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Thread: Hodge Dual and Wedge Product

  1. #1 Hodge Dual and Wedge Product 
    Moderator Moderator Markus Hanke's Avatar
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    Calling all mathematicians on this forum - could someone explain to me please in simple enough terms the definitions for

    1. The Hodge Dual ( Hodge Star operator )
    2. The wedge product

    Never really got my head around this. If an actual example for a calculation could be provided for each of these, that would be perfect - understanding the actual meaning behind it is more important to me than mathematical rigour.
    Also, is there a physical meaning to the Hodge Star in particular, i.e. what is the physical meaning of an expression like this



    as appears in the Chern-Pontryagin scalar, a curvature invariant in GR.

    Any help on this will be much appreciated ! Not going anywhere with this, just trying to further understand this mathematical concept.


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  3. #2  
    Forum Professor river_rat's Avatar
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    Hi Markus

    Are you after motivation or definition here?


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Markus, you might like to checkout this ancient thread. You will notice I made some errors, and also that there is a lot more detail in there that you are after. Skim to find the bits that answer your question!
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  5. #4  
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    Quote Originally Posted by river_rat View Post
    Hi Markus

    Are you after motivation or definition here?
    I am after motivation and meaning, not so much rigorous definition. I basically want to understand the meaning of these operations, particularly in regards to their application in modern physics, i.e. GR.

    Markus, you might like to checkout this ancient thread. You will notice I made some errors, and also that there is a lot more detail in there that you are after. Skim to find the bits that answer your question!
    Thanks Guitarist, I will have a read through this.
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  6. #5  
    Forum Professor river_rat's Avatar
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    Ah ok, I guess the place to start would be to ask if you are ok with the reason for using forms as integrands and what they are trying to capture?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  7. #6  
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    Quote Originally Posted by river_rat View Post
    Ah ok, I guess the place to start would be to ask if you are ok with the reason for using forms as integrands and what they are trying to capture?
    I always think of forms as infinitesimal measurements in the geometric sense - that way I get a "point" element ( 0-form ), a line element ( 1-form ), a surface element ( 2-form ), a volume element ( 3-form ) and so on. The integral to me is then nothing other than an infinite sum of these forms, which hopefully converges, so I get a function, a curve, a surface, etc etc.
    This is probably not mathematically rigorous, and perhaps that is where my issue lies...?
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  8. #7  
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    In a nutshell yes, forms take things represented by vectors and give us numbers. We have a problem though, geometric concepts typically involve vectors while integration needs numbers. So forms are the ideal thing to use for integration of geometric concepts, as by definition they map vectors to numbers but more importantly do it in such a way that the linear properties of the integral are preserved.

    Now the next step is how we use these forms on vectors to extend this idea further. We have something that can take a single vector and give us a number but not all geometric ideas are represented by a single vector, and we need a way to build these higher dimensional objects from things we understand. This is what the exterior product helps with, it is the correct way of joining vectors together so that the properties we want to hold for a nice integral to be defined work.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
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    Quote Originally Posted by river_rat View Post
    Now the next step is how we use these forms on vectors to extend this idea further. We have something that can take a single vector and give us a number but not all geometric ideas are represented by a single vector, and we need a way to build these higher dimensional objects from things we understand. This is what the exterior product helps with, it is the correct way of joining vectors together so that the properties we want to hold for a nice integral to be defined work.
    So, in essence any higher dimensional element is thus the exterior ( wedge ) product, such as



    Is this basic idea correct ?
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  10. #9  
    Forum Professor river_rat's Avatar
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    Yep pretty much, the wedge product is just the right thing to use if you are worried about signed areas etc (which is what you are worried about when working out work done by a force, or flux through a surface). It is also co-ordinate free, so the same sort of construction works no matter what the actual co-ordinate system you are using is - which is one of the reasons we believe physical laws are actually written in terms of exterior products etc - it basically encodes our idea of relativity into the law.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  11. #10  
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    Quote Originally Posted by river_rat View Post
    Yep pretty much, the wedge product is just the right thing to use if you are worried about signed areas etc (which is what you are worried about when working out work done by a force, or flux through a surface). It is also co-ordinate free, so the same sort of construction works no matter what the actual co-ordinate system you are using is - which is one of the reasons we believe physical laws are actually written in terms of exterior products etc - it basically encodes our idea of relativity into the law.
    Nice, thanks river_rat. I got the idea now.
    So then, what's the deal with the Hodge dual ?
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  12. #11  
    Forum Professor river_rat's Avatar
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    mmm, at what level do you want a motivation for the Hodge Dual?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  13. #12  
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    Quote Originally Posted by river_rat View Post
    mmm, at what level do you want a motivation for the Hodge Dual?
    Geared towards application in physics, so at what ever level it needs to be at to achieve this. I have read the Wikipedia article on the Hodge Dual, and to be honest it is a bit too abstract for me to grasp the main idea behind it, so if it can be explained in a more straightforward way that would be great
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  14. #13  
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    Then I am lost! I linked you to a reasonably comprehensive post on p-vectors, differential forms and the Hodge dual,

    I also showed in that link how the exterior derivative combined with the Hodge operator (a couple of times over) can be used to extract the Laplacian,

    If you don't think this is "an application in physics" it might have been wiser not to include this operator in your GR thread.
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  15. #14  
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    Quote Originally Posted by Guitarist View Post
    If you don't think this is "an application in physics" it might have been wiser not to include this operator in your GR thread.
    I didn't say that. I think I just need to go through that thread in a little more detail one more time, because the first time I read it, it didn't really "click" with me.
    I'll go through it again
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  16. #15  
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    If I assume that not "clicking" means not understanding, then the fault was in my presentation.

    But ya know - you can always ask questions..........
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  17. #16  
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    Quote Originally Posted by Guitarist View Post
    If I assume that not "clicking" means not understanding, then the fault was in my presentation.
    But ya know - you can always ask questions..........
    I think your presentation was perfectly fine, just don't forget that I never studied any maths beyond college level, so for me some concepts take longer to sink in. That is not any reflection on your presentation, but rather on my lack of a suitable background.
    I am currently going through that thread in detail, and yes, I will be asking questions in the end. Purely conceptually I understand the hodge dual being a mapping from choice space to remainder space, and that makes sense to me. I also get the algebra linking it and the exterior derivative to grad, div, curl and Laplace operators; with that a formulation such as



    and



    makes perfect sense, and is very elegant. What I do not get at the moment is how the exterior derivative and the hodge dual are independent of the choice of coordinates, and also how the hodge star can be applied to any arbitrary tensor - there would need to be some connection with the metric tensor in such an operation.
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  18. #17  
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    Markus: First I apologize for hassling you - I had spent the day dealing with morons (and I should have known you are not one!)
    Quote Originally Posted by Markus Hanke View Post
    What I do not get at the moment is how the exterior derivative and the hodge dual are independent of the choice of coordinates
    Actually the proofs I know of are rather messy, and to be honest I just take it for granted. Let's see if I can cook up a simple one.......,
    and also how the hodge star can be applied to any arbitrary tensor - there would need to be some connection with the metric tensor in such an operation.
    Beware of equating p-vectors with tensors - usually it's fine, but from time-to-time it might lead to problems. But OK, consider this......

    Recall the cross (vector) product of vectors. Recall also this is ONLY defined on a 3-space with an inner product. So that with the usual metric. The definition of the cross product obviously requires us to know the angle between u and v and their relative lengths, and these are given by the inner product on a vector space.

    Let's use our fancy terminology and re-write this as , it being understood this simply means that for any choice of 2 vectors we can find another vector such that the vector (cross) product properties hold.

    But notice this is a mapping to a which is just how I defined the Hodge dual and the Star operator.

    Obviously the vector cross product is only valid in a metric 3-space. The Hodge Star is quite simply a generalization of the cross product operator, but if this generalization is to make sense, the metric (inner product) is still required
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  19. #18  
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    Markus: First I apologize for hassling you
    No problem, I totally understand. That sort of thing can be incredibly frustrating - I know from experience.

    The Hodge Star is quite simply a generalization of the cross product operator, but if this generalization is to make sense, the metric (inner product) is still required
    Wow, that makes sense ! Thanks Guitarist.

    Let's see if I can cook up a simple one.......,
    That would be good. If you can't, maybe just give a simple example of an exterior derivative and a hodge dual calculation in some coordinate system other than a cartesian one, so that I can see what happens. Whichever is easier for you.

    Thanks again for all your help. It is truly appreciated
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  20. #19  
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    Ok, after having gone through the thread referenced earlier in detail, and also having studied a few other sources, I think I kind of get the basic ideas now. Please excuse any misuse of notation in the following.
    If we have a p-form A in d dimensions, then this can always be written in terms if basis vectors such as



    Furthermore the volume form can be generalized as



    This immediately provides a connection to the metric tensor, and shows that this is quite independent of the choice of coordinates. The wedge product also naturally takes care of orientation and the appropriate signs because



    So far so good. The exterior derivative then maps p-forms to (p+1)-forms, wheres the hodge dual maps p-forms to (d-p)-forms. One can construct explicit relations similar to the first one above for both of these operators, however I am too lazy to type those out. Putting it all together it is then possible to generalize the three classic operators grad, div and curl as follows :

    Gradient - exterior derivate on a 0-form :

    Divergence - *d* applied to 1-form :

    Curl - *d applied to 1-form :

    Again, written this way everything is independent of choice of coordinates as well as how many dimensions there are - this is now obvious to me, since the above relations provide a connection to the metric tensor. This whole exercise is thus really just a way to write physical laws in an arbitrary number of dimensions, and independent of any particular coordinate system.

    As a practical example I consider a vector field in 3 dimensions :



    which is precisely a mapping from a 2-form to a 1-form ( and vice versa ), and which is really pretty neat

    Am I getting all this approximately right ??
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  21. #20  
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    Yeah, everything looks fine until we get here
    Quote Originally Posted by Markus Hanke View Post
    Gradient - exterior derivate on a 0-form :
    I do not understand your second equality - I am not saying it is wrong, simply that I don't understand it. Please show how you arrive at it


    Curl - *d applied to 1-form :
    Naughty boy! You haven't told us what is or what is


    ]Am I getting all this approximately right ??
    Possibly - when I say I don;t understand, or need symbols defined, I am willing to accept this as a fault in me, not that you are wrong
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  22. #21  
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    Please show how you arrive at it
    By sitting down with a pen and a piece of paper and actually doing the algebra in three different coordinate systems ( cartesian, polar, spherical ), because it was my express aim to figure out the relation to the metric tensor. When I wrote down the elements of the metric tensor in those coordinate systems, I noticed that the end result ( the gradient vector ) was just the matrix multiplication between the tensor and a vector with only the derivatives with respect to the new coordinates in it. Hence the above expression. And it seemed to work in each of the three coordinate systems, so I figured there must be something to it. Due to my lack of a mathematical background I often sit down and try around until things fit, and then write down what I think is the general relation from that.

    Is this relation not correct ? If not, what is the correct relation to the metric tensor ? If it is correct, how do you derive it more elegantly ?

    Naughty boy! You haven't told us what is or what is
    The Levi-Civita symbol and the covariant derivate, respectively. You are right, I should have stipulated that.
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  23. #22  
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    Quote Originally Posted by Markus Hanke View Post
    Gradient - exterior derivate on a 0-form :
    OK, walking the dogs last night I think I saw the problem. I warn you, this will be a lengthy post......

    In what follows I use the so-called "summation convention" throughout - just as you did.

    By definition, an element of an arbitrary vector space is written as where the are scalars and the are basis vectors.

    Likewise an element in the dual to this space, is written as with the same general meaning.

    Now if is an inner product space, we have that such that for any there is some where , which is quite simply a non-negative number. One calls a metric for this inner product space - it is a type (0,2) tensor.

    Recall from the "raising and lowering indices" thread that

    Now suppose be a differential manifold, with local coordinates for some point I may define a vector space all of whose elements are tangent to which one denotes by . Now the basis vectors for this space are the operators so by analogy with the above I seek some scalar multipliers.

    Now since our manifold is differentiable, then will do as well as another other, provided ONLY that is a scalar valued function, since then is a "genuine" scalar. So avector in may be written as (Remember I am assuming summation over like indices)

    So, again by analogy, at each point I may have a dual space called whose basis vectors are . Once again I choose the scalar multipliers to be , so a covector, or one-form in is just

    Since we may not assume that our manifold is globally "flat" then the metric tensor may well be different at each point , so let's write it in component form as, say to acknowledge this fact

    Putting this all together (if anyone out there hasn't lost the will to live) we will have that



    Which is not quite exactly as Markus wrote - but I concede it was a valiant attempt
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  24. #23  
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    Quote Originally Posted by Guitarist View Post
    OK, walking the dogs last night I think I saw the problem. I warn you, this will be a lengthy post......

    In what follows I use the so-called "summation convention" throughout - just as you did.

    By definition, an element of an arbitrary vector space is written as where the are scalars and the are basis vectors.

    Likewise an element in the dual to this space, is written as with the same general meaning.

    Now if is an inner product space, we have that such that for any there is some where , which is quite simply a non-negative number. One calls a metric for this inner product space - it is a type (0,2) tensor.

    Recall from the "raising and lowering indices" thread that

    Now suppose be a differential manifold, with local coordinates for some point I may define a vector space all of whose elements are tangent to which one denotes by . Now the basis vectors for this space are the operators so by analogy with the above I seek some scalar multipliers.

    Now since our manifold is differentiable, then will do as well as another other, provided ONLY that is a scalar valued function, since then is a "genuine" scalar. So avector in may be written as (Remember I am assuming summation over like indices)

    So, again by analogy, at each point I may have a dual space called whose basis vectors are . Once again I choose the scalar multipliers to be , so a covector, or one-form in is just

    Since we may not assume that our manifold is globally "flat" then the metric tensor may well be different at each point , so let's write it in component form as, say to acknowledge this fact

    Putting this all together (if anyone out there hasn't lost the will to live) we will have that



    Which is not quite exactly as Markus wrote - but I concede it was a valiant attempt
    Wow, thanks Guitarist. I can follow you and understand it, even though I wouldn't have been able to do this myself. I figured it had something to do with inner products, since that is how the metric tensor is defined, yet I couldn't quite figure out the inner product of what exactly. Of course I didn't think of dual spaces - that is where my lack of a mathematical background unfortunately fails me. My notation was also crap since I was left with a tensor index which wasn't summed over ( I took the remaining index to be a vector index, since the gradient should be a vector, but nonetheless the notation was of course wrong ).

    Once thing though I am not clear about - why is ? Shouldn't that be , since we are summing over like indices ?
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  25. #24  
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    Quote Originally Posted by Markus Hanke View Post
    Once thing though I am not clear about - why is ? Shouldn't that be , since we are summing over like indices ?
    Oops-a-daisy, that was stupid of me. Yes you are quite correct - thanks for pointing that out.
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  26. #25  
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    Quote Originally Posted by Guitarist View Post
    Quote Originally Posted by Markus Hanke View Post
    Once thing though I am not clear about - why is ? Shouldn't that be , since we are summing over like indices ?
    Oops-a-daisy, that was stupid of me. Yes you are quite correct - thanks for pointing that out.
    Excellent - I think I understand all of this much better now.

    I wish to genuinely thank you, Guitarist ( and also river_rat ), for taking your time to explain this stuff, and I really mean that. I am not a mathematician, and thus far away from a total understanding of this subject, but I feel I have learned a lot about differential forms here. Much appreciated

    Just one more quick question, to take this full circle - sorry, this may be a really stupid question, but I don't quite understand the notation used for the Euler scalar :



    In what sequence would one calculate such an expression ? And what exactly does the hodge star to the right of the tensor, but before the summation indices mean ? The notation is not really clear to me.
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    Markus - it is not at all a stupid question, but I have no idea what the answer might be. Ever the pedant, I would need to have the notation explained before I could even begin to think about it
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  28. #27  
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    Quote Originally Posted by Guitarist View Post
    Markus - it is not at all a stupid question, but I have no idea what the answer might be. Ever the pedant, I would need to have the notation explained before I could even begin to think about it
    No problem at all, don't worry about it. I read it here :

    Curvature invariant (general relativity) - Wikipedia, the free encyclopedia

    and was wondering about how to actually do the calculation. Not really important though.
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  29. #28  
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    Ok, I think I got this figured out now. First thing is that whoever wrote that Wikipedia article cheekily omitted a very important bit of notation; this should really have read



    I then did some further research and found out that there is actually a left hodge dual operator and a right hodge dual operator, defined like so :



    and



    which of course explains the notation for the Euler scalar

    Source : http://opus.bath.ac.uk/14074/1/Vedad...hesisFinal.pdf
    Dave Wilson likes this.
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