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A scalar triple product identity

39 pointsby luualmost 2 years ago

7 comments

jiggawattsalmost 2 years ago
There&#x27;s a fascinating history of how out of the four founding fathers of modern multi-dimensional mathematics, three preferred geometric algebra, and only one preferred vector and matrix algebra. The three all left academia or died young, the one dissenter lived to a ripe old age and was a vocal proponent of his preferred approach -- because it&#x27;s more useful for statistics. He wasn&#x27;t a physicist.<p>All of these weird &quot;triple products&quot;, limitations to 3D space, etc... <i>vanish</i> in geometric algebra. You can use the exact same formulas in 2D space, 3D space, 4D spacetime, or 18 dimensional spaces with degenerate dimensions if you please.<p>There&#x27;s this obstinate refusal to just <i>admit</i> that the maths that&#x27;s ideally suited to solving statistical problems may not be ideal for physics, robotics, or optics. Instead, physicists insist on re-inventing the good stuff over and over, badly, with different names, each time.<p>All of the following are just geometric algebra in disguise, or various &quot;subsets&quot; of a geometric algebra, or a geometric algebra operation that got renamed:<p><pre><code> - Cross products - Triple product - Exterior algebra - Complex numbers - Quaternions - Octonions - Pauli matrices - Gamma matrices </code></pre> And on, and on, and on.<p>As a random example: the way we represent 3D rotations using a vector in 3D doesn&#x27;t work in 2D, because a 2D rotation vector would <i>point out of the plane</i>. It also doesn&#x27;t work in 4D or higher dimensions. It just happens to &quot;work&quot; in 3D not because 3D is natural, preferred, or special, but because the broken maths happens to not be completely broken in <i>one case</i> due to a simple coincidence.<p>In geometric algebra, instead of using a vector, a <i>bivector</i> is used, which is like a surface patch. Notice that you can have a surface in 2D, so rotations in GA work in 2D. You can also have a surface in 3D, so rotations work in 3D. And in 4D (including both 4D space and 3+1D space-time!), and 5D... and <i>all of rest</i>, with the same formula!<p>PS: Geometric algebra also prevents gimbal lock, doesn&#x27;t store redundant values, and has better numerical precision. Its transformations can be interpolated unlike (famously!) matrix transformations, which can&#x27;t. In robotics or computer graphics this means everyone uses quaternions instead, which are... <i>drumroll</i>... the &quot;even subset&quot; of a geometric algebra.
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adrian_balmost 2 years ago
This so-called &quot;scalar&quot; triple product is not scalar.<p>It is a pseudo-scalar, which means that its value depends on the system of coordinates that is used. More precisely, its sign changes between right and left systems of coordinates.<p>In physics it is very important to understand the differences between the different kinds of quantities, e.g. scalars, vectors a.k.a. polar vectors, pseudo-vectors a.k.a. axial vectors, pseudo-scalars, symmetric tensors and so on, otherwise it is easy to make mistakes.
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ultrafilteralmost 2 years ago
Near the end, the author claims that if a,b,c form a basis of R3, then the cross products a x b, b x c, and c x a are orthogonal.<p>This is false. E.g., a=(1,1,0), b=(0,1,1), c=(1,0,1) implies<p>a x b = (+1,-1,+1), b x c = (+1,+1,-1), c x a = (-1,+1,+1).<p>However, this &quot;dual basis&quot; still serves its claimed purpose, which essentially is to precompute part of Cramer&#x27;s rule.
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mijoharasalmost 2 years ago
Alright, I got nerd sniped halfway through when he said:<p><pre><code> Scalar triple products only exist in 3D, because they involve cross products which only exist in 3D (please keep the “they exist in 7D too” comments for some other time). </code></pre> There isn&#x27;t a reference, and me googling &quot;higher dimensional cross product&quot; seems to return results generalising to any number of dimensions.<p>Can anyone explain what he&#x27;s talking about or provide references? Off the top of my head I can&#x27;t see why you can&#x27;t generalise cross products to D &gt; 3. Or if you can&#x27;t I don&#x27;t see what&#x27;s special about dimensions 3 and 7...
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amlutoalmost 2 years ago
Somehow, these topics often seem to devolve into discussions of how cross products are evil and (insert alternative) is better or into math full of jargon that’s incomprehensible without studying the math in question. [0]<p>I’ll try to say something intelligent about this identity. This comes from a class I took on different geometry, but I’ll leave out the differential geometry :)<p>First, u, v, w, etc are n-dimensional vectors [1]. n = 3 here, and this is important later.<p>The object [u v w] is a function of three vectors. You feed it three vectors and it spits out a number. It has two interesting properties:<p>1. It’s multilinear. This means that [u+v w x] = [u w x] + [v w x] and [a·u w x] = a·[u w x]. The same holds for the other two parameter. This just means that, if you fix all but one parameter, you’re left with a linear function of the parameter you didn’t fix.<p>2. It’s totally antisymmetric. If you swap any two parameters, you negate the result. So [u v w] = -[u w v] = [v w u]. Play with this — it’s fun. With three parameters, the cyclic permutations are positive. With any number of parameters, the <i>even</i> permissions are positive.<p>These properties are pretty easy to prove from the definition of the triple product.<p>Now on to the messy function in the article:<p>f(u,v,w,x) = u[v w x] – v[w x u] + w[x u v] – x[u v w]<p>f is multilinear: each term is fairly trivially multilinear, and the sum of multilinear functions is multilinear.<p>f is antisymmetric. Try it — swapping any two arguments negates it! [2]<p>Now for the fun: In n dimensions, if you have a k-parameter antisymmetric multilinear function, then, if k&gt;n, your function is always zero!<p>Lots of other cases are interesting, too:<p>k=1: these are just linear functions of vectors. One might call them “dual vectors” or “covectors” or whatever. If you are a bit sloppy, you can thing of them as vectors, and they are vectors, but they are not the same type of vector as the type of their input. If you <i>cough</i> erase that type difference, you end up in the rabbit hole that leads to cross products making sense in 3D but not otherwise.<p>k=n: This is a “volume element”. In high school calculus, x·dx or f(x,y)·dx·dy are things that live inside an integral and you aren’t supposed to do too much in the way of removing the integral sign. In 1D, x·dx is a function mapping x (a <i>point in space</i>) to dx, a volume element. If you have a mapping from points to volume elements, you can integrate it and get a scalar! In Euclidean space, you can get away with integrating something like a volume element (e.g. dx·dy) and ignoring the point part. In non-Euclidean space, a vector here and a vector there are different things — a vector (say, the direction and ant is crawling if it’s at a certain spot on a balloon) is <i>not a vector elsewhere* — an ant crawling that direction elsewhere on the balloon whirl be crawling off the surface!<p>k=0: scalar. It’s a scalar-valued function of zero parameters. That would be a scalar.<p>Other goodies from high school geometry and calculus are hiding in here, too, with appropriate k and n.<p>[0] For example, “Algebra” means something to most people who went to high school. It means something rather different in fancy math. I don’t know the history of how this came to be.<p>[1] If you are working with a non-Euclidian manifold, then they’re all vectors originating at the same point. Yes, this all generalizes to any differentiable manifold.<p>[2] If you feel fancy, the “mess” is the exterior product of the identity on one vector with the scalar triple product, times plus or minus 1 (I didn’t check). And wow, the Wikipedia article on the exterior product is a mess. You can read several pages, try to remember a bunch of math you haven’t used in a while, and still have no idea how to compute the thing. It’s really not that bad once you get past the jargon.</i>
Scene_Cast2almost 2 years ago
That operator looks eerily similar to the one in Clifford Algebra. And C.A. generalizes nicely* out of 3D, too.<p>* I found that current Clifford Algebra Python libraries don&#x27;t handle high dimensional (even 32D, fairly primitive by ML standards) spaces.
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eternauta3kalmost 2 years ago
Get familiar with the Levi Civita tensor[1], never look back.<p>[1] <a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Levi-Civita_symbol" rel="nofollow noreferrer">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Levi-Civita_symbol</a>