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Assertion: With any simplex $P$ with vertices $V_{0..N+1} \subset R^N$, it is possible to compute the sign of the oriented volume using the expression $det(V_1-V_0, V_2-V_0, ..., V_{N-1}-V_0, V_{N}-V_0)$, where the vector differences are represented as a list of column vectors.

Further, it is possible to permute the indices of the vertices, so as to "rotate" them $k$ times so that V'_i := V_{((i + k) \mod (N+1))}, and the determinant stays the same.

However, this does not seem to hold with simplices in $R^1$ (where the simplex is a line segment) and it prevents some simple tests from working in the general case. What am I leaving out, or how is my assertion incorrect? My gut is telling me there's something special about the first dimension that's missing from my rotation operation.

Can anyone explain this special case?

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    Thanks Robert. You are correct, every simplex is convex, so I should reword the question to be more concerned with the indices of the vertices, and their effect on the determinant.2011-04-29

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It is simply wrong that a "rotation" of the vertices does not change the orientation (i.e. sign of the determinant).

A "rotation" is a cyclic permutation which is odd if and only if the number of permuted elements is even. So, in the corrected version, the rotation will preserve orientation for even $N$ and change orientation for odd $N$.

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    Thanks @user9325, this is exactly what I was unable to see.2011-04-29
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I think one of the main problems is the language you are using. If the $V_i$ are the vertices of an $N$-simplex in $\mathbb{R}^N$, then \begin{equation} \det (V_1-V_0,\ldots, V_N-V_0) \end{equation} is equal to $N!$ times the oriented volume of the simplex. This is discussed on the Wikipedia page. You are using "convex" when you mean positively oriented I think. (I think you also say "rotate the vertices" when you mean permute the vertices.)

I do not see the problem you are having in $\mathbb{R}^1$, but that may be because you have edited the question in response to comments without realizing that it fixed the problem. A 1-simplex has two vertices $V_0$, and $V_1$. In this case $\det(V_1-V_0) = V_1-V_0$ since it is a $1\times 1$ matrix. It is the length of your segment, and the orientation is positive if $V_1 \gt V_0$.

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    Thanks! This answer is on the right track, and I will correct the question so that the language is more obvious. I suppose I wasn't really asking about convexity, as you point out, I did in fact mean "positively oriented." The trick that @user9325 pointed out was the difference between "rotation" permutations in odd or even dimensions.2011-04-29