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Dual spaces and the meaning of "Dual"

  1. May 26, 2015 #1
    So, I understand that the dual space, [itex]V^{\ast}[/itex] of a vector space [itex]V[/itex] over a scalar field [itex]\mathbb{F}[/itex] is the set of all linear functionals [itex]f^{\ast}:V\rightarrow\mathbb{F}[/itex] that map the vectors in [itex]V[/itex] to the scalar field [itex]\mathbb{F}[/itex], but I'm confused as to the meaning of the term dual?!
    Is it just that the two vector spaces form a dual pair, such that there is a pairing between them [itex]\langle\cdot , \cdot\rangle : V^{\ast}\times V\rightarrow\mathbb{F}[/itex], that constructs a unique relation between them (i.e. each vector space has only one dual space), or is there some other reasoning to it?
     
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  3. May 26, 2015 #2

    jedishrfu

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  4. May 26, 2015 #3
    Thanks for the link. Unfortunately I've read it already and it didn't fully clear it up for me. Would what I put in my first post be correct at all?
     
  5. May 26, 2015 #4

    atyy

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    "Dual" in "dual space" need not have a precise meaning independent of "dual space". In other words, the meaning of "dual" in "dual space" is defined by the definition of "dual space".

    From the Wikipedia link, you can see that "dual" may be part of a more general concept, but it is not necessarily so. Terminology in mathematics is not completely uniform.
     
  6. May 26, 2015 #5
    yes, I understand that to an extent, but the term as a whole must have some meaning, right? It's referring to a specific connection between two vector spaces.
     
  7. May 27, 2015 #6

    HallsofIvy

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    There is a theorem, for finite dimensional spaces, that if B is the dual space to A, then A is (isomorphic to) the dual space of B. That's the reason for the word "dual"- there are always two things in a "dual" set.

    (That theorem is not true for infinite dimensional spaces
     
  8. May 27, 2015 #7

    micromass

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    The term comes from projective geometry. Do you know anything about that?
     
  9. May 27, 2015 #8
    No I don't really, sorry.
     
  10. May 27, 2015 #9

    mathwonk

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    well it refers to the reciprocal nature of some of the correspondences set up by the pairing you mentioned VxV*-->F. (assume V has finite dimension.) Then a single vector x in V is annihilated by a vector subspace of V* of codimension one, and more generally a subspace W of dimension r in V is annihilated by a subspace Wperp of codimension one in V*.

    This gives rise to what micromass is referring to in projective geometry, i.e. a correspondence between hyperplanes H of the projective space PV, and points of the projective space PV*, where s is a linear functional vanishing precisely on the linear subspace defining the hyperplane H.

    As Halls said, this duality goes both ways, since V is naturally isomorphic to (V*)*. I.e. fixing an element x of V, the pairing {x} x V*-->F lets us evaluate x on elements of V*, so x is a linear functional on V*. This mapping V-->(V*)* is injective between spaces of equal finite dimensions, hence is an isomorphism.

    So, V* is the space of linear functionals on V, and dually V is the space of inear functionals on V*. So V and V* have the same relation to each other, i.e. they are dual.

    by the way why is the software dissing me with these lineouts?
     
    Last edited: May 27, 2015
  11. May 27, 2015 #10

    mathwonk

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    the projective duality micromass is referring to is related but can be stated without mentioning linear functionals. consider a 3 dimensional vector space only, for simplicity. lines of this space are considered as points of the associated projetive space, and planes in this space are considered as projective lines.

    then any two distinct lines (i.e. two distinct points of the associated projective space) span a unique plane (i.e. span a line in the projective space). Dually, any two distinct planes (or lines in the projective space) intersect in a unique line (or point of the projective space).

    thus two distinct points of the projective plane always give a line, and 2 distinct lines always give a point. So since the reciprocal relations are the same, you could define the dual of a projective plane P to be the space P* whose points correspond to lines of the original plane P. Then 2 points of P* (lines of P) would give a line in P* (point of P), and 2 lines of P* (points of P), would give a point of P* (line of P).

    the two spaces are said to be dual to each other. This projective setup of course can be modeled on the linear one as above, using the language of linear functionals, since a line in a projective plane is just a plane in a linear 3 space, which is the kernel of a linear functional on that linear 3 space.


    if you are confused, join the club! i have never ceased to be confused by duality.
     
  12. May 29, 2015 #11

    Fredrik

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    When you type [S], everything that follows will be lined out. If the editor is in rich text mode, you can't see the [S], but it will show up if you click on the "paper" button in the upper right corner.
     
  13. May 29, 2015 #12
    Thanks for your input mathwonk. I have to admit, like you I have always been confused by the concept.

    Does the concept of duality in terms of vector spaces have anything to do with the fact that the linear functionals that map a vector space into its underlying scalar field are unique and hence the two vector spaces are dual, as they are only related to one another (and not any other vector spaces). Also, while we're on the subject, I read a proof of existence of linear functionals between vector spaces in which the author put "Consider a linear functional [itex]f:V\rightarrow W[/itex] between two vector spaces and let [itex]\lbrace\mathbf{v_{i}}\rbrace[/itex] be a basis for [itex]V[/itex] and [itex]\lbrace\mathbf{w_{i}}\rbrace[/itex] be any set of vectors in [itex]W[/itex]. Now, let [itex]\mathbf{v}\;\in V[/itex], then we define [itex]f[/itex] such that [tex]f(\mathbf{v})=\sum_{i=1}^{n}a^{i}\mathbf{w}_{i}[/tex] He then claims that this is well-defined as the linear combination [itex]\mathbf{v}=\sum_{i=1}^{n}a^{i}\mathbf{v}_{i}[/itex] is unique. By this, does he mean that each vector is mapped to a unique linear combination of vectors [itex]\lbrace\mathbf{w_{i}}\rbrace\;\in W[/itex] by [itex]f[/itex], as the set of coefficients [itex]\lbrace a^{i}\rbrace[/itex] for each vector (with respect to the given basis) is unique. Thus the functional [itex]f[/itex] has one and only one output for each input vector (and thus is well-defined)?! Is it correct to deduce from this, that as [itex]f[/itex] is linear, [itex]f(\mathbf{v}_{i})=\mathbf{w}_{i}[/itex]?
     
  14. May 29, 2015 #13

    Fredrik

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    I don't fully understand what you're saying or asking here. All finite-dimensional vector spaces are isomorphic to each other, so V and V* will certainly have some relationship with other spaces.

    I think the main reason why the word "dual" is appropriate is that V can be said to be associated with V* in the same way that V* is associated with V. What I mean by that is that V** (the dual space of the dual space of V) is isomorphic to V. Also, this isomorphism is different from say the isomorphisms we can define between V and V* in that it doesn't require an inner product, a preferred basis, or anything like that. Because of this isomorphism, you can say that if you take some vector space V and apply the dual "operation" twice, you get the original space back. (There are infinite-dimensional spaces for which this isn't true. The word "dual" is used there too, but I guess it just seemed pointless to think of a new term).

    This terminology is non-standard, I think. If the map is from a vector space into a vector space, it's usually called a linear transformation or a linear operator, not a linear functional. (Some authors reserve the term linear operator for linear transformations ##L:X\to Y## such that X=Y). A linear functional is usually defined as a linear map from a vector space to its underlying field. Many physics books also require the vector space to be a set of functions, but I have always found that very odd.

    So he's saying right at the start that the f he's talking about is linear.

    OK.

    Huh? He said earlier that ##f## is some linear transformation and that ##v## is some vector in V. So the notation ##f(v)## is already reserved for some vector in ##W##, and can't be defined by this statement. Another reason why this can't be a definition of ##f(v)## is that you haven't said what the ##a_i## are. If this equality defines something, it's the ##a_i##. But it only works as a definition if ##\{w_1,\dots,w_n\}## is a basis for W.

    Are you sure that you quoted the book right? The above doesn't make sense.
     
  15. May 29, 2015 #14

    Fredrik

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    I think this is what the author meant:

    Let V and W be two finite-dimensional vector spaces. Define n=dim V. Let ##E=(e_1,\dots,e_n)## be an ordered basis for V. Let ##w_1,\dots,w_n## be elements of W. For each ##v\in V##, let ##(v_1,\dots,v_n)## be the n-tuple of components of ##v## with respect to ##E## (i.e. the unique n-tuple ##(v_1,\dots,v_n)## of real numbers such that ##v=\sum_{i=1}^n v_i e_i##). Now define ##f:V\to W## by
    $$f(v)=\sum_{i=1}^n v_i w_i$$ for all ##v\in V##. (It's essential that the ##v_i## are the components of ##v##). This f is linear because for all ##x,y\in V## and all ##a,b\in\mathbb R##, we have
    $$f(ax+by)=\sum_{i=1}^n (ax+by)_i w_i =a\sum_{i=1}^n x_iw_i +b\sum_{i=1}^n y_iw_i =af(x)+bf(y).$$
     
  16. May 29, 2015 #15
    It seemed to make sense to me why it's called a "dual" from the notion of a dual pair (c.f. my original post), but I'm unsure as to whether I'm conflating two different concepts with this.

    Is the idea, that due to this isomorphism one can describe vectors etc. either via their components (with respect to the dual basis) or by their decomposition with respect to a basis in the "original" vector space (thus there are "dual" descriptions of the same object)?



    Ah, ok thanks. Sorry for the mix-up of terminology in the previous posts, just been talking about functionals a lot recently and went into "auto-pilot mode".
    So is this well defined because the n-tuple [itex](v_{1},\ldots , v_{n})[/itex] (describing a vector in [itex]V[/itex] with respect to a particular basis) is unique, and therefore each vector [itex]\mathbf{v}\in V[/itex] is mapped to a unique element of [itex]W[/itex] by the linear map [itex]f[/itex]?
     
  17. May 29, 2015 #16

    Fredrik

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    The isomorphism I'm talking about is between V** and V, not between V* and V. I don't think the two descriptions of an element of V* that you're talking about are very useful.

    Because it's uniquely determined by v, yes.
     
  18. May 29, 2015 #17

    mathwonk

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    i think the word dual has maybe two meanings. the algebraic dual of a linear space V is the space V* of linear functions V-->R. but two spaces V,W are dual to each other, if there is a pairing VxW-->R which is perfect in the sense that this pairing gives an isomorphism of V with W* and also of W with V*.

    Thus for finite dimensional vector spaces V, V and V* are dual to each other in that sense, i.e. both senses, because the natural evaluation pairing VxV*-->V does make V isomorphic to (V*)* as well as of course V* isomorphic to V*. But an infinite dimensional space V can have a "dual" space V*, but the two are not "dual" to each other! i.e. the pairing does not make both spaces the dual of each other.

    So this makes it more confusing. Infinite dimensional spaces V for which there is some definition of a "dual" space V* with some continuity conditions, and such that in these definitions V is isomorphic to (V*)*, are called "reflexive".

    In topology, "Poincare duality" is a pairing between homology and cohomology of oriented manifolds making each the dual of the other. I.e. on a manifold, a homology class can be considered as a cohomology class, by letting it act on homology classes of complementary dimension, via intersecting with them. I.e. there is a perfect pairing between homology in dim r and homology in dim n-r. oho! so here actually homology is dual to itself!

    I.e. cohomology is always dual to homology, and the theorem is that here cohomology and homology are isomorphic, so homology is duakl to itself. similarly (or dually!) Serre duality for complex algebraic varieties is a duality of (coherent sheaf) cohomology and itself. then there is a Grothendieck duality, Verdier duality,.....all of which are presumably pairings of some cohomology with other cohomology.
     
    Last edited: May 29, 2015
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