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I am reading the short paper due to Erdös and Bollobás "On a Ramsey-Turán type problem", where they obtain a lower bound for the number of edges on an $n$-graph without $K_4$ as a subgraph and small independence number (namely, $o(n)$).

The construction is geometric and uses that for given $k$, there is a sufficiently large non-negative integer $n$ such that the sphere $S^{k+1}=\{x\in R^{k+2}: \|x\|=1\}$ can be divided into $n$ sets 1) having equal measure and 2) small diameter.

This recall me Borsuk's conjecture, but not sure how to get this result, as here the number of parts is larger and we need that all the pieces have the same measure.

I assume that this is a folklore result, but I do not know where to find a reference for that.

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This is too long for a comment. If you want a single $n=n(k,\varepsilon)$ and do not care about the dependency on the parameters ($\varepsilon$ is the largest diameter of the pieces), then here is one way to do it. This is only a half-answer, since it does not provide a well-known reference.

The result is obvious for $k=0$, so we assume $k\geq1$. Choose some $r\geq100k$.

  1. Cut the sphere along half-hyperplanes $$\{-\sin(\theta_i)x_1+\cos(\theta_i)x_2=0,\cos(\theta_i)x_1+\sin(\theta_i)x_2\geq0\}$$ for $\theta_i$ very close to $i\pi/r$ (say $\pi/4r$-close), $1\leq i\leq2r$, in such a way that the volumes of the pieces $\{C_i\}$ you created are rational fractions of the total volume of the sphere. These are convex pieces, with respect to the spherical geodesics. (This is a bit ridiculous since we can just choose $\theta_i=i\pi/r$, but bear with me.)
  2. Cut every $C_i$ along half-hyperplanes $$\{-\sin(\theta_{ij})x_1+\cos(\theta_{ij})x_3=0,\cos(\theta_{ij})x_1+\sin(\theta_{ij})x_3\geq0\}$$ for $\theta_{ij}$ very close to $j\pi/r$, so that all the pieces $\{C_{ij}\}$ thus created still have rational volume. This is possible since the projection of the interior of $C_i$ on these components is connected ($C_i$ is convex). Iterate for all pairs of dimensions.
  1. Find a common denominator $q$ for all these ratios, i.e. every piece $C_{i_{11},\ldots,i_{k+2,k+2}}$ has volume $p_{i_{11},\ldots,i_{k+2,k+2}}\operatorname{vol}(S^{k+1})/q$. For each of them, consider an isometry $\phi:R^{k+2}\to R^{k+2}$ such that $\phi^{-1}(0)$ belongs to its interior, and cut the piece along half-hyperplanes of the form $$\{-\sin(\theta)\phi(x)_1+\cos(\theta)\phi(x)_2=0,\cos(\theta)\phi(x)_1+\sin(\theta)\phi(x)_2\geq0\}$$ so that every piece has volume $\operatorname{vol}(S^{k+1})/q$.

These $q$ pieces all have the same volume. We can bound their diameter as follows. Let $x,y$ be two elements of the same piece, and suppose they are at (spherical) distance at most $\pi/2$. Since they are unit, the norm $|x\wedge y|$ of their exterior product is the sine of the angles between the two, i.e. $|x\wedge y|=\sin(d(x,y))$ for $d$ the spherical distance. Moreover, $$ |x\wedge y|^2 = \sum_{i<j}(x_iy_j-x_jy_i)^2 < \binom{k+2}2\sin\left(\frac{3\pi}{2r}\right)^2. $$ We find a crude estimate $d(x,y)\leq4(k+3)/r$. This is smaller than $\pi/2$, so by convexity the diameter has to be at most $4(k+3)/r$, which can be made arbitrarily small by choosing $r$ very large.

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    $\begingroup$ By the way, this is not really a topology result: it is true for any closed manifold with an absolutely continuous measure with full support, by choosing instead of the half-planes some $\{f_i(x)=\alpha\}$ such that the $f_i$ are Morse functions whose differentials collectively generate the cotangent space at every point. One has to be a bit more careful about connectedness though. $\endgroup$
    – Pierre PC
    Commented May 11, 2023 at 18:35
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The OP asks for a reference. One such result is Lemma 21 in On the optimality of the random hyperplane rounding technique for MAX CUT by Feige and Schechtman. It says that "For each $0 < \gamma < \pi/2$ the sphere $S^{d-1}$ can be partitioned into $N = (O(1)/\gamma)^d$ equal volume cells, each of diameter at most $\gamma$." The proof involves Voronoi cells.

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