Xerxes

Right. By a "continuous number system" I assume you mean a complete number system. Which would include \(\mathbb{R}\) but not include a countable subset, as I had said. So my question is this (and it's not rhetorical!); if time is continuous, as I beleive the spectra of some QM operators also are, and if the rational numbers are not sufficient to describe continuous phenomena, what meaning can be attached to a measurement of, say, \(e\) or \(\sqrt{2}\) or, for that matter, of \(\pi\) which as you say is a ration, nota measurement

May we say that a measurement is an injection from a set of observations to a countable subset of \(\mathbb{R}\), and this injection may be composed of many linked injections? If so, then it might appear that it is time that is the outlier, not count.

But surely, if one uses a 4 coordinate system, and sets \(t=ict\), then the coordinate transformation \(C \to C'\) is orthoganal i.e. metric-preserving?

That's because you don't understand mathematics. "Relational Geometry" - your term. The word geometry implies a metric (-metry ; metric). Specifically, the geometry of a concrete space comes entirely from its metric. Different metric, different geometry. No metric, no geometry. Others may have more patience with this, but I'm out

Ha! Serves me right for trying to be sarcastic on an international forum. But the fact remains that both your theory and category theory are deeply non-constructivist, by which is meant they assume the existence of objects they are not equipped to construct. Category Theory is unashamedly like this, as it is not designed to dispense with what it might regard as "lower level constructions" rather to build upon them as a sort of meta-mathematics Your theory, on the other hand, arrogantly declares it has no need of the usual sorts gadgets we use in differential geometry, while implicitly using them all. Hence studiot's very relevent questions.

Wow! I'm impressed! Last week, when I casually remarked that your ideas reminded me somewhat of category theory, you admitted you knew very little about the subject. And now you have become sufficiently proficient in the subject to lecture us on one of its more subtle points. Well done you! However, although it is obvious that spacetime is an object in the category of smooth manifolds, it is not obvious to me in which category you think energy belongs. Is it the category of natural numbers, or some other category? And can you please describe in detail the functors between these categories?

Is anybody else here struck by the similarities between Anton's ideas and Category Theory? recall that this theory, which is 80 yers old and highly respectable, seeks to unify all mathematics in a sort of meta-mathematiccs and where the different "objects" in the different categories (of sets, of rings, of groups etc) are of no real interest, the relations between them ("morphisms") of passing interest only, the main focus being on the relations between categories and thus the relations between maps between different categorical object, ("maps of maps") I can can see how, in outline, it might be applied to both the special and general theories, but having no real knowledge of physics, cannot see the details. Maybe that's where the devil lies? I beleive that John Baez (Riverside, U. Cal??) has an interest in applying category theory to physics problems. Might be worth Anton's while to look up is work

Well you guys are no fun. But as this is a subject that interests me, I shall have fun on my own. Recall that for every vector space \(V\), there exists a companion space \(V^*\), this being the space of all linear maps \(V^* \to \mathbb{R}\), so that for any \(\varphi \in V^*\) and all \(v \in V\) we will have that \(\varphi(v) = \alpha \in \mathbb{R}\). Now it is a fact that if and only if there exists a metric in \(V\), with an inner product \((v,w) \in \mathbb{R}\) then there will be some particular \(\phi in V^*\) such that \(\phi(w) = (v,w) in \mathbb{R}\). I will temporarily write this as \(\phi_v\) to emphasize this fact. This cooincides very nicely with a useful and interseting property of the metric tensor, which some people (who really are old enough to know better) call "index raising" and "index lowering". So recall that the metric tensor is a bilinear gadget, that is it takes 2 vectors as argument and returns a scalar, say \(g(v,w) \in \mathbb{R}\). So if we feed this beast just a single vector as argument it's not going to work unless it has had already another vector to feed on. Like \(g(v, -) (w) \in \mathbb{R}\). \(g(v, -)\) is therefore a dual vector So with an elegant sleight of hand, I make the "equivalence" \(\phi_v \equiv g(v, -)\). And for those that insist on using index notation for tensors (they really shouldn't, since in so doing they are actually referring to tensor components, which are scalars of sorts, not tensors at all), this becomes \(g_{j,k}A^k= B_j\), which is index lowering. So there.

Very nice studiot. An ineresting thread you started What's even nicer is that every metric space is automaticaly a topological space with the Hausdorff property. This means, as studiot showed, that if there are no open sets \(a\in A\) and \(b\in B\) such that \(A \cap B =\emptyset\), then we must have that \(a = b \) Furthermore, if the metric on every subset \(S\) is the Euclidean metric, then we may have a homeomorphism (i.e. a sort of isomorphism) \(h: S \to R^n\), which is essentially the definition of a manifold.

First let me apologize for my last post- it came out way more symbolic than I had intended. More like a student tutorial as MigL said. Not really what you want in a discussion forum. See if I can do better...... We have our metric tensor field (that is, if you think I was making any sort of sense). Let's assume, as Einstein did, that gravitation and acceleration are in some sense equivalent. Let's further assume, again as Einstein did, that within a small enough spacetime interval, an accelerating body is in inertial (unaccelerated) motion at each point in its trajectory. So that at each point the Special Theory applies, and the different coordinates there are related by a local orthogonal tranformation (a Lorentz transformatin). Now as you know, orthogonal transformations are characterized be the fact that they preserve the metric, which is very nice locally, but doesn't apply for the different inertial systems at different points. So if we wish to do relativity, we may not assume that transformations between them are orthogonal (metric conserving). The solution is to allow the line element that I pedanticaly showed you (or do I mean didacticaly) to vary from point to point. So we will now have a non-constant metric field, which by the assumed equivalence of acceleration and gravitation, applies in the presence of gravitating bodies. Which is basically the General Theory

This is an interesting and important topic, so if I may be allowed to pick it up...... First recall that for every vector space \(V\) there exists a dual space \(V^*\), this being the vector space of all linear functions \(V \to \mathbb{R}\), so that for any \(v \in V\) and any \(\varphi \in V^*\) then \(\varphi(v) = \alpha \in \mathbb{R}\). Suppose now an arbitrary manifold \(M\). Here our vector space may be taken to be, in some vague sense, "tangent" to it at some defined point, say \(p\). Accordingly, it is called the "tangent space" and is written as \(T_p(M\) and its dual as \(T^*(M)\). But still we will have that \(\phi(v) = \alpha \in \mathbb{R}\). Now every vector space must have a basis that allows us to write, for any vector, \(v=\sum_{nolimits j}\beta^j e_j\) where as beforthe set \({\beta^j\) are scalars and the set \({e_J}\) are the basis vectors. in the specific case of our tangent dual space, the basis is the set \({dx^k}\), which I can prove but it would take too long (even if any one were intereste Let's now introduce a new operation \(T^*_p(M) \otimes T^*_p(M): T_p(M) \times T_p(M) \to \mathbb{R}\) so that in our specific case, \(\varphi \otimes \phi = \sum_{nolimits jk} g_j g_k dx^j\otimes dx^k\) By convention, the producct \(g_jg_k\) is usually written as \(g_{jk}\) and since the product \(dx^dx^k\) ican be taken to be the infinitessemal coordinate "distance", the construction \(g_{jk}(dx^j dx^k\) = ds^2\) is called the line element, and the product \(g_{jk}\) is referred to as the metric tensor. A Riemann manifold is simply one that boasts a field of such tensors. When we recall that, in any field, not every element need to be the same (though it might be), we can start to see how this beast is so important in the relativistic theory of gravitation.

Of course it wasn't my objective, Why would you think such a thing? Surely you're not just trying to be offensive?

So, it seems that both you and your source use the term "unbound vector" to refer to what most mathematicians would call a whole equivalence class of vectors. And since congruence in general is a broad (though not especially deep) subject, that is why I advised the opening poster not to think of vectors as directed line segments. But since she, or he, seems to have lost interest in the subject, it's really not worth falling out over

Well, the terms "bound vector" and "unbound vector" are most certainly non-standard in mathematics. Maybe you could explain what you mean. Also, recall that the "length" of a vector, and the angle between two vectors requires an inner product. And a vector space equipped with an inner product is called a metric space. But not every vector space is a metric space. (Of course, those spaces whithout a metric are of no interest to physicists or enigineers!)

First I would caution against referring to a "normal" vector as a tensor. Yes it is true that tensors are elements in a vector space, and therefore vectors, but I doubt that's what you mean. It is also true that, when talking about tensors in general, it is customary to refer to vectors as type 1,0 tensors but that is a harmless notational abuse which I advise you not to adopt for now. To answer your question: yes, in the naive view of vecyors as arrows in a plane, they can have "heads" or "tails" anywhere, but to understand why you are never told this would require you to understand equivalence classes and stuff like that. Why bother? Why not just use the grown-up definition of a vector? Namely a vector space comprises a set called /(V/) equipped with a closed binary operation called addition (closed just means that \(x+y\) is in \(V\) whenever \(x\) and \(y\) are), together with an associated scalar field which acts by multiplication on elements of the space, this also being a closed operation. When feeling pompous, we talk of a vector space [i]over[/i] a certain field, and may even write it as \(V(\mathbb{F}\) (thankfully this is rare) Don't even think about tensors until you are comfortable with all there is to know about vectors - there is quite a lot more

Well, I believe that the pull-back, like the push-forward, is a more general construction. Here's my thinkinig..... Suppose the maps (\g:V \to \mathbb{F}\} and (\h:W \to \mathbb{F}\). These are, of course elements of the dual spaces (\V^*\) and (\W^*\}, repectively. Then the pull-back of the map (\f:V \to W\) maps (\W^* \) onto (\V^*\),which is not quite what I was looking for

Suppose \(V, W\) are vector spaces overthe field \(\mathbb{F}\) and that \(L:V\to W\) is a linear transformation. We know that the dual spaces \(V^*, W^*\) exist. What would be the dual transformation i.e. \(V^* \to W^*\)? I'm tempted to suggest it has something to do with pullbacks, but I can't seem tp get it to work