# Signed volume of $n$-simplex

Given an ordered sequence of vertices for an $n$-simplex, $(v_1, v_2, \cdots, v_{n+1})$, embedded in $\mathbb{R}^{m}$ for $m \geq n$, is there any compact way to compute the signed volume?

I am aware of using the following formula, based on the Cayley-Menger Determinant, for the absolute volume $V_n$ of an $n$-simplex:

\begin{align} V_n &= \sqrt{(-1)^{n+1} \frac{\det{(B)}}{2^n (n!)^2}}\\ \text{where}& \\ B &= \begin{pmatrix} 0 & \boldsymbol{1}^T \\ \boldsymbol{1} & \beta \end{pmatrix}\\ \beta_{ik} &= \lVert v_i - v_k\rVert_2^2 \end{align}

Is there anything similar to this for the signed volume case?

• For $m>n$ you don't really have a "signed" volume. Jan 9, 2018 at 6:51
• @LordSharktheUnknown I'm thinking of it in terms of, say, the example of a triangle with vertices in $\mathbb{R}^3$. While the vertices are in $\mathbb{R}^3$, you could technically project them onto a plane and compute the signed area of the triangle in that coordinate frame. So in that sense, I am after the "signed volume" of an $n$-simplex. Does that make sense? I'm hoping I don't have to explicitly compute hyperplanes for some $n$-simplex. Jan 9, 2018 at 7:01
• I think you may be seeking the concept of exterior products. Jan 9, 2018 at 7:08
• @LordSharktheUnknown I'll look into it and see if that fits what I'm after, thanks. Jan 9, 2018 at 7:17
• @LordSharktheUnknown Exterior products definitely look like something I should look into long term, though it will take some learning on my end. That said, I ended up just projecting the vertices from $\mathbb{R}^m$ to $\mathbb{R}^n$, for $m \gt n$, using modified Gram-Schmidt and then computing the signed volume as $V^{(s)}_{n} = \frac{1}{n!}\det(\hat{v}_2 - \hat{v}_1, \cdots, \hat{v}_{n+1} - \hat{v}_1)$, where $\hat{v}_k$ is the projected version of $v_k$. Since I am dealing with $n \leq 4$ for my research, this is not bad to compute. Jan 9, 2018 at 16:34