# The volume of an $n$-dimensional unit hypercube intersected with the plane $\sum_{i=1}^n x_i=0$.

Let $$\mathcal{C}$$ be the unit cube in $$\mathbb{R}^n$$ centred at $$\mathbf{0}$$, for some integer $$n>1$$. Consider the following $$(n-1)$$-dimensional subspace:

$$V=\left\{(x_1,x_2,\dots,x_n) \in \mathbb{R}^n: \sum_{i=1}^n x_i=0\right\}.$$

I want to calculate the volume of the polytope $$\text{Vol}(\mathcal{C} \cap V)$$, where by volume we mean the $$(n-1)$$-dimensional volume as measured in the space $$V$$ (that is, the volume of the $$(n-1)$$-dimensional polytope projected onto the plane $$V$$).

So far, I have determined that such a shape must have vertices whose coordinates are either $$0$$, $$-1/2$$ or $$1/2$$ and such that the coordinates sum to zero. However, beyond this I am uncertain how to calculate the volume of such a shape, and was wondering if anyone had any idea of how to go about solving this problem, or whether it is possible to determine the exact volume of this shape at all for general $$n$$.

• The intersection is a convex hull of $\binom{n}{n/2}$ points of distance $\sqrt n/2$ from the origin when $n$ even and $n\binom{n-1}{(n-1)/2}$ points with distance $\sqrt{n-1}/2$ from the origin when $n$ is odd. Not sure how that helps. Commented Jan 4, 2023 at 19:12
• In 1D, $C\cap V$ is the origin and has "0D-volume" 1. In 2D, it is a diagonal of a square and has length $\sqrt2$. In 3d, it is a "diagonal" plane passing through four vertices (in particular, it is the plane orthogonal to the vector $(1,1,1)$) and has area $\sqrt2$ since it is a rectangle with one side an edge of the cube and one side the diagonal of a face. Notice how the 3D case reduces to the 2D case on a face of the cube. You can probably prove by induction that the answer is either always $\sqrt2$ with the 1D case exceptional, or that it's something like $(\sqrt2)^{\lceil(n-1)/2\rceil}$. Commented Jan 4, 2023 at 19:20
• If I didn't make any mistake, it is $\frac{\sqrt{n}}{(n-1)!}\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^k \binom{n}{k}\left(\frac{n}{2}-k\right)^{n-1}$. The basic idea is compute the hypervolume of the intersection of the unit cube $[-\frac12,\frac12]^n$ with the halfspace $\sum_i x_i < t$ and taking derivative of $t$ to get the hyperarea. Look at my answer for a similar problem in 3d for the general idea.... Commented Jan 4, 2023 at 19:28
• A simple formula for $n=2,3,4$ is $\frac{n\sqrt{n}}{2(n-1)}$... Commented Jan 4, 2023 at 20:23
• Thank you all for your helpful answers! @NicholasTodoroff, unfortunately I don't think that your answer is correct - take for example the 3-dimensional case. Then the unit cube actually traces out a 2D regular hexagon on the plane's surface, as you can see in this lovely geogebra graphic: imgur.com/a/ulhAv9S The hexagon has side-length equal to $1/\sqrt{2}$, and so the area of the shape generated is $\frac{3}{4}\sqrt{3}$, unless I am mistaken. Commented Jan 4, 2023 at 21:01

To simplify derivation, we will translate

• unit hypercube $$\mathcal{C}$$ to $$\mathcal{C}' = [0,1]^n$$.
• plane $$V$$ to $$V' = \left\{ (x_1,\ldots,x_n) \in \mathbb{R}^n : \sum\limits_{i=1}^n x_i = \frac{n}{2} \right\}$$.

Before we start, let's adopt some conventions.

• For any $$p \in \mathbb{R}^n$$, we will use $$p_1,\ldots,p_n$$ to denote its components. ie. $$p = (p_1,p_2,\ldots,p_n)$$.
• For any $$p, q \in \mathbb{R}^n$$, we will use $$p \ge q$$ as a shorthand for the relation, $$p_i \ge q_i$$ for all $$i = 1,\ldots, n$$.
• For any $$p\in \mathbb{R}^n$$, let $$C_p$$ be the cone $$\{ x \in \mathbb{R}^n : x \ge p \}$$.
• Let $$\mathcal{V}$$ be vertices of $$\mathcal{C}'$$ and for $$v \in \mathcal{V}$$, we will use $$(-1)^v$$ be a shorthand of $$(-1)^{\sum_{i=1}^n v_i}$$.
• For any $$U \subset \mathbb{R}^n$$, we will use $$\mathbb{1}_U$$ to denote its indicator function.
• For any expression $$(\cdots)$$, we will use $$(\cdots)_{+}$$ to denote its positive part, ie. $$\max((\cdots),0)$$.

Let $$m \in \mathbb{R}^n$$ be an unit vector with all components positive. For any $$t \in \mathbb{R}$$, let

• $$H_t$$ be the half-space $$\{ x \in \mathbb{R}^n : m \cdot x \le t \}$$.
• $$P_t$$ be the hyperplane $$\{ x \in \mathbb{R}^n : m \cdot x = t \}$$.

Let $$\mathcal{S}$$ be the collection of geometric shapes $$\mathbb{R}^n$$ that can be formed by finite intersection of open/closed half-spaces. For any $$S \in \mathcal{S}$$, let $$\mu_{n}(S)$$ and $$\mu_{n-1}(S)$$ be the $$n$$ and $$n-1$$ dimensional measure of $$S$$ whenever it make sense.

When $$S$$ is bounded, aside from finitely many $$t$$ where $$P_t$$ contains a facet of closure of $$S$$, the hyper area of its intersection with $$P_t$$ is related to the hyper volume of its intersection with $$H_t$$ by the relation:

$$\mu_{n-1}(S \cap P_t) = \frac{d}{dt}\mu_n(S \cap H_t)$$

Let $$\mathcal{C}'' = [0,1)^n$$, notice its indicator function is related to those of $$C_v$$ be the relation.

$$\mathbb{1}_{\mathcal{C}''} = \sum_{v \in \mathcal{V}} (-1)^v \mathbb{1}_{C_v}$$

We find \begin{align} \mu_n(\mathcal{C}' \cap H_t) =\mu_n(\mathcal{C}'' \cap H_t) &= \sum_{v \in \mathcal{V}} (-1)^v \mu_n( C_v \cap H_t )\\ &= \sum_{v\in \mathcal{V}} (-1)^v \mu_n( C_0 \cap H_{t-m\cdot v} )\\ &= \frac{1}{n!\prod_{i=1}^n m_i}\sum_{v\in \mathcal{V}} (-1)^v (t - m\cdot v)_{+}^n \end{align} This leads to $$\mu_{n-1}(\mathcal{C}' \cap P_t) = \frac{d}{dt} \mu_n(\mathcal{C}' \cap H_t) = \frac{1}{(n-1)!\prod_{i=1}^n m_i}\sum_{v\in \mathcal{V}}(-1)^k (t - m\cdot v)_{+}^{n-1}$$ For the problem at hand, we can take all $$m_i = \frac1{\sqrt{n}}$$ and $$P_{\frac{\sqrt{n}}{2}}$$ coincides with hyperplane $$V'$$. This means,

$${\rm Vol}(\mathcal{C} \cap V) = {\rm Vol}(\mathcal{C}' \cap V') = \mu_{n-1}(\mathcal{C}' \cap P_{\frac{\sqrt{n}}{2}} )$$

Notice $$m\cdot v = \frac{1}{\sqrt{n}}\sum\limits_{i=1}^n v_i$$ and for $$k = 0,\ldots, n$$, there are $$\binom{n}{k}$$ vertices with $$\sum\limits_{i=1}^n v_i = k$$. At the end, we find:

\begin{align} {\rm Vol}(\mathcal{C},V) &= \frac{(\sqrt{n})^n}{(n-1)!}\sum_{k=0}^n (-1)^k \binom{n}{k}\left(\frac{\frac{n}{2} - k}{\sqrt{n}}\right)_{+}^{n-1}\\ &= \frac{\sqrt{n}}{(n-1)!}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} (-1)^k \binom{n}{k}\left(\frac{n}{2} - k\right)^{n-1}\tag{*1} \end{align}

As a double check, let's look at the special case $$n = 3$$.

$$V$$ will be a regular hexagon will side $$\frac{1}{\sqrt{2}}$$. So its area will be $$\frac{3\sqrt{3}}{2}\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{3\sqrt{3}}{4}$$.

For comparison, the formula $$(*1)$$ reproduces $$\frac{\sqrt{3}}{2!}\left(\left(\frac32\right)^2 - 3\left(\frac12\right)^2\right) = \frac{3\sqrt{3}}{4}$$ as expected.

• This is absolutely beautiful! Thank you so much for this answer. In fact, this problem constitutes a part of a lemma in a paper that I'm currently writing up, so can I credit you in some way ? Maybe add you in the acknowledgements? (Of course I'll cite the post). Commented Jan 5, 2023 at 11:49
• @Chris there is no need to credit me explicitly, citing a post is enough. Commented Jan 5, 2023 at 14:09
• Can this language be translated to an integral? Commented Jan 5, 2023 at 16:57

This can be written as a graph $$x_n= -\sum_1^{n-1} x_i$$ over a subset $$D$$ of the corresponding $$n-1$$ dimensional hypercube, $$D$$ to be determined. If you happen to know that the area of the graph of a function $$f: D\rightarrow \mathbb{R}$$ can be calculated as $$A(f, D) = \int_D \sqrt{1+ ||\nabla f||^2}d\bar x$$ (with the notation $$\bar x=(x_1,\ldots, x_{n-1})\in D \subset \mathbb{R}^{n-1}$$) you should be able to calculate the area of the surface in question once you figure out to which subset of the $$n-1$$ dimensional cube the hyperplane projects.

• Hi Thomas, thanks for your answer. Maybe I have got this wrong, but when I try to use this method using $f(x_1,x_2,\dots,x_{n-1})=-\sum_{i=1}^{n-1}x_i$, this gives me $\sqrt{1+\|\nabla f\|^2}=n$ and so letting $D$ be the unit cube, this gives $A=\sqrt{n}$, which would be incorrect when $n=3$. But, I assume I must be misunderstanding your method... Commented Jan 5, 2023 at 0:03
• @Chris the misunderstanding (and an error in the first version of my answer) is that $D$ is assumed to be the unit cube. If you check (e.g. in three space) at a corner (e.g. $x_1 = x_2 = - \frac{1}{2}$) you will see that $x_3 = -x_1 - x-2 = 1$ is way outside of the bounding cube in $\mathbb{R}^3$. I will leave a corrected but, at this time, incomplete answer undeleted, just in case anyone wants to see the (correct) formula for the surface area. Commented Jan 5, 2023 at 16:42
• I see- thanks for the information @Thomas, my bad! Commented Jan 5, 2023 at 16:53