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*}


\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.

  • 1
    $\begingroup$ 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 ... $\endgroup$ Nov 21, 2017 at 17:20
  • 1
    $\begingroup$ I mean the latter case, where $\delta$ does not depend on $n$ and $d$. I have updated the phrasing to reflect this. $\endgroup$ Nov 21, 2017 at 17:45
  • 1
    $\begingroup$ Probably not relevant, but it’s not true over the reals. $\endgroup$ Nov 22, 2017 at 2:11
  • 1
    $\begingroup$ 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? $\endgroup$ Nov 22, 2017 at 6:43
  • 1
    $\begingroup$ 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 $A_d = B_d$. $\endgroup$ Nov 22, 2017 at 16:43

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.