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Friday, July 20, 2012

Excellent polynomial mappings

This  answers  a question of Lior Bary-Soroker on Mathoverflow.  \newcommand{\bR}{\mathbb{R}} \newcommand{\pa}{\partial}

Suppose that f:\bR^n\to \bR is a polynomial. For any \vec{a}\in\bR^n we define

f_{\vec{a}}:\bR^n\to \bR,\;\; f_{\vec{a}}(\vec{x})=f(\vec{x})-\vec{a}\cdot\vec{x}.

We want to prove that  for generic \vec{a} the function f_{\vec{a}} is an excellent Morse function, i.e., all critical points are nondegenerate, and no two of them  are  on the same level set of f_{\vec{a}}.  We set

C(\vec{a})=\lbrace \vec{x}\in\bR^n;\;\;df_{\vec{a}}(\vec{x})=0\;\rbrace=\lbrace\;\vec{x}\in\bR^n;\;\;df(\vec{x})=\vec{a}\;\rbrace.

We say that a set S\subset\bR^n is generic (in semialgebraic sense) if its complement is a semialgebraic set of dimension <n.  Sard's theorem implies that for generic \vec{a} the critical set C(\vec{a}) is discrete. Being semialgebraic this implies that it is also finite.

Define \newcommand{\eZ}{\mathscr{Z}}

\eZ:=\bigl\lbrace\; (\vec{x},\vec{a})\in\bR^n\times\bR^n;\;\;df(\vec{x})=\vec{a}\;\bigr\rbrace.

The set \eZ is semialgebraic. We denote by \pi:\eZ\to\bR^n the projection

\eZ\ni (\vec{x},\vec{a})\to \vec{a}\in \bR^n.

For any S\subset \bR^n we set \eZ(S):=\pi^{-1}(S). There exists a generic set G\subset \bR^n such that for any connected component  A of G the  induced  map \pi:\eZ(A)\to A is a twice differentiable   covering.   Thus there exists a positive integer  m=m(A) and  twice differentiable maps

\vec{u}_1,\dotsc,\vec{u}_m:A\to \bR^n

such that

\vec{u}_i(\vec{a})\neq\vec{u}_j(\vec{a}), \;\;\forall i\neq j,  \;\;\vec{a}\in A, \tag{1}

  and the  set \eZ(A) is the union of the graphs  of the maps \vec{u}_i. In other words, for any \vec{a}\in A we have

C(\vec{a})=\lbrace\; \vec{u}_1(\vec{a}),\dotsc,\vec{u}_m(\vec{a})\;\rbrace.

We claim that f_{\vec{a}} is excellent for generic  \vec{a}\in A. We argue by contradiction. Suppose that this is not the case. Then there exists a nonempty, connected  open subset A_*\subset A and indices i\neq j such that

f( \vec{u}_i(\vec{a}) -f(\vec{u}_j(\vec{a}))=\vec{a}\cdot\bigl(\vec{u}_i(\vec{a})-\vec{u}_j(\vec{a})\;\bigr),\;\;\forall \vec{a}\in A_*. \tag{2}

We denote by \pa_k f the k-th partial derivative of f and we write

\vec{a}= (a^1,\dotsc, a^n),\;\;\vec{u}_i=(u_i^1,\dotsc, u_i^n).

Note that

\pa_kf(\vec{u}_i(\vec{a}))=a^k=\pa_kf(\vec{u}_j(\vec{a})),\;\;\forall k=1,\dotsc, n.\tag{3}

 Derivating (2) with respect to the variable a^\ell  using the equality  (3) we deduce

\sum_k a^k\pa_{a^\ell}\bigl(u_i^k-u_j^k\bigr)= u^\ell_i(\vec{a})-u^\ell_j(\vec{a}) +\sum_k a^k\pa_{a^\ell}\bigl(u_i^k-u_j^k),\;\;\vec{a}\in A_*.


We deduce  that

u^\ell_i(\vec{a})=u^\ell_j(\vec{a}),\;\;\forall\ell=1,\dotsc, n,\;\forall\vec{a}\in A_*.

This contradicts (1).



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