Talk:Simplex

in the plane or under it?
The article now says:


 * The standard n-simplex is the subset of Rn+1 given by
 * $$\Delta^n = \left\{(t_0,\cdots,t_n)\in\mathbb{R}^{n+1}\mid\Sigma_{i}{t_i} \leq 1 \mbox{ and } t_i \ge 0 \mbox{ for all } i\right\}$$

Shouldn't $$\Sigma_{i}{t_i} \leq 1$$ be $$\Sigma_{i}{t_i} = 1$$? —Tamfang (talk) 17:59, 28 June 2009 (UTC)


 * No, I think it's correct as written. This definition uses $$n$$ variables to define a set in $$\mathbb{R}^{n+1}$$, and that's why the condition is $$\Sigma_{i=1}^n {t_i} \leq 1$$.  If you wanted, you could define an extra variable $$t_{n+1} = 1 - \Sigma_{i=1}^n {t_i}\,$$, and then the condition would be $$\Sigma_{i=1}^{n+1}{t_i} = 1$$.  I don't think this is necessary, though, since the simplex is only defined by $$n$$ independent variables. --Apocralyptic (talk) 19:13, 15 July 2009 (UTC)


 * Are we looking at the same thing? As I read it, the definition uses $$n+1$$ variables ($$t_0,\cdots,t_n$$), not n variables, to define a set in $$\mathbb{R}^{n+1}$$.  If the intention is to define something in an n-dimensional subspace, it ought to be more like a surface than a volume, thus somewhere there ought to be a strict equality, not only a series of inequalities.  Take n=2: the definition given is equivalent to { (x,y,z) | x+y+z ≤ 1 and x,y,z ≥ 0 }, which defines a (right) tetrahedron, not a triangle. —Tamfang (talk) 05:16, 16 July 2009 (UTC)


 * I agree with you now... I think I was confused by the variables starting with t0 instead of t1. What I said before made no sense.  The correct definition should be
 * $$\Delta^n = \left\{(t_1,\cdots,t_{n+1})\in\mathbb{R}^{n+1}\mid\Sigma_{i=1}^{n+1}{t_i} = 1 \mbox{ and } t_i \ge 0 \mbox{ for all } i\right\}$$. --Apocralyptic (talk) 14:40, 16 July 2009 (UTC)


 * Now, if in the above definition you lower all subscripts by 1, you'll be onto something.Daqu (talk) 01:01, 7 July 2010 (UTC)


 * It amazes me a bit that I've rarely seen an edit war over that choice of convention. —Tamfang (talk) 20:12, 25 October 2019 (UTC)

Simplices in functional analysis
I think that a reference to simplices in functional analysis is missing here. Terminologies such as "Choquet simplex" or "Bauer simplex" are often met. As far as I understand, they are infinite-dimensional extensions of the notion of simplex discussed in that article. — Preceding unsigned comment added by Gamesou (talk • contribs) 14:52, 6 December 2019 (UTC)


 * Which article? I don't see any reference to Choquet simplex or Bauer simplex in the functional analysis article.—Anita5192 (talk) 17:22, 6 December 2019 (UTC)

Normal vectors and dihedral angles
It appears that the normal vectors in the dihedral angles sections are wrong: https://en.wikipedia.org/wiki/Simplex#Dihedral_angles_of_the_regular_n-simplex

For example, in 2d you would get (0, 1) and (1, 0), which can not be the correct normal vectors of an equilateral triangle. Further, calculating the angle between two such vectors leads to arccos(1/(n-1)), which is not correct.

Both of those issues appear if the normals are instead given by (-n, 1, ..., 1) and permutations, I think that's the correct answer. — Preceding unsigned comment added by 2001:16B8:A0C4:1800:7500:BF95:344F:B5AC (talk) 09:01, 13 October 2020 (UTC)


 * I did the calculations and got the same result as you, so I changed it.—Anita5192 (talk) 21:19, 13 October 2020 (UTC)

Incoherent sentence
The section Geometric properties contains this incoherent sentence:

"If P is a general parallelotope, the same assertions hold except that it is no longer true, in dimension > 2, that the simplexes need to be pairwise congruent; yet their volumes remain equal, because the n-parallelotope is the image of the unit n-hypercube by the linear isomorphism that sends the canonical basis of $$\mathbf R^n$$ to $$e_1,\ldots, e_n$$."

This is way too confusing.

Much better: State exactly what does hold true, instead of trying to use a previous paragraph with modifications.01:51, 10 November 2020 (UTC)

Questionable "Applications" entry
I removed this bullet that reads like original research, but I got reverted twice by someone who disagrees. The part about methane sounds particularly problematic. –LaundryPizza03 ( d c̄ ) 06:16, 5 January 2021 (UTC)
 * In chemistry, the hydrides of most elements in the p-block can resemble a simplex if one is to connect each atom. Neon does not react with hydrogen and as such is a point, fluorine bonds with one hydrogen atom and forms a line segment, oxygen bonds with two hydrogen atoms in a bent fashion resembling a triangle, nitrogen reacts to form a tetrahedron, and carbon forms a structure resembling a Schlegel diagram of the 5-cell. This trend continues for the heavier analogues of each element, as well as if the hydrogen atom is replaced by a halogen atom.


 * This is definitely not original research. The well-known molecular geometry in this example can be found in most any first-year college chemistry textbook and also at Molecular geometry.  See Molecular geometry.  In particular, the geometry of methane, CH4, is shown in that table, and can be compared to the Schlegel diagram of the 5-simplex in File:5-simplex verf.png.—Anita5192 (talk) 06:59, 5 January 2021 (UTC)

Volume formula in more than n dimensions
There is a question about the volume formula that is contention. I am hoping that someone has a textbook handy that shows this formula and thus we can add a citation to the article and keep the formula. Specifically, until a few days ago the article has indicated, correctly IMHO, that

\mathrm{Volume} = {1\over n!} \left(\det \left[ \begin{pmatrix} v_0^T & 1 \\ v_1^T & 1 \\ \vdots & \vdots \\ v_n^T & 1 \end{pmatrix} \begin{pmatrix} v_0 & v_1 & \cdots & v_n \\ 1  & 1   & \cdots & 1 \end{pmatrix} \right]\right)^{1/2}\,, $$ and, in particular, that this expression works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions. Because has questioned whether this formula applies in a Euclidean space with more than n dimensions, I bring the topic to this discussion. — Q uantling (talk &#124; contribs) 20:02, 20 February 2022 (UTC)


 * Especially if this formula is includable under WP:CALC then the question of the truth of the statement is highly relevant. In particular, we say that $L: Rn → Rm$ is a mapping that preserves distances and angles (and hence preserves volumes) if $u &middot; v = Lu &middot; Lv$ for every $u$ and $v$.  Another way of saying that is that $uTv = uTLTLv$.  Another way of saying that is $LTL = I$, the identity matrix.  The above formula includes $MTM$ and the question is whether $(LM)T(LM)$ gives the same value.  The answer is yes, because that expression equals $MTLTLM$ and we can use $LTL = I$ to immediately get $MTLTLM = MTM$.


 * Or putting it another way,

M^T M = \begin{pmatrix} v_0^T & 1 \\ v_1^T & 1 \\ \vdots & \vdots \\ v_n^T & 1 \end{pmatrix} \begin{pmatrix} v_0 & v_1 & \cdots & v_n \\ 1  & 1   & \cdots & 1 \end{pmatrix} = \begin{pmatrix} v_0 \cdot v_0 + 1 & v_0 \cdot v_1 + 1 & \cdots & v_0 \cdot v_n + 1 \\ v_1 \cdot v_0 + 1 & v_1 \cdot v_1 + 1 & \cdots & v_1 \cdot v_n + 1 \\ \vdots & \vdots & \ddots & \vdots \\ v_n \cdot v_0 + 1 & v_n \cdot v_1 + 1 & \cdots & v_n \cdot v_n + 1 \end{pmatrix} $$
 * The right hand side depends only upon the dot products, which means that it depends only upon the vector lengths and the angles between the vectors. In particular, it does not depend upon the number of dimensions.  — Q uantling (talk &#124; contribs) 20:26, 20 February 2022 (UTC)


 * If there is a counterexample where the formula $(1/n!) det1/2 MTM$ fails to compute the correct volume then that would help me to understand the objection to the formula. — Q uantling (talk &#124; contribs) 20:34, 20 February 2022 (UTC)


 * Dot products between vectors from the origin to the vertices are not translation-invariant; only dot products between vectors from vertices to other vertices are. It’s quite easy to see that the formula is wrong: for example, it gives an area of ½√(z² + 1) for the 3D triangle with vertices (0, 0, z), (1, 0, z), and (0, 1, z), which should be ½ independent of z. —Anders Kaseorg (talk) 22:31, 20 February 2022 (UTC)


 * Thank you — Q uantling (talk &#124; contribs) 03:22, 21 February 2022 (UTC)

Another volume formula
et al.: Is it reasonable to add the following volume formula? The expression $$\mathrm{Volume} = \frac{1}{n!} \sqrt{ \det \left[ \begin{pmatrix} v_0^T & 1 \\ \vdots & \vdots \\ v_n^T & 1 \end{pmatrix} \begin{pmatrix} v_0 & \ldots & v_n \\ 1 & \ldots & 1 \end{pmatrix} \right] - \det \left[ \begin{pmatrix} v_0^T \\ \vdots \\ v_n^T \end{pmatrix} \begin{pmatrix} v_0 & \ldots & v_n \end{pmatrix} \right]}\,,$$ is symmetric in the vertices and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions.

Alternatively, Let
 * $$M = \begin{pmatrix}

v_0^T \\ \vdots \\ v_n^T \end{pmatrix} \begin{pmatrix} v_0 & \ldots & v_n \end{pmatrix} = \begin{pmatrix} v_0 \cdot v_0 & v_0 \cdot v_1 & \cdots & v_0 \cdot v_n \\ v_1 \cdot v_0 & v_1 \cdot v_1 & \cdots & v_1 \cdot v_n \\ \vdots & \vdots & \ddots & \vdots \\ v_n \cdot v_0 & v_n \cdot v_1 & \cdots & v_n \cdot v_n \end{pmatrix}, \qquad U = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}\,. $$ Then $$\mathrm{Volume} = \frac{1}{n!} \sqrt{\det(M+U) - \det(M)}$$ is symmetric in the vertices and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions. Or mix and match somehow. Thank you — Q uantling (talk &#124; contribs) 14:59, 21 February 2022 (UTC)


 * That’s cute but probably WP:OR. Another cute formula: $\mathrm{Volume} = \tfrac1{n!} \sqrt{\sum_{i,j}(\operatorname{adj} M)_{ij}}$ . —Anders Kaseorg (talk) 20:13, 21 February 2022 (UTC)


 * I believe that it is the square root of the sum of the (un-squared) elements of $adj M$ rather than the Frobenius norm that goes in that formula. If by original research you mean that no one has ever discovered and published this before now, ... I very much doubt that.  But if you mean that my library isn't big enough that I could find the citation myself ... then I agree!  I will let it sit here for a while and hope that someone has a better familiarity with the literature than I.  Thank you — Q uantling (talk &#124; contribs) 22:22, 21 February 2022 (UTC)


 * Right about that formula, corrected above. —Anders Kaseorg (talk) 00:38, 22 February 2022 (UTC)


 * If you know of a citation that supports the notability of $\mathrm{Volume} = \tfrac1{n!} \sqrt{\sum_{i,j}(\operatorname{adj} M)_{ij}}$, I'd want to add that formula to the article. — Q uantling (talk &#124; contribs) 02:51, 23 February 2022 (UTC)
 * Only a little off topic ... do you know of an algorithm that computes $\sum_{i,j} (\mathrm{adj~} M)_{ij}$, or any of these other formulations for volume, asymptotically more quickly than the running time of a matrix inversion? — Q uantling (talk &#124; contribs) 15:46, 9 May 2024 (UTC)

Signs

 * $$\mp$$ should indeed be changed to $$\pm$$. Just look at the case where $$n=1$$, for example. 2601:547:500:E930:944C:F88D:FBB3:E92F (talk) 02:30, 23 March 2022 (UTC)


 * I agree with you now. Thank you for being persistent.  — Q uantling (talk &#124; contribs) 01:22, 24 March 2022 (UTC)

Dimensionality of space containing the simplex
I am moving a discussion with to this talk page. The Simplex article is a little inconsistent about the dimensionality of the space containing a simplex. Specifically, if a $k$-simplex is non-degenerate, we'll need at least $k$ dimensions for placing its $k+1$ vertices; such as the 3 vertices of a triangle (2-simplex) fit well in a 2-dimensional plane. But one of our main examples is placing a $k$-simplex into $k+1$ dimensions; such as a triangle with vertices at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. So, how should we modify the article to clarify this seeming inconsistency? As far as I am concerned, anyone with ideas should go ahead and make edits. If we find that there is contention over those edits we can then continue the discussion here. Thanks! — Q uantling (talk &#124; contribs) 16:46, 6 February 2023 (UTC)


 * Ha I edited before I saw this comment, so I'm glad you endorsed that :). In the lead section, I don't see any reason to specify the ambient space in general -- it is equally true that three affinely independent points span a triangle in $$\mathbb{R}^{17}$$ as in $$\mathbb{R}^{2}$$ and $$\mathbb{R}^{3}$$.  Does this inconsistency extend to the body, as well?  --JBL (talk) 20:16, 6 February 2023 (UTC)
 * I think this makes sense, the definitions of linear independence and affine independence do not specify dimensions. Dukeleimao (talk) 16:35, 7 February 2023 (UTC)

The article's history section states that Clifford "wrote about these shapes in 1886" but that was seven years after his death. Perhaps 1886 was when that writing was published but I haven't been able to verify that reference. In this reference about the earliest known uses of some mathematical terms, it states that Clifford wrote about "prime confines" in the January 1886 issue of Educational Times, but yesterday I spent a couple hours scanning through this archive that includes the January 1886 issue of Educational Times and (although it was interesting) I found no reference to Clifford writing about prime confines in that January 1886 issue. My best guess is that the 1886 reference year is a typo and it was never checked by any of the article's editors. Gj7 (talk) 16:48, 2 June 2023 (UTC)

Notation error in volume formula?
The volume is shown as |det...| but shouldn't that be either det[...] or |...| for the short form of the determinant? I didn't know the formula, so I don't want to change it myself, but it looks like a notational error. It is confusing (to me), however. — Preceding unsigned comment added by 2A02:58:157:9B00:70B:DBE4:BC88:CA19 (talk) 10:41, 28 October 2023 (UTC)


 * In the first formula, the determinant could come out negative. It is the absolute value of that possibly negative value that is the actual volume, which is what the $|det(...)|$ is indicating.  — Q uantling (talk &#124; contribs) 14:44, 28 October 2023 (UTC)