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What exactly is differential form?

  1. Jun 30, 2014 #1
    I'm reading about multlinear algebra and I'm stuck at differential form and outer product. The definitions involve tensor product and quotient set and I really cannot grab the concrete idea of the differential form.

    Can someone explain in a somewhat layman term what is differential form? (like: "a 2_tensor is a n*n matrix)

    Thank you!
     
  2. jcsd
  3. Jun 30, 2014 #2

    jedishrfu

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  4. Jun 30, 2014 #3
    Ok it's an alternative, but what is it? I mean, how can I imagine a differential form in my head?
     
  5. Jun 30, 2014 #4

    Matterwave

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    A differential n-form can simply be regarded as a completely anti-symmetric rank (0,n) tensor. A 0-form is therefore simply a scalar function. A 1-form can be regarded as linear operators of vectors into numbers. In other words, a one form ##\tilde{\omega}## produces a real number when it acts on an arbitrary vector ##\vec{q}##:

    $$\tilde{\omega}(\vec{q})=\sum_a \omega_a q^a$$

    It is linear in that, for real numbers ##a## and ##b##, and for all vectors ##\vec{q}## and ##\vec{p}##:

    $$\tilde{\omega}(a\vec{q}+b\vec{p})=a\tilde{\omega}(\vec{q})+b \tilde{\omega}(\vec{p})$$


    An n-form is then a multi-linear (linear in each argument) operator of n-vectors into real numbers:

    $$\tilde{\omega}(\vec{q},\vec{p},...,\vec{t})=\sum_{a,b,...,n}\omega_{ab...n}q^ap^b...t^n$$

    With the added condition that it is totally anti-symmetric under interchange of any 2 vectors:

    $$\tilde{\omega}(\vec{q},\vec{p},...,\vec{t})=-\tilde{\omega}(\vec{p},\vec{q},...,\vec{t})$$

    Forms are useful because they generalize the notion of "volume" to an arbitrary differential manifold that otherwise would not have any rigidity. The most important result in the theory of differential forms is probably Stoke's theorem, that for an n-form ##\tilde{\omega}## defined on a region ##\Omega##, its integral over the (closed) boundary of ##\Omega## which we call ##\partial\Omega## is equal to the integral of its exterior derivative ##d\tilde{\omega}## over the region ##\Omega##:

    $$\oint_{\partial\Omega} \tilde{\omega} = \int_\Omega d\tilde{\omega}$$

    This theorem can be thought of as a generalization of the fundamental theorem of calculus, and from this single result we can derive all of the "divergence theorems" of ordinary vector calculus (including the Green's theorem, and the vector calculus Kelvin-Stoke's theorem).
     
  6. Jun 30, 2014 #5
    It seems like there are many ways to imagine a differential form.
    I have a feeling that a k_form is simply something have the form of (a,b,....,k) that is linear-in-each-argument and alternating. A sum, an operator,...
    How can we integrate a somewhat mysterical object like that? (I want to imagine the process of integrate a k_form)
     
  7. Jul 1, 2014 #6

    HallsofIvy

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    I think this would be better placed in "Differential Geometry".
     
  8. Jul 1, 2014 #7

    Matterwave

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    A form is not really mysterious, it's a specific mathematical object. It might seem mysterious at first, but once you understand it, it really isn't (like many mathematical objects).

    There are many ways to imagine a k-form. One is from Misner Thorne and Wheeler who have you imagining a one-form as a series of surfaces (infinitesimal) through which a vector pierces. Each time the vector pierces a surface there is a bong of a bell and the number produced is the number of bongs of the bell. A 2-form is then a mesh of these surfaces like an egg-crate structure (and so on for higher dimensions). But of course, these are just visualizations. This can be found in Misner Thorne and Wheeler chapter 4.

    Integration over a form, or how a form can be thought of as a volume element, requires us to understand first that an n-form on an n dimensional manifold has only 1 degree of freedom (one independent component). You are probably familiar with this fact from the connection with the Levi-Civita totally anti-symmetric symbol ##\epsilon_{ijk}## which has only 1 independent component in 3-space, and is a 3-form in regular Euclidean 3-space. So, given an n-form ##\omega## on an n-dimensional manifold, all other n-forms on this manifold can be expressed as ##f\omega## where ##f## is some scalar function. So, if we introduce a coordinate patch on an open region of our manifold, call it ##\{x^n\}##, then we can express any n-form as:

    $$\omega = f(x^1,...,x^n)dx^1\wedge dx^2\wedge...\wedge dx^n$$

    We can then define on our open region the integral of any n-form ##\omega## as:

    $$\int_U \omega = \int_U f(x^1,...,x^n)dx^1\wedge dx^2\wedge...\wedge dx^n\equiv \int_U f(x^1,...,x^n)dx^1 dx^2...dx^n$$

    Where the last expression is simply the regular Riemann integral.

    To extend the integration over our whole manifold (assuming it is orientiable), we have to create what is called a partition of unity (so as not to over count overlap regions). The details get a little more complicated there.
     
  9. Jul 1, 2014 #8
    For some inspiration for the idea of a 2-form, you can think about the concept of flux through a parallelogram. You take two vectors that span the parallelogram, and then maybe you have some fluid flowing through it or some other field, so you measure the flux. So, you put in two vectors and you get out a number. If you switch the order of the vectors, you change the orientation of the parallelogram, so you get a minus sign in front, so that illustrates why it's alternating. So, differential forms just generalize this to higher dimensions.

    If you want to know why a 2-form is something that you can integrate over a surface, you can break up the surface into little parallelograms. 2-forms take in parallelograms and give you a number. So, just apply the 2-form to each parallelogram and add it all up. Then, of course, you'd have to take a limit of small parallelograms to get the integral. And you can play the same game in higher dimensions.

    If you wanted to take the exterior derivative of a 2-form, that would be a 3-form that you could get as follows. You want to get something that takes in 3 vectors and spits out a a number, and somehow you want to build it out of the 2-form. The 3 vectors will span a parallelopiped. So how do you go down one dimension? Integrate the 2-form over the parallelopided (in the limit as its size goes to zero). That's essentially the meaning of the exterior derivative.

    And morally speaking, this also indicates where Stokes theorem comes from. For example, say you are integrating a 1-form over the boundary of a surface. Intuitively, what the integral is doing is breaking up the boundary into little vectors, which get plugged into the 1-form at their base points and then you add it all up. You can also break up the surface into parallelograms. All the middle stuff will cancel out due to orientation, so the integral of the 1-form is equal to the integral of the 1-form over each one of those tiny parallelograms, in other words, its exterior derivative. Same idea for higher dimensions.

    This may not be the cleanest way to actually prove these theorems, even though, to my mind, it's the best motivation, so unfortunately, a lot of this intuition can be hard to find when you read more formal treatments.
     
  10. Jul 1, 2014 #9
    I haven't thought that we will come back to Riemann integral to define the integral of a k_form.

    It starts to make sense now. Thanks you guys!
     
  11. Jul 2, 2014 #10

    jedishrfu

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  12. Jul 3, 2014 #11

    WWGD

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    In the most general sense, a k-form is an element of the n-dual of a vector space V, i.e., an element of :

    (VxVx...xV)*--->F ,

    Where F is the base field. To simplify, assume for now that F=Reals.

    A 1-form is just any element of the dual space of a vector space.

    Example: Say V=Reals over the Reals. Choose a basis {##v_1,..,v_n##} and write ## v=c_1v_1+...c_nv_n ## . Then the map ## v-\in V-> c_1+c_2+...+c_n ##is an element of V*. An n-form is then an n-linear: bilinear, trilinear, etc. map.



    Maybe in the most basic sense, in differential geometry, a 1-form w is a linear map from each tangent vector, as an element of T_p M , as elements (usually) to the Reals (most-often the base field of the 1-D tangent plane), i.e., w is an element of (T_pM)*,
    the dual of the tangent space, more generally, an n-form is an element of ##(T_pM \times T_p M \times .. \times T_pM )^*-->\mathbb R ##
    i.e., a multilinear map in what is called the exterior algebra of ##T_pM ##. A 2-tensor you may be familiar with is the inner-product, which is representable as a matrix.

    BUT, a form is a subset (I think actually a subspace) of the n-th dual, consisting of linear maps that are alternating, meaning n-linear maps that invariant under coordinate changes (up to parity of the changes of coordinates). A 1-form is a linear assignment of a Real number at each tangent space, a k-form is an assignment of an alternating n-linear map to each vector in ##(T_p M \times T_p M \times.. \times T_p M) ##.

    More formally, a k-form w_k is an element of ##TM^k:= (T_p M)^*:= ( |_| T^*p M) ##, where ##TM^k ## is the k-th power of the cotangent bundle.

    Maybe you can start by reading on the Tensor Algebra : http://en.wikipedia.org/wiki/Tensor_algebra

    Like you said, a 1-tensor is just a linear map, a 2-tensor is a quadratic form so that B(x,y): ##x^T MY ##, for a matrix ## M ## ,and 3- and higher- tensors are not (AFAIK) describable with matrices, at least not in a standard form.

    The exterior algebrar is a so-cold (specially in winter ;) ) graded algebra, where the n-th grade consists of the n-linear maps, and you can go between different grades to obtain a (k+l)-linear map ( as long as (k+l) is less than the total dimension ) by tensoring maps, or doing a tensoring between graded subspaces.

    When you restrict the exterior algebra to alternating forms, you get the tensor algebra.

    About tensor products, a tensor product of two ( or more) spaces allow you to have a single space that simplifies your maps in a specific sense: in the case of modules, or vector spaces, the tensor product allows you to construct a new space in which every n-linear map on ##(V_1 \times V_2 \times..\times V_n) ## becomes a linear map on the space ## V_1 \otimes... \otimes V_n ##; so that, e.g., for the case of two vector spaces ( over the same field ) ## V,W## , the tensor space ## V \otimes W ## allows you to describe every bilinear map on ## V \times W \rightarrow \mathbb R## as a linear map on ## V \otimes W \rightarrow \mathbb R ##. This construction (meaning the justification for why it is possible to construct a space where every map factors through) ultimately comes down to properties that allow certain maps to "factor through " other maps, i.e., the map f is said to factor through the map g , if there is a map h with ## f=g \circ h ##, so you see how this seems similar to factoring numbers as n= rs . These properties have to see mostly with kernels and normal subgroups. Maybe you can look up "conditions for maps to factor through ".

    This idea of the tensor product can be defined for other types of objects.
     
    Last edited: Jul 3, 2014
  13. Jul 4, 2014 #12

    mathwonk

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    do you know what a vector space is? next question, do you know what a dual of a vector space is?

    if so, then you can understand a differential form.


    given a smooth geometric object, called a "manifold", the fundamental vector space is a tangent space at some point. then the dual space at that point is the "cotangent space".

    a tangent vector is a vector in some tamgent space. a covector is a dual vector in some cotangent space.

    a "vector field" is a family of tangent vectors, one at each point, and a covector field is a family of covectors, one at each point.

    at last! a differential 1-form is a covector field, i.e. a CHOICE of a dual tangent vector at each point of your space. These objects can then be differentiated over parametrized curves.

    probably more useful, a differential form is something you can integrate over a parametrized curve or piece of surface. I.e. a differential n form can be integrated over a paramnetrized piece of n dimensional surface.

    thank you, homeomorphic! i just read your excellent answer!

    I was trying to keep this simple, but.....ok a 1 - form sees a vector and spits out a number.

    a 2-form sees an ordered pair of vectors and spits out a number. AND we require that it spit out minus that number if the order of the two vectors is reversed.

    e.g.a typical 2 form projects the two vectors onto a coordinate plane and takes the oriented area of the projected parallelogram. i.e. it takes the 2x2 determinant of say the 2x2 submatrix of the first two entries of the 2xn diml matrix of the two n diml vectors.

    etc for k forms. i.e. a typical k form assigns to an ordered set of k, n-diml vectors, the determinant of the kxk submatrix formed by choosing k of the columns of the matrix having those vectors as rows.
     
    Last edited: Jul 4, 2014
  14. Jul 4, 2014 #13
    Technically, that would just be a 1-form, but you can build other forms out of those using the wedge product and linear combinations.
     
  15. Jul 4, 2014 #14

    WWGD

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    It all comes down to multilinear algebra , and figuring out expressions like ##(T_p M \times...\times T_p M)^* ## which is isomorphic to ##(T_p M (+)+....(+) T_p M)^* "="(T_pM)^* (+)...(+)(T_pM)^* ##. The collection of all tangent spaces is itself a manifold, called the tangent bundle and so is the union of the duals of the tangent spaces, called the cotangent bundle. And then you get into more linear algebra with, e.g., quotient bundles, Riemannian metrics, etc.

    Here the "=" means algebraically-isomorphic.

    And then you have multilinear maps , a.k.a., tensors, that take arguments both in V and in ## V^*##. One thing that is good to understand here is the trace isomorphism that gives a justification to the index contractions used; this is a good result to remember , or, if you can, try to prove, because it is the basis of index contractions: ## Hom(V,V) "=" T^1_1(V) ## , where Hom(V,V) is the vector spaces of homeomorphisms of V to itself and ## T_1^1(V) ## is the space of multilinear maps with arguments in V, ## V^*## respectfully. The generalization of this result is the justification for contracting indices.

    And like Mathwonk said, k-forms are defined in such a way not only to be integrated over k-manifolds, but to be integrated in such a way that the integral is independent of coordinates.

    It is amazing to see the amount of (multiinear) algebra that comes into play in DG, specially considering the fact that there is no analysis in pure algebra (unless, e.g., you talk about topological groups).
     
    Last edited: Jul 4, 2014
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