I wanted to say a few words about manifolds, even though I would hardly call myself an expert. Why? Because to give two examples; spacetime, which has been frequently talked of here, is in fact a manifold. Perhaps, surprisingly, the Lie groups \(SO(n)\) and \(SU(n)\) that Ben is fond of, are also manifolds. So it seems that manifolds are important to physicists. They are also interesting in their own right, I dare say. Anyway, all contributions, positive or negative are welcome. So a manifold is a topological space with certain additional properties, so I may as well begin my outline with what a topological space is. To start us off, I'll define the power set. Let \(S\) be a point set. The power set on \(S\) is the set formed from all possible subsets of \(S\). So if, for instance, \(S=\{a,b,c\}\) then the power set \(\mathcal{P}(S) = \{\{a\},\; \{b\},\; \{c\},\; \{a,b\},\; \{a,c\},\; \{b,c\},\; \{a,b,c\},\;\emptyset\}\). You may note three things from this: elements in \(\mathcal{P}(S)\) are sets; \(S\) and \(\emptyset\) are always subsets of \(S\); \(\{a\}\) is called the singleton set and is not the same as \(a\).This is so, because, whereas \(a \in \{a\}\) is always true, \(a \in a\) is meaningless. So a topology \(T\) on \(S\) is a subset of \(\mathcal{P}(S), \;\;T\subseteq \mathcal{P}(S)\) that satisfies the following; finite intersection of elements (sets, recall) of \(T\) are in \(T\); arbitrary union of elements in \(T\) are in \(T\); \(\emptyset \in T\); \(S \in T\). The ensemble \( S,T\) is called a topological space. I'll give a couple of examples straight away, If \( T = \{\emptyset, S\}\) then \(S,T\) is said to have the trivial, concrete or indiscrete topology. In contrast, if \(T= \mathcal{P}(S)\), then \(S,T \) is said to have the discrete topologogy. Elements in \(T\) are called the open sets in \(S,T\). I'll get to closed sets in a bit. Suppose for some arbitrary set \(X\) I have \(Y\) as a proper subset, that is \(Y \subset X\). Then the set of elements in \(X \) not in \(Y\) is referred to as the complement of \(Y\) in \(X\), and is written \(Y^c \subset X\). OK, so the closed sets in \(S,T\) are simply those sets in the complement of \(T\) in \(S,T\). Note that the notion of openness and closedness is not exclusive: sets can also be both open and closed or neither. For example, the \(\emptyset\) is open, as it is always in \(T\), likewise \(S\). But \(S\) is the complement of \(\emptyset\) and \(\emptyset\) is the complement of \(S\), so these sets are both open and closed. Just in case this way of defining open and closed sets is making anyone feel queasy, let me say this. Elements in what's called the standard topology on \(\mathbb{R}\) are intervals of the form \((a,b)\). These are the open sets in \(\mathbb{R},T\), which you can easily check satisfy the axioms above; we also recognize these as open intervals. In particular, note that, by the axiom of arbitrary union, I must have \((-\infty, \infty) \in T\), which we recognize as the real line. Often one doesn't really care what the precise topology on a set is, so It is usual to slightly abuse notation and write simply "let \(X\) be a topological space". So I guess that's enough definitions to be going on with. There may be more to follow, if you want, but don't worry; once we have an understanding of topological spaces, manifolds, in their simplest form, follow quite easily, I think. Mmm. Looking back on this, I wonder if it isn't too dense to be comprehensible - I can promise you, though, that it is correct
Quark--- Thank you for starting yet another informative thread. It looks like you are starting with point-set topology. I think that everything I am familiar with falls under algebraic topology, and I don't know how the two are related. Unless you have a strong opinion, I will follow along untill the end, then give a few examples of the things that I know.
Yes, Ben, I am doing point-set topology. The algebraic variety I know less about, having only covered homotopy theory. But it would be fun to discuss this later. Now, when I first heard of the definition of open and closed sets given in my last post, I went into a kind of mental tail-spin, so I'll say a bit more about this, which may calm a few nerves. It requires some more definitions, though, I'm afraid. But don't panic! In most cases, your visual intuition will be a perfectly reliable guide. Let \(X\) be a topological space. For any subset \(A \subset X\), the interior of \(A\), written as \(A^{\circ}\) is the largest open set contained in \(A\). The closure of \(A\), written as \(\overline{A}\) is the smallest closed set containing \(A\). We then have the nice construction: the boundary of \(A,\;\partial A\) is simply the set such that \( \partial A = \overline{A} - A^{\circ}\). Obviously, if \(A\) is open, then \(A = A^{\circ}\), and if \(A\) is closed, then \(A=\overline{A}\) and moreover, \( \partial A \subset \overline{A}\), and \(\partial A \cap A^{\circ} = \emptyset\), always. I hope this is some comfort. Ah - but note the rather unfortunate but quite standard notation: the boundary \(\partial A\) is not a derivative, it's just a label! So let \(p \in X\), where \(p\) is some point in \(X\). If there is an open set \(U\) containing \(p\), I will call \(U\) a neighbourhood of the point \(p\). This is not hard to see, but it will become important, so hang on to it! Now let's give our topological space some properties. The topological space \(X\) is said to be connected iff it cannot be written as the union of any two non-empty disjoint sets. We recognize two "levels" of connectedness; if a space is path-connected, I mean that any two points in \(X\) can be connected by some continuous path, even though there may be a "hole" in \(X\). If I say that \(X\) is arc-wise-connected, I mean that any path between two points can be continuously deformed to another such path - therefore "holes" are not permitted. The space \(X\) is said to be a Hausdorff space iff, for any two points \(p,\;q \; p \ne q \in X\) there are neighbourhoods \(U \ni p,\; V \ni q\) such that \(U \cap V = \emptyset\). Essentially, this one several possible criteria which allow me to say whether two points are the same or different in my topological space. There is another important topological property I need to address - compactness - but l "have" to go a drinks party now. If I get back later, I will be er umm, how shall I put it? Informal?
OK, compactness. Let \(X\) be a topological space. If every convergent sequence in \(X\) has a limit in \(X\) then this space is said to be compact. This is not the definition you will find in topology texts, but it will suit our purposes, I think. Very loosely, this means that \(X\) is no bigger than it needs to be, or, you might say that \(X\) is bounded. This more-or-less coincides with our intuition about compactness. Good. Suppose \(X,\;Y\) are topological spaces. The function \(f:X \to Y\) will be continuous at \(x \in X\)iff, for each neighbourhood \(U \in X\) of \(x\) and each image neighbourhood \(V \in Y\) of \(f(x)\), the pre-image set \(f^{-1} (V) \subseteq U \in X\) is a neighbourhood of (open set containing) \( x \in X\). (This rather cumbersome definition actually coincides with that which we learn in elementary real analysis; simply set \(V = \epsilon \in Y,\; \delta = U \in X\)) Suppose now there is some continuous function (as defined above) \(g:Y \to X\). In the case that for all \(U \in X,\; V \in Y\) that \(g^{-1}(U) = V\) and \(f^{-1}(V) = U\) then \(f,\;g\) are seen to be mutual inverses, and one says that \( g = f^{-1}\). Under this circumstance, the function \(f:X \to Y\) is said to be a homeomorphism, and \(X\) and \(Y\) are homeomorphic. (This is the topological version of isomorphism). So, we are now ready to give a definition of a manifold. Later for that!
Ya know what? I'm losing heart here. Anyone want me to continue, or would you all rather have "hummor to a very difficulty subject" at my expense?
Quark--- Please continue, at least for mePlease Register or Log in to view the hidden image! Note to all: Please keep discussions in the SR threads strictly on topic. Thank you.
First let me apologize to all, and especially Billy T, for my infantile outburst last night. Sorry. OK. What is a manifold. Very loosely speaking, it is a geometric object whose geometry is locally Euclidean. I can make this more precise as follows: Let \(M\) be a topological space that is a Hausdorff space, is path-connected and compact. Then \(M\) will be called a topological manifold if; a) each point \(p \in M\) is contained in some open subset (neighbourhood) \(U\) of \(M\); b) all functions in and out of \(M\) are continuous; c) for each neighbourhood \(U \in M\) there is a continuous, invertible map (a homeomorphism) \(h: U \to R^n\) for some \( n \in N\). So, that's your most basic definition, and you will see straight away that \(R^n,\; \forall n \in N\) is trivially a manifold. In a bit we'll put bells and whistles on this definition, but let's see what my set-up implies. By the connectedness property, I must be able to find a collection \(\mathcal{C}\) of open sets such that \(M\) is the union of all \(U \in \mathcal{C}\), and there is no \(U \in \mathcal{C}\) for which \(U \cap V= \emptyset, \;\forall V \in \mathcal{C}\). This is simply to re-assert that \(M\) cannot be written as the union of two disjoint non-empty open sets. Now recall that the elements of any \(R^n\) are a "system" of \(n\)real numbers \(x^i, \; i= 1, 2, ..., n\), called an n-tuple. So, like elements in \(R\) are of the form \(x^1 = x\), elements in \(R^2\) are of the form \((x^1,x^2)\) and so on. (The superscripts are not, of course, exponents, merely labels). So my definition (c) above implies that, for each \(p \in M\), and each \(U \ni p\), there corresponds an n-tuple \(x^i \in R^n\), given by the homeomorphism \(h: U \to R^n, \;h(p) = x^i\); one says that the neighbourhood \(U\) of \(p\) has assigned to it the real coordinates \(x^i\). But note this crucial point: My definition of a manifold does not allow me to infer that, for some \(V \ni q\) the above homeomorphism still applies: rather I must allow that there may have to be some other homeomorphism \(g:V \to R^n, \;g(q) = x^j\) My definition (a) of a manifold, together with the connectedness property, implies that for each \(p \in M\) there are open sets \(U,\;V\) such that \(p \in U \cap V\). We now have the following situation: there is a homeomorphism \(h: U \to R^n\) and \(g: V \to R^n\), and \(p \in U \cap V\). A schizophrenic dilemma for poor old \(p\) - am I best described by the n-tuple (coordinates) given by \(h: U \to R^n,\;h(p) =x^i\) or by the coordinates \(g:V \to R^n, \; g(p) = x^j\)? What a cliff-hanger!
Now this looks like really tough stuff, right? Actually, you all know the basics already, here..... Consider the problem of representing a map of Earth on a 2-plane, i.e. a sheet of paper. It can't be done without introducing unwelcome discontinuities, at, say the poles and at some meridian (usually on the Bering Straight). But the solution is simple, as you all know: For each arbitrarily small area on Earth we can devise a local map, each of which has a coordinate set, often called a grid, and each of which overlaps (usually) four adjacent maps. You will have, I dare say, a book of such maps in your car; perhaps you call it a road atlas? We are doing nothing more fancy than this, conceptually. Each page in your road atlas I am calling a coordinate neighbourhood of the manifold Earth, and I am merely asking the following question: If my house is in the overlap between pages 4 and 5 in this book of maps, and if it is grid reference 4D8 on one page and grid reference 5A1 on another, how do I relate these grid reference to each other? The basic idea here is really no more more profound than this, though some of the arithmetic is quite challenging. But here's the thing; coordinate neighbourhoods of a manifold are referred to as charts and the collection of all charts that cover a manifold is called an atlas. Geddit?
Not sure. Perhaps it is the advanced notation, that is blocking me. For example, I have forgotten what the inverted A means (Is it "there exists") i.e. it would be helpful at least to me if you were to offer a "dictionary" (or a link to one.) then perhaps I might have an on-thead comment or two. Based on your last post with the "2D map of globe" example, it would seem that the scheme used to define the "local domains" contains the answer to your main question (stated in that post also.)
OK, Billy, consider it done! In fact I propose a stand-alone thread, give me a few moments. But first this. Recall I included in the definition of a manifold that there are continuous invertible maps of the form \(h: U \to R^n\) where \(U \ni p\) is an open set containing (aka "neighbourhood" of) \(p\). Note that, by our topological definition of continuity, this implies that the function \(h\) has an open subset of \(R^n\) as its codomain. So far, so good I think. Let's be posh, and say that a continuous function is a function of class \(C^0\). Recall also that we required that \(\forall p \in M\) (M is our manifold), we must have \(p \in U \cap V\). So the following must be so; since \(h: U \to R^n\) and \(g: V \to R^n\) are invertible, then I may have \(g^{-1}\;\circ\; h: U \to V\) and \(h^{-1}\; \circ\; g: V \to U\). So let's now return to the existential crisis of \(p \in U \cap V\), where we have the \(x^i\) as the coordinates on \(U\) and \( x^j\) as the coordinates on \(V\). Since we may assume that \(p \in U,\;p \in V\) are the same point, we must be able to specify \(p\) by reference to either the \(x^i\) or the \(x^j\) in such a way as to leave \(p\) unchanged; the argument above tells us this can bo done by a transformation \(U \to V\), and specifically by a coordinate transformation \(x^i \to x'^j\) and vice versa, where I have introduced the prime to indicate this fact. So can I write, say \(g^{-1}\; \circ\; h(x^i) = x'^j\)? No, because this implies that \(g^{-1}\; \circ \;h(x^1) = x'^1,\;g^{-1} \;\circ\; h(x^2) = x'^2\) and so on, and this cannot be right - each \(x'^j\) must depend on all the \(x^i\). So I write; \(x'^1 = f^1(x^1, x^2, ......, x^n)\) \(x'^2 = f^2(x^1, x^2,......., x^n)\) .............................................. \( x'^n = f^n(x^1, x^2, ....., x^n)\) where, obviously, the \(f^n\) are \(n\) distinct functions, each in \(n\) variables. Let's write this succinctly as \(x'^j = f^j(x^i),\; i,j = 1, 2, ..., n\). And note that as, from the above, we see there is strict one-to-one correspondence between the \(x'^j\) and the \(f^j\), so with an ever so slight laxity in notation I may reformulate the above as \(x'^j = x'^j(x^i)\;i,j = 1, 2,...n\) Hmm....I've a ghastly feeling this is far too detailed for most tastes. Anyway, I'm outta puff for now!
Thanks I will look for it (The "dicitionary") but your text above, despite not being sure of \(h: U \to R^n\) symbols seems to remind me of what in an earlier era was called "1 to 1 conformal mapping." However; unlike my old conformal maping concepts, I think you only want that the possible discrete elements of both U an R be "path connected" -i.e. keeping open the possibility of being only like the caulk lines of a tile floor, and not the entire floor. If you think it too much work to bring me (and others, equally ignorant) up to speed so we can follow you, that is OK. I may drop out anyway as I will never have need of any of this, so perhaps wise for you to only put a translation of each new symbol when you use if first time in separate line after the first use instead of full dictionary. Your right and left sort of € like signs remind me of unions of sets but I do not think that correct and do not know what the right vs left version indicate. Sorry I am so ignorant of your notation. Does "open set" mean that the elements of the set are not countable or that it can be extened as much as one likes or both? Let me try to put into english what I am guessing your notation is stating (more to give you an idea of what I need than to expose my ignorance, but it does both.) Assume a function, h, that can transform all of the members of the set U into region R (of dimention n, if need be) at least for all elements of U which are within some (not sure how) defined neighborhood of element p (of the set U, I think, not R)
The inverted A means "for all", and € like signs mean "contained in" or "contains" depending on where the the fork heading.
Thanks that is exactly the type of thing I need to read QuarkHead's post. (Understanding them is a separate question, that comes later.) Also again thanks QuarkHead. At least the notation is reasonable - perhaps and Old timer, like me, can even remember it with out frequent fer to you new thread. Now I will try to read to see if I can understand your posts.
Billy T doubts and questions raised by post 1: With these 4 conditions, especially the first two (intersections and unions), how can topology \(T\) on \(S\) be less than the “power set” of \(S\) ? It would seem to me (but surely I am wrong) to be identical to “power set” of \(S\) Perhaps what your are stating is there is a sub set of the power set (not all of it) which with unions and intersections added is “expandable” to the full power set of \(S\)? For example with sub set, Ss = {{a,b}, {b,c}, {c,a}, \(\emptyset \\)} of the power set of your set S = {a,b,c} various intersections will give me {{a},{b},{c}} and the "tripple union" gives {a,b,c} and unions of the sub sets of Ss, especailly with \(\emptyset \\) and itself, if needed, reproduces itself and the other three sub sets of Ss for the full "power set of S." How does an “ensemble” differ from the union S & T? Probably not yet seeing this as did not understand how the topology conditions could produce anything other than the power set. I will not ask more now in detail now, or comment, but note that I do not know meaning of following two symbols: \(\mathbb{R},T\) but assume it is an “ensemble” as you separate with a comma and as the “after comma part” is a topology, probably this is also “topological space.” \(\mathbb{R}\) And not sure what “intervals” of the form (a,b) are. I doubt you are restricting this to be “real numbers” between a & b. Power set, and complement sets concepts seem very clear to me. Not much problem with “open and closed” sets either, but since they are not always mutual exclusive, hard for me at this stage to anticipate what these adjective concepts might be useful for. The probability of my “dropping out” is growing, but I will stick with it a little more as would like to be able to at least read with understanding what you are saying.
Perhaps it is worth mentioning that for any set S, T={emptyset,S} is a topology (called the trivial, indiscrete, or antidiscrete topology). You can check this T satisfies the 4 conditions (axioms). I think this answers your first question. ps: Quark, I am also learning differential geometry and love it. I hope you don't mind if I occasionally pitch in and try to answer some questions.
Welll bloody analyzed!!!!!!!!!!!! If anyone denies this work, they are intetipnally being dogmatic. Please Register or Log in to view the hidden image!
''In particular, note that, by the axiom of arbitrary union, I must have , which we recognize as the real line. '' Just too add, if you don't mind, that the mentioning of: ''In particular, note that, by the axiom of arbitrary union, I must have , which we recognize as the real line.'' can be seen in rotational Lorantzian Coordinated math. r' and r.
temur: Great! Any assistance you can give would be very welcome. (I thought I had mentioned the indiscrete topology. If not, I should have - pathological it may be, but an important example nonetheless) Not quite sure what you mean. First note, as temur said, the powerset \(\mathcal{P}, (S)\) cannot be derived from the topology \(T = \{\emptyset, S}\). Second, note the key phrase finite intersection. Quite obviously, in general, infinite intersection always descends to singletons, which are closed in all topologies, and also open in only a minority. Third, note that the property of finite intersection an arbitrary union are just that - properties. They are not how the topology is arrived in the first instance. Fourth, note that \(\{\{a,b\},\; \{b, c\},\; \{c,a\},\;\emptyset, \;\{a, b, c\}\}\) is not a topology. \(\{a, b\} \cap \{b, c\} = \{b\}\), for example, which is not in your topology. Hugely! S is a set of points, T is a set whose elements are subsets of S. I can't convince myself that the "union" of a set with one of its elements makes sense. And even if it did (any thoughts, temur?) \( S \cup T = T\) a topology on \(S\), not a topological space. Just think of the ensemble as describing a certain structure - a topological structure! - on S. EDIT to add: Let's say you have a set \(B\) of bricks. Then a house \(H=B, T\) is (let's say) the set \( B\) endowed with certain structure. Bingo! That's the general idea, yes. Oh but I am! That's what the set \(\mathbb{R}\) is, a set of real numbers. Just to remind you; an "interval" only makes sense in a metric space, which \(\mathbb{R}\) is, when considered as a topological space (some other sorts of spaces too, but that's of no relevance here) Please don't drop out - it's good to see you thinking about this
You can take union of S and T formally, but then it would be just a set where points and sets of points "mixed" together. I think you need some way to differentiate which is the base set and which is the set of subsets that defines the topology. So they need to be separate.
