Well, using the literature I have in front of me, the collection of functions \(fi_1i_2...i_k\) where the latter identity function \(i_k\) with the subscript \(k\) is a type of label to map out the summing over all possible values of \(i\) on something like \(dx^{i_1} \wedge dx^{i_2}... \wedge dx^{i_k}\). Is what is happening in the case of (k+n) a differential form called the exterior derivative?
A friend has just alerted me to the fact that my post history appears as a subtext in this thread. Mister Reiku, drop it: as I have not so far been a participant, it is like talking about someone behind their back. It has nothing to do with this thread, or your ability to even half understand the answers that may be given here.
Why should I? I HAVE BEEN BANNED MUCH LONGER THAN YOU HAVEN'T BEEN AROUND! AND YET MY NAME HAD STILL BEEN RASIED IN DEBATES WITH OTHER MEMBERS NOT LINKED TO ME TO SLAG THEM OFF! What makes you think you are so special? Why should I give you any leniency anyway? Any chance you could get to you slagged me off!!!
And the person who raised my name behind my back was AN. Secondly, it has everything to do with this thread, if the likes of Guest who are coming in slagging me off, alphanumeric too, but you were shown some kind of pardon to alphanumeric whilst guest couldn't stand you in the slightest.
And in case AN forgets what I am referring to, whilst speaking to likes of Pincho, magnetman ect, he's often corrected them on things with added statements like ''you show all the attributes of a previous member here, REIKU under the dilusion, blah blah blah'' Get real QH! Life's cruel eh? Well go and reflect on how you treated me along with your other crony pals.
Mister Reiku: Either you want your question answered, or you would prefer to engage in a slanging match with virtually all members here. Decide what you want. I repeat, are you sure that you are equipped to understand the answer to your original question? There seems little point in pretending if you are not.
''Either you want your question answered, or you would prefer to engage in a slanging match'' does engaging take away the fact that four or five members pounced on the discussion of this thread to hope for a slanging match? answer that question. Most of these people are your friends, minus one. They are to blame as much as the fish bites. Do as you preach in future, not as you act!
I think it's time for this thread to take another break. Comments and criticisms to the usual address.
The thread is now reopened. Please do not let it descend into a flame war and stick to the topic, which is the Hodge star. This thread has already been locked and unlocked twice, which in my view has been extremely lenient. Should there be another closure there will not be a reopening.
It is a map on differential forms on N dimensional space. It maps p forms to N-p forms. I did ask Mister for an explanation but he fell short. I could go into a lot of detail but I might get accused of showing off by Mister... Its a very interesting area though, you can phrase pretty much all major physics jn terms of differential forms so its something any theoretical physicist should at least be aware of.
I already told you that I don't have a clue what a hodge star is - other than what AlphaNumeric wrote about it above - and I'm not sure I understand all that either. As to your reply: As I understand it, there is a difference between a function and a functional. The definition you cut and pasted earlier referred to functionals, not functions. From this, I take it that you have no idea what a functional is. Which makes me wonder how you hope to be able to phrase a relevant question about the Hodge star for which you will comprehend the answer. You didn't answer my question. From this I take it that you don't know what makes a function linear, let alone a functional. This doesn't answer my question, because it doesn't distinguish between functions and functionals - obviously. I'm asking you, not the other way around. If you're asking about the Hodge star, I would assume you have the necessary background to understand the answers you receive. It appears to me that you do not. From (5) it appears to me that you don't know the difference between a vector and a form. Why are you asking about the Hodge star if you don't know that? From (7) it appears to me that you don't know what a vector space is. I understand enough of the background here to appreciate that you don't have clue what you're talking about or asking about. That doesn't mean I know all about this stuff; I have already said I do not. My purpose there was to get you to see that multiplying two square matrices always results in a matrix and never in a single number. I hoped that I could get you to see this by asking you to consider a general case.
Even with the qualifier that it was easy (and the bad form of replying to a banned member's post) I can't let this go without comment. You seriously say that you had an assignment teaching you to draw, on paper, functions like \(y=ax+b\)? In college? You realise that this is stuff taught to 10 year olds?
Then you should be quaking in your boots. As I am not one to bear a grudge, let's see if we can help Mister Reiku. But I warn him/her there will exercises along the way. Failure to even attempt them will see your help withdrawn, at least by me Suppose first the n-dimensional linear vector space \(V_n\). Exercise: what exactly does this statement mean? Now suppose the vector space \(\Lambda^p(V_n)\) of dimension \(\begin{pmatrix}n\\ \\p\end{pmatrix}\) which very roughly speaking means from n choose p. This is called the space of p-vectors. Don't fret about the notation - it is just that and arbitrary, like all notation The definition above of the dimension of this space has a name, widely used in mathematics. Exercise: what is it? Exercise: find another way to write this. Hint - it involves a bunch of factorials. So the Hodge operator, as Alpha said is the mapping \(\ast: \Lambda^p(V_n) \to \Lambda^{n-p}(V_n)\). Very roughly speaking again, you may think of this latter space as that of those vectors which are "left over" having chosen p from n in the former. Exercise: What is the dimension of this vector space? If you do this and the previous exercise right, you might (just might) be a bit surprised, given the notation. You may find some of these exercises hard, I dunno, but if you have no clue where to start with them, I suggest you hit those books and come back with some more elementary questions.
Well, you're talking mince in that case. Nothing I have recited is anything about functionals, yet as usual, you try and make this look more dubious than it really is! h t t p ://mathworld.wolfram.com/HodgeStar.html Had to find it to show you are talking mince again. No mention of functionals here apart from my cloudy mind to trying to recall something which I named wrong. So exactly where have I presented, ''a cut and paste'' of functionals might I ask? You didn't answer my question. From this I take it that you don't know what makes a function linear, let alone a functional. I am not really to concerned about functionals. I already said to you, in a previous post I mixed up my terminology, so why in Gods name would you bring it up again? I am quite aware of what a bloody linear function is. As I have explained, I did very basic examples as a cover of what we could do in College. And I mean it was very brisk! (To much of fudge muffins laughter and anothers) This doesn't answer my question, because it doesn't distinguish between functions and functionals - obviously. James, are you trolling? Considering my very first reply was: ''1) Did I say functional, I should have said, function, maybe then it would have been clearer? '' You are clearly trolling. After I have written about myself mistaking the terminology, you continue to repeat it, over and over again. This reminds me of when alphanumeric said ''Just admit when you are wrong and nothing else would be said about it.'' I said something along the lines of ''who you trying kid, I know fine well that will never be the conduct towards me.'' James, you are clearly showing the same behaviour. And in light of this, I have every reason to ignore your post now. You are clearly ignoring the fact that I've admitted not the word functional, but function. (plus no where have I posted about functionals, not in your vocabulary anyway!) I understand enough of the background here to appreciate that you don't have clue what you're talking about or asking about. That doesn't mean I know all about this stuff; I have already said I do not. Well, paint me surprised, that almost sounded a little hypocritical James! Atleast I know some mathematics to ask my question with, you come into my thread accuse me of not knowing what I am asking, but equally you admit you've never even heard of a hodge star!!!!!!! Get real! Did you also completely ignore my OP? Anyone with a brain will see I came across some math work I was doing which was involving the use of the Hodge dual. I admitted in the OP I knew next to nothing about the hodge dual, but explained what I did know, yet you in a completely disingenuous attitude come in here and proclaim I don't know what I am talking about, yet equally admitting you know s**t as well! lol So ultimately... what are you blaming or accusing me of? Admitting that I have said I know not much about hodge duals? Because if you had my way, you'd rather ignore that part and just state the obvious anyway? :/ QH Ok, for now, bygones be bygones. Let's hope our discussion will be more fruitful. Suppose first the n-dimensional linear vector space . Exercise: what exactly does this statement mean? Well, I know of a very typical case in physics which explains an n-dimensional vector space, that being \(R^n\). It is a little offensive to my general knowledge by some here that I don't even know what a linear vector space is, I mean this stuff is pretty much damn easy as I have explained before. In the most simplest notation that I know of, an n-dimensional vector space is given as \((v_1,v_2... v_n)\) The definition above of the dimension of this space has a name, widely used in mathematics. Exercise: what is it? It's a binomial coefficient. Exercise: find another way to write this. Hint - it involves a bunch of factorials. This has me a little stumped. Are you perhaps wanting a polynomial expansion? Or is it something more technical? Exercise: What is the dimension of this vector space? If you do this and the previous exercise right, you might (just might) be a bit surprised, given the notation. Your terminology has me confused. I don't know exactly what you mean by ''dimension''. My use of the word covers a lot of things, like the dimensions of an equation, the dimension of single quantity ect. Something tells me you are talking about a Hamel Dimension, and that is out of my league concerning mappings. Anyway, I am sure no one really needs to delve into mappings to explain a Hogde star. I have seen plenty explanations which don't even think about going near mappings. Anyway, despite all that, I couldn't wait any longer to reply to this. Being banned a second time will be my third and last. I have no intentions ever coming here again. I am not even going to give a big farewell note to the cliques right now. You know who you are and you've made me terribly uncomfortable here the last week or so. Anyway, AN, prom ect, it don't matter. I am sending a message to james saying that I want my Mister account perminently banned and that it was generous of him to let me back for some time, even though I've had no peace. Nearly every one of my threads under the Mister account has been locked and that is no reflection on me in general, it's how some here want to treat me and if that is an ''ok'' policy here, then I'd suit myself better to just never return. Bye
Right, let's go through Mister's comments and beat them with a stick. Differential forms, specifically 1-forms, are functionals on the space of vectors. To follow on from stuff others have said given a vector space V you can define the space of functionals, let's call it V*, such that if v is in V and X is in V* then X(v) is in some field, typically the Reals or Complex numbers. Now let's just show off because I can.... In the usual construction of differential forms we start with some differentiable manifold M, which has a tangent bundle TM. Each point \(p \in M\) has a tangent space \(T_{p}M\) (such that \(\bigcup_{p}T_{p}M = TM\)) and we can define the dual to this tangent space, \(T_{p}^{\ast}M\), and union all of them into a co-tangent bundle \(T^{\ast}M\). This, once you crunch through the details (see Chapter 5 of Nakahara's Geometry, Topology and Physics for an excellent overview. Actually the whole damn book is awesome), is the space of 1-forms. Since, by some fundamental linear algebra V and V* have the same dimensionality we can thus conclude that the dimensionality of the co-tangent bundle is the same as the tangent bundle. In fact, since we know the dimensionality of the tangent space at p is equal to the dimensionality of M (by definition almost, call it N) we know the dimensionality of the tangent and co-tangent bundles, 2N. Thus at any point we have basis vectors \(\partial_{i}\) and basis 1-forms \(\textrm{d}x^{j}\) such that \(\textrm{d}x^{j}(\partial_{i}) = \delta_{i}^{j}\). Thus we can form linear combinations, \(\textrm{d}\lambda = \lambda_{i}\textrm{d}x^{i}\) or \(v = v^{j}\partial_{j}\) and we can use the slick notation \(\textrm{d}\lambda(v) = v(\lambda) = v^{j}\partial_{j}(\lambda_{i}x^{i}) = v^{j}\lambda_{i}\partial_{j}(x^{i}) = v^{j}\lambda_{i}\delta_{j}^{i} = v^{j}\lambda_{j}\). Indeed as we'd require from the dual nature! Now to make p-forms you build antisymmetric products of the 1-forms. So if I'm building p-forms how many different ones can I build? Well I'm picking p from N, not caring about the order. That's N-choose-p, ie \(\frac{N!}{(N-p)!p!}\). If I consider the entire ring (look that up) of forms then I have a total of \(\sum_{p=0}^{N}\frac{N!}{(N-p)!p!} = 2^{N}\) (who has a proof of that sum? It's really nice). I can now act these on objects with multiple vector structures, giving me a HUGE space of functionals on an even larger vector space of objects! You don't even know the meaning of the word. You should be concerned if you want to understand differential forms. There's a function in the sense of y=mx+c and a linear operator in the sense f(ax+by) = af(x) + bf(y) for a,b scalar and x,y vector space elements. That's the sort of linear operator differential forms are. y=mx+c has nothing to do with this stuff. Because you never admitted you were wrong. How can we respond to a behaviour you never do? Differential forms are functionals. You obviously missed that when you got your degree from the University of Copy and Paste. Nothing you could possibly be doing would involve the Hodge star in a valid way. You are simply too poorly educated to be doing anything related to it. It's not even taught during the Cambridge degree! You have to wait till you do their Masters course before you learn differential forms and the Hodge star! I didn't take that course and ended up learning it during my PhD and my thesis uses tons of differential geometry. You are simply lying. Everyone here sees it and you simply can't accept no one is swallowing your nonsense. No, you lied and claimed you know more than you do. You cannot accuse me of 'knowing s**t as well', I have published work based on generalised Hodge star maps and cohomologies applied to spaces which don't even have proper names yet! I am firmly saying you don't know anything about this which couldn't be gleaned from an afternoon on Google and Wikipedia. All you've shown if you can Wiki 'vector space'. No, that's perhaps an example of a basis. A vector space is not a set of n vectors, it is a topological space with additional structure. Feel free to Wiki 'topological space'. It's essential in the definition of a manifold too. You just failed a very simple test. The dimensional stuff I mentioned above is what was being referred to. The fact that N-choose-p and N-choose-(N-p) have the same dimensionality means that there could be an map between them which is an isomorphism. The Hodge star is such a map. This is an example of you showing how little you understand. You've clearly Googled for the word and found it's used in loads of ways. The context makes it obvious to anyone who understands it. Many textbooks make comments like "We will abuse notation and say..... except in instantances where it might be conflicting. In all other cases it'll be clear from the context". What they mean is it'll be clear if you follow WTF the book is on about. You don't. Nice try, throw out a random Google result and see if you hit on the right one. It's a ****in' map from the space of p-forms to the space of (N-p)-forms!! It's clear from the definition, which you couldn't give! My spidey sense is telling me this is a lie. I'm sorry you don't like people pointing out you're a liar and a fraud. Might I suggest you lie less in future? It'll serve you well in life if you follow that advice. My god, learn to take some damn responsibility. From your very first posts, before I suspected you were Reiku, you were spouting BS and I asked you not to. You are an obvious and terrible liar and you need to learn to deal with what is quite clearly a problem you have.
Moderator action : Trolling posts by trolling now banned poster removed For those who cannot see deleted posts (sorry lesser mortals) my spidey sense that Mister would return was correct. Hands up who is surprised? ... Thought as much...
\( { N \choose k } = \left{ { \frac{N!}{k! \, (N-k)!} \quad \quad \quad \textrm{if} \, 0 \le k \le N \\ \quad \\ 0 \quad \quad \quad \textrm{otherwise} } \right. \) Proof via Pascal's Triangle: \({ N \choose k - 1 } + {N \choose k } = \left{ { 0 + \frac{N!}{0! \, (N-0)!} = \frac{(N+1)!}{0! \, (N + 1 - 0)!} \quad \quad \quad \textrm{if} \, k = 0 \\ \quad \\ \frac{N!}{(k-1)! \, (N+1-k)!} + \frac{N!}{(k)! \, (N-k)!} = \frac{ N! \, (k)! \, (N-k)! + N! (k-1)! \, (N+1-k)! }{(k-1)! (k)! (N+1-k)! (N-k)!} = \frac{ N! k + N! (N + 1 - k) }{ (k)! (N+1-k)!} = \frac{ (N + 1)! }{ (k)! (N+1-k)!} \quad \quad \quad \textrm{if} \, 1 \le k \le N \\ \quad \\ 0 \quad \quad \quad \textrm{otherwise} } \right} = {N + 1 \choose k }\) \(\sum_{k=0}^{0} {0 \choose k } = { 0 \choose 0 } = 1 = 2^0\) \(\sum_{k=0}^{N + 1} {N + 1 \choose k} = \sum_{k=0}^{N+1} \left( { N \choose k - 1 } + {N \choose k } \right) = {N \choose -1} + \sum_{k=0}^{N} {N \choose k} + \sum_{k=0}^{N} {N \choose k} + {N \choose N+1} = 2 \sum_{k=0}^{N} {N \choose k} \) Then it follows by induction: \(N \in \mathbb{N}_0 \quad \Rightarrow \quad \sum_{k=0}^{N} {N \choose k} = 2^N\) Alternately, proof via Binomial Theorem: \((x + y)^N = \sum_{k=0}^{N} {N \choose k} x^{N-k} y^k \quad \Rightarrow \quad 2^N = (1 + 1)^N = \sum_{k=0}^{N} {N \choose k} 1^N = \sum_{k=0}^{N} {N \choose k}\)
