Let $A,B \subset M$ be two transversal submanifolds of a compact manifold $M$. It seems rather intuitive that if $A$ and $B$ are deformed (say smoothly) in a way that they remain transversal to each other, the homotopy type of the intersection remains unchanged during the deformation.
I would like to know if there is a way to write such a homotopy in the following situations:
- $A = f^{-1}(0)$ is the zero level set of some smooth function $f : M \to \mathbb{R}$, and $B = B_s$ is a smooth family of submanifolds such that $A$ and $B_s$ are always transverse. Is it true that the homotopy type of $A \cap B_s$ is independent of $s$ ?
- $A_s := f_s^{-1}(0)$ is the zero set of a smooth function $f_s : M \to \mathbb{R}$, and the family $A_s$ is always transversal to a fixed submanifold $B$. Is it true that the homotopy type of $A_s \cap B$ is independent of $s$ ?
Any help will be appreciated. Thanks in advance.
I'm editing my question with an idea regarding the two points above. I'll be happy to get feedback on this.
- Let us assume that $B_s = \rho^{-1}(s)$ is a family of regular level sets of a smooth function $\rho : M \to \mathbb{R}$. Since $F^{-1}(0)$ is assumed to be transversal to $\rho^{-1}(s)$ for all $s \in [0,1]$, the restriction $$ \rho_{|F^{-1}(0)} : F^{-1}(0) \to \mathbb{R} $$ has only regular values on the interval $[0,1]$. We can then proceed as in Morse's deformation lemma: note that there exists $\epsilon > 0$ such that $\rho$ admits only regular values on $[-\epsilon, 1 + \epsilon]$. Let $\chi : \mathbb{R} \to \mathbb{R}_{\geq 0}$ be a smooth function such that $$ \chi(x) = \begin{cases} 1 \text{ on } [0,1]\\ 0 \text{ on } (-\infty , -\epsilon] \cup [1 + \epsilon, \infty). \end{cases} $$ For any given Riemannian metric on $M$, we consider the vector field $$ X(x) := \frac{\nabla_x \rho}{|| \nabla_x \rho||^2} \chi(\rho(x)). $$ Then $X$ is well-defined, and if $\phi_s$ denotes its flow, and $\phi_s(x) \in \rho^{-1}(0)$, then $$ \frac{d}{d s} \rho(\phi_s(x)) = 1. $$ Thus, $\rho(\phi_s(x)) = s$, and $\phi_1$ sends $\rho^{-1}(1)$ to $\rho^{-1}(0)$.
I think that this answers the question in the special case where $B_s$ are regular levels of a smooth function. However, I would like to extend this to any family of submanifolds.
- For the second point, it might be possible to apply a similar argument as above: consider the family of restrictions $f_{s|B} : B \to \mathbb{R}$. By assumption, $0$ is a regular value of $f_{s|B}$, for all $s \in [0,1]$. In particular, (by compactness of [0,1]), there exists $\epsilon > 0$ such that for all $s$, $f_{s|B}$ has only regular values in the interval $[-\epsilon, \epsilon]$. Define a smooth function $\chi : \mathbb{R} \to \mathbb{R}_{\geq 0}$ satisfying $$ \chi(x) = \begin{cases} 1 \text{ on } [-\frac{\epsilon}{3}, \frac{\epsilon}{3}]\\ 0 \text{ on } (-\infty, -\frac{\epsilon}{2}] \cup [\frac{\epsilon}{2}, \infty). \end{cases} $$ For any Riemannian metric on $B$, one can then define a time-dependent vector field $X$ on $B$ by $$ X_s(x) := - \chi(f_{s|B}(x)) \frac{\nabla_x f_{s|B}}{||\nabla_x f_{s|B}||^2} \frac{\partial}{\partial s} f_{s|B}(x). $$ As above, this vector field is well-defined, and if $\phi_s$ denotes its flow, and $f_{s|B}(\phi_s(x)) \in [-\frac{\epsilon}{3}, \frac{\epsilon}{3}]$, then $$ \frac{d}{ds} f_{s|B}(\phi_s(x)) = 0. $$ In particular, for any $x \in f_{0|B}^{-1}(0)$, the (continuous) function $s \mapsto f_{s|B}(\phi_s(x))$ is constant equal to $0$ in a neighbourhood of $0 \in [0,1]$. But then, by continuity, it must be constant equal to $0$ on the whole interval $[0,1]$. In particular, $$ \phi_s(f_{0|B}^{-1}(0)) = f_{s|B}^{-1}(0). $$ Now I guess that it should be possible to extend $\phi_s$ to an isotopy of $M$, by the homotopy extension theorem, or something close to it.
Does this seem ok ? Thanks again for your help.
