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Calculus on Manifolds

  1. May 31, 2015 #1
    So this is beginning to feel like the beginning of the 4th movement of Beethoven's Ninth: it is all coming together.
    Manifolds,Lie Algebra, Lie Groups and Exterior Algebra.

    And now I have another simple question that is more linguistic in nature.

    What does one mean by "Calculus on Manifolds"? (Yes,I have seen other posts on this topic in this forum, but please allow me to state it in my terms.)

    I now feel I was severely mis-educated as an engineer. And perhaps I do not really appreciate calculus. Thus...

    Can you give me an physical (but engineering) example of calculus being done on manifolds?
    Why do I want to do calculus on a manifold?
    What does DOING calculus mean?
    (FORGIVE ME for this one, but... you have a function of variables, you take derivatives... what's the big deal? what is the DOING part of the calculus on manifolds?
    What are examples of it?
    What can I do in R3 that I CANNOT do in.... another place... unless that place is like a manifold?

    Assuming that a Lie group is a continuous manifold, how does one DO calculus on a Lie Group?
    I assume one uses Lie Algebraic structures.
    How is that related... IN WORDS... not equations... to the Lie Derivative.

    And regarding differential forms... it now seems that topic matters to me ONLY if I want to understand the distinction between vectors and covectors... but if my TARGET is dynamics, then I need not spend so much time with differential forms and exterior algebra as much as I should spend with manifolds, Lie groups and Lie algebras. yes?
     
    Last edited: May 31, 2015
  2. jcsd
  3. May 31, 2015 #2

    MarcusAgrippa

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    Mmm. I see confusion. Let's start with "calculus". Generally, a calculus is a slick, stylized system for calculation. The differential calculus and the integral calculus are two examples. Another is the exterior calculus.

    Your question appears to be about differential calculi. Let's start with the simplest one: differentiation of a function of one variable. The derivative is defined as the limit of a differential quotient as the denominator is made smaller and smaller. Were you to leave the matter there, you would spend all your life calculating these limits. A great waste of time! How do we get around this? First, we prove some general results. Most elementary functions we encounter are sums, products or quotients of a limited set of elementary functions. So we prove 1. The derivative of a sum is the sum of the derivatives, 2. The product rule, 3. The quotient rule. These are all proved using the definition of the derivative as the limit of a differential quotient. Then we use the same definition to find the derivative of a power of x, the derivatives of trigonometric and hyperbolic functions, the exponential an log functions - anything else? Now we learn these results and turn them into rules. These rules enable us to find the derivative of most functions. Occasionally we come across something that cannot be differentiated by these rules and we need to return to the definition as a limit to get the answer.

    The set of results and rules obtained enable us to bypass the original definition of the derivative and to calculate derivatives quickly, slickly and easily: behold, we have devised a calculus. This is called the differential calculus.

    Got it?

    This calculus does not allow you to differentiate vector valued functions of one variable. Why? Because your definition was only for scalar valued functions of a single variable. So a new definition is needed. Luckily, vectors can be subtracted, so we can again form differential quotients. We again prove results from the definition - the quotient rule no longer makes sense, but the others do. We next show that by decomposing the vectors into components, we can reduce the differentiation of vector valued functions of a single variable to differentiation of the scalar valued components. We have now invented a new calculus, the calculus of vector valued functions of one variable.

    Are you getting the idea? Every new kind of function requires the invention of a new kind of calculus. In particular, functions with a domain that is not Rn, I.e. Whose domain is a manifold, requires a definition of a derivative, and then a new calculus to ease the process of calculation.
     
    Last edited: May 31, 2015
  4. May 31, 2015 #3

    mathwonk

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    the difference between a manifold and R^n is that in R^n we have a unique way to assign numbers to each point, i.e. coordinates. On a manifold, we may have several different ways to choose coordinates for each point. thus we need to know how the results we get from taking derivatives in terms of one set of coordinates relate to what happens in the other coordinates. I.e. since there are several ways to assign coordinates, we need to understand when our calculations depend on those choices and when they do not. E.g. in physics,i.e. the physical world, even if you believe that world is euclidean, there is no special choice of origin, so you need to know when your calculations have physical meaning independent of the coordnates you chose to make them. more generally if you believe the world is not euclidean, maybe the universe is a sphere, then you need to understand how to choose coordinates locally in different parts of that sphere and interpret the calculations made in each system.

    a very simple example is to understand what is a vector. if you choose an origin, then every point has a "position vector", the arrow from the origin to that point. But this is an artificial sort of "vector" and physically it is n it a vector at all. a real vector is velocity or acceleration. I.e. when you move the origin, the position vector changes, but when you move the origin, the velocity vector stays the same. probably you understand this better than I. the same sort of thing obtains in the case of manifolds, i.e. you want to understand which measurem,ents give intrinsic results and which ones are artifacts of the coordinate system. usually on a manifold, it is physically meaningful to say a derivative is non zero, but not so much to say it equals 4.
     
  5. Jun 1, 2015 #4
    Marcus,

    I am extremely grateful to you and to others who are helping me through this process.
    As a mechanical engineer (and an older one) I was not taught well. And I have also forgotten a lot.
    And I am also cursed with a need for understanding WHY people say things the way they do.
    So these questions, I know, are trivial. But you have helped me immensely.
    Thank you for your patience.

    With that, may I push my luck?

    Exterior Calculus... Well, I now understand exterior alegebra and its various rules: wedge, hodge. And I see they can create the generalized Stoke's Theorem and that is a generalization of first fundamental theorem, Greens, Curl, etc. (with functions, gradients, curls and divergences following as forms of increasing orders)

    1. But what do you mean by Exterior Calculus?
    2. And why is the word EXTERIOR... Why not... oh.. I don't know... supercalifragilisticexpialidocious calculus? I am NOT being facetious here. I can find NO EXPLANATION of why the word is exterior.
     
    Last edited: Jun 1, 2015
  6. Jun 1, 2015 #5
    And the same to you, too, mathwonk (what I wrote to Marcus)

    That has really helped me a LOT. Thank you...

    Your example with the vectors is extremely helpful.
    And recognizing points and coordinates too.
     
    Last edited: Jun 1, 2015
  7. Jun 1, 2015 #6

    MarcusAgrippa

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    First, what are forms?

    A 1-form is a covariant vector. A two form is an antisymmetric tensor of rank 2. A p-form is a completely antisymmetric tensor of rank p.

    In tensor algebra, one of the fundamental operations is that by which we take two tensors, one of rank r and other of rank s, and we construct a new tensor of rank r+s. This operation is called the tensor product. Tensors of rank r form a vector space (or, linear space). Those of rank s form a completely different vector space. The result of taking the tensor product produces a new animal, not contained in either of the original spaces: it belongs to a completely different space. In this general sense, it is an "external" or "exterior" operation, sometimes also called an outer product. Products that produce tensors of smaller rank are called "interior products", such as contraction of tensors, or the dot-product of elementary vectors.

    Suppose now that we take a fully antisymmetric tensor of rank r, and another of rank s, and we form their tensor product. This product produces a tensor of rank r+s, bot not an antisymmetric tensor of rank r+s. To extract its fully antisymmetric part, we need to antisymmetrise it in a separate operation. In many applications, it is important to extract the antisymmetric part of the product tensor. To do this, Elie Cartan invented an efficient new operation which does both things at once, thus saving a great deal of labour. The new product takes an r-form and an s-form and produces an (r+s)-form, i.e. a fully antisymmetric tensor of rank r+s. This product he called the exterior product - exterior here is a general word with a specialised meaning. It is exterior because it produces another object of the same kind which is or a higher rank and lives in a different vector space.

    That probably answers both questions.
     
    Last edited: Jun 1, 2015
  8. Jun 1, 2015 #7




    (Ode to Joy excerpt now playing loudly in my head.)

    Wow... that was great...
    You probably think this is silly and, in time, I may too. But right now, I need to hear someone say what you just said.
    It takes a really kind and smart person to know where someone is being dense and how not to insult them in the process of shining the light.
     
  9. Jun 1, 2015 #8

    MarcusAgrippa

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    Thank you for your kind words.

    If I may be so bold, your question is a probing one and requires someone with a global view of the subject to answer it. Few folk pause to think about the foundations and significance of a subject - they are too busy trying to make their careers by pushing out papers that require minimal manipulative knowledge of their field. One cannot criticise this practice - it is what the current system of funding and promotion encourages.
     
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