Well, no, not really, just a new look at old ones. It's fun. Suppose \(U,\;V,\;W\) are vector spaces over the field \(\mathbb{F}\). Let \(f\) and \(g\) be linear maps (transformations, operators) \(U \to V\) and let \(a,\;b \in \mathbb{F}\) be scalar. Then we can define a linear vector space of such maps (over the same field) by \((f+g)(u) = f(u) + g(u)\) and \((af)(u) = a(f(u))\) I will denote these spaces by \(L(U,V)\) etc., whose elements (vectors) are precisely these maps. A natural question is; what do linear maps on these spaces look like? Right. Returning to the vanilla case, consider \(f:U \to V\) and \(g:V \to W\). Then obviously, since the codomain of \(f\) is the domain of \(g\), I can compose these as \(U \longrightarrow^f V \longrightarrow^g W = U \longrightarrow^{g \cdot f} W\) where I follow the usual convention of composing right-to-left. But writing this in terms of our new friends I find this implies \(g:L(U,V) \to L(U,W), \;g(f) = g \cdot f\), since \(f \in L(U,V),\; g \cdot f\in L(U,W)\) This cannot be right, since the domain of \(g\) is \(V\) and not \(L(U,V)\). So I need a new map which is obviously, by the immediately preceding, related to \(g\). I can call this \(L(U,g)\) which nicely captures the idea, but, for a reason I hope to make clear, I will use the other convention and use \(g_*\) such that \(g_*(f) = g \cdot f\). This map is called the "pushforward" of \(f \) along, or on, \(g\). Now consider the linear space \(L(U,\mathbb{F})\), which we immediately recognize as the dual space \(U^*\). Suppose now we have \(\mathbb{F} \longleftarrow^h U \longrightarrow^f V\longrightarrow^k\mathbb{F}\). How am I going to compose these? If the following formatting works.... \(\;U \longrightarrow^f V \\ h \searrow \; \swarrow k \\ \;\;\;\;\; \mathbb{F}\\) it's easy to see that \(k(f)=k \cdot f = h \in L(U,\mathbb{F})\). But again, since \(k \in L(V,\mathbb{F})\) this implies that \(f: L(V,\mathbb{F}) \to L(U,\mathbb{F})\), which makes even less sense than before, since \(f: U \to V\)! So is there a map \(L(V,\mathbb{F}) \to L(U, \mathbb{F})\) based on \(f: U \to V\)? Yes, for we can do this: define the map \( f^*: L(V,\mathbb{F}) \to L(U, \mathbb{F})\) by \(f^*(k) = k \cdot f\). This is called the "pullback" of \(k\) on \(f\), for the obvious reason. And here's the punch-line: Since \(L(V,\mathbb{F}) =V^*\) we can write \(f^*: L(V,\mathbb{F}) \to L(U, \mathbb{F})\) as \(f^*: V^* \to U^*\). This is nice, don't you think? It has an immediate, and familiar, consequence when our vector spaces are inner product spaces. It also provides us with a brand new gadget, which I.m keeping secret for now!
Ya, well - as always I appear to be talking to myself (maybe if I insulted a few members I might get more response......?) Any, leaving out the obvious: recall we have a map \(f: U \to V\) and another map \(f^*: V^* \to U^*\). This implies the existence of some gizmo, call it *, that sends objects in one class of vector spaces (\(U,\;V,\;...\)) to objects in a different class of vector spaces (\(U^*,\;V^*\;...\)) and maps in the first class (\(f,\;...\)) to maps in the second class \(f^*,\;...\)) with certain structure-preserving constraints This is a new sort of operation, and has very wide application: functions send objects to objects, operators send maps to maps. The gizmo that does both simultaneously is called a functor. Now, who can I insult.......
Somehow, for bringing up the memories of my Matrix Theory classes, you have insulted me... naw, just kiddin. I can't remember everything from that class, but doesn't this property already exist? Granted you may have proven it, but I'm pretty sure I've seen this particular axiom before. Then again, I didn't take very much from that class, I'm an engineering major, not a mathmatician, so practicality tends to be what sticks to me, but just a thought!
What you call pull-back seems to be the transpose of a linear transformation. Pretty fact: If V and W are finite dimensional linear spaces over a field F, and T:V->W is a linear transformation and T(t):V*->W* is its transpose then the annihilator of the nullspace of T is the range of T(t).
What about compactification, and covers? Can you extend the discussion of your space into these, or have you already? That's called "embedding", right? Like Mr Lisi thought he could do with F8 and QCD? Cause I don't understand all the terminology in adv. algebras and tensor calculus (maybe one or two others here have this "gap", as well).
Whoa guys. Either we are at cross purpose or I am terribly confused. If \(f: V \to W\) is a linear transformation, then the (non-conjugate) transpose operator is defined to be \(f^t: W^* \to V^*\) by \( f^t(\phi) = \phi \cdot f \; \forall \phi \in W^*\). It is therefore synonymous with the pullback. In the case that \(V,\;W\) are inner product spaces, we may have \(f^t: W \to V\) given by \(g_V(v, f^t(w)) = g_W(f(v), w)\) This we may identify with the adjoint operator only if \(V = W\) and real. Surely this is right? In the latter case, we also may have, for \(h:V \to V,\;u,\;v \in V\) and \(\psi \in V^*\) that \(\psi_v(u) = g_V(u,v)\) and therefore \(h^t(\psi_v)(u) = g_V(u,h^t(v)) =g_V(h(v),u) \)
I think you are in the wrong domain of discourse here. Although I don't know what "compactification" means exactly (and can't be bothered to look it up), I have seen it it. But I do know that the notion of a cover is an entirely topological concept, and I also know that compactness is also. If you want, I can explain, but I doubt you will be interested But this discussion, such as it is, is entirely to do with common-or-garden vector spaces. There are such things as topological vector spaces, maybe that's what you are referring to? But that is not what this thread is about. or maybe you're a bit muddled (fruddled??)?
Constantly. A vector space is topological, I think I got that bit. The complex plane (e^st) can be used to represent certain kinds of "time-domain" variables (like in circuit analysis, I know a bit about circuits and electronics). It's hyperbolic, too, or something (at least I can remember using sinh, and cosh). There's a "need" to define and discriminate different kinds of spaces, and a need to connect the differences together in some sort of a model...? Your post appears to be using conjugate pairs and binary operations. Reimannian manifolds are a general kind of vector space (but not "common-or-garden")? I thought closure and boundaries etc, were sort of important to the idea of vectors and spaces...? But, no biggie. Is the space you're discussing regular, semiregular, or just "common or garden"?
So I was confused adjoint and transpose, I may have seen in my dreams adjoint defined as transpose by your definition.
Hey! Don't beat yourself up, you weren't as far wrong as you seem to think. As you will know, the adjoint of an operator is defined to be the matrix transpose of its complex conjugates, usually written \(f^{\dagger}\). Since conjugation is a redundant operation on real numbers, the adjoint reduces to the matrix transpose, aka the pullback. We saw all that yesterday. There are other adjoints in representation theory which I touched on in another thread, which tie in much better with the pull-back. Maybe if I get time later I'll come back to that.
No. A vector space is not a topological space, they are two quite different concepts. I could give you an everyday example of an algebraic structure that has the compatible properties of a vector space and a topological space, but I shan't, as this would fruddle you even more; but even here, as elsewhere, one needs to be sure which axioms one is calling upon when describing the properties of such spaces. So no, in general, a vector space is not "topological". Once again you are fruddled. Riemannian manifolds are just that - manifolds, i.e. topological spaces with additional structure. They are most emphatically not vector spaces.
The complex plane is the set of all Real x and y put into the form x+iy where i*i=-1. e^(st) is not a standard way of representing a general complex number. You mention you knwo a bit about circuits and electronics. I would guess e^(st) relates to a Laplace transform in your mind. There's a lot more to the complex plane than Laplace transform. Never, ever, think your engineering textbook is a good source of rigorous maths! 'Hyperbolic' relates to entirely different notions. sinh and cosh (and tanh) are hyperbolic functions. The Poincare disk is a hyperbolic space, because it ha a metric defined on it. Simply talking about the complex numbers doesn't automatically imply it's a hyperbolic space. You need to notion of a metric to say such a thing and while you can define a metric or a norm a space of complex numbers, the complex numbers themselves do not inherently have a notion of distance. .... must...not....bash....head....into....keyboard.... Frud, don't take this the wrong way but stop throwing in words you don't know the meaning of. Far better to be honest and work your way up from the bottom than try to fool people who know these things that you know it too. It'll result in less posts correcting you and you'll benefit/learn so much more that way. The first step in learning something is admitting you don't know it.
Thanks for the tip, but actually an EE textbook is a good source of applied math, and explanations of behaviour--of things like semiconductors, electrons, quasiparticles, time and frequency dependent "reactive" elements. Electronics is a big field and covers a lot of physics and math. Perhaps you mean to imply something like: I shouldn't try to compose a concerto before I learn how to play the piano? There's this bit too: I think I understand that the complex plane isn't actually something that only circuit designers use. It's a mathematical "structure", yes? Involves, and also connects ideas of geometry and algebra (there's that Euler constant in there, and sines, cosines and tangents, and hyperbolics). I learned about geometry and algebra being a part of math some time before I started the EE course. After you take something for the headache: are manifolds (Reimann manifolds too) a general kind of space? Or is it that Reimann spaces can't be said to be a general space or surface? I thought they were just a class of manifolds (I know what a manifold is, I have an idea of what a space is, and what a surface is). Don't take this the wrong way, but what gives you the idea that "being honest" or "trying to fool people", is better/worse, than someone with an opinion, like your opinion, that there are "corrections" that you, or any other opinionated type has to make? Or at least, imply? What's wrong with saying "the wrong thing"? You've just said a whole heap of stuff that looks a bit opininated, in my opinion. You should stop assuming things (maybe you have a need to believe something other than the stuff you appear to be attached to just now). At least that's my opinion. Does a (vector) space have a requirement that it has to be regular, or can it be "irregular"? Semiregular is like "not fully regular", or something? How many kinds of "regularity" are there? Algebra terminology sometimes uses what looks like contradictory meaning, but you have to see a deeper meaning; often the terminology describes some dual relation, or operation, and the words are constrained in meaning by the relation, so you have to understand the relation and how it works, or what it's made out of.
Well first, I don't think it's nice to call a Riemannian manifold a "space"; I prefer to think of it as a geometry on a space, in this case a topological space. But it's a matter of taste, perhaps. But more importantly; Riemannian manifolds are not a generalized anything, in fact they are highly specialized bits of kit. Um... let's try this "order of specialization" Set > topological space > manifold > differentiable manifold > Riemannian manifold. (and the first person to tell me that a set is not greater than a topological space gets a slap. OK?) I have to say, though, that I tend to agree with Alpha; from what I have seen, your math skills are not highly developed. That's OK, there are things we all need to learn, so this comment is not a put-down. But you do have a tendency to parade your lack of knowledge with fists flailing, even when faced with a bunch of heavyweights. It is probably a struggle for them to remain courteous under these circumstances. Like Alpha said, you would get a far better response if you candidly admitted ignorance and, in a friendly (not necessarily importunate) way, asked for advice. That way you can learn stuff Admitting ignorance is no sign of weakness, in fact it's a sign of strength of character.
Acting like an "ignorant" questioner, is sometimes a more rewarding kind of tactic to use. There's far more stuff about math, just for starters, that I don't understand than what I can say I do understand. Saying things like "numbers have a shape", or "you can hear numbers", is either dumb and ignorant, or thought provoking. I agree that it's largely a matter of taste. (And I don't "flail" anything, except a lot of "dumb" words around, and also opinions that look like they could use a good thrashing) OK. This implies that there are different, discrete kinds of spaces, and at least two members of "the set of spaces", are 1) vectors and 2) topological "whatsits" (these must be geometrical, because that's a synonym for "topological"). Can you explain "why" a vector isn't topological/geometrical? I'm having trouble visualising this property of vectors. This is maybe one of the contradictory-looking things that you advanced mathematicians seem to be able to brew up.
