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 Infinite-dimensional space bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2006-12-16, 16:42 (This post was last modified: 2006-12-16, 16:44 by bo198214.) quickfur Wrote:Even with independent-plane rotations? Maybe "principal rotation" is the wrong term... my question was whether this accounts for all combinations of plane rotations as well (there may be an infinite number of combined plane rotations, since we have infinite possible independent pairs of dimensions to rotate in). Oh, now I see what you mean. For dimensions $\le 3$ every rotation can be given by just 1 angle. And for higher dimensions maybe one needs more angles, or in our case maybe infinite angles. So I must admit I dont know the finite dimensional representation of rotations (though that should be contained in some standard theory), so I can not speak about it (and matrices were anyway always a red rag for me). The $A^{-1} R A$ are not even closed by multiplication which rotations should be (and here even infinite products of rotations should be a rotation again). Quote:Really? I think we're doing exactly the same thing here... maybe the base case is different? I start with the points (-1) and (1), in one dimension. Maybe your base case is different. What I did was to construct an n-simplex from an (n-1)-simplex by translating the latter downwards by an appropriate amount, and adding a new vertex. The new vertex is of the form $(0,\ 0,\ \ldots,\ P_n)$, and $P_n$ converges to $\sqrt{2}$ as n goes to infinity. Ok then I see the link. My simplices all have side length 1 and your simplices have all the side length 2 (if you start with (-1,0..), (+1,...) and I have $\pm\frac{1}{2}$ at step 2. So yours is simply modified by 2 and it should roughly be $2 d_{n-1} = P_n$ and hence $2\frac{1}{\sqrt{2}} = \sqrt{2}$ which is at least true *g*. But I would suspect that our formulas not completely match, in my formula there is this $\frac{n+1}{n}$ factor contained ... can you just check? Quote:In fact, our definition of rotation should include maps from these vertices into any other equal-normed vector. That should not be that difficult. We have the rotation plane (spanned by both vectors) and the angle already. And exactly to that case was my $A^{-1} R(\theta) A$ tailored. We just project a point on the rotation plane, rotate with $R(\theta)$ in the plane coordinates, and project it back. Without going into the details I think it is just expressable as given above. Quote:Anyway, I'm starting to get the suspicion that if we want a consistent geometry for infinite-dimensional vectors, we will have to allow transfinite coordinates and work with a transfinite superset of the reals ... Excellent observation! We need to extend $\mathbb{R}$ to a number system, where additionally arbitrary (countable?) sums *of positive summands* always have a limit/are defined. We need only positive summands if we want to take things like $\sqrt{x_1^2+x_2^2+...}$. And it also saves us from the sum rearrangement problem in $\mathbb{R}$. If all the summands are positive and the limit exists in $\mathbb{R}$ then the series absolutely convergent and hence reordering does not matter. And in our to be built extension, the real series with real limit should have the same limit as in $\mathbb{R}$. (A reordering of positive summands would destroy all our efforts of extension, because then we wouldnt have a unique limit in $\mathbb{R}$ already). Because such an extension is a topic for its own, I will start a new thread. quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2009-09-04, 04:34 (This post was last modified: 2009-09-04, 04:38 by quickfur.) This is a very belated reply, but I just wanted to say that I'm starting to be convinced that perhaps the idea of an infinite-dimensional space that preserves nice geometrical properties of finite dimensions simply isn't possible. The next nail in the coffin involves the matter of compositions of rotations. Intuitively speaking, every rotation can be decomposed into a composition of principle plane rotations. Now, assuming that we are able to rotate the vector <1,1,1,...> into , where C is the transfinite norm of <1,1,1,...>. If such a rotation were indeed possible, it must also be possible to decompose it into a countable sequence of plane rotations. In other words, for the transformation from <1,1,1,...> into to be a valid rotation, rotations should be closed under infinite composition (provided the rotations in the sequence take place in orthogonal 2-planes). Such a property, however, spells trouble: take now the sequence of rotations that map the first coordinate axis to the second, and then the second to the third, then the third to the fourth, and so on. Since rotations are closed under infinite composition, this infinite sequence of rotations should qualify as a "linear transformation", R, in infinite-dimensional space. Now what happens if we apply R to the vector <1,0,0,0,...>? Answer: there is no such vector, because the limit of the "partial compositions" in the infinite composition (the first n such rotations, for each n), applied to the vector <1,0,0,0,...>, is the zero vector, which doesn't make sense. That first coordinate of 1 has vanished into thin air! To recover from this situation, we have to introduce the idea of transfinite dimensions, because after an infinite sequence of rotations, if the number of coordinates is transfinite, then there is somewhere for the 1 to end up after an infinite number of rotations. However, this brings with it a host of other even nastier problems (now we have a transfinite number of coordinates, so we must also allow a transfinite number of independent rotations in order to map the transfinite hypercube vertex to the vertical axis). So the only other solution is to concede that infinite composition of rotations doesn't make sense, which in turn implies that the infinite-dimensional hypercube cannot be oriented such that its diagonal is parallel to the vertical axis, so that the circumsphere of the infinite-dimensional hypercube is not a complete hypersphere, since it doesn't intersect any of the infinite number of coordinate axes. Two opposite vertices of the infinite-dimensional cross polytope also cannot be oriented to be parallel to the diagonal of the infini-cube, because their coordinates would just shrink to zero (because they would be of the form k*<1,1,1,1,...>, with the condition that the norm is a finite number). In other words, infinite-dimensional polytopes have very counterintuitive properties that make them very unlike the finite-dimensional polytopes that we know and love. Retreating back to L^2 Hilbert space doesn't help, either, because the measure polytope does not exist in that space, so the familiar polytopes are absent. bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2009-10-29, 17:40 (2009-09-04, 04:34)quickfur Wrote: This is a very belated reply, but I just wanted to say that I'm starting to be convinced that perhaps the idea of an infinite-dimensional space that preserves nice geometrical properties of finite dimensions simply isn't possible. *Wears, like quickfur, the mourning clothes* « Next Oldest | Next Newest »

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