# The simplex value *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This lesson computes \(R_d\) for simplices, following OpenAI [OpenAI-PB, Section 3]. By Cauchy's formula (Proposition 2.3 of [Projection bodies of polytopes](projection-bodies-of-polytopes.md)), the projection body of the standard simplex is the sum of \(d+1\) segments. Up to a translation and a factor, it is the unit cube plus one diagonal segment, and its volume follows from a fibre count. All simplices of one dimension have the same value, by affine invariance. Throughout, \(d\ge2\), \(e_1,\dots,e_d\) is the standard basis, \(w=e_1+\dots+e_d\), and \[ \begin{aligned} T_d=\{x\in\mathbb R^d:{}&x_i\ge0,\\ &x_1+\dots+x_d\le1\}. \end{aligned} \] Measure-theoretic facts are cited from the text of Fremlin [Fremlin] by number. ## 1. Volume and facets of the simplex **Lemma 1.1.** \(|T_d|=1/d!\), and the \((d-1)\)-dimensional simplex \(\{x\in\mathbb R^{d-1}:x_i\ge0,\ \sum x_i\le1\}\) has volume \(1/(d-1)!\). *Proof.* For \(d=1\), \(T_1=[0,1]\). For \(d\ge2\), the section of \(T_d\) at \(x_d=t\in[0,1]\) is \((1-t)T_{d-1}\), of volume \((1-t)^{d-1}|T_{d-1}|\) [Fremlin, 263A]. By Fubini's theorem [Fremlin, 251N and 252B], \(|T_d|=|T_{d-1}|\int_0^1(1-t)^{d-1}dt=|T_{d-1}|/d\). \(\square\) **Lemma 1.2.** \(T_d\) is a polytope in the sense of [Projection bodies of polytopes](projection-bodies-of-polytopes.md), with \(d+1\) facets: \[ \begin{gathered} F_i=T_d\cap\{x_i=0\}\quad(1\le i\le d),\\ F_0=T_d\cap\{x_1+\dots+x_d=1\}. \end{gathered} \] The outward unit normal of \(F_i\) is \(-e_i\) and its area is \(1/(d-1)!\); the outward unit normal of \(F_0\) is \(w/\sqrt d\) and its area is \(\sqrt d/(d-1)!\). So the vectors \(s_ja_j\) are \[ -\frac{e_i}{(d-1)!}\quad(1\le i\le d),\qquad\frac w{(d-1)!}. \] *Proof.* The inequalities are \(\langle-e_i,x\rangle\le0\) and \(\langle w/\sqrt d,x\rangle\le1/\sqrt d\). Each \(F_i\) is a copy of the \((d-1)\)-dimensional simplex in the coordinate hyperplane, of area \(1/(d-1)!\) by Lemma 1.1. None of the inequalities can be omitted: each set \(F_j\) contains a relatively open subset of its hyperplane, hence a point outside the other hyperplanes, where the other inequalities are strict; moving from it a little in the direction of the outward normal violates inequality \(j\) only. The projection of \(F_0\) along \(e_d\) onto the hyperplane \(\{x_d=0\}\) is the \((d-1)\)-dimensional simplex. By Lemma 2.2 of [Projection bodies of polytopes](projection-bodies-of-polytopes.md), its area \(1/(d-1)!\) equals the area of \(F_0\) times \(|\langle w/\sqrt d,e_d\rangle|=1/\sqrt d\). \(\square\) ## 2. The projection body of the simplex By Proposition 2.3 of [Projection bodies of polytopes](projection-bodies-of-polytopes.md) and Lemma 1.2, \[ \begin{aligned} \Pi T_d=\frac1{(d-1)!}\Bigl(&\sum_{i=1}^d\bigl[-\tfrac{e_i}2,\tfrac{e_i}2\bigr]\\ &+\bigl[-\tfrac w2,\tfrac w2\bigr]\Bigr), \end{aligned} \] because a segment \([-\frac v2,\frac v2]\) does not change when \(v\) is replaced by \(-v\). Since \([-\frac v2,\frac v2]=[0,v]-\frac v2\), this is a translate of \[ \frac1{(d-1)!}\bigl(Q+[0,w]\bigr),\qquad Q=[0,1]^d . \] **Lemma 2.1** (extrusion). Let \(C\subseteq\mathbb R^d\) be compact and convex and \(v\ne0\). Then, with \(\pi_v\) the orthogonal projection onto \(v^\perp\), \[ |C+[0,v]|=|C|+|v|\operatorname{vol}_{d-1}(\pi_v[C]). \] *Proof.* After an orthogonal change of coordinates [Fremlin, 263A], \(v=|v|e_d\). For \(y\in\mathbb R^{d-1}\) let \(C_y=\{t:(y,t)\in C\}\), a compact interval or empty. A point \((y,t)\) lies in \(C+[0,v]\) exactly when \(t=t'+\theta|v|\) with \(t'\in C_y\) and \(\theta\in[0,1]\), since adding a multiple of \(v\) does not change \(y\). So the fibre of \(C+[0,v]\) over \(y\) is \(C_y+[0,|v|]\), of length \(|C_y|+|v|\) when \(C_y\) is nonempty, that is, when \(y\in\pi_v[C]\), and empty otherwise. Fubini's theorem [Fremlin, 251N and 252B] gives \[ \begin{aligned} &|C+[0,v]|\\ &\quad=\int\bigl(|C_y|+|v|\,\mathbf 1_{\pi_v[C]}(y)\bigr)\,dy\\ &\quad=|C|+|v|\operatorname{vol}_{d-1}(\pi_v[C]).\qquad\square \end{aligned} \] **Proposition 2.2** (OpenAI). For \(d\ge2\), \[ \begin{gathered} |\Pi T_d|=\frac{d+1}{((d-1)!)^d},\\ R_d(T_d)=c_d=\frac{(d+1)\,d^d}{d!} . \end{gathered} \] Every \(d\)-dimensional simplex has the same value \(c_d\). *Proof.* By Lemma 2.1 with \(C=Q\) and \(v=w\), and by Proposition 2.3 of [Projection bodies of polytopes](projection-bodies-of-polytopes.md) with homogeneity, \[ \begin{aligned} |Q+[0,w]|&=1+|w|\operatorname{vol}_{d-1}(\pi_w[Q])\\ &=1+h_{\Pi Q}(w). \end{aligned} \] The cube has two facets with normals \(\pm e_i\) and area \(1\) for each \(i\), so \(h_{\Pi Q}(w)=\frac12\sum_i2|\langle e_i,w\rangle|=d\), and \(|Q+[0,w]|=d+1\). Scaling by \(1/(d-1)!\) multiplies volume by \(((d-1)!)^{-d}\) [Fremlin, 263A], and translation does not change it [Fremlin, 134A]. Then \[ \begin{aligned} R_d(T_d)&=\frac{d+1}{((d-1)!)^d}\,(d!)^{d-1}\\ &=\frac{(d+1)\,d^{d-1}}{(d-1)!}=\frac{(d+1)\,d^d}{d!}. \end{aligned} \] A \(d\)-dimensional simplex, the convex hull of \(d+1\) affinely independent points \(p_0,\dots,p_d\), is \(LT_d+p_0\) with \(L\) the matrix of columns \(p_i-p_0\). Proposition 3.1 of [Projection bodies of polytopes](projection-bodies-of-polytopes.md) gives \(R_d(LT_d+p_0)=R_d(T_d)\). \(\square\) The first values are \(c_2=6\), \(c_3=18\), \(c_4=\frac{160}3\) and \(c_{10}=\frac{17\,187\,500}{567}\approx30\,313.5\). In dimension three the bound \(R_3\le18\) holds for all convex bodies [CFLXX, Theorem 1.2]. ## 3. Exercises **3.1** (easy). Draw \(\Pi T_2\): show that it is a hexagon of area \(3\), and that \(R_2(T_2)=6\). **3.2** (medium). Prove the volume formula for zonotopes: for vectors \(g_1,\dots,g_k\in\mathbb R^d\), \[ \Bigl|\sum_{i=1}^k[0,g_i]\Bigr|=\sum_{S}|\det(g_i)_{i\in S}|, \] the sum over the \(d\)-element subsets \(S\) of \(\{1,\dots,k\}\). Use it to recompute \(|Q+[0,w]|=d+1\). **3.3** (easy). Show that \(c_d<(d+1)e^d\). ## 4. Solutions **3.1.** \(\Pi T_2=[-\frac{e_1}2,\frac{e_1}2]+[-\frac{e_2}2,\frac{e_2}2]+[-\frac w2,\frac w2]\), the hexagon with vertices \(\pm(1,1)\), \(\pm(1,0)\) and \(\pm(0,1)\); as a translate of \([0,1]^2+[0,(1,1)]\), its area is \(3\) by Proposition 2.2. With \(|T_2|=\frac12\), \(R_2(T_2)=3/\frac12=6\). **3.2.** Induction on \(d\), and for fixed \(d\) on \(k\). For \(d=1\) both sides are \(\sum_i|g_i|\). For \(d\ge2\) and \(k