Problem ContributionArnold Mathematical Journal

Received: 5 July 2015 / Accepted: 28 July 2015

Disconjugacy and the Secant Conjecture

Alexandre Eremenko Supported by NSF grant DMS-1361836.Department of Mathematics Purdue University West Lafayette IN 47907 USA

We discuss the so-called secant conjecture in real algebraic geometry, and show that it follows from another interesting conjecture, about disconjugacy of vector spaces of real polynomials in one variable.

Disconjugacy, Wronskian, Schubert calculus

Let $V$ be a real vector space of dimension $n$ whose elements are real functions on an interval $[a,b]$. The space $V$ is called disconjugate if one of the following equivalent conditions is satisfied:

  1. (a)

    Every $f\in V\backslash\{0\}$ has at most $n-1$ zeros, or

  2. (b)

    For every $n$ distinct points $x_{1},\ldots,x_{n}$ on $[a,b]$ and every basis $f_{1},\ldots,f_{n}$ of $V$ we have $\det(f_{i}(x_{j}))\neq 0$.
     One can replace “every basis” by “some basis” in (b) and obtain an equivalent condition.
     If $V$ is disconjugate then the determinant in (b) has constant sign which depends only on the ordering of $x_{j}$ and on the choice of the basis.
     A space of real functions on an open interval is called disconjugate if it is disconjugate on every closed subinterval.
     We are only interested here in spaces $V$ consisting of polynomials.
     Suppose that a positive integer $d$ is given, and $V$ consists of polynomials of degree a most $d$. Then every basis $f_{1},\ldots,f_{n}$ of $V$ defines a real rational curve $\mathbf{RP}^{1}\to\mathbf{RP}^{n-1}$ of degree $d$. Indeed, we can replace every $f_{j}(x)$ by a homogeneous polynomial $f^{*}_{j}(x_{0},x_{1})$ of two variables of degree $d$, such that $f_{j}(x)=f^{*}_{j}(1,x)$, and then $f^{*}_{1},\ldots,f^{*}_{n}$ define a map $f:\mathbf{RP}^{1}\to\mathbf{RP}^{n-1}$ (if polynomials have a common root, divide it out).
     Then the geometric interpretation of disconjugacy is:

  1. (c)

    The curve $f$ constructed from a basis in $V$ is convex, that is intersects every hyperplane at most $n-1$ times.

For every basis $f_{1},\ldots,f_{n}$ in $V$ we can consider its Wronski determinant $W=W(f_{1},\ldots,f_{n})$. Changing the basis results in multiplication of $W$ by a non-zero constant, so the roots of $W$ only depend on $V$.

Conjecture 1.

Suppose that all roots of $W$ are real. Then $V$ is disconjugate on every interval that does not contain these roots.

This is known for $n=2$ with arbitrary $d$ (see below), and for $n=3,d\leq 5$ by direct verification with a computer.

This conjecture arises in real enumerative geometry (Schubert calculus), and we explain the connection. The problem of enumerative geometry we are interested in is the following:

Let $m\geq 2$ and $p\geq 2$ be given integers. Suppose that $mp$ linear subspaces of dimension $p$ in general position in $\mathbf{C}^{m+p}$ are given. How many linear subspaces of dimension $m$ intersect all of them non-trivially?

The answer was obtained by Schubert in 1886 and it is

$\displaystyle d(m,p)=\frac{1!2!\ldots(p-1)!(mp)!}{m!(m+1)!\ldots(m+p-1)!}.$

Now suppose that all those given subspaces are real. Does it follow that all $p$-subspaces that intersect all of them non-trivially are real? The answer is negative, and we are interested in finding a geometric condition on the given $p$-subspaces that ensure that all $d(m,p)$ subspaces of dimension $m$ that intersect all the given $p$-subspaces non-trivially are real.

One such condition was proposed by B. and M. Shapiro. Let $F(x)=(1,x,\ldots,x^{d}),{}d=m+p-1$ be a rational normal curve, a. k. a. moment curve. Suppose that the given $p$-spaces are osculating $F$ at some real points $F(x_{j})$. This means that subspaces $X_{j}$ are spanned by the (row)-vectors $F(x_{j}),F^{\prime}(x_{j}),\ldots,F^{(p-1)}(x_{j})$ for some real $x_{j},1\leq j\leq mp$. Then all $m$-subspaces that intersect all $X_{j}$ non-trivially are real.

This was conjectured by B. and M. Shapiro and proved by Mukhin, Tarasov and Varchenko (MTV) [Mukhin et al.2009]. Earlier it was known for $n=2$ [Eremenko and Gabrielov2002], and in [Eremenko and Gabrielov2011] a simplified elementary proof for the case $n=2$ was given.

We are interested in the following generalization of this result.

Secant Conjecture. Suppose that each of the $mp$ subspaces $X_{j},1\leq j\leq mp$ is spanned by $p$ distinct real vectors $F(x_{j,k}),0\leq k\leq p-1$, and that the sets of points $\{x_{j,k}:0\leq k\leq p-1\}$ are separated, that is $x_{j,k}\in I_{j}$, where $I_{j}\subset\mathbf{RP}^{1}$ are disjoint intervals. Then all $m$-subspaces which intersect all $X_{j}$ non-trivially are real.

This is known when $p=2$, [Eremenko et al.2006] and has been tested on a computer for $p=3$ and small $m$, [Hillar et al.2010], [Garcia-Puente et al.2012]. The special case when the groups $\{x_{j,k}\}_{k=0}^{p-1}$ form arithmetic progressions, $x_{j,k}=x_{j,0}+kh$ has been established [Mukhin et al.2009].

Next we show how the Secant Conjecture follows from Conjecture 1 and the results of MTV.

Let us represent an $m$-subspace $Y$ that intersects all subspaces $X_{j}$ as the zero set of $p$ linear forms, and use the coefficients of these forms as coefficients of $p$ polynomials $f_{0},\ldots,f_{p-1}$. Then the condition that $Y$ intersects some $X_{j}$ is equivalent to linear dependence of the $p$ vectors $f_{i}(x_{j,m})_{m=0}^{p-1},\;i=0,\ldots,p-1$. That is


These equations for $j=1,\ldots,mp$ define the subspaces $Y$, and we have to prove that all solutions are real.

Let $I_{j}$ be the intervals with disjoint closures which contain the $x_{j,k}$. We may assume without loss of generality that $\infty\notin I_{j}$. We place on each $I_{j}$ a point $y_{j}$, and consider the $d(m,p)$ real rational curves $\mathbf R\to\mathbf{RP}^{p}$ with inflection points at $y_{j}$. These curves exist by the MTV theorem, and they depend continuously on the $y_{j}$.

Let $f=(f_{0}\ldots,f_{p-1})$ be one of these curves. Fix $k\in\{1,\ldots,mp\}$. For all $j\neq k$, fix all $y_{j}\in I_{j}$. When $y_{k}$ moves in $I_{k}$ from the left end to the right end, the determinant $\det(f_{i}(x_{k,m}))_{i,m=0}^{p-1}$ must change sign, in view of Conjecture 1. So this determinant is 0 for some position of $y_{k}$ on $I_{k}$.

Then it follows by a well-known topological argument that one can choose all $y_{j}\in I_{j}$ in such a way that $\det(f_{i}(x_{j,m}))=0$ for all $j$.

Thus we have constructed $d(m,p)$ real solutions of the secant problem. As the total number of solutions is also $d(m,p)$, for generic data, we obtain the result.

Proof of Conjecture 1 for $n=2$. We have two real polynomials $f_{1}$ and $f_{2}$, such that $f_{1}^{\prime}f_{2}-f_{1}f_{2}^{\prime}$ has only real zeros. This means that the rational function $F=f_{1}/f_{2}:\mathbf{\overline{C}}\to\mathbf{\overline{C}}$ is real and all its critical points are real. Let $I\subset\mathbf R$ be a closed interval without critical points. Then $F$ is a local homeomorphism on $I$, so $F(I+i\epsilon)$ belongs to one of the half-planes $\mathbf{C}\backslash\mathbf R$, for all sufficiently small $\epsilon>0$. Suppose without loss of generality that it belongs to the upper half-plane $H$. Let $D$ be the component of $F^{-1}(H)$ that contains $I+i\epsilon$. Then $D$ is a region in $H$ with piecewise analytic boundary, and $I\subset\partial D$. The map $F:D\to H$ is a covering because it is proper and has no critical points. As $H$ is simply connected, $D$ must be simply connected and $F:D\to H$ must be a conformal homeomorphism. Then $F^{-1}:H\to D$ is a conformal homeomorphism. As $\partial D$ is locally connected, this homeomorphism extends to $F^{-1}:\overline{H}\to\overline{D}$. This last map must be injective because this is a left inverse of a function. Thus $F^{-1}:\overline{H}\to\overline{D}$ is a homeomorphism. Then $F:\overline{D}\to\overline{H}$ must be also a homeomorphism, in particular $F$ is injective on $I$.

This implies that the linear span of $f_{1},f_{2}$ is disconjugate.


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