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Exterior product

  1. Aug 8, 2007 #1
    I'm trying to understand what an exterior product is. If I understood the wikipedia's article correctly, it is an equivalence class

    [tex]
    x\wedge y = \{a\otimes b\;|\;\exists z\;:\; a\otimes b - x\otimes y = z\otimes z\}
    [/tex]

    But I don't understand, how to generalize this to products of more than two vectors, like this [itex]x\wedge y\wedge z[/itex]. The recursive definition doesn't seem to be a correct way because

    [tex]
    (x\wedge y)\wedge z = x\wedge (y\wedge z)
    [/tex]

    is not true literally.
     
  2. jcsd
  3. Aug 8, 2007 #2
    Never mind all that nonsense. Try this page to find out what the exterior product actually is.
     
  4. Aug 8, 2007 #3
    I don't understand that either. What is that geometric product supposed to be?
     
  5. Aug 8, 2007 #4

    quasar987

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    I'm learning about the exterior product at the moment, but the kind I'm studying is perhaps not the most general case. Nevertheless, it is probably the easiest case to understand and justify, so this is where one should begin imo.

    A k-tensor over a real vector space V is a k-linear map from V^k to R. The set of all k-tensors, together with the natural way to define addition and multiplication by a scalar is itself a vector space over R. The tensor product is an operation sending a k-tensor and an l-tensor to a (k+l)-tensor in the most natural way possible:

    [tex](T\otimes S)(v_1,...,v_{k+l})=T(v_1,...,v_k)\cdot S(v_{k+1},...,v_{k+l})[/tex]

    A tensor T is alternating or completely antisymmetric if interchanging any two of the k vectors in the argument changes the sign of T. I.e. if

    [tex]T(v_1,...,v_i,...,v_j,...,v_k) = -T(v_1,...,v_j,...,v_i,...,v_k)[/tex]

    There is a natural way to take a k-tensor and make it into an alternating tensor. We denote this operation Alt(T).

    The set of all alternating k-tensor is a subspace of the space of all k-tensor, but if you take the tensor product of an alternating k-tensor and an alternating l-tensor, the result is an (k+l)-tensor, but it is not necessarily alternating. The exterior product or wedge product is the most natural way of taking a k-tensor and an l-tensor and making it into an alternating (k+l)-tensor:

    [tex]T\wedge S = \mbox{Alt}(T\otimes S)[/tex]

    As a consequence of these definitions, if {v_1,...,v_n} is a basis for V and {f_1,...,f_n} is its dual basis, then any alternating (k+l) tensor can be written as a linear combination of all the ways to wedge k of the n {f_i} together.

    (And yes, it is not evident at first sight, but [itex]\wedge[/itex] is associative: [tex](x\wedge y)\wedge z = x\wedge (y\wedge z)[/tex])
     
    Last edited: Aug 8, 2007
  6. Aug 10, 2007 #5

    Hurkyl

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    The geometric product is something different -- it's for a Clifford algebra, not for the exterior algebra. The two are related; they have the same underlying vector space, and you can convert back and forth between the two.
     
    Last edited: Aug 10, 2007
  7. Aug 11, 2007 #6

    mathwonk

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    you have to know what space you want to take the product in. there are two dual concepts. for any vector space M, you can define another vector space M^M, and a product from MxM-->M^M, or MtensorM-->M^M, if you like, simply formally, by sending v,w to the symbol v^w, subject to the rules that v^w = -w^v or equivaklently, that v^v = 0. thus vtensw and vtensw + vtensv go to the same element namely v^w. This is what wikimnopedia is trying to say. It generalizes by saying that v^v^w = 0, and v^v^w^r = 0, etc...

    This map MxM-->M^M is bilinear and alternating by construction. Thus composing it, i.el following it, by a linear function M^M-->R, where R is the real numbers, gives a bilinear alternating function on MxM.

    Thus to avoid defining M^M, some authiors instead focus on its dual space, namely linear functions on M^M, which can be described more easily as bilinear alternating functions (or tensors) on MxM. This is the version Quasar is talking about.

    However, one still needs to define the multiplication map,

    (M^M)^M-->M^M^M, which in the first or formal case is easy, just write down the symbols and juxtapoose them.

    In Quasars case however, one must define a way to combine two alternating functions into an alternating function. The big secret is there is onloy one way to do this, namely the determinant. thats all an alternating or exterior product is, a determinant.

    namely, if you want the tensor product of the basic coordinate functions x and y on R^2, you just multiply them and get xy. Buy if you want an alternatinbg product of x and y you take their determinant xy-yx, i.e. the alternating sum of their products in all possible orders.

    More generally, if f,g are any two linear functions on M, their exterior product in (M^M)* is fg-gf. the product of f,g,h in (M^M^M)* is fgh-gfh+-....you get it.

    I.e. if you know what a determinant is, you have seen all the alternating or exterior products in the world. (There are also some normalization factors stuck in front of these products to make them come out = 1 when acting on the standard basis vectors. See Spivak, Calculus on maniofolds.)
     
    Last edited: Aug 11, 2007
  8. Aug 11, 2007 #7

    mathwonk

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    so up to a constant multiple, f1^f2^...^fs acting on v1^v2^...^vs, is just the determinant of the sxs matrix [fi(vj)].

    since determinants are complicated gadgets, one can make this seem as complicated as desired by discussing it in explicit formulas, instead of concepts.

    then further, since determinants measure volumes, one can also give a geometric version, by considering the wedge product of some vectors to be the oriented parallelipiped or "block" they span. or the wedge product of their volume functions to be the volume function of the associated block construction.

    you pays your money and takes your choice.
     
    Last edited: Aug 11, 2007
  9. Aug 11, 2007 #8

    mathwonk

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    well there may be a real difference between clifford and exterior algebras, i believe you hurkyl, but it is not visible to me from the discussion linked above. that is apparently nothing but exterior algebra.

    indeed mathematically, all forms of "products" are encoded in the tensor product, i.e. every multilinear function in the world is an element of the tensor product with some extra conditions. if those conditions are just alternayingness, then by definition you have an exterior product. so equations 11 and 12 in the linked reference seem to prove that the product discussed there is abstractly just the exterior product. at least on "vectors".

    i would look in artins little book on geometric algebra, or the appendix to the books of bott or atiyah on k theory.

    ok from artins book the clifford algebra does not appear to satisfy the alternating property except on orthogonal elements, i.e. xoy = -yox if x,y, are orthogonal, but xox = x^2 is not nec zero.

    i do not see this in the article above, and i am also confused since the article above uses wedge notation, unlike artin. maybe the exterior algebra is contained somehow in the clifford algebra? (they also seem to have the same dimension, 2^n = the sum of the binomial coefficients of total degree n.)

    after rereading the article linked above, i will go out on a limb here and say that the author is not actually discussing the clifford algebra at all. but i could easily be wrong, since i admit i myself have never seen them before today.

    but i did read artin's definition, and it apparently aint always alternating. also there is an orthogonal structure involved, not mentioned in the somewhat imprecise link. maybe it is taken for granted there.

    can you say more about what gives here, hurkyl??
     
    Last edited: Aug 11, 2007
  10. Aug 11, 2007 #9

    mathwonk

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    well it is very clearly stated in wikipedia:

    "Given a vector space V one can construct the exterior algebra Λ(V), whose definition is independent of any quadratic form on V. It turns out that if F does not have characteristic 2 then there is a natural isomorphism between Λ(V) and Cℓ(V,Q) considered as vector spaces (and there exists an isomorphism in characteristic two, which may not be natural). This is an algebra isomorphism if and only if Q = 0. One can thus consider the Clifford algebra Cℓ(V,Q) as an enrichment (or more precisely, a quantization, cf. the Introduction) of the exterior algebra on V with a multiplication that depends on Q (one can still define the exterior product independent of Q)."

    so the exterior algebra is the special case of the clifford algebra constructed from the trivial quadratic form, where any two vectors are considered orthogonal. so indeed it appears the link above concerns not the general clifford algebra at all, but the exterior algebra, i.e. the trivial cliffiord algebra.
     
    Last edited: Aug 11, 2007
  11. Aug 11, 2007 #10

    mathwonk

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    i recommend the wiki articles on both clifford and exterior algebra.
     
  12. Aug 12, 2007 #11

    mathwonk

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    have i at last clarified the exterior product so well in posts 6 and 7 that there are no more questions! yes! yes! yes!
     
  13. Aug 12, 2007 #12

    quasar987

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    I'm sure this is not true literally, so what do you mean by it?

    Surely, Alt(T) is the "most natural" way to make T alternating because

    i. if T is already alternating, then Alt does not modify T.

    ii. if T is completely symetric, then Alt(T)=0 (there was no hope for T to begin with :frown:)

    But one could also spot a tensor that is alternating, and let Alt indiscriminately map everyone to it. Anyway, I just want to hear you talk about this, so here's the mic.
     
  14. Aug 12, 2007 #13

    mathwonk

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    alt is nothing but the definition of a determinant.

    a determinant is a multilinear function that alternates when you interchange arguments. hence the most natural way to define it to take the alternating sum of the products of all permuted values. this also the definition of alt.

    notice i said above that the exterior product of f and g is fg-gf, note also that this is the same definition as alt of their tensor product.

    i.e. ALT of f1...fn applied to v1,...vn is just the detrrminant of the matrix fi(vj).

    here i have simplified the alt business a little by only applying alt to a product of 1- tensors. that makes it more obvious why they are really all determinants. i.e. the definiton of ALT(ftensorg) is a little more complicated looking if f and g are r and s tensors so I write only decomposable ones, and consider only r tensors of form f1tens...tensfr.

    that does determine the action on all tensors however since these span.
     
    Last edited: Aug 13, 2007
  15. Aug 12, 2007 #14

    mathwonk

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    OK TO BE PRECISE, in post 4, line -4, that definition of T^S = ALT(TtensorS), in case T is T1...Tr and S is Tr+1...Tr+s = n, acting on v1,...vn,

    is just the determinant of the matrix [Ti(vj)].
     
    Last edited: Aug 13, 2007
  16. Aug 12, 2007 #15

    mathwonk

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    the basic idea in a simpler form is the theorem on page 51 of my web notes for math 845-3. it says that every r multilinear function of an n dimensional space is a linear combination of the ones obtained by taking rxr subdeterminants of the rxn matrix of entries of the r argument vectors. that of course is why the dimension of that space is the binomial coefficient n choose r.

    unfortunately i did not discuss the wedge product in these simple terms explicitly in those notes. i.e. i discussed the vector space structure of the exterior algebra there in the vector space case but not the algebra structure. i discussed the algebra structure abstractly later in general, but then it is sort of trivial.

    But what p.51 of my notes is saying is how to take the wedge product of any collection of one forms, which is sufficient for computing any wedge product. i.e. that theorem says that basis for r multiinear functions is obtained by taking all r fold wedge products of a basis of linear functions.

    it follows that since all functions are sums of products of linear ones, that you only need to know how to multiply linear ones, i.e. to take determinants, in order to know how to take any wedge product.

    so maybe i am saying the general ones are sums of determinants.
     
    Last edited: Aug 13, 2007
  17. Aug 12, 2007 #16

    mathwonk

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    yes, the wedge product of one forms is a determinant.

    every r form is a sum of products of one forms,

    hence every wedge product is a sum of determinants.


    I think that says it.

    (I was listening to an Elvis special on PBS when I wrote this last night.)


    here's the mic back.
     
    Last edited: Aug 13, 2007
  18. Aug 15, 2007 #17

    quasar987

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    So to summarize, an alternating n-tensor on an n-dimensional space is literally the determinant (up to a multiplicative constant), and an alternating k-tensor on an n-dimensional space is a linear combination of all the ways (=n choose k) to take a k x k subdeterminant out of the k x n matrix obtained by lining up the k arguments.

    let's hear it for the math wonk ladies and gents!
     
  19. Apr 30, 2009 #18
  20. Apr 28, 2010 #19
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