Complexity of Sparse Polynomial Solving: Homotopy on Toric Varieties and the Condition Metric

Abstract

This paper investigates the cost of solving systems of sparse polynomial equations by homotopy continuation. First, a space of systems of n-variate polynomial equations is specified through n monomial bases. The natural locus for the roots of those systems is known to be a certain toric variety. This variety is a compactification of $(\mathbb {C}\setminus \{0\})^n$, dependent on the monomial bases. A toric Newton operator is defined on that toric variety. Smale’s alpha theory is generalized to provide criteria of quadratic convergence. Two condition numbers are defined, and a higher derivative estimate is obtained in this setting. The Newton operator and related condition numbers turn out to be invariant through a group action related to the momentum map. A homotopy algorithm is given and is proved to terminate after a number of Newton steps which is linear on the condition length of the lifted homotopy path. This generalizes a result from Shub (Found Comput Math 9(2):171–178, 2009. https://doi.org/10.1007/s10208-007-9017-6).

Appendix A: Proof of Lemma 2.5.3

We start with a real version of Lemma 2.5.3. This will be used to recover the complex version. The notation $\langle \cdot . \cdot \rangle$ stands for the canonical Hermitian inner product in $\mathbb {C}^n$, and $\langle \cdot . \cdot \rangle _{\mathbb {R}^n}$ is the real canonical inner product. Identifying $\mathbb {C}^n$ to $\mathbb {R}^{2n}$ we can write

$$\begin{aligned} \mathfrak {R}\left( \langle \cdot . \cdot \rangle \right) = \langle \cdot . \cdot \rangle _{\mathbb {R}^{2n}} . \end{aligned}$$

Since the same norm arises from those two inner products, we use the notation $\Vert \cdot \Vert$ for it. Here is the real lemma:

Lemma A.1

Suppose that $\mathbf {x}, \mathbf {y}, \varvec{\zeta }\in \mathbb {R}^{n+1}$ with $\varvec{\zeta }-\mathbf {x} \perp \mathbf {x}$, $\mathbf {y}-\mathbf {x} \perp \mathbf {x}$ and $\Vert \mathbf {y}-\varvec{\zeta }\Vert \le \Vert \mathbf {x}-\varvec{\zeta }\Vert$. Then,

$$\begin{aligned} \frac{\Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \varvec{\zeta }\Vert } \le \frac{\Vert \mathbf {y}-\varvec{\zeta }\Vert }{\Vert \mathbf {x}\Vert }, \end{aligned}$$

where $\pi _{\mathbb {R}}(\mathbf {y}) = \frac{\Vert \varvec{\zeta }\Vert ^2}{\langle \mathbf {y},\varvec{\zeta }\rangle _{\mathbb {R}^{n+1}}} \mathbf {y}$ is the radial projection onto the real affine plane $\varvec{\zeta }+\varvec{\zeta }^{\perp }$.

Proof

Rescaling the three vectors $\mathbf {x}, \mathbf {y}$ and $\varvec{\zeta }$ simultaneously we can assume that $\Vert \mathbf {x}\Vert =1$. Then we can choose an orthonormal basis $(\mathbf e_0, \dots , \mathbf e_n)$ so that $\mathbf {x} = \mathbf e_0$, $\varvec{\zeta }$ is in the span of $\mathbf e_0$ and $\mathbf e_1$ and y is in the span of $\mathbf e_0, \mathbf e_1$ and $\mathbf e_2$. In coordinates,

$$\begin{aligned} \mathbf {x} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix} \text {, } \quad \varvec{\zeta }= \begin{pmatrix} 1 \\ t \\ 0 \\ 0 \\ \vdots \end{pmatrix} \quad \text { and } \quad \mathbf {y} = \begin{pmatrix} 1 \\ s \\ r \\ 0 \\ \vdots \end{pmatrix}. \end{aligned}$$

We can further assume that $t \ge 0$ and $r \ge 0$. Squaring both sides of the hypothesis $\Vert \mathbf {y}-\varvec{\zeta }\Vert \le \Vert \mathbf {x}-\varvec{\zeta }\Vert$ we obtain

$$\begin{aligned} r^2 + (s-t)^2 \le t^2 \end{aligned}$$

that is

$$\begin{aligned} r^2 \le 2 st - s^2 \end{aligned}$$

(14)

which implies $s \ge 0$.

We claim first that

$$\begin{aligned} \frac{ \Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \mathbf {y}-\varvec{\zeta }\Vert } \le \frac{ \Vert \pi _{\mathbb {R}}(\mathbf {x})-\varvec{\zeta }\Vert }{\Vert \mathbf {x}-\varvec{\zeta }\Vert } . \end{aligned}$$

(15)

We compute

$$\begin{aligned} \Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert ^2= & {} {{\left( t^2+1\right) \,\left( r^2\,t^2+t^2-2\,s\,t+s^2+r^2\right) }\over {\left( s\,t+1\right) ^2}} \\ \Vert \mathbf {y}-\varvec{\zeta }\Vert ^2= & {} t^2-2\,s\,t+s^2+r^2 \\ \Vert \pi _{\mathbb {R}}(\mathbf {x})-\varvec{\zeta }\Vert ^2= & {} t^2\,\left( t^2+1\right) \\ \Vert \mathbf {x}-\varvec{\zeta }\Vert ^2= & {} t^2. \end{aligned}$$

To show in Eq. (15), we just need to verify that

$$\begin{aligned} K = \Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert ^2 \Vert \mathbf {x}-\varvec{\zeta }\Vert ^2 - \Vert \pi _{\mathbb {R}}(\mathbf {x})-\varvec{\zeta }\Vert ^2 \Vert \mathbf {y}-\varvec{\zeta }\Vert ^2 \le 0. \end{aligned}$$

Using the Maxima computer algebra system [27],

$$\begin{aligned} K = - \frac{t^3 (t^2+1)}{(st+1)^2} \left( A r^2 + B\right) \end{aligned}$$

with

$$\begin{aligned} A=(s^2-1)t + 2s \quad \text { and } \quad B=s (t-s)^2 (st+2). \end{aligned}$$

From the factorization above, K is negative if and only if $A r^2 + B \ge 0$. Clearly, $B \ge 0$. If $A \ge 0$ we are done, so assume $A<0$. Then multiplying both sides of (14) by A, one obtains

$$\begin{aligned} A r^2 \ge 2 A st - A s^2 \end{aligned}$$

and

$$\begin{aligned} A r^2 +B \ge s^2 t(1+t^2) \ge 0 . \end{aligned}$$

This shows (15). Also,

$$\begin{aligned} \frac{ \Vert \pi _{\mathbb {R}}(\mathbf {x})-\varvec{\zeta }\Vert ^2}{\Vert \mathbf {x}-\varvec{\zeta }\Vert ^2} = 1+t^2 = \frac{\Vert \zeta \Vert ^2}{\Vert x\Vert ^2} . \end{aligned}$$

Taking square roots and combining with (15),

$$\begin{aligned} \frac{\Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \varvec{\zeta }\Vert } \le \frac{\Vert \mathbf {y}-\varvec{\zeta }\Vert }{\Vert \mathbf {x}\Vert } . \end{aligned}$$

$\square$

Lemma 2.5.3Suppose that$\mathbf {x}, \mathbf {y}, \varvec{\zeta }\in \mathbb {C}^{n+1}$with$\varvec{\zeta }-\mathbf {x} \perp \mathbf {x}$, $\mathbf {y}-\mathbf {x} \perp \mathbf {x}$and$\Vert \mathbf {y}-\varvec{\zeta }\Vert \le \Vert \mathbf {x}-\varvec{\zeta }\Vert$. Then,

$$\begin{aligned} \frac{\Vert \pi (\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \varvec{\zeta }\Vert } \le \frac{\Vert \mathbf {y}-\varvec{\zeta }\Vert }{\Vert \mathbf {x}\Vert }, \end{aligned}$$

where$\pi (\mathbf {y}) = { \frac{\Vert \varvec{\zeta }\Vert ^2}{\langle \mathbf {y},\varvec{\zeta }\rangle } \mathbf {y}}$is the radial projection onto the affine plane$\varvec{\zeta }+\varvec{\zeta }^{\perp }$.

Proof

We identify $\mathbb {C}^{n+1}$ with $\mathbb {R}^{2n+2}$ and claim that

$$\begin{aligned} \Vert \pi (\mathbf {y})-\varvec{\zeta }\Vert \le \Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert . \end{aligned}$$

(16)

Since complex orthogonal vectors are also real orthogonal, Eq. (16) and Lemma A.1 imply

$$\begin{aligned} \frac{\Vert \pi (\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \varvec{\zeta }\Vert } \le \frac{\Vert \pi _{\mathbb {R}}(\mathbf {y})-\varvec{\zeta }\Vert }{\Vert \varvec{\zeta }\Vert } \le \frac{\Vert \mathbf {y}-\varvec{\zeta }\Vert }{\Vert \mathbf {x}\Vert } . \end{aligned}$$

To show (16) we choose coordinates so that

$$\begin{aligned} \varvec{\zeta }= \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \end{pmatrix} \quad \text { and } \quad \mathbf {y} = \begin{pmatrix} a+b i \\ c \\ 0 \\ \vdots \end{pmatrix} \end{aligned}$$

with $c \ge 0$. A straightforward computation gives

$$\begin{aligned} \pi (y) = \begin{pmatrix} 1 \\ \frac{c}{a+bi} \\ 0 \\ \vdots \end{pmatrix} \quad \text { and } \quad \pi _{\mathbb {R}} (\mathbf {y}) = \begin{pmatrix} 1+\frac{b}{a} i \\ \frac{c}{a} \\ 0 \\ \vdots \end{pmatrix}. \end{aligned}$$

We have

$$\begin{aligned} \Vert \pi (\mathbf {y})-\varvec{\zeta }\Vert ^2 = \frac{c^2}{a^2+b^2} \le \frac{b^2}{a^2} + \frac{c^2}{a^2} = \Vert \pi _{\mathbb {R}} (\mathbf {y})-\varvec{\zeta }\Vert ^2 \end{aligned}$$

with equality if $b=0$. This finishes the proof. $\square$