I have the following question which is in some way related to an application of Randell Isotopy Theorem to complex hyperplane arrangements.

Let $h,k\geq1$ be integer numbers and let $F_{1},\ldots,F_{h}$ be elements of the polynomial ring $\mathbb{C}\left[t_{1},\ldots,t_{k}\right].$ Now, consider the space $$X=\lbrace P\in\mathbb{C}^{k}\mid F_{j}(P)=0\text{ for }1\leq j\leq h\rbrace$$ where $F_{j}(P)$ simply denotes the evaluation of the polynomials $F_{j}$ at the point $P.$

Let us assume $X$ is endowed with the $\textit{classical}$ topology. In general $X$ is not a topological manifold. However, I have the following question regarding the path connectivity of the connected components of $X.$ In fact, $X$ a locally path connected space, so that its connected components coincide with its path connected components. To be more precise, my question regards the smoothness of paths connecting points in $X.$

$\textbf{Definition:}$ A path $\gamma:\left(a,b\right)\longrightarrow\mathbb{C}^{k}$ is called $\textit{smooth}$ if every derivative of any order are continuous.

$\textbf{Remark:}$ Pay attention that this definition of smoothness is different from the one of $\textit{regularity}$ of a parametrized curve. I am NOT asking the tangent vector $\gamma'(t)$ is not vanishing.

$\textbf{QUESTION:}$ Let $P,Q\in X$ be two points of $X$ which are in the same connected component. Then, there exists $\epsilon>0$ and a smooth path $\gamma:\left(-\epsilon,1+\epsilon\right)\longrightarrow\mathbb{C}^{k}$ with $\gamma(t)\in X$ for any $t\in\left(-\epsilon,1+\epsilon\right)$ and such that $\gamma(0)=P$ and $\gamma(1)=Q.$


2 Answers 2


This is true. It is easy to find a piecewise smooth path by following the smooth strata. Then reparameterize the path so that it becomes smooth by stopping of infinite order at each point where it it is not smooth (use pieces $t\mapsto e^{-1/t^2}$).

  • $\begingroup$ Thank you very much for this suggestion! I was pretty sure that a smart use of the exponential map solve my problem. However, your answer confirm my first impressions. With this result in mind, I am now able to build a smooth lattice isotopy in my moduli space of arrangements. Thank you again! $\endgroup$
    – snaleimath
    May 12, 2015 at 11:48

Edit. Peter Michor showed that the answer to the question is yes. By contrast, the following example shows that it is in general no when one replaces smooth path with regular path (I at first assumed that smooth and regular were used here as synonims, as sometimes happens in the literature, see the discussion in the comments below).

Take the connected affine space $X:=\{z_1z_2=0 \} \subset \mathbb{C}^2$. Under the identification $\mathbb{C}^2 = \mathbb{R}^4$ given by $$(x_1+x_2i, \, x_3 + x_4 i) \mapsto (x_1, \, x_2, \, x_3, \, x_4),$$ the underlying real space of $X$ is the connected union of two planes $$\Pi_1 \cup \Pi_2,$$ where $\Pi_1=\{x_1=x_2=0\}, \quad \Pi_2 =\{x_3 = x_4 =0\}$.

Let us take two points $p_1 \in \Pi_1$ and $p_2 \in \Pi_2$. Then any continuous path $\gamma(t)$ such that $\gamma(0)=p_1$ and $\gamma(1)= p_2$ must contain the origin $\mathbf{0}=(0, \, 0, \, 0, \, 0)=\Pi_1 \cap \Pi_2$, so there exists $t_0 \in (0, \, 1)$ such that $\gamma(t_0)=\mathbf{0}$.

Now, the path $\gamma(t)$ is clearly not regular at $t_0$. In fact, we have $$\lim_{t \to t_0^-} \gamma'(t) \in \Pi_1, \quad \lim_{t \to t_0^+} \gamma'(t) \in \Pi_2,$$

so if $\gamma(t)$ is smooth at $t_0$ it follows $\gamma'(t_0)=\mathbf{0}.$

  • $\begingroup$ Are you sure that there is no way to find a path such that $\operatorname{lim}_{t\rightarrow t_{0}^{-}}\gamma'(t)=\operatorname{lim}_{t\rightarrow t_{0}^{+}}\gamma'(t)=0$? $\endgroup$
    – snaleimath
    May 12, 2015 at 10:10
  • $\begingroup$ In this case $\gamma'(t_0)=0$ and so $\gamma(t)$ is not regular at $t_0$. $\endgroup$ May 12, 2015 at 10:12
  • $\begingroup$ But I am not looking at regular paths, just smooth paths. $\endgroup$
    – snaleimath
    May 12, 2015 at 10:14
  • $\begingroup$ Sorry, what is your definition of smooth? For me (and for the literature) is equivalent to $C^{\infty}$ parametrization, or at least $C^1$ parametrization. This means in particular that $\gamma'$ must be continuous and $\gamma'(t) \neq 0$ for all $t \in (0, \, 1)$. The tangent vector must be nonzero at every point! $\endgroup$ May 12, 2015 at 10:18
  • 2
    $\begingroup$ I am sorry. I use the word $\textit{regular}$ for your $\textit{smooth}.$ To me, and as far I know before, $\textit{smooth}$ just means that all derivatives of any order are continuous. $\endgroup$
    – snaleimath
    May 12, 2015 at 10:21

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