[Paper Review] Multiple points of immersions
This paper constructs a complete obstruction in a bordism group for determining whether a regular homotopy class of immersions $V^n \looparrowright M^m$ contains a $k$-immersion (i.e., an immersion without $k$-fold points), under the dimension condition $(k+1)(n+1) \leq km$. The obstruction vanishes if and only if such a $k$-immersion exists, generalizing Haefliger's result for $k=2$ using bordism-theoretic methods and equivariant techniques.
Given smooth manifolds $V^n$ and $M^m$, an integer $k$, and an immersion $f:V\looparrowright M$, we have constructed an obstruction for existence of regular homotopy of $f$ to an immersion $f':V\looparrowright M$ without $k$-fold points. This obstruction takes values in certain framed bordism group, and for $(k+1)(n+1)\leq km$ turns out to be complete.
Motivation & Objective
- To determine when a regular homotopy class of immersions $V^n \looparrowright M^m$ contains a $k$-immersion, i.e., an immersion without $k$-fold points.
- To generalize Haefliger's result for $k=2$ to higher $k$ using bordism theory and equivariant methods.
- To construct a complete obstruction in a bordism group that detects the existence of $k$-immersions under the dimension condition $(k+1)(n+1) \leq km$.
- To extend the theory to the case of multiple immersions $f_1, \dots, f_k$ with common $k$-fold intersection points, providing a criterion for eliminating such intersections via regular homotopy.
Proposed method
- Define the manifold of $k$-fold points $\overline{\pitchfork}(f^{(k)})$ for a generic immersion $f: V^n \looparrowright M^m$, which is an immersed submanifold of dimension $kn - (k-1)m$.
- Associate to $f$ an element $b(f^{(k)})$ in a certain equivariant bordism group $\Omega_{kn - (k-1)m}(E_k; \xi(f))$, which depends only on the regular homotopy class of $f$.
- Use the action of the symmetric group $\Sigma_k$ on the $k$-fold covering $\widehat{\overline{\pitchfork}(f^{(k)})}$ to define a $\Sigma_k$-equivariant bordism class.
- Construct a $\Sigma_k$-equivariant bordism $J: \widehat{W} \to \widehat{E_k}$ between $\widehat{\overline{\pitchfork}(f^{(k)})}$ and a representative $N$ of the same bordism class.
- Apply general position arguments to perturb the map $B: (\widehat{W} \times T)/\Sigma_k \to M$ to ensure transversality and disjointness from self-intersections of $f(V)$.
- Construct a $\Sigma_k$-invariant immersion $H: \widehat{W}^+ \times T^+ \to M$ that induces a regular homotopy eliminating $k$-fold points when the bordism class vanishes.
Experimental results
Research questions
- RQ1Under what dimensional conditions can a regular homotopy class of immersions $V^n \looparrowright M^m$ contain a $k$-immersion?
- RQ2Is there a complete bordism-theoretic obstruction to eliminating $k$-fold points in immersions?
- RQ3Can the method be extended to eliminate common $k$-fold intersection points of $k$ distinct immersions $f_1, \dots, f_k$?
- RQ4How does the sign or mod 2 count of $k$-fold points relate to the existence of $k$-immersions in special cases?
Key findings
- The obstruction $b(f^{(k)}) \in \Omega_{kn - (k-1)m}(E_k; \xi(f))$ is complete for detecting the existence of a $k$-immersion when $(k+1)(n+1) \leq km$.
- If $b(f^{(k)}) = 0$ in the bordism group, then there exists a regular homotopy of $f$ to a $k$-immersion.
- For $V^{(k-1)r} \looparrowright M^{kr}$ with $r > k$, a generic immersion is regularly homotopic to a $k$-immersion if and only if the signed count $I(f) = 0$.
- In the case $M = \mathbb{R}^{kr}$ and $r > k$ odd, any immersion $V^{(k-1)r} \looparrowright \mathbb{R}^{kr}$ is regularly homotopic to a $k$-immersion.
- For $k$ immersions $f_1, \dots, f_k$, they can be regularly homotoped to have no common $k$-fold intersection if and only if their images can be continuously homotoped to have empty common intersection.
- When $V_i$ are $p$-connected and $M$ is $(p+1)$-connected with $p \in \{4,5,12\}$, any $k$ immersions can be regularly homotoped to have no common intersection.
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This review was created by AI and reviewed by human editors.