2. Preliminaries

Homeomorphs of $S^1$ are just cycle graphs and so for this reason the Turán number for simplicial 1-spheres is trivially $n - 1$ . For homeomorphs of $S^2$ , Brown, Erdős, and Sós [Reference Sós, Erdős and Brown23] give a double counting argument to show the $O(n^{5/2})$ upper bound on $ex(n, S^2)$ . We induct on this argument to prove Proposition 3. For our lower bound, we follow the strategy of Linial’s proof of $ex(n, S^2) \geq \Omega (n^{5/2})$ . Let us start with a review of that argument. A fundamental part of Linial’s randomised construction is an enumeration result of Tutte.

Moreover, it is well known by Steinitz’ Theorem that every simplicial 2-sphere is the boundary of a 3-polytope, see Chapter 4 of [Reference Ziegler25]. From here, the proof for the lower bound for $ex(n, S^2)$ follows by a standard probabilistic argument. More specifically it uses the alterations method; we initially construct a complex that has the right number of facets, but still a few $d$ -spheres, and then delete a face from each of those spheres. Take $Y_d(n, p)$ to denote the Linial–Mesulam random $d$ -complex model, that is, each face of dimension $(d - 1)$ is present, and the $d$ -faces are included independently with probability $p$ . Recall that for random simplicial complexes (including random graphs), a property is said to hold with high probability if the property holds with probability tending to one as the number of vertices goes to infinity.

For the question of the extremal construction for the 2-sphere one takes $Y \sim Y_2(n, p)$ with $p = {\varepsilon }/{\sqrt {n}}$ , $\varepsilon \gt 0$ a small constant. As the number of triangles in $Y$ follows a binomial distribution with $\binom {n}{3}$ trials and success probability $p$ , it follows that for $p = \varepsilon /\sqrt {n}$ with high probability the number of triangles in $Y_2(n, p)$ is $\Theta (n^{5/2})$ .

On the other hand, by the Euler characteristic a simplicial 2-sphere with $k$ vertices has $2k - 4$ triangles, so by the linearity of expectation and Tutte’s result the expected number of embedded 2-spheres is at most

\begin{eqnarray*} \sum _{k = 4}^{\infty } \binom {n}{k} k! C^k \left (\frac {\varepsilon }{\sqrt {n}} \right )^{2k - 4} &\leq & O(n^2), \end{eqnarray*}

for $\varepsilon \lt 1/C$ . With high probability, there are therefore $o(n^{5/2})$ embedded 2-spheres, and after simply deleting a triangle from each one, there are still $\Theta (n^{5/2})$ triangles, but no 2-spheres.

Our strategy to prove Theorem2 is to adapt this randomised argument to higher dimensions. In place of Tutte’s result, we assume that we work with classes of spheres which have exponential size in terms of the number of facets. As discussed in the introduction, this is a reasonable assumption as there are many natural classes of simplicial $d$ -spheres for which such a bound provably holds, and, as far as anyone currently knows, such a bound may hold for all $d$ -spheres. Classes of spheres which are known to have exponential size include all simplicial spheres in the following hierarchy:

\begin{equation*}\{\text{polytopal spheres}\} \subseteq \{\text{shellable spheres}\} \subseteq \{\text{LC spheres}\} \subseteq \{\text{2-LC spheres}\},\end{equation*}

as well as spheres that have bounded discrete Morse vector.

Since we do not know whether the collection of all $d$ -spheres has exponential size, it could be that the true value of $ex(n, S^d)$ is smaller than our conditional lower bound from Theorem2. However, if that is the case, a proof of a better upper bound would have to be exceptionally difficult, as it would necessarily involve finding a simplicial sphere that does not belong to any of the classes listed above. On the other hand, the proof of our upper bound in Proposition 3 finds an iterated suspension of a cycle in every sufficiently dense $d$ -complex, which in particular is a polytopal sphere.

For context regarding the common types of spheres for which our lower bound holds unconditionally, we review some standard definitions.

We remind the reader that a simplicial complex is pure if all facets are of the same dimension. All polytopes are shellable [Reference Bruggesser and Mani4], but not all shellable spheres are polytopal and not all simplicial spheres are shellable; see the discussion in Chapter 8 of [Reference Ziegler25]. With respect to counting by vertices there is an upper bound of Goodman and Pollack of $n^{(d+1)(d+2)n}$ polytopal $d$ -spheres on $n$ labelled vertices, while Kalai showed there are at least $\exp (\Theta (n^{\lfloor d/2 \rfloor }))$ simplicial $d$ -spheres on $n$ labelled vertices [Reference Kalai9]; Lee [Reference Lee13] showed that the spheres constructed in [Reference Kalai9] are shellable.

Broadening out from shellable spheres we have LC-spheres introduced by Durhuus and Jónsson [Reference Durhuus and Jónsson5], and more generally $t$ -LC spheres introduced by Benedetti and the second author [Reference Benedetti and Pavelka2]. When $t=1$ , the $1$ -LC class is precisely the LC class. Of particular interest here is the case $t = 2$ as these are the largest class in the $t$ -LC hierarchy for which there is a proof of exponential size.

A tree of $d$ -simplicies in the above definition refers to a tringulation of the $d$ -ball whose dual graph is a tree. Regarding enumeration we have the following bound, which holds also for all 2-LC manifolds.

Beyond the exponential-size assumption, another key result used in our proof is a lower bound theorem of Goff, Klee, and Novik [Reference Goff, Klee and Novik6] for balanced spheres. Specifically, this lower bound theorem is on the $f$ -vector of a balanced $d$ -sphere. The $f$ -vector for an arbitrary $d$ -dimensional simplicial complex is the vector $(f_0, f_1, \ldots , f_d)$ where $f_i$ is the number of faces of dimension $i$ . A simplicial $d$ -sphere, or more generally a $d$ -complex, is balanced if it is possible to colour the vertices with exactly $d + 1$ colours so that each facet is rainbow, that is contains exactly one vertex of each colour. A graph is balanced if it is bipartite and non-bipartite graphs have quadradic (ordinary) Turán number. In the same way, a non-balanced $d$ -complex $X$ has ordinary Turán number $ex(n, X) = \Theta (n^{d + 1})$ as we can simply take the complete $(d + 1)$ -partite hypergraph with $n/(d + 1)$ vertices in each part and have an $X$ -free $d$ -complex with $\Omega (n^{d + 1})$ facets. If we can construct a complex $Y$ that contains no balanced triangulation of $S^d$ then by colouring each vertex of $Y$ uniformly at random from a set of $d + 1$ colours and keeping only the rainbow facets, we end up with a complex $Y'$ that has no $d$ -spheres at all and $f_d(Y')/f_d(Y)$ very close to $\frac {(d + 1)!}{(d + 1)^{d + 1}}$ with high probability, so we lose only a constant factor in the number of facets. In other words, to study the Turán number of all triangulations of spheres, it suffices to restrict to the case of balanced triangulations.

3. Upper bound

The proof for the upper bound is a straightforward induction from the standard proof for the upper bound in the $d = 2$ case; however, it does not seem to appear in the literature, so we include it here. For this upper bound, we recall the definition of a link in a simplicial complex. For $X$ a simplicial complex and $\sigma$ a face of $X$ , the link of $\sigma$ in $X$ is defined as

\begin{equation*}{lk}(\sigma ) = \{\tau \in X \mid \sigma \cap \tau = \emptyset \text{ and } \sigma \cup \tau \in X \}.\end{equation*}

Proof of Proposition 3. By induction on $d$ . The $d = 1$ case is trivial, as it is just the Turán number for a cycle which is exactly $n - 1$ . For $d \gt 1$ suppose that we have a $d$ -complex $X$ with at least $C n^{d + 1 - 1/2^{d - 1}}$ facets for some large constant $C$ that depends on $d$ . Now for $u$ and $v$ vertices, let $\mathrm{d}(u, v)$ denote the number of $(d - 1)$ faces in ${lk}(u) \cap {lk}(v)$ and take the following sum.

\begin{eqnarray*} \sum _{\{u, v\} \in \binom {[n]}{2}} \mathrm{d}(u, v) &=& \sum _{\sigma \in \mathcal{F}_{d - 1}(X)} \binom {\deg (\sigma )}{2}, \end{eqnarray*}

where $\mathcal{F}_{i}(X)$ denotes the set of $i$ -dimensional faces, so $f_i = |\mathcal{F}_{i}(X)|$ , and the degree of a face is the number of faces of one dimension larger that contain it. We continue by estimating the above expression from below using Jensen’s inequality.

\begin{eqnarray*} \sum _{\sigma \in \mathcal{F}_{d - 1}} \binom {\deg (\sigma )}{2} &\geq & |\mathcal{F}_{d - 1}(X)| \binom {\frac {1}{f_{d-1}}\sum _{\sigma \in \mathcal{F}_{d - 1}} \deg (\sigma )}{2} \\ &\geq & \frac {1}{4} \frac { \left (\sum _{\sigma \in \mathcal{F}_{d - 1}} \deg (\sigma )\right )^2}{f_{d - 1}} \\ &\geq & \frac {1}{4} \frac { ((d + 1) f_d)^2 }{n^d} \\ &\geq & \Omega \left ( \frac {n^{2d + 2 - \frac {1}{2^{d - 2}}}}{n^d} \right ) \\ &=& \Omega \left (n^{d + 2 - \frac {1}{2^{d - 2}}}\right ) . \end{eqnarray*}

As the sum was taken over all pairs of vertices this means that there is a pair of vertices with at least $\Omega (n^{d - 1/2^{d - 2}})$ -many $(d - 1)$ -faces in the common link. Thus by induction, there are two vertices $u$ and $v$ so that ${lk}(u) \cap {lk}(v)$ contains a $(d - 1)$ -sphere $S$ . Now, let $S_u$ be the subcomplex obtained by adding $u$ to each $(d- 1)$ -face of $S$ and $S_v$ be obtained by adding $v$ to each $(d - 1)$ -face of $S$ . Then, the facets (of dimension $d$ ), $S_u \cup S_v$ form a $d$ -sphere in $X$ , specifically, this is the suspension of the $(d - 1)$ -sphere $S$ .

4. Lower bound

Here, we prove Theorem2. As discussed we will use a lower bound theorem for balanced spheres due to Goff, Klee, and Novik:

Theorem 9. (From Theorem 5.3 of Goff–Klee–Novik [Reference Goff, Klee and Novik6]). Let $X$ be a balanced, simplicial $d$ -sphere with $n$ vertices. Then

\begin{equation*}f_d(X) \geq \frac {2^{d + 1} - 2}{d + 1} n - 2^{d + 1} + 4.\end{equation*}

With this result, we prove the following enumeration lemma that will be critical for our randomised construction. For a collection $\mathcal{S}$ of $d$ -dimensional simplicial complexes, we denote by $\mathcal{S}_m$ the subcollection of $\mathcal{S}$ of complexes having exactly $m$ facets.

Lemma 10. Let $\mathcal{S}$ be a collection of unlabelled, balanced simplicial $d$ -spheres that has exponential size, say $|\mathcal{S}_m| \leq C^m$ . Then, the number of labelled copies of elements of $\mathcal{S}_m$ in the simplex $\Delta _{n - 1}$ is at most

\begin{equation*}C^m n^{(d + 1)m/(2^{d + 1} - 2)} n^{(d + 1)(2^{d + 1} - 4)/(2^{d + 1} - 2)}.\end{equation*}

Proof. For an unlabelled, balanced simplicial $d$ -sphere $M \in \mathcal{S}_m$ the number of labelled copies of $M$ in $\Delta _{n - 1}$ is at most the number of ways to label the vertices of $M$ with elements from $[n]$ . The number of such labelings is at most

\begin{equation*}n^{f_0(M)}.\end{equation*}

By Theorem9,

\begin{equation*}m = f_d(M) \geq \frac {2^{d + 1} - 2}{d + 1} f_0(M) - 2^{d + 1} + 4\end{equation*}

and so

\begin{equation*}f_0(M) \leq \frac {(m + 2^{d + 1} - 4)(d + 1)}{2^{d + 1} - 2}. \end{equation*}

Thus,

\begin{equation*}n^{f_0(M)} \leq n^{\frac {(m + 2^{d + 1} - 4)(d + 1)}{2^{d + 1} - 2}}.\end{equation*}

The result follows as there are at most $C^m$ choices for $M$ .

We can now prove Theorem2 via a standard alterations argument.

Proof of Theorem 2. Let $\mathcal{S}$ be a family of unlabelled $d$ -spheres that has exponential size. Take $Y \sim Y_d(n, \varepsilon n^{-(d + 1)/(2^{d + 1} - 2)})$ for $\varepsilon$ a small constant to be set later. By a usual application of the Chernoff bound, with high probability $Y$ has at least $\Omega (n^{d + 1 - (d + 1)/(2^{d + 1} - 2)})$ facets. By Lemma 10, the expected number of balanced complexes in $\mathcal{S}$ that are included in $Y$ is at most

\begin{eqnarray*} &&\sum _{m = 2^{d + 1}}^{\infty } C^m n^{(d + 1) m/(2^{d + 1} - 2)} n^{(d + 1)(2^{d + 1} - 4)/(2^{d + 1} - 2)} \left (\frac {\varepsilon }{n^{(d + 1)/(2^{d + 1} - 2)}} \right )^m \\ &=& n^{(d + 1)(2^{d + 1} - 4)/(2^{d + 1} - 2)}\sum _{m = 2^{d + 1}}^{\infty } (C\varepsilon )^m . \end{eqnarray*}

For $\varepsilon \lt 1/C$ the above is $O(n^{(d + 1)(2^{d + 1} - 4)/(2^{d + 1} - 2)})$ . Thus for any constant $\delta$ , the probability that there are more than

\begin{equation*}\delta n^{d + 1 - (d + 1)/(2^{d + 1} - 2)}\end{equation*}

balanced complexes from $\mathcal{S}$ in $Y$ is, by Markov’s Inequality:

\begin{equation*}O\left (\frac {n^{(d + 1)(2^{d + 1} - 4)/(2^{d + 1} - 2)}}{n^{d + 1 - (d + 1)/(2^{d + 1} - 2)}} \right ) = O\left (\frac {1}{n^{(d + 1)/(2^{d + 1} - 2)}}\right ) = o(1).\end{equation*}

Thus, with high probability, we may remove an arbitrarily small fraction of the facets from $Y$ to end up with a complex $\widetilde {Y}$ still with at least $\Omega (n^{d + 1 - (d + 1)/(2^{d + 1} - 2)})$ facets, but without balanced members of $\mathcal{S}$ . For the unbalanced members of $\mathcal{S}$ , we simply colour each vertex of $[n]$ independently and uniformly from the set of $d + 1$ colours. Now, let $Z$ be the subcomplex of $\widetilde {Y}$ generated by the rainbow facets under this colouring. With high probability, $Z$ will contain about a $\frac {(d + 1)!}{(d + 1)^{d + 1}}$ fraction of the facets of $\widetilde {Y}$ , but it will not contain any complexes of $\mathcal{S}$ .