. Oka Mixed hypersurface singularities

Here ∂ g = ( ∂ g ∂ z 1 , . . . , ∂ g ∂ z n ) and ¯ ∂ g = ( ∂ g z 1 , . . . , ∂ g z n ). As g is a real ∂ ¯ ∂ ¯ valued polynomial, using the equality ∂ g ( x , y ) = ¯ ∂ g ( x , y ) we get z ) = 1 − ti ¯ ∂ f ( z , ¯ ∂ f ( z , ¯ z ) . 1 + ti Thus it is enough to take α = 1 − ti 1+ ti . z ) = α ¯ Conversely assume that ∂ f ( z , ¯ ∂ f ( z , ¯ z ) for some α = a + bi with a 2 + b 2 = 1. Using the notations d x g = ( ∂ g , . . . , ∂ g d y g = ( ∂ g , . . . , ∂ g ) , ) , etc , ∂ x 1 ∂ x n ∂ y 1 ∂ y n we get (1 − a ) d x g + b d y g = − b d x h − (1 + a ) d y h − b d x g + (1 − a ) d y g = ( a + 1) d x h − b d y h . M. Oka Mixed hypersurface singularities

Solving these equations assuming a ̸ = 1, we get − 2 b d R g = ( d x g , d y g ) = (1 − a ) 2 + b 2 d R h which proves the assertion. If a = 1, the above equations implies that dh = 0 and the linear dependence is obvious. M. Oka Mixed hypersurface singularities

Transversality We assume again f ( z , ¯ z ) is a mixed weighted homogeneous polynomial as before. First we observe that Proposition 0.2 f − 1 ( t ) is mixed non-singular for any t ̸ = 0 and z ∈ f − 1 ( t ) as df z : T z f − 1 ( t ) → T t C is surjective. Let V = f − 1 ( 0 ). Assume that the radial weight q j > 0 for any j . Then V is contractible to the origin 0 . If further 0 is an isolated mixed singularity of V , V \{ 0 } is smooth. We omit the proof of the first assertion. A canonical deformation retract β t : V → V is given as β t ( z ) = t ◦ z , 0 ≤ t ≤ 1. (More precisely, the action is defined for t > 0 but it is easy to see β 0 ( z ) = lim t → 0 β t ( z ) = 0 .) Then β 1 = id V and β 0 is the contraction to 0 . Assume that z ∈ V \{ 0 } is a singular point. Consider the decomposition into real analytic functions f ( z ) = g ( x , y ) + ih ( x , y ). M. Oka Mixed hypersurface singularities

Using the radial R + -action, we see that g ( r ◦ ( x , y )) = r m r g ( x , y ) , h ( r ◦ ( x , y )) = r m r h ( x , y ) . (2) This implies that g ( x , y ) , h ( x , y ) are weighted homogeneous polynomials of ( x , y ) and the Euler equality can be restated as ( ) = ∑ n ∂ g ∂ g m r g ( x , y ) j =1 p j x j ∂ x j ( x , y ) + y j ∂ y j ( x , y ) ( ) = ∑ n x j ∂ h ∂ x j ( x , y ) + y j ∂ h m r h ( x , y ) j =1 p j ∂ y j ( x , y ) . Differentiating the equalities (2) in r , we get ∂ g ( r ◦ ( x , y )) = r m r − q j ∂ g ∂ h ( r ◦ ( x , y )) = r m r − q j ∂ h ( x , y ) , ( x , y ) . ∂ x j ∂ x j ∂ x j ∂ x j This implies that these differentials are also weighted homogeneous polynomials of degree m r − q j . M. Oka Mixed hypersurface singularities

Thus the jacobian matrix ( ∂ ( g , h ) ) ∂ ( x i , y i )( r ◦ ( x , y )) is the same with the jacobian matrix at z = ( x , y ) up to scalar multiplications in the column vectors by r m r − q 1 , . . . , r m r − q n , r m r − q 1 , . . . , r m r − q n respectively. Assume that ( x , y ) ∈ V \ { 0 } is a mixed singular point. Then any points of the orbit r ◦ ( x , y ) , r > 0 are singular points of V . This is a contradiction to the assumption that 0 is an isolated singular point of V , as lim r → 0 r ◦ ( x , y ) = 0 . M. Oka Mixed hypersurface singularities