Embedding products of graphs into Euclidean spaces

We need the following two simple results:

(*) If a compact polyhedron $K\hookrightarrow\mathbb{R}^{d}$ and $d>0$, then $K\times I$, $K\times S^{1}\hookrightarrow\mathbb{R}^{d+1}$ (it is sufficient to prove this for $K=I^{d}$, for which this is trivial).

(**) For any compact $d$-dimensional polyhedron $K$, the cylinder $K\times I\hookrightarrow\mathbb{R}^{2d+1}$ [RSS95].

Denote $G=G_{1}\times\dots\times G_{n}$. By general position, $G\hookrightarrow\mathbb{R}^{2n+1}$. If $i\neq 0$, then by (**) $G\times I\hookrightarrow\mathbb{R}^{2n+1}$. If, say, $G_{1}$ is planar, that is, $G_{1}\hookrightarrow I^{2}$, then by (**) and (*) we get $I^{2}\times G_{2}\times\dots\times G_{n}\hookrightarrow\mathbb{R}^{2n}$, whence $G\hookrightarrow\mathbb{R}^{2n}$. Applying (*) repeatedly, we get the embeddability assertion in all cases considered. ∎

Note that any connected graph, not homemorphic to a point, $I$, and $S^{1}$, contains a triod $Y$. So it suffices to prove that $Y^{n}\times I^{s+i}\not\hookrightarrow\mathbb{R}^{2n+s+i-1}$. Since $CK\times CL\cong C(K*L)$ and $K*\sigma^{0}_{0}=CK$ for any polyhedra $K$ and $L$, it follows that

If a polyhedron $K\not\hookrightarrow S^{d}$ then the cone $CK\not\hookrightarrow\mathbb{R}^{d+1}$ (because we work in piecewise linear category). So the non-embeddability in case (1) follows from $(\sigma^{0}_{2})^{*n}\not\hookrightarrow S^{2n-2}$, which is proved in [Kam32] (or alternatively from $Y^{n}\not\hookrightarrow S^{2n-1}$, which is proved in [Sko07]). ∎

First we reduce the theorem to the particular case $s=0$ analogously to Example 3. Indeed, assume that $s>0$ and Theorem 1 does not hold in case (2) for a product $G\times(S^{1})^{s}=G_{1}\times\dots\times G_{n}\times(S^{1})^{s}$, i.e., $G\times(S^{1})^{s}\hookrightarrow\mathbb{R}^{2n+s}$. Then by assertion (*) from the proof of the embeddability in Theorem 1 it follows that

The composition of the two embeddings is an embedding of a product containing no factors homeomorphic to $S^{1}$. The existence of the latter embedding contradicts to the case $s=0$ of Theorem 1 because the theorem gives the dimension $d=2n+2s+1$ for the product $G\times(K_{5})^{s}$. The obtained contradiction reduces the theorem to the particular case $s=0$, which is considered now.

By the Kuratowsky graph planarity criterion any nonplanar graph contains a subgraph homeomorhic to $K_{5}$ or $K_{3,3}$. So we may assume that each $G_{k}$ is either $K_{5}$ or $K_{3,3}$. Now we are going to replace all the graphs $K_{3,3}$ by $K_{5}$-s.

Note that $K_{5}$ is almost embeddable to $K_{3,3}$ (Fig. 1). Indeed, map a vertex of $K_{5}$ into the midpoint of an edge of $K_{3,3}$ and map the remaining four vertices bijectively onto the four vertices of $K_{3,3}$ not belonging to this edge. Then map each edge $e$ of $K_{5}$ onto the shortest (with respect to the number of vertices) arc in $K_{3,3}$, joining the images of the endpoints of $e$, and the almost embedding is constructed.

A product of almost embeddings is again an almost embedding, thus we get an almost embedding $(K_{5})^{n}\to G_{1}\times\dots\times G_{n}$. Assume that there is an embedding $G_{1}\times\dots\times G_{n}\to\mathbb{R}^{2n}$. The composition of the almost embedding and the embedding is an almost embedding $(K_{5})^{n}\to\mathbb{R}^{2n}$, which contradicts to Lemma 2. Thus the non-embeddability in case (2) of Theorem 1 follows from Lemma 2. ∎

Take triangulations of $A$ and $B$ and identify pairs of linear maps $\bar{f}:A\to\mathbb{R}^{2n}$, $\bar{g}:B\to\mathbb{R}^{2n}$ with points in $\mathbb{R}^{2nN}$. For a pair of simplices $a\subset A$, $b\subset B$ and almost all $(\bar{f},\bar{g})\in\mathbb{R}^{2nN}$, the set $\bar{f}a\cap\bar{g}b$ has at most one point, if $\dim a=\dim b=n$, and is empty, if $\dim a+\dim b=2n-1$. Indeed, otherwise $\bar{f}a$ and $\bar{g}b$ lie in one hyperplane in $\mathbb{R}^{2n}$, which is only possible for $(\bar{f},\bar{g})$ on certain algebraic hypersurface in $\mathbb{R}^{2nN}$. Analogously, for almost all $(\bar{f},\bar{g})\in\mathbb{R}^{2nN}$, the set $\bar{f}A\cap\bar{g}B$ is in bijection with the set $S$ of pairs of simplices $a\subset A,b\subset B$ such that $\bar{f}a\cap\bar{g}b\neq\emptyset$. In what follows we often omit the phrase “for almost all $(\bar{f},\bar{g})\in\mathbb{R}^{2nN}$”.

It remains to prove that $S$ has an even number of elements. Extend $\bar{f}:A\to\mathbb{R}^{2n}$ linearly to the cone $CA$ so that the cone vertex is mapped to the origin of $\mathbb{R}^{2n}$. The extension is still denoted by $\bar{f}$. Analogously to the above, for each pair of simplices $a\subset A$, $b\subset B$, the set $\bar{f}Ca\cap\bar{g}b$, if nonempty, is a single straight line segment when $\dim a=\dim b=n$, and a single point when $\dim a+\dim b=2n-1$; it is empty when $\dim a+\dim b<2n-1$. Construct the following graph. The set of vertices is the set of pairs of simplices $a\subset CA,b\subset B$ such that $\dim a+\dim b=2n$ and $\bar{f}a\cap\bar{g}b\neq\emptyset$. Two distinct vertices $(a_{1},b_{1})$ and $(a_{2},b_{2})$ are joined by an edge, if there are top-dimensional simplices $a\subset CA$ and $b\subset B$ such that $a\supset a_{1},a_{2}$ and $b\supset b_{1},b_{2}$. In this case, $\bar{f}a\cap\bar{g}b$ is the segment joining $\bar{f}a_{1}\cap\bar{g}b_{1}$ and $\bar{f}a_{2}\cap\bar{g}b_{2}$; in particular, the latter points are distinct (otherwise, either $\bar{f}\left|{}_{a}\right.$ is non-injective and $1=\dim\bar{f}a\cap\bar{g}b=\dim\bar{f}\partial a\cap\bar{g}b=0$, or $\bar{g}\left|{}_{b}\right.$ is non-injective, or $\bar{f}(a_{1}\cap a_{2})\cap\bar{g}(b_{1}\cap b_{2})\neq\emptyset$). There is no other vertex $(a_{3},b_{3})$ with $a_{3}\subset a,b_{3}\subset b$ because the segment has just two endpoints. Thus the set of edges is in a bijection with the set of pairs of top-dimensional simplices $a\subset CA$ and $b\subset B$ such that $\bar{f}a\cap\bar{g}b\neq\emptyset$. Hence the degree of a vertex $(a,b)$ is odd if and only if $a\subset A$, i.e., $(a,b)\in S$ (because $\partial(CA)=A\cup C\partial A$ and $\partial A=\partial B=\emptyset$). Thus $S$ has an even number of elements. ∎

Eq. (3) is proved analogously to Lemma 3. Indeed, take triangulations of $A\times I$ and $B\times I$ and identify pairs of linear maps $\bar{F}:A\times I\to\mathbb{R}^{2n}$, $\bar{G}:B\times I\to\mathbb{R}^{2n}$ with points in $\mathbb{R}^{2nN}$. For a pair of simplices $a\subset A\times I$, $b\subset B\times I$ and almost all $(\bar{F},\bar{G})\in\mathbb{R}^{2nN}$, the intersection $\bar{F}^{\uparrow}\mathaccent 28695{a}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}$ is a single open interval of a straight line or empty when $\dim a=\dim b=n+1$, a single point or empty when $\dim a+\dim b=2n+1$ or $\dim a+\dim b=2n$ with $a\subset A\times\{t\}$, $b\subset B\times\{t\}$ for some $t\in I$; and the intersection is empty otherwise. For almost all $(\bar{F},\bar{G})\in\mathbb{R}^{2nN}$, we have $\left|\bigcup_{t\in I}(\bar{F}_{t}\partial A\cap\bar{G}_{t}B)\right|=\left|\bar{F}^{\uparrow}(\partial A\times I)\cap\bar{G}^{\uparrow}(B\times I)\right|$. Take $(\bar{F},\bar{G})$ close enough to $(F,G)$ so that $\bar{F}_{t}\partial A\cap\bar{G}_{t}\partial B=\bar{F}_{\tau}\partial A\cap\bar{G}_{\tau}B=\bar{F}_{\tau}A\cap\bar{G}_{\tau}\partial B=\emptyset$ for each $t\in I,\tau\in\{0,1\}$. Then Eq. (3) is equivalent to the set $\left(\bar{F}^{\uparrow}\partial(A\times I)\cap\bar{G}^{\uparrow}(B\times I)\right)\cup\left(\bar{F}^{\uparrow}(A\times I)\cap\bar{G}^{\uparrow}\partial(B\times I)\right)$ having an even number of elements. We may assume that the latter set is in bijection with the set $S$ of pairs of simplices $a\subset A\times I,b\subset B\times I$, where $a\subset\partial(A\times I)$ or $b\subset\partial(B\times I)$, such that $\bar{F}^{\uparrow}\mathaccent 28695{a}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}\neq\emptyset$. In what follows, we may impose assumptions, which are automatic for almost all $(\bar{F},\bar{G})\in\mathbb{R}^{2nN}$, without listing them explicitly.

To show that $|S|$ is even, construct the following graph. The set of vertices is the set of pairs of simplices $a\subset A\times I,b\subset B\times I$ such that $\bar{F}^{\uparrow}\mathaccent 28695{a}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}$ is a single point. Two distinct vertices $(a_{1},b_{1})$ and $(a_{2},b_{2})$ are joined by an edge, if there are top-dimensional simplices $a\subset A\times I$, $b\subset B\times I$ such that $a\supset a_{1},a_{2}$ and $b\supset b_{1},b_{2}$. In this case, $\bar{F}^{\uparrow}a\cap\bar{G}^{\uparrow}b$ is the segment joining $\bar{F}^{\uparrow}\mathaccent 28695{a}_{1}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}_{1}$ and $\bar{F}^{\uparrow}\mathaccent 28695{a}_{2}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}_{2}$ (in particular, the latter points are distinct because the images of the interiors of faces of $a$ are disjoint unless $\dim\bar{F}^{\uparrow}a\leq n$). Thus the set of edges is in a bijection with the set of pairs of top-dimensional simplices $a\subset A\times I$, $b\subset B\times I$ such that $\bar{F}^{\uparrow}\mathaccent 28695{a}\cap\bar{G}^{\uparrow}\mathaccent 28695{b}\neq\emptyset$. Thus the set of vertices of odd degree is precisely $S$. Thus $|S|$ is even, and (3) holds.

Eq. (3) implies that the intersection index $fA\cap gB$ is well-defined (if $f:A\to\mathbb{R}^{2n}$ and $g:B\to\mathbb{R}^{2n}$ satisfy $f\partial A\cap gB=fA\cap g\partial B=\emptyset$). Indeed, let $F(x,t)=f(x)$ and $G(x,t)=g(x)$ be constant homotopies. Take two triangulations of $A$ such that $f$ is linear on both and two maps $\bar{F}_{0}:A\to\mathbb{R}^{2n}$ and $\bar{F}_{1}:A\to\mathbb{R}^{2n}$ linear on the first and the second triangulation respectively. Extend the two triangulations to a triangulation of $A\times I$ such that $F$ is linear and the two maps to a linear map $\bar{F}:A\times I\to\mathbb{R}^{2n}$. Construct a triangulation of $B\times I$ and maps $G,\bar{G}:B\times I\to\mathbb{R}^{2n}$ analogously. All pairs $(\bar{F},\bar{G})$ linear on our triangulations can be obtained by this construction. If $(\bar{F},\bar{G})$ is close to $(F,G)$, then the right side of (3) vanishes, thus for some $\varepsilon>0$ the number $|\bar{F}_{0}A\cap\bar{G}_{0}B|$ has the same parity for all triangulations of $A$ and $B$ and almost all $(\bar{F}_{0},\bar{G}_{0})$, $\varepsilon$-close to $(f,g)$.

As a consequence, if linear maps $f:A\to\mathbb{R}^{2n}$ and $g:B\to\mathbb{R}^{2n}$ intersect ’transversely’, i.e. each point of $fA\cap gB$ has a single $f$-preimage and a single $g$-preimage, both lying in the interior of top-dimensional simplices, then the intersection index $fA\cap gB=|fA\cap gB|\mod 2$.

Finally, the intersection index of homotopies $\bigcup_{t\in I}F_{t}A\cap G_{t}B$ is well-defined (if $F:A\times I\to\mathbb{R}^{2n}$ and $G:B\times I\to\mathbb{R}^{2n}$ satisfy the assumptions in its definition; in particular, now $\dim B=n-1$ or $B=\emptyset$) because

for a general position pair $(\bar{F},\bar{G})$ close to $(F,G)$. Here the first equality is the definition. The second one is obvious, where we denote $C:=C(B\times I)\cong B\times I^{2}/B\times I\times\{0\}$ and $\bar{G}^{\Uparrow}(y,t,s):=(2s\bar{G}^{\uparrow}(y,t),1-2s)\subset\mathbb{R}^{2n+1}\times\mathbb{R}$ for all $y\in B$ and $t,s\in I$. The third one follows from the previous paragraph. The last equality is obtained by applying (3) to the rectilinear homotopy between $(\bar{F}^{\uparrow},\bar{G}^{\Uparrow})$ and $(F^{\uparrow},G^{\Uparrow})$. ∎

Assume that there exists an almost embedding $f:K=K_{5}\times\dots\times K_{5}\to\mathbb{R}^{2n}$. Let $O=O_{1}\times\dots\times O_{n}$ be a vertex of $K$. By the well-known formulae, we get

Let $\alpha,\beta\subset\operatorname{Lk}O$ be a pair of $(n-1)$-spheres given by Lemma $1^{\prime}$. Identify $\operatorname{Lk}O$ and $\operatorname{Lk}O_{1}*\dots*\operatorname{Lk}O_{n}$. Since $\alpha$ and $\beta$ are disjoint, it follows that for each $k=1,\dots,n$ the sets $\alpha\cap\operatorname{Lk}O_{k}$ and $\beta\cap\operatorname{Lk}O_{k}$ are disjoint. Each of the sets $\alpha\cap\operatorname{Lk}O_{k}$ and $\beta\cap\operatorname{Lk}O_{k}$ contains more than $1$ point because one of the spheres $\alpha$ and $\beta$ would be a cone otherwise. Thus each of the sets consists of exactly 2 points. By definition, put $\{A_{k},C_{k}\}:=\alpha\cap\operatorname{Lk}O_{k}$ and $\{B_{k},D_{k}\}:=\beta\cap\operatorname{Lk}O_{k}$. Consider two $n$-tori

contained in $K$.

Clearly, $T_{\alpha}\supset C\alpha$, $T_{\beta}\supset C\beta$ and $T_{\alpha}\cap T_{\beta}=O$. Since $f$ is an almost embedding, it follows that the intersection index $fT_{\alpha}\cap fT_{\beta}=fC\alpha\cap fC\beta$. So $fT_{\alpha}\cap fT_{\beta}$ is odd by the choice of $\alpha$ and $\beta$. By Parity Lemma 3 we obtain a contradiction, hence $K$ is not almost embeddable into $\mathbb{R}^{2n}$. ∎

The proof is similar to that of Conway–Gordon–Sachs theorem and applies the idea of [Kam32], with a slightly more refined obstruction. The reader can restrict attention to the case when $n=2$ and obtain an alternative proof of the Sachs theorem. (The proof for $n>2$ is analogous to that for $n=2$.)

We show that for any $(n-1)$-simplex $c$ of $L$ and any almost embedding $f:CL\to\mathbb{R}^{2n}$ there exist a pair of disjoint $(n-1)$-spheres $\alpha,\beta\subset L$ such that $\alpha\supset c$ and the intersection index $fC\alpha\cap fC\beta$ is odd.

For an almost embedding $f:CL\to\mathbb{R}^{2n}$, let

be the Van Kampen obstruction to linkless embeddability. Here the sum is over all pairs of disjoint $(n-1)$-spheres $\alpha,\beta\subset L$ such that $c\subset\alpha$. It suffices to prove that $v(f)=1$. Our proof is in 2 steps: first we show that $v(f)$ does not depend on $f$; then we calculate $v(f)$ for certain ‘standard’ embedding $f:CL\to\mathbb{R}^{2n}$.

Let us prove that $v(f)$ does not depend on $f$ [cf. Kam32, CG83]. Take any two almost embeddings $F_{0},F_{1}:CL\to\mathbb{R}^{2n}$. By general position in piecewise linear category, there exists a piecewise linear homotopy $F:CL\times I\to\mathbb{R}^{2n}$ between them such that

1) there is a finite number of singular moments $t$, i.e., $t\in I$ such that $F_{t}$ is not an almost embedding;

2) for a singular $t$, there is a unique pair of disjoint $(n-1)$-simplices $a,b\subset L$ such that $F_{t}Ca\cap F_{t}b\neq\emptyset$;

3) the intersection $F_{t}Ca\cap F_{t}b$ is ’transversal in time’, i.e., $F_{t}^{-1}(F_{t}Ca\cap F_{t}b)\times t$ consists of exactly two points, and $\left.F\right|_{Ca\times I}$ and $\left.F\right|_{b\times I}$ are smooth at those points.

Consider a singular moment $t$ and the pair $a,b$ of simplices given by condition 2). Conditions 3) and 1) imply that the intersection index of homotopies $\bigcup_{\tau\in[t-\varepsilon,t+\varepsilon]}F_{\tau}Ca\cap F_{\tau}b$ is odd for small enough $\varepsilon>0$. Then by Lemma $3^{\prime}$, for a pair of disjoint $(n-1)$-spheres $\alpha,\beta\subset L$, the intersection index $F_{t}C\alpha\cap F_{t}C\beta$ changes with the increasing of $t$ if and only if either $\alpha\supset a$, $\beta\supset b$ or $\alpha\supset b$, $\beta\supset a$. Such pairs $(\alpha,\beta)$ satisfying the condition $\alpha\supset c$ are called critical. If $c\cap(a\cup b)=\emptyset$, then there are exactly $2$ critical pairs. Indeed, we have either $\alpha\supset a\cup c$ or $\alpha\supset b\cup c$. Each of these two conditions determines a unique critical pair. If $c\cap(a\cup b)\neq\emptyset$, then there are two distinct vertices $v,w\in L-(a\cup b\cup c)$ belonging to the same copy of $\sigma^{0}_{3}$. Then there is an involution without fixed points on the set of critical pairs. Indeed, $\mathbb{Z}/2\mathbb{Z}$ acts on the set of vertices of $L$ by interchanging $v$ and $w$, and it also acts on the set of critical pairs, because $v,w\notin a\cup b\cup c$. So the number of critical pairs is even, thus $v(F_{0})=v(F_{1})$.

Fig. 2.

Now let us prove that $v(f)=1$ for certain ‘standard’ embedding $f:CL\to\mathbb{R}^{2n}$ (Fig. 2). To define the standard embedding $f:CL\to\mathbb{R}^{2n}$, take a general position collection of $n$ lines in $\mathbb{R}^{2n-1}\subset\mathbb{R}^{2n}$. For each $k=1,\dots,n$ take a quadruple $\sigma_{k}$ of distinct points on $k$-th line. Taking the join of all $\sigma_{k}$, we obtain an embedding $L\to\mathbb{R}^{2n-1}$. The standard embedding $f:CL\to\mathbb{R}^{2n}$ is defined to be the cone over this embedding. Further we omit $f$ from the notation of $f$-images.

Clearly, for a pair of disjoint $(n-1)$-spheres $\alpha,\beta\subset L$, the intersection index $fC\alpha\cap fC\beta$ has the same parity as the linking number $\operatorname{lk}(\alpha,\beta)$ in $\mathbb{R}^{2n-1}$. Indeed, let $H$ be the half-space bounded by $\mathbb{R}^{2n-1}$ and containing the cone vertex. Take a pair of triangulations of $C\alpha$ and $C\beta$ and a pair of linear embeddings $(\bar{f}\colon C\alpha\to H,\bar{g}\colon C\beta\to H)$ close enough to $(\left.f\right|_{C\alpha},\left.f\right|_{C\beta})$ such that $\bar{f}\alpha=\alpha$, $\bar{g}\beta=\beta$, and $\bar{f}a\cap\bar{g}b=\emptyset$ for each pair of simplices $a\subset C\alpha,b\subset C\beta$ unless $\dim a=\dim b=n$. Applying Lemma $3^{\prime}$ to the rectilinear homotopy between $(\bar{f},\bar{g})$ and $(\left.f\right|_{C\alpha},\left.f\right|_{C\beta})$, we get $fC\alpha\cap fC\beta=\bar{f}C\alpha\cap\bar{g}C\beta=|\bar{f}C\alpha\cap\bar{g}C\beta|\mod 2$. Passing to a subdivision, if necessary, we guarantee that for each simplex $b\subset C\beta$ there is at most one simplex $a\subset C\alpha$ such that $\bar{f}a\cap\bar{g}b\neq\emptyset$. Then $\beta$ is homologous in $H-\bar{f}C\alpha$ to the sum of the boundaries $\partial b$ of all simplices $b\subset C\beta$ such that $\bar{f}a\cap\bar{g}b\neq\emptyset$ for some simplex $a\subset C\alpha$. A simple computation shows that each such $\partial b$ contributes $1$ to $\operatorname{lk}(\alpha,\beta)\mod 2$, hence $|\bar{f}C\alpha\cap\bar{g}C\beta|=\operatorname{lk}(\alpha,\beta)\mod 2$.