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Consider a simplicial complex $K$. A piecewise linear map $f: K \to \mathbb{R}^n$ is an almost-embedding if $f(\sigma) \cap f(\tau) = \emptyset$ for any two disjoint simplices $\sigma,\tau$ in $K$.

Question 1: Is every almost-embedding an embedding in the case where $K$ is a simplicial 2-manifold, and $n=3$?

There is some related discussion in Segal, Skopenkov, Spieiz, "Embeddings of polyhedra in $\mathbb{R}^n$ and the deleted product obstruction", but they specifically leave this case open (2-manifolds into $\mathbb{R}^3$).

Alternatively, I would be happy with an answer to the following variant:

Question 2: Suppose $K$ is a simplicial 2-manifold where every vertex has degree at least 4, and $f: K \to \mathbb{R}^3$ is a piecewise linear map such that $f(\sigma) \cap f(\tau) = \emptyset$ for any two disjoint 2-simplices $\sigma,\tau$ in $K$. Is $f$ necessarily an embedding?

Here, triangulations with degree-3 vertices are excluded since otherwise there are some obvious counterexamples, e.g., the tetrahedron with all four vertices coplanar. (This case is already excluded from Question 1, since any vertex and the complementary triangle are disjoint.)

Here are some not-quite-counterexamples to consider:

Two examples that fail to be almost-embeddings.

In both cases, $K$ is a simplicial 2-sphere. On the left there is an intersection between triangles $t_1$ and $t_2$ (which share an edge $e_1$), but this map is not quite an almost-embedding because the images of disjoint edges $e_2$ and $e_3$ intersect at the point $p$. On the right, there is an intersection between triangle $t_1$, with vertices $i,k,l$, and triangle $t_2$, with vertices $i,j,l$, but this map is not quite an almost-embedding because the images of disjoint triangles $t_3$ and $t_4$ intersect at $f(k)$.

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    $\begingroup$ Just another reference concerning almost-embeddings: Skopenkov, Arkadiy, and Martin Tancer. "Hardness of almost embedding simplicial complexes in $\mathbb{R}^d$." Discrete & Computational Geometry 61 (2019): 452-463. $\endgroup$ Mar 11, 2023 at 15:16

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"Yes" for Question 1.

We need to show that interiors of distinct smplexes $\alpha$ and $\beta$ do not intersect. Assume $\mathring \alpha\cap\mathring \beta\ne\varnothing$. Then $\alpha$ and $\beta$ share a simplex, say $\sigma$. Note that $\sigma\in\partial (\alpha\cap\beta)$. Choose a point $z\in \alpha\cap\beta$ that maximize the distance to $\sigma$. Denote by $\alpha'$ and $\beta'$ the smallest subsimplexes of $\alpha$ and $\beta$ that contain $z$. Observe that $\alpha'\cap\beta'=\varnothing$.

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  • $\begingroup$ Sorry, I don't completely follow your argument (though it's quite possible I'm misinterpreting your notation). Suppose $\alpha$ and $\beta$ are coplanar triangles, neither containing the other, and sharing an edge $\sigma$, as in this image: i.stack.imgur.com/vz3z7.png. Then following your construction I can pick $x$ and $y$ such that $\alpha^\prime = \alpha$ and $\beta^\prime$ is an edge sharing a vertex with $\alpha^\prime$. But in this case $\alpha^\prime$ and $\beta^\prime$ are not disjoint, hence there is no immediate contradiction. $\endgroup$
    – Omega Tree
    Mar 12, 2023 at 1:32
  • $\begingroup$ Oh --- there was a mistake, I will fix it in a minute. $\endgroup$ Mar 12, 2023 at 18:46
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Sorry, my previous answer is based on the impression that the "manifold" can mean "manifold with boundary. In that case, the countercase can be constructed easily (see my previous answer). But for true manifold, here is my answer:

Question 1: Yes, except the simplicial 1-manifold embedded to 2/3D. But I think this case can always be viewed as a slicing of the 2 manifolds.

Question 2: I actually believe this is true. I even drafted a proof:

  1. First let's filter out all the degenerate cases. There are 3 degenerate cases, as shown in the following figure. (a,b) correspond's to the rank of $\nabla f$ being 1; (c) corresponds to the rank of $\nabla f$ being 0. Both violate the almost-embedding definition (a: green vertex and the bottom edge; b: green vertex and the red vertex; c: all three ). enter image description here

  2. Then we can actually check whether there can be a local intersection that satisfies the almost-embedding definition. We only need to find 2 2-simplex (triangle): $t_1$ and $t_2$ from the embedding of the manifold, s.t., $t_1 \cap t_2\neq \emptyset$, and an emersion $f$, s.t., $f(t_1/(t_1\cap t_2))\cap f(t_2/(t_1\cap t_2))\neq\emptyset$, and $f$ is causing no intersection of non-adjacent faces.

  3. $t_1 \cap t_2\neq \emptyset$ includes two cases: they can share 1 or 2 vertices (but not 3). as shown in the following figure. We can break it down to different cases and prove they all violate the definition of almost embedding. enter image description here

  4. If $t_1 $ and $ t_2$ shares 2 vertices and $f(t_1/(t_1\cap t_2))\cap f(t_2/(t_1\cap t_2))\neq\emptyset$, $f(t_1)$ and $f(t_2)$ must be coplanar, then we can just enumerate all the intersection cases of $f(t_1)$ and $f(t_2)$, there are totally 4 cases (mode out the symmetry). All of them are not almost embedding, violating parts are marked by the red cross. enter image description here We can guarantee this is an exhaustive list, because there are local intersections if an only if the yellow vertex is mapped to one of the regions 1,2,3,4 (or the boundary between them) in the following figure. Algebraic analysis can prove the intersection condition and the categorization of the cases. enter image description here

  5. Case 2: $t_1$ and $t_2$ only share one vertex. In this case if $f(t_1/(t_1\cap t_2))\cap f(t_2/(t_1\cap t_2))\neq\emptyset$ we can further categorize it into 2 subcases:

5.1. Edge intersections: when $f(e_1)^\circ \cap f(e_2)^\circ \neq \emptyset$. where $e_1, e_2$ are edges, $e_1 \in t_1$ and $e_2 \in t_2$ and $e_1 \cap e_2=$shared vertex. There are 3 cases, as shown in the figure. None of them are almost embedding. enter image description here

5.2. $f(t_1)^\circ \cap f(e_2)^\circ \neq \emptyset$, where $e_2 \in t_2$ and contains the shared vertex. There are 3 cases, and non of them are almost embedding. enter image description here

5.3. Every case that is neither 5.1 nor 5.2. Then e mush $f(t_1)^\circ \cap f(t_2)^\circ \neq \emptyset$. This is a very intersting case, actually, it can be handled similarly to 5.1. Denote $e_1$, $e_2$ as the edge of $t_1$ and $t_2$ opposite to the shared vertex. In this case, eigher $f(e_1) \cap f(t_2)^\circ \neq \emptyset$, or $f(e_2) \cap f(t_1)^\circ \neq \emptyset$, or or $f(e_1) \cap f(e_2) \neq \emptyset$. Either way, it's not an almost embedding. enter image description here

We've proved that in every possible case, the local intersection cannot be an almost embedding. Thus question 2 is proven.


Below are my old answers. Have you considered the simplicial 3-manifold cases?

For Question 1: I think an almost-embedding can also be a 3D simplicial complex (tetrahedral mesh) mapped to $\mathcal{R}^3$ (for example, an identity map).

For Question 2: You might have degenerate cases or flipped cases, like this $\mathcal{R}^2 \mapsto \mathcal{R}^2 \times \{0\}$ case shown in the figure. Because the bijectivity of already adjacent simplexes is not controlled by the definition. (the website won't let me directly post images) Inversion map causes local intersecction

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  • $\begingroup$ I kind of see your point; if the 2-manifold is a closed surface, then things become more complicated, and my countercase won't work; please allow me to think about this, I will get back to it later. $\endgroup$
    – Anka Chen
    Mar 11, 2023 at 19:59
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As stated, the answer to both questions is no. A counterexample is: Take any surface PL-embedded in $\mathbb{R}^3$. This can be for example boundary of the regular octahedron in order to satisfy the conditions of Question 2.

Then take any embedded $2$-simplex $\sigma$ and locally form a self-intersection in the interior of $\sigma$. One can for example `pull a finger' from the interior and then to let it intersect with other part of the interior in a circle. (The picture below shows an analogous move for 1-manifolds in the plane.) If this is performed locally, then no other simplices are affected. Therefore we have an almost embedding but not an embedding.

enter image description here

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  • $\begingroup$ You have to remain PL. Is it obvious that it can be made an almost-embedding in the PL way? Only images of non-disjoint simplices are allowed to additionally intersect. $\endgroup$ Mar 23, 2023 at 7:53
  • $\begingroup$ PL is not a problem. One way to phrase this is to take a subdivision of $\sigma$ containing in the interior some other 2-simplex $\tau$ and a disjoint vertex $v$. Then the construction that I have described is essentially van Kampen's finger move from $\tau$ around $v$. These finger moves are PL. Everything can be performed locally in close neighborhood of $\sigma$ thus no additional intersections are introduced. $\endgroup$ Mar 23, 2023 at 8:08
  • $\begingroup$ Should not the answer include this argument? Not that I do follow it well enough... $\endgroup$ Mar 23, 2023 at 8:12
  • $\begingroup$ I have added a simple 1-dimensional picture which hopefully clarifies the move. (I couldn't make it immediately when first posting the answer.) The main point is that PL maps allow subdivision of simplices. Then it is very easy to make self intersections and there are many ways how to get them. $\endgroup$ Mar 23, 2023 at 11:01
  • $\begingroup$ Does not the question presume that PL in this case means linear on each simplex? Because if one interprets your example in this way, you will have intersecting images of disjoint simplices, no? $\endgroup$ Mar 23, 2023 at 13:12

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