# Proximity of solutions to system of degree two polynomials

Everything in this post is over the complex numbers. I would like to know if for every $\epsilon > 0$ there exists $\delta > 0$ such that the following holds for every $n$ and every $d$ which is sufficiently large depending on $n$.

Let $p_1,\ldots,p_n$ be a set of homogeneous degree two polynomials in the variables $x_1,\ldots,x_d$. Assume that all coefficients in the $p_k$ are $1$.

Consider the systems

\begin{equation*} \mathcal{S}_1 = \begin{cases} p_1(x_1,\ldots,x_d) = a_1 \\ p_2(x_1,\ldots,x_d) = a_2 \\ \hspace{0.5in} \vdots \\ p_n(x_1,\ldots,x_d) = a_n \end{cases} \end{equation*}

and

\begin{equation*} \mathcal{S}_2 = \begin{cases} p_1(x_1,\ldots,x_d) = b_1 \\ p_2(x_1,\ldots,x_d) = b_2 \\ \hspace{0.5in} \vdots \\ p_n(x_1,\ldots,x_d) = b_n \end{cases} \end{equation*}

where $a_k$ and $b_k$ have magnitude bounded by one. Suppose there exist solutions $\overline{\alpha} = (\alpha_1,\ldots,\alpha_d)$ to $\mathcal{S}_1$ and $\overline{\beta} = (\beta_1,\ldots,\beta_d)$ to $\mathcal{S}_2$, both in the unit ball, such that $||\overline{\alpha} - \overline{\beta}||_2 \leq \delta$. Then for every solution $\overline{\alpha}'$ to $\mathcal{S}_1$ in the unit ball there exists a solution $\overline{\beta}'$ to $\mathcal{S}_2$ such that $||\overline{\alpha}' - \overline{\beta}'||_2 \leq \epsilon$.

My intuition is that since the left sides of $\mathcal{S}_1$ and $\mathcal{S}_2$ are the same, the varieties they define should be in some sense parallel, as happens in the linear case.

• I'm slightly confused about the quantifiers for n and d. Are n and d fixed? Or is it like this: For every epsilon, there is a delta, so that for every n, the following holds: for every d sufficiently large ... – Dmitri Gekhtman Nov 21 '17 at 17:20
• I mean the latter case, where $\delta$ does not depend on $n$ and $d$. I have updated the phrasing to reflect this. – burtonpeterj Nov 21 '17 at 17:45
• Probably not relevant, but it’s not true over the reals. – Dmitri Gekhtman Nov 22 '17 at 2:11
• Consider for each $d$ the polynomial $p_d(x_1,\ldots, x_d) = \sum_{i=1}^d\sum_{j=1}^d x_ix_j$. This polynomial sends the vector with all entries $1/d$ to 1. Let $A_d$ be the intersection of the unit ball with $p_d^{-1}(0)$. Let $B_d$ be the intersection of the unit ball with $p_d^{-1}(0)$. Your conjecture would imply that $\max_{v\in B_d} \text{Distance}(v,A_d)$ goes to 0 as $d\rightarrow \infty$. Is this more specific thing true? – Dmitri Gekhtman Nov 22 '17 at 6:43
• It's interesting that it's not true over the reals. Why is that the case? I'm not sure what you mean by the second comment, with what you have written $A_d = B_d$. – burtonpeterj Nov 22 '17 at 16:43

I posted something slightly wrong previously. Let's try this again:

The thing you're asking about doesn't hold over $\mathbb{R}$. It does hold over $\mathbb{C}$ if we restrict to $n=1$ and look only at varieties defined by single polynomials. I'm not sure what happens in the general situation.

Counterexample over $\mathbb{R}$: For each positive integer $d$, let $p_d(x_1,\ldots, x_d)= \sum_{i\leq j}x_ix_j$. The matrix of the corresponding symmetric bilinear form has each diagonal entry 1 and each off-diagonal entry $\frac{1}{2}$. The matrix has two eigenvalues:

$\frac{d+1}{2}$ with eigenspace spanned by the vector $(1,\ldots, 1)$

$\frac{1}{2}$ with eigenspace equal to the orthocomplement of $(1,\ldots, 1)$.

So after an orthogonal change of coordinates, $p_d$ takes the form $\frac{d+1}{2} y_1^2+ \frac{1}{2}\sum_{i=2}^{d}y_i^2$. The zero set $p_d^{-1}(0)$ is always 0. The locus $p_d^{-1}(1)$ is a badly squashed ellipsoid containing vectors $\sqrt{\frac{2}{d+1}}\cdot \vec{e}_1$ and $\sqrt{2}\cdot\vec{e}_2$.

This doesn't give a counterexample over $\mathbb{C}$. Indeed, given a point of $p_d^{-1}(1)$, there's always a nearby point of $p_d^{-1}(0)$ obtained by slightly varying the $y_1$ component.

If we restrict to $n=1$, your conjecture does hold over $\mathbb{C}$, and the example above suggests a strategy for proving it. Suppose there is a sequence $p^i$ of degree 2 polynomials that yield a counterexample to the conjecture.

So there are $a,b, \overline{\alpha}_i,\overline{\beta_i}$ so that $$p^i(\overline{\alpha}_i) = a$$ $$p^i(\overline{\beta_i}) = b$$ with $$\|\overline{\alpha}_i - \overline{\beta}_i \|\rightarrow 0,$$ but there's $\overline{\alpha}'_i$ in the $a$ level set of each $p^i$ which is distance $>\epsilon$ from the $b$ level set.

If the eigenvalues of the matrices corresponding to the $p^i$ are all uniformly bounded, then the norm of the derivative $dp^i$ is uniformly bounded on the unit ball, so $\|\overline{\alpha}_i - \overline{\beta}_i \|\rightarrow 0$ implies $p^i(\overline{\alpha}_i) - p^i(\overline{\beta_i})\rightarrow 0$, which is a contradiction.

If the eigenvalues of the matrices corresponding to the $p^i$ are not uniformly bounded, then you can get from any point of the $a$ level set of $p^i$ to the $b$ level set by slightly varying in the eigendirection of a big eigenvalue. Again, there's a contradiction.