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I think I might be misunderstanding the concept of a simplex and its volume.

Take the 2-dimensional simplex (a triangle) embedded in 3-dimensional space with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. When I calculate its area, I get $\frac{\sqrt{3}}{2}$. However, I've come across information suggesting that the volume (area in this case) of this probability simplex should be $\frac{1}{2}$.

Similarly, the 3-dimensional tetrahedron (a 3-dimensional simplex) embedded in 4-dimensional space with vertices $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, and $(0,0,0,1)$. When I calculate its volume, I obtain $\frac{1}{3}$, but I've read that the volume of this 4-dimensional probability simplex is $\frac{1}{6}$.

Can someone explain the meaning of this discrepancy in the volume?

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  • $\begingroup$ Maybe you're thinking of $\{(x, y) \,:\, 0\leq x \leq y \leq 1\}$, which has volume 1/2, and $\{(x,y,z)\in\mathbb{R}^3\,:\,0\leq x \leq y \leq z \leq 1\}$, which has volume 1/6. $\endgroup$
    – Dennis
    Commented Oct 17, 2023 at 15:43

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The area of the 2-simplex is indeed $\sqrt{3}/2$. There's a different 2-simplex, with vertices $(0,0), (1, 0), (0, 1)$ in the plane whose area is $1/2$; the corresponding 3-simplex in 3-space (with vertices at the origin and at all points with exactly one non-zero coordinate, which is 1, i.e. $(1,0,0), (0,1,0), (0,0,1)$) has volume $1/6$. In dimension $n$, that simplex has volume $1/n!$.

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