aa r X i v : . [ m a t h . S G ] O c t Morse-Darboux lemma for surfaces with boundary
Ilia Kirillov ∗ Lomonosov Moscow State University
Abstract
We formulate and prove an analog of the classical Morse-Darboux lemma forthe case of a surface with boundary.
Throughout this paper the word smooth means C ∞ smooth. The aim of this paper isto prove the following theorem. Theorem 1.
Let M be a 2D surface with an area form ω , and let f : M → R bea smooth function. Let also O ∈ ∂M be a regular point for f and a non-degeneratecritical point for f | ∂M . Then there exists a chart ( p, q ) centered at O such that we have q ≥ wherever q is defined, the boundary ∂M satisfies the equation q = 0 , ω = dp ∧ dq, and f = α ◦ S , where S = q + p or S = q − p (See Figure 1). The function α of onevariable is smooth in the neighborhood of the origin ∈ R and α ′ (0) = 0 . Theorem 1 is closely related to the classical Morse-Darboux lemma. Let us recallthe statement of that lemma.
Theorem 2.
Let M be a 2D surface with an area form ω , and let f : M → R be asmooth function. Let also O ∈ M \ ∂M be a non-degenerate critical point for f. Thenthere exists a chart ( p, q ) centered at O such that ω = dp ∧ dq, and f = α ◦ S , where S = pq or S = p + q . The function α of one variable is smooth in the neighborhoodof the origin ∈ R and α ′ (0) = 0 . The Morse-Darboux lemma is a particular case of Le lemme de Morse isochore,see [1], and also a particular case of Eliasson’s theorem on the normal form for anintegrable Hamiltonian system near a non-degenerate critical point, see [2, 3]. TheMorse-Darboux lemma is an important tool in topological hydrodynamics, see [4], andtheory of integrable systems, see [5]. ∗ e-mail: [email protected] q (a) Case S = q + p . pq (b) Case S = q − p . Figure 1: Level sets of f . The horizontal axis is the boundary of M. We expect that the result of the present paper will also be useful in 2D fluid dynam-ics. In particular, it gives a partial answer to Problem 5 . ′ whichis equivalent to Theorem 1 . The proof of Theorem 1 ′ is given in Section 4. Section 3contains several lemmas useful for the proof of Theorem 1 ′ . Theorem 1 ′ . Let ω = ω ( x, y ) dx ∧ dy be an area form on R , and f = f ( x, y ) be asmooth function such that f x (0 ,
0) = 0 , f y (0 , > and f xx (0 , > . Then there existsa chart ( p, q ) centered at (0 , such that ω = dp ∧ dq, f ( p, q ) = α ( p + q ) , and q = 0 ifand only if y = 0 . The function α of one variable is smooth in the neighborhood of theorigin ∈ R and α ′ (0) > . Proposition 1.
Theorem follows from Theorem ′ . Proof.
Let us choose a chart ( x, y ) centered at O in ∂M such that P ∈ ∂M if and only if y ( P ) = 0 . The function f ( x, y ) and the form ω ( x, y ) dx ∧ dy can be smoothly extended onsome neighborhood of (0 , ,
0) is non-degenerate critical point for f | ∂M we have f x (0 ,
0) = 0, f y (0 , = 0, f xx (0 , = 0 . To fulfil conditions f y (0 , > f xx (0 , > , we may need some of the following transformations: f → − f, y → − y. Now, weobtain the chart ( p, q ) from Theorem 1 ′ . If q ≤ q → − q, p → − p. It remains to resctrict the chart ( p, q ) to the upper half plane.
In this section we assume that conditions of Theorem 1 ′ hold. Also from now on wewill assume that f (0 ,
0) = 0 . This will simplify notation.First of all, we want to prove an analog of the classical Morse Lemma for a surfacewith boundary.
Lemma 1.
There exists a chart (ˆ x, ˆ y ) centered at (0 , such that . ˆ x (0 ,
0) = ˆ y (0 ,
0) = 0 ;2. f (ˆ x, ˆ y ) = ˆ x + ˆ y ;3. ˆ y ( x, y ) = 0 if and only if y = 0 . Proof.
Hadamard’s lemma implies that f ( x, y ) = f ( x, y ) x + f ( x, y ) y, where f and f are smooth functions, and f (0 ,
0) = f x (0 , f (0 ,
0) = f y (0 , . Since f x (0 ,
0) = 0 Hadamard’s lemma similarly implies that f ( x, y ) = ( f ( x, y ) x + f ( x, y ) y ) x + f ( x, y ) y = f ( x, y ) x + f ( x, y ) xy + f ( x, y ) y = ( x p f ( x, y )) + y ( f ( x, y ) x + f ( x, y )) . Recall that f xx (0 , > f (0 ,
0) = f xx (0 , x ( x, y ) = p f ( x, y ) x ˆ y ( x, y ) = y ( f ( x, y ) x + f ( x, y )) . The Jacobian determinant of this transformation at the point (0 ,
0) is equal to p f (0 , f (0 , > . It follows from the inverse function theorem that functions ˆ x and ˆ y form a chart centered at (0 , . By construction f (ˆ x, ˆ y ) = ˆ x + ˆ y, and ˆ y ( x, y ) = y ( f ( x, y ) x + f ( x, y )) = 0 if and only if y = 0 . Remark . It follows from Lemma 1 that without loss of generality it can be assumedin Theorem 1 ′ that in the chart ( x, y ) we have f ( x, y ) = x + y. So from now on we willforget about the chart (ˆ x, ˆ y ) . Corollary 1.
Let D ( f, ε ) := { ( x, y ) ∈ R | f ( x, y ) ≤ ε and y ≥ } . Than the function A f ( ε ) := Z D ( f,ε ) ω ( x, y ) dx ∧ dy is well-defined if ε > is small enough (to use Lemma 1). Using the chart ( x, y ) thefunction A f ( ε ) can be expressed as A f ( ε ) = √ ε Z −√ ε dx ε − x Z ω ( x, y ) dy. emark . The function A f gives us an invariant of a pair ( f, ω ) . It will play a crucialrole in the proof of Theorem 1 ′ . Example 1.
Consider the upper half-plane H with an area form ω = dp ∧ dq and afunction f = α ( p + q ) , where α ′ (0) > . Then the function A f can be expressed as A f ( α ( ε )) = Z D ( α ( p + q ) ,α ( ε )) dp ∧ dq = Z D ( p + q,ε ) dp ∧ dq = √ ε Z −√ ε dp ε − p Z dq = √ ε Z −√ ε ( ε − p ) dx = 43 ε √ ε = 43 ε / , so [ A f ◦ α ]( ε ) = 4 / ε / and α ( ε ) = A − f (4 / ε / ) or α − ( ε ) = [3 / A f ( ε )] / . So we know how to determine the function α from Theorem ′ . Now we want to provethat α is a smooth function. Lemma 2.
The function ˜ A ( ε ) := A f ( ε ) / is smooth in some neighborhood of zero.Proof. Let u ( x, ε ) := ε − x Z ω ( x, y ) dy. Note that u is a smooth function of two variables. Further, A f ( ε ) = √ ε Z −√ ε u ( x, ε ) dx. Introducing a new variable δ = √ ε we obtain A f ( δ ) = δ Z − δ u ( x, δ ) dx. This function is smooth and odd. Let us find the third order Taylor polynomial of4 f ( δ ) : A f ( δ ) = δ Z − δ dx δ − x Z ω ( x, y ) dy = δ Z − δ dx δ − x Z [ ω (0 ,
0) + O ( x ) + O ( y )] dy = ω (0 , δ Z − δ dx δ − x Z dy + δ Z − δ dx δ − x Z O ( x ) dy + δ Z − δ dx δ − x Z O ( y ) dy = ω (0 ,
0) 43( δ ) / + O ( δ Z − δ dx δ − x Z xdy ) + O ( δ Z − δ dx δ − x Z ydy )= ω (0 ,
0) 43 δ + O ( δ ) + O ( δ ) = ω (0 ,
0) 43 δ + O ( δ ) . It means, that A f ( δ ) = δ B ( δ ), where the function B ( δ ) is smooth, even, and B (0) = 0. So, A f ( ε ) = ε / B ( √ ε ) and ˜ A ( ε ) = ε [ B ( √ ε )] / . Remark . The function ˜ A is defined only if ε >
0. But it extends to a smooth functionon a neighborhood of zero.Further in this section we will try to do things in the same way as in the proof ofthe classical Darboux Lemma (see [7], p. 230).
Definition 1.
Recall that one-forms on a surface M with a fixed area form ω maybe identified with vector fields, and every smooth function f : M → R determines aunique vector field X f , called the Hamiltonian vector field with the
Hamiltonian f, byrequiring that for every vector field Y on M the identity df ( Y ) = ω ( Y, X f ) holds. Letalso P f be the flow ( hamiltonian flow ) corresponding to the vector field X f . Definition 2.
Recall that in the chart ( x, y ) we have f ( x, y ) = x + y (see Remark 1).Let t f ( ε ) be the time necessary to go from ( −√ ε,
0) to the point ( √ ε,
0) under theaction of P f , i.e. t f ( ε ) is defined by P t f ( ε ) f ( −√ ε,
0) = ( √ ε, . Definition 3.
The curve γ ( ε ) := P t f ( ε ) f ( −√ ε, ε > bisector . Lemma 3.
The bisector is smooth and transversal to the boundary { y = 0 } . Proof.
Let us introduce a new coordinate system ( x, z ) , where z ( x, y ) := f ( x, y ) = x + y y (a) Chart ( x, y ) . xz (b) Chart ( x, z ) . Figure 2: Level sets of the function f in charts ( x, y ) and ( x, z ). The thick curve is theboundary of M. (see Figure 2). Then in these new coordinates f ( x, z ) = z , ω = ω ( x, z ) dx ∧ dz , y = 0 ifand only if z = x , and X f = ( − ω ( x,z ) , t f . Note that − ω ( x, z ) dx = dt (1)Integrating (1) over the horizontal segment between the points ( −√ z, z ) and ( √ z, z ),we get t f ( z ) = − √ z Z −√ z ω ( τ, z ) dτ. In the same way we obtain equations for the bisector ( s ( z ) , z ) s ( z ) Z −√ z ω ( τ, z ) dτ = 12 √ z Z −√ z ω ( τ, z ) dτ. (2)Introducing a new variable w = √ z we obtain an equation for the function ˆ s ( w ) := s ( w ): ˆ s ( w ) Z − w ω ( τ, w ) dτ = 12 w Z − w ω ( τ, w ) dτ, (3)Equation (3) allows us to define ˆ s ( w ) even if w < . We claim that ˆ s is a smoothfunction and ˆ s ( − w ) = ˆ s ( w ) . Partial derivative of (3) with respect to ˆ s is ω (ˆ s ( w ) , w ). For any ( x, z ) we have ω ( x, z ) = 0 . It follows from the implicit function theorem that ˆ s ( w ) depends smoothlyon w. w → − w. We obtain: ˆ s ( − w ) Z − ( − w ) ω ( τ, ( − w ) ) dτ = 12 − w Z − ( − w ) ω ( τ, ( − w ) ) dτ ⇐⇒ − w Z ˆ s ( − w ) ω ( τ, w ) dτ = − w Z − w ω ( τ, w ) dτ ⇐⇒ w Z ˆ s ( − w ) ω ( τ, w ) dτ = 12 w Z − w ω ( τ, w ) dτ ⇐⇒ ˆ s ( − w ) Z − w ω ( τ, w ) dτ = 12 w Z − w ω ( τ, w ) dτ ⇐⇒ ˆ s ( − w ) = ˆ s ( w ) . It means that equation (3) defines ˆ s as an even function of w . s ( z ) = s ( √ z ) = ˆ s ( √ z ) , so s is a smooth function of z. Now it is clear that the bisector is transversal to theboundary { z = x } . Remark . It follows from the proof of Lemma 3 that the bisector can be smoothlyextended to the lower half plane.
Definition 4.
Let T f ( x, y ) be be the time necessary to go from the bisector to thepoint ( x, y ) under the action of P f . Remark . In the chart ( x, z ), we have: T f ( x, z ) = x Z s ( z ) − ω ( τ, z ) dτ = s ( z ) Z x ω ( τ, z ) dτ, where the function s is defined in Lemma 3. Now it is clear that T f is a smooth function.Also note that since P f is the flow of the vector field X f , it follows that dT f ( X f ) = 1 . Lemma 4. ω = df ∧ dT f . Proof.
Using that dT f ( X f ) = 1 (see Remark 4), we get i X f df ∧ dT f = df ( X f ) dT f − df dT f ( X f ) = − dT f ( X f ) df = − df = i X f ω, so i X f ( df ∧ dT f − ω ) = 0 , and, since the ambient surface is 2-dimensional and X f = 0, it follows that ω = df ∧ dT f . emma 5. ddε A f ( ε ) = | t f ( ε ) | . Proof.
To proof this, let us use the chart ( x, z ) from Lemma 3 . Remind that in thischart f ( x, z ) = z. Now it follows from the definition of A f and from Lemma 4 that A f ( ε + δ ) − A f ( ε ) = | √ ε Z −√ ε ε + δ Z ε dz ∧ dT f | + o ( δ ) = | √ ε Z −√ ε dT f ε + δ Z ε dz | + o ( δ ) == δ | T f ( √ ε, − T f ( −√ ε, | + o ( δ ) = δ | t f ( ε ) | + o ( δ ) . So ddε A f ( ε ) = | t f ( ε ) | . Lemma 6.
Suppose that after a coordinate transformation ( x, y ) → ( p, q ) the followingconditions hold:1. f ( p, q ) = p + q. ω = dp ∧ dq.
3. The equation p = 0 describes the bisector.4. A f ( ε ) = ε √ ε. Then y ( p, q ) = 0 if and only if q = 0 .Proof. First of all, by the Condition 4 the function A ′ f ( ε ) can be computed as: ddε / ε √ ε = 2 √ ε. (4)Let us check that y = 0 if and only if q = 0 . It is follows from Lemma 3 that thecurve { y = 0 } is transversal to the bisector { p = 0 } . So, the curve y = 0 is a graphof some function q = r ( p ) (see Figure 3). It follows from the definition of bisector that r ( x ) = r ( − x ) . Let us proof that r ( x ) ≡
0. Assume that there exists some p such that q := r ( p ) > q < A ′ f ( q + p ) = [by equation (4)] = 2 q q + p > | p | = [by conditions (1),(2),(3) and the definition of t f ] = | t f ( q + p ) | = [by Lemma 5] = A ′ f ( q + p ) . This contradiction concludes the proof. 8 q pq r ( p )Figure 3: Illustration to the proof of Lemma 5. Proof.
Consider the function α ( ε ) := A − f ( 43 ε √ ε ) . It follows from Lemma 2 that α is a smooth function. Let also H ( x, y ) := [ α − ◦ f ]( x, y ) p ( x, y ) := − T H ( x, y ) q ( x, y ) := H − p ( x, y ) . Then dp ∧ dq = − dT H ∧ d ( H − T H ) == − dT H ∧ dH + dT H ∧ T H dT H = dH ∧ dT H = [by Lemma 4] = ω, so dp and dq are linearly independent. Further, in the chart ( p, q ), we have1. H ( p, q ) = p + q and f ( p, q ) = α ( p + q ) . ω = dp ∧ dq.
3. The equation p = 0 describes the bisector, because p ( x, y ) = 0 if and only if T ( x, y ) = 0 , while the latter means that the point ( x, y ) belongs to the bisector.4. A H ( ε ) = A f ( α ( ε )) = A f ( A − f ( ε √ ε )) = ε √ ε. So the chart ( p, q ) fulfils all conditions of Lemma 6. And now it follows fromLemma 6 that the chart ( p, q ) satisfies all conditions of Theorem 1 ′ . The author is grateful to A.M. Izosimov and A.A. Oshemkov for useful comments anddiscussions. This research is supported in part by the Russian Foundation for BasicResearch (grant No. 16-01-00378-a), the program Leading Scientific Schools (grant no.NSh-6399.2018.1) and the Simons Foundation.9 eferences [1] Y. C. de Verdi`ere and J. Vey, “Le lemme de Morse isochore,”
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