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quickfur · 2006-12-02, 23:49

How far can we go in extending a geometric interpretation to infinite-dimensional space $\Re^{\infty}$ ?

An infinite-dimensional vector is essentially a function $v:N\rightarrow \Re$ . Given two such functions, we can easily define addition, subtraction, scalar multiplication, and equality. How much further can we go?

Here are some ideas: we cannot easily generalize vector norm, because the norm of an infinite-dimensional vector would be non-finite in general. But it is still possible to compare the lengths of two vectors. Since for finite vectors the norm is the square root of the sum of the squares of each component, we simply do the comparison of two infinite dimensional vectors v and w component-wise: we say that $||v||=||w||$ iff $\sum_{i=0}^{\infty}\left(v(i)^2 - w(i)^2\right) = 0$ . Similarly, we can define inequalities of norms indirectly in this way.

For dot products, again it is not possible to assign a finite value to the dot product of two infinite-dimensional vectors in general; however, we can define the angle between them as:
$\cos\theta = \frac{\sum_{i=0}^{\infty}v(i) w(i)}{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}(v(i) w(j))^2}$ .
(from an analogous construction with finite-dimensional vectors.)

My conjecture is that this sum always converges (proof, anyone?). A cursory glance at it also reveals that there is a high probability that any two random infinite-dimensional vectors would be orthogonal (the sum converges to 0), which intuitively makes sense if you consider that there are $\aleph_0$ possible dimensions that are orthogonal to any given vector.

Further vector operations that would be nice to generalize: rotation in any 2-plane, taking unit vectors of any given vector, etc.

Eventually, I'd like to see if it's possible to have infinite-dimensional polytopes. $Smile$ Will an infinite-dimensional simplex have itself as facets? (After all, it seems that there should be a rotation that transforms a facet into the simplex itself.)

DirkU · 2006-12-07, 22:38

Hm, I am not a geometer, but I guess to have something like a 'nice geometry in inifinite many dimensions' it would be preferable to have at least a vector space with an inner product $<.,.>$ (like a Hilbert space). At least such a space is much more 'geometry-like' than a general normed vector space. For instance, you can get angles via $\cos(angle(x,y)):= <x,y>/(\sqrt{<x,x>}\sqrt{<y,y>})$ .

Dirk

***bo198214*** · (This post was last modified: 2006-12-08 11:36 by bo198214.)

@quickfur

What you are looking for is called ${l^2$ space. The solution is too simply regard only those vectors for which the norm is finite.

DirkU Wrote:it would be preferable to have at least a vector space with an inner product $<.,.>$ (like a Hilbert space). ... For instance, you can get angles via $\cos(\angle(x,y)):= \frac{<x,y>}{\sqrt{<x,x>}\sqrt{<y,y>}}$ .

In the wikipedia article is also mentioned:

Wikipedia Wrote:The space $l^2$ is the only of the spaces $l^p$ that is a Hilbert space ...

So this would be one possibility for such an infinite dimensional extension (, i.e. allowing only finitely normed vectors).

If we however stick to including all vectors the inner product $\sum_{i=1}^\infty v_iw_i$ would also often fail to be finite.
So the idea to compute with angles (which are always finite) is quite appealing though I never heard of that.

To see that $\frac{a_n}{b_n}:=\frac{\sum_{i=0}^n v_iw_i}{\sqrt{\sum_{i=0}^n v_i^2}\sqrt{\sum_{i=0}^n w_i^2}}$ (I love TeX $Wink$ ) indeed converges, we first observe from finite dimensional vector spaces that
$\sum_{i=0}^n \frac{|v_i||w_i|}{b_n}\le 1$ ,
especially it converges. If we now split the nominator sum into positive and negative part (let $p(x) = 1$ for $x\ge 0$ and 0 otherwise):
$\sum_{i=0}^n \frac{v_iw_i}{b_n}=\sum_{i=0}^n\frac{{v_iw_ip(v_iw_i)}}{b_n}-\sum_{i=0}^n\frac{{-v_iw_ip(-v_iw_i)}}{b_n}$
both parts of the difference only have positive summands and are also restricted $\le 1$ hence they converge and hence $a_n/b_n$ converges.

So what rules would we postulate for our operation $\angle:\ (V\setminus\{\mathbf{0}\})\times (V\setminus\{\mathbf{0}\})\to [0,\pi]$ to make V an "angled" vector space?

$\angle(\mathbf{x},\mathbf{x})=0$
$\angle(\mathbf{x},\mathbf{y})=\angle(\mathbf{y},\mathbf{x})$
$\angle(\alpha \mathbf{x},\beta \mathbf{y}) = \angle(\mathbf{x},\mathbf{y})$ for $\alpha,\beta > 0$
$\angle(-\mathbf{x},\mathbf{y})=\pi - \angle(\mathbf{x},\mathbf{y})$
$\angle(\mathbf{x},\mathbf{y})=\angle(\mathbf{x}+\mathbf{z}, \mathbf{y} + \mathbf{z} )$
$\angle(\mathbf{x},\mathbf{z}) \le \angle(\mathbf{x},\mathbf{y}) + \angle(\mathbf{y},\mathbf{z})$

Or the corresponding cosined forms $c:\ (V\setminus\{\mathbf{0}\})^2 \to [-1,1]$ :

$c(\mathbf{x},\mathbf{x})=1$
$c(\mathbf{x},\mathbf{y})=c(\mathbf{y},\mathbf{x})$
$c(\alpha \mathbf{x},\beta \mathbf{y}) = c(\mathbf{x},\mathbf{y})$ for $\alpha,\beta > 0$
$c(-\mathbf{x},\mathbf{y})=c(\mathbf{x},-\mathbf{y})=-c(\mathbf{x},\mathbf{y})$
$c(\mathbf{x},\mathbf{y})=c(\mathbf{x}+\mathbf{z},\mathbf{y}+\mathbf{z})$
$c(\mathbf{x},\mathbf{z})\le c(\mathbf{x},\mathbf{y})+c(\mathbf{y},\mathbf{z})$

Note that if we would only allow to measure angles between 0 and $\pi/2$ we could drop the condition 4 and allow in condition 3 also negative scalars.

Also one would be tempted in analogy to a metric, to add: $\angle(\mathbf{x},\mathbf{y})=0$ if and only if there exists an (positive) $\alpha$ such that $\mathbf{x}=\alpha \mathbf{y}$ . So we can consider the space of the corresponding equivalence classes.

But this construct (for $\alpha$ not restricted to positive values) is also already known as Projective Space P(V) $Smile$

quickfur · 2006-12-10, 23:26

bo198214 Wrote:@quickfur

What you are looking for is called ${l^2$ space. The solution is too simply regard only those vectors for which the norm is finite.

No, I already know about this space. But what I'm looking for is to include the entire set of infinite-dimensional vectors. Obviously, there's no way to do this with finite-valued norms, hence I tried to explore how far we can get with working around having an actual value for things like the dot product.

Quote:[...]If we however stick to including all vectors the inner product $\sum_{i=1}^\infty v_iw_i$ would also often fail to be finite.
So the idea to compute with angles (which are always finite) is quite appealing though I never heard of that.
[...snipped proof of convergence...]

Cool, so the ratio does indeed converge.

What about the set of equivalence classes of equal-length vectors? I wonder what kind of structure that will have... it will be some superset of the finite reals to include distinct infinite numbers, I suspect. (This is somewhat related to my magnitudes idea: instead of grappling with the intricacies of functions over the reals, we can immediately get the infinite component just by considering countable sequences of reals as infinite-dimensional vectors and defining a comparison on them. This, however, doesn't let us get beyond the cardinality of the reals.)

***bo198214*** · (This post was last modified: 2006-12-11 00:48 by bo198214.)

quickfur Wrote:What about the set of equivalence classes of equal-length vectors? I wonder what kind of structure that will have... it will be some superset of the finite reals to include distinct infinite numbers, I suspect. (This is somewhat related to my magnitudes idea: instead of grappling with the intricacies of functions over the reals, we can immediately get the infinite component just by considering countable sequences of reals as infinite-dimensional vectors and defining a comparison on them. This, however, doesn't let us get beyond the cardinality of the reals.)

Not sure what you mean. You get no more information about infinite real numbers just by putting behind an infinite dimensional vector space. We have finite real lengths/sums and infinity, and to know this we dont need a vector space or geometrical interpretation. And whether we regard sums, sequences or functions with respect to their behaviour at infinity is not that different (except that only with functions we can have inverse functions $Wink$ ).

The operation that works always on (infinite dimensional) $\mathbb{R}$ -vector spaces is the (cos)angle. The equivalence classes of equal angle correspond to the equivalence classes $\mathbf{v}\equiv\mathbf{w}\Leftrightarrow\mathbf{v}=\alpha\mathbf{w}$ .

Or do you mean by $a$ has equal length to $b$ , that $a$ is a rotation of $b$ , by some angle? But then you come to the question of determinants of infinite dimensional matrices (because for a rotation must $\det{A}=1$ ). Not sure whether this is possible, or if it depends on the order of summations/multiplications, that surely must be taken, whether it converges (to 1) or not.

quickfur · 2006-12-11, 06:44

bo198214 Wrote:[...]
Or do you mean by $a$ has equal length to $b$ , that $a$ is a rotation of $b$ , by some angle? But then you come to the question of determinants of infinite dimensional matrices (because for a rotation must $\det{A}=1$ ). Not sure whether this is possible, or if it depends on the order of summations/multiplications, that surely must be taken, whether it converges (to 1) or not.

What I meant is the definition I gave: $a=b$ iff $\sum_{i=1}^{\infty} (a_i^2 - b_i^2) = 0$ . This is an equivalence relation that obviously transcends finite norms; I'm just wondering what kind of structure it will give us. Obviously, if the sum is positive or diverges to $+\infty$ , then we may say $||a||>||b||$ , and if the sum is negative, or diverges to $-\infty$ , we may say $||a||<||b||$ (even though $||a||$ and $||b||$ are generally non-finite). So this gives us a linear ordering. I wonder what it looks like? (On that note, I wonder if the triangle inequality holds?)

I was thinking something along the lines of, take some simple infinite vector as $(1,\ 1,\ 1,\ \ldots)$ , and give its norm a name, say $\Delta$ , and then compare other vectors to it to see what kind of structure we can uncover. For example, $||(2,\ 2,\ 2,\ \ldots)||$ could probably be reasonably called $2\Delta$ , since it is the scalar multiplication of the former vector by 2. Then we could have another vector, say $(1,\ 2,\ 3,\ \ldots)$ , which probably has a larger norm than $\Delta$ . But is this true for all multiples of $\Delta$ or only up to a certain factor? Etc..

Now, rotation is a very interesting topic. Is it provable that if $a=b$ (according to the above definition), then there must be a way of rotating $a$ into $b$ ? That would certainly involve infinite matrices, which I'm not sure how to handle. Or maybe we can work around this problem by defining rotation as any transformation that preserves norm: $||T\cdot x|| = ||x||$ . But you still have the problem of how to define a transformation, since it will involve infinite matrices.

***bo198214*** · 2006-12-12, 23:28

quickfur Wrote:What I meant is the definition I gave: $a=b$ iff $\sum_{i=1}^{\infty} (a_i^2 - b_i^2) = 0$ . This is an equivalence relation that obviously transcends finite norms; I'm just wondering what kind of structure it will give us. Obviously, if the sum is positive or diverges to $+\infty$ , then we may say $||a||>||b||$ , and if the sum is negative, or diverges to $-\infty$ , we may say $||a||<||b||$ (even though $||a||$ and $||b||$ are generally non-finite). So this gives us a linear ordering.

No this gives no linear ordering because there are much more choices than converge or converge to $\pm\infty$ , for example diverge $Wink$ . Especially if we have a subseries of partial sums greater 0 and one smaller 0.

Thats for example the reason, why in the construction of the reals, Cauchy sequences are taken. You could allow arbitrary sequences, with the same equivalence relation (i.e. $(a_n)\equiv(b_n)\Longleftrightarrow (a_n-b_n)\to 0$ ). But you could not linearly order it in any intended way. Cauchy sequences however you can linearly order by $(a_n)<(b_n)\Longleftrightarrow (b_n-a_n)>0$ and $(a_n) > 0\Longleftrightarrow \exists\varepsilon\in\mathbb{Q}_+\exists N\forall n>N: a_n > \varepsilon$

Quote:Now, rotation is a very interesting topic. Is it provable that if $a=b$ (according to the above definition), then there must be a way of rotating $a$ into $b$ ? That would certainly involve infinite matrices, which I'm not sure how to handle. Or maybe we can work around this problem by defining rotation as any transformation that preserves norm: $||T\cdot x|| = ||x||$ .

I could also think of having an infinite matrix $R(\theta)$ which is mostly 0, but 1 at the diagonal and the typical 2x2-cos-sin-rotation matrix in the left upper corner. The operation of this matrix would be well-defined (even without limits) by only affecting the first two indexes of the vector. Then we would call a transformation T a rotation if there is an A such that $A^{-1}TA=R(\theta)$ . It would also satisfy your $||T\cdot x|| = ||x||$ . In your definition you have at least also mirrorings.

Quote:But you still have the problem of how to define a transformation, since it will involve infinite matrices.

No, a transformation is simply a linear mapping.

quickfur · 2006-12-13, 00:19

bo198214 Wrote:
quickfur Wrote:What I meant is the definition I gave: $a=b$ iff $\sum_{i=1}^{\infty} (a_i^2 - b_i^2) = 0$ . This is an equivalence relation that obviously transcends finite norms; I'm just wondering what kind of structure it will give us. Obviously, if the sum is positive or diverges to $+\infty$ , then we may say $||a||>||b||$ , and if the sum is negative, or diverges to $-\infty$ , we may say $||a||<||b||$ (even though $||a||$ and $||b||$ are generally non-finite). So this gives us a linear ordering.
No this gives no linear ordering because there are much more choices than converge or converge to $\pm\infty$ , for example diverge $Wink$ . Especially if we have a subseries of partial sums greater 0 and one smaller 0.

Hmm, that's true. I also realized that it's probably possible to get conditionally convergent series (by suitable definitions of the vector coordinates) that converge to different numbers depending on the arrangement of the coordinates. This would be bad, since it means that the norm of a vector is not preserved under reflections (i.e. coordinate rearrangements).

I wonder if there's a way to massage the definition so that we will always get series that either converge or diverge?

Quote:Thats for example the reason, why in the construction of the reals, Cauchy sequences are taken. You could allow arbitrary sequences, with the same equivalence relation (i.e. $(a_n)\equiv(b_n)\Longleftrightarrow (a_n-b_n)\to 0$ ). But you could not linearly order it in any intended way. Cauchy sequences however you can linearly order by $(a_n)<(b_n)\Longleftrightarrow (b_n-a_n)>0$ and $(a_n) > 0\Longleftrightarrow \exists\varepsilon\in\mathbb{Q}_+\exists N\forall n>N: a_n > \varepsilon$

Good point.

Quote:
Quote:Now, rotation is a very interesting topic. Is it provable that if $a=b$ (according to the above definition), then there must be a way of rotating $a$ into $b$ ? That would certainly involve infinite matrices, which I'm not sure how to handle. Or maybe we can work around this problem by defining rotation as any transformation that preserves norm: $||T\cdot x|| = ||x||$ .
I could also think of having an infinite matrix $R(\theta)$ which is mostly 0, but 1 at the diagonal and the typical 2x2-cos-sin-rotation matrix in the left upper corner. The operation of this matrix would be well-defined (even without limits) by only affecting the first two indexes of the vector. Then we would call a transformation T a rotation if there is an A such that $A^{-1}TA=R(\theta)$ . It would also satisfy your $||T\cdot x|| = ||x||$ . In your definition you have at least also mirrorings.

This only accounts for principal rotations. Maybe we can make it more powerful by starting with $R(\theta)$ and any transposition $A$ (transposition = swapping of coordinates), and define rotation to be the closure of $R$ and $A$ under matrix multiplication. This will also include non-principal rotations.

However, I'm not sure if this is sufficient. Consider for example, a (hypothetical) infinite-dimensional simplex, where the first vertex is $(1,\ 0,\ 0,\ 0,\ \ldots)$ and all the other vertices have the first coordinate equal to 0. Then it must be possible to rotate the simplex on the hyperplane defined by all except the first vertex. Then the question is, is our definition of rotation strong enough to include all possible such rotations?

Quote:
Quote:But you still have the problem of how to define a transformation, since it will involve infinite matrices.
No, a transformation is simply a linear mapping.

OK.

***bo198214*** · (This post was last modified: 2006-12-14 23:29 by bo198214.)

quickfur Wrote:I also realized that it's probably possible to get conditionally convergent series (by suitable definitions of the vector coordinates) that converge to different numbers depending on the arrangement of the coordinates. This would be bad, since it means that the norm of a vector is not preserved under reflections (i.e. coordinate rearrangements).

I could live with only rearranging finitely many coordinates.

Quote:I wonder if there's a way to massage the definition so that we will always get series that either converge or diverge?

I have absolutely no idea, your task $Wink$

Quote:
Quote:I could also think of having an infinite matrix $R(\theta)$ which is mostly 0, but 1 at the diagonal and the typical 2x2-cos-sin-rotation matrix in the left upper corner. The operation of this matrix would be well-defined (even without limits) by only affecting the first two indexes of the vector. Then we would call a transformation T a rotation if there is an A such that $A^{-1}TA=R(\theta)$ . It would also satisfy your $||T\cdot x|| = ||x||$ . In your definition you have at least also mirrorings.
This only accounts for principal rotations.

Nonono $R(\theta)$ is a principal rotation, but $AR(\theta)A^{-1}$ gets all rotations (I think). For example if A is a coordinate swapping, we get a similar thing you described. But also note that not all coordinate swappings are rotations, some are also mirrorings.

Quote:Consider for example, a (hypothetical) infinite-dimensional simplex, where the first vertex is $(1,\ 0,\ 0,\ 0,\ \ldots)$ and all the other vertices have the first coordinate equal to 0.

Yeah, interesting. But what are the coordinates of the other points?
Assume we have already constructed the centered n-simplex with vertices $p_1,...,p_n$ with each $p_i$ having n coordinates (and the others 0). We can then place the new vertex $q$ at $(\underbrace{0,...,0}_{n},q_{n+1})$ , getting an offcentered n+1-simplex in $\mathbb{R}^{n+1}$ . To compute $x$ , we use $q$ having the distance 1 to all other $p_i$ . Which is automaticly given when it has distance 1 to $p_1$ , so $q_{n+1}=\sqrt{1-|p_1|^2}$ .

The center of the new simplex is then also along $(0,...,0,c)$ given by
$\frac{p_1+...+p_n+q}{n+1}$
so $c=\frac{\sqrt{1-|p_1|^2}}{n+1}$
After centering the new simplex is then
$q_{n+1}=\sqrt{1-|p_1|^2}(1-\frac{1}{n+1})$ which is also the distance of all other vertices to 0.
(Performing this I wondering whether we already discussed a similar thing at tetraspace ... here ... but I get different formulas ...).
Hence the distance of the vertices from 0 following the following law
$\begin{eqnarray} d_1 &=& 0\\ d_{n+1} &=& \sqrt{1-d_n^2}(1-\frac{1}{n+1}) = \frac{n}{n+1}\sqrt{1-d_n^2}\\ \end{eqnarray}$
The limit should be $\frac{1}{sqrt{2}}$ , at least it does not converge to 0 (what would mean there is no infinite simplex ... all vertices 0) which I just wanted to verify. So the vertices of an infinite simplex lie on a (infinite dimensional) sphere with radius $\frac{1}{\sqrt{2}}$

But didnt you figure such a nifty closed formula for the recurrance? In the moment I am unable to relate both threads though the limit was always something with $\sqrt{2}$ $Wink$ , can you do this?

Quote: Then it must be possible to rotate the simplex on the hyperplane defined by all except the first vertex. Then the question is, is our definition of rotation strong enough to include all possible such rotations?

In the moment I have also to take a bit sleep so I will finish with the coordinates formula (at least by my humble derivations ...)
$ \begin{eqnarray} p_1 &=& (-\frac{1}{1}d_2,-\frac{1}{2}d_3,-\frac{1}{3}d_4,....)\\ p_2 &=& (d_2,-\frac{1}{2}d_3,-\frac{1}{3}d_4,....)\\ p_3 &=& (0,d_3,-\frac{1}{3}d_4,-\frac{1}{4}d_5,....)\\ p_n &=& (\underbrace{0,...,0}_{n-2},d_n,-\frac{1}{n}d_{n+1},....) \end{eqnarray} $

Rotations hopefully next time.

quickfur · 2006-12-15, 21:53

bo198214 Wrote:
quickfur Wrote:I also realized that it's probably possible to get conditionally convergent series (by suitable definitions of the vector coordinates) that converge to different numbers depending on the arrangement of the coordinates. This would be bad, since it means that the norm of a vector is not preserved under reflections (i.e. coordinate rearrangements).
I could live with only rearranging finitely many coordinates.

That would work around this particular problem, although I'm not sure if it will break anything else (e.g. rotations in $(\infty-1)$ hyperplane).

Quote:
Quote:I wonder if there's a way to massage the definition so that we will always get series that either converge or diverge?
I have absolutely no idea, your task $Wink$

OK, I'll have to work on this some other time.

Quote:
Quote:
Quote:I could also think of having an infinite matrix $R(\theta)$ which is mostly 0, but 1 at the diagonal and the typical 2x2-cos-sin-rotation matrix in the left upper corner. The operation of this matrix would be well-defined (even without limits) by only affecting the first two indexes of the vector. Then we would call a transformation T a rotation if there is an A such that $A^{-1}TA=R(\theta)$ . It would also satisfy your $||T\cdot x|| = ||x||$ . In your definition you have at least also mirrorings.
This only accounts for principal rotations.
Nonono $R(\theta)$ is a principal rotation, but $AR(\theta)A^{-1}$ gets all rotations (I think).[...]

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).

Quote:
Quote:Consider for example, a (hypothetical) infinite-dimensional simplex, where the first vertex is $(1,\ 0,\ 0,\ 0,\ \ldots)$ and all the other vertices have the first coordinate equal to 0.
Yeah, interesting. But what are the coordinates of the other points?
[...]
(Performing this I wondering whether we already discussed a similar thing at tetraspace ... here ... but I get different formulas ...).

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.

Quote:[...]
The limit should be $\frac{1}{sqrt{2}}$ , at least it does not converge to 0 (what would mean there is no infinite simplex ... all vertices 0) which I just wanted to verify. So the vertices of an infinite simplex lie on a (infinite dimensional) sphere with radius $\frac{1}{\sqrt{2}}$

But didnt you figure such a nifty closed formula for the recurrance? In the moment I am unable to relate both threads though the limit was always something with $\sqrt{2}$ $Wink$ , can you do this?

Heh, $\sqrt{2}$ seems to show up everywhere, doesn't it? $Smile$ I think it comes from the type of recurrence you get when you construct an n-simplex in this way.

Quote:
Quote: Then it must be possible to rotate the simplex on the hyperplane defined by all except the first vertex. Then the question is, is our definition of rotation strong enough to include all possible such rotations?

In the moment I have also to take a bit sleep so I will finish with the coordinates formula (at least by my humble derivations ...)
$ \begin{eqnarray} p_1 &=& (-\frac{1}{1}d_2,-\frac{1}{2}d_3,-\frac{1}{3}d_4,....)\\ p_2 &=& (d_2,-\frac{1}{2}d_3,-\frac{1}{3}d_4,....)\\ p_3 &=& (0,d_3,-\frac{1}{3}d_4,-\frac{1}{4}d_5,....)\\ p_n &=& (\underbrace{0,...,0}_{n-2},d_n,-\frac{1}{n}d_{n+1},....) \end{eqnarray} $

Rotations hopefully next time.

Well, it doesn't have to be vertices of a simplex, really. I just used it 'cos it was the first thing that came to my mind. I think what I really wanted to say was, if you take the set S of all infinite vectors with equal norms (as per my definition of norm equality), in other words, all the points on an infinite-dimensional hypersphere of some radius R, then given any two vectors $v_1,\ v_2 \in S$ , a linear transformation that maps $v_1$ to $v_2$ should be included in the definition of a rotation.

When R is finite, this isn't too hard, but ideally, we want to account for equal-norm vectors with transfinite norms as well. For example, the vertices of the infinite hypercube (all possible sign permutations of $(1,1,1,\ldots)$ ) are equal-norm to each other, and our definition of rotation should include all transformations that map these vertices into each other. In fact, our definition of rotation should include maps from these vertices into any other equal-normed vector.

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, in order for operations like the dot product and norms to be closed in our number set. For example, if you take the diagonal vector (1,1,1,...,1) in n dimensions (for finite n), it is possible to rotate it such that it is of the form (C,0,0,0,...,0), where C is its norm. We can't do this in $\Re^{\infty}$ because the norm of (1,1,1,...) is not finite, which means that you cannot rotate an infinite-dimensional hypercube so that two opposite vertices lie along one of the coordinate axes! It also means that the hypersphere defined by the set of all vectors equal norm with some basis vector v is incomplete if the norm of v is not finite. For example, the set of all vectors equal-normed with the cube vertex (1,1,1,...) describes the circumsphere of the cube, but it has many "holes", because it doesn't intersect with any of the coordinate axes! So it seems that we must have a transfinite number system to express coordinates in, so that these operations are closed. This number system should probably be linearly-ordered, since otherwise we will get very strange "un-geometric" behaviour of the vectors.

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