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What is the meaning of the term "dual"?

  1. Dec 3, 2015 #1
    Apologies if this is a really trivial question, but I've never been quite sure as to the usage of the terminology dual space. I get that given a vector space ##V## we can construct a set of linear functionals that map ##V## into its underlying field and that these linear functionals themselves form a vector space ##V^{\ast}##, but what is meant by calling it dual to ##V##? Is it simply that given one we can construct the other and so they are intricately related to one another? For example, given a vector ##\mathbf{v}\in V##, and a basis ##\mathcal{B}=\lbrace\mathbf{e}_{i}\rbrace##, such that ##\mathbf{v}=a^{i}\mathbf{e}_{i}##, then we can define a unique map ##f\in V^{\ast}## such that ##f(\mathbf{v})=a^{i}##. As this map is linear it follows that $$f(\mathbf{v})=a^{i}f(e_{i})$$ and so ##f## is determined uniquely by its action on the basis vectors ##\lbrace\mathbf{e}_{i}\rbrace##. Is this what is meant by dual in that if we no the form of one then we can determine the other?
     
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  3. Dec 3, 2015 #2

    jedishrfu

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    With respect to mathematical duality in general:

    https://en.wikipedia.org/wiki/Duality_(mathematics)

    and here's a list of dualities:

    https://en.wikipedia.org/wiki/List_of_dualities

    My understanding was that a a problem that is difficult to solve under one system of math could be mapped to a dual system and more easily solved and then converted back to get the answer you were looking for.

    The one I'm somewhat familiar with is the dual vector:

    https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors

    https://en.wikipedia.org/wiki/Dual_space
     
  4. Dec 3, 2015 #3

    Krylov

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    Yes, I agree that
    Otherwise, it is really just a definition.

    Incidentally, it's maybe good to know that when ##\text{dim}\,V = \infty## and ##V## is endowed with a norm (or, more generally, a vector space topology), then in contrast with the finite dimensional case there always exists at least one discontinuous linear functional on ##V##. Hence one needs to distinguish between the algebraic dual of ##V## (the set you defined in your OP) and the topological dual of ##V##, which is the smaller vector space consisting of all continuous linear functionals. The norm in ##V## induces a natural norm on the topological dual.
     
  5. Dec 3, 2015 #4
    I like this way of thinking about it, makes sense to me :-)
     
  6. Dec 3, 2015 #5

    Krylov

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    Sometimes, amusingly, it also works the other way around: You start with, say, a normed space and ask whether this space is itself the (topological) dual space of some other normed space. (This is not always true.) For example, you can sometimes study linear problems on spaces of measures (hard) by studying the pre-adjoint problem on a space of continuous functions.
     
  7. Dec 3, 2015 #6

    mathwonk

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    i kind of like what my prof Lynn Loomis said once; consider a function f(x) acting on a point x. Which one do you think is the function, f or x? well usually we think f is, and sends x to a number, but if you want, you can notice that given f, plugging in x sends f to a number too. so x is a function on the set of f's! this is the dual way of looking at it. The point is you have two objects f and x, and combining them gives a number, so if you fix either one, you get a function on the other variable. i.e. ( )(x) is a function in the f variable, and f( ) is a function in the x variable. That's the basic example of duality. To make the point more visually he wrote f(x) as x(f), when x is thought of as the function, and that simple reversal blew my young mind, since I had never before considered it.

    In the case where f is linear, the two points of view have similar properties: i.e. f(x+y) = f(x) + f(y), but also (f+g)(x) = f(x) + g(x),

    This also illustrates the principle of double duality. i.e. if V* is the space of all linear functions on V, then if f is such a function, we have just seen that a vector x in V gives such a function on f, namely "evaluation at x". so elements of V give linear functions on V*, i.e. there is a natural map from V to linear functions on V*, i.e. to (V*)* = V**, and in finite dimensions it is even an isomorphism V ≈ V**.

    That wikipedia article linked above looks pretty good to me, by the way, and the first few sentences inspired my thinking.

    i am also led to remark that sometimes spaces of functions have better properties than geometric spaces they act on, e.g. it is easy to multiply two functions f and g, but not so easy to multiply two points. This gives the dual object, in the sense of the one whose elements are functions, more structure sometimes which can be very useful. In this sense cohomology is dual to homology in topology, but is stronger for m aking some proofs. I.e. there are topological spaces whose homology vector spaces are isomorphic, but whose cohomology spaces can be distinguished by their multiplicative structures, so cohomology can bne used to prove the spaces are not homeomorphic but their homology cannot do this.
     
    Last edited: Dec 3, 2015
  8. Dec 3, 2015 #7

    Krylov

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    Nice explanation, I like it a lot.
     
  9. Dec 3, 2015 #8

    Mark44

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    The dual concept also shows up in Linear Programming, where the problem in standard form is
    Maximize ##c_1x_1 + c_2x_2 + \dots + c_nx_n## -- the objective function
    Subject to the constraints
    ##a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n \le b_1##
    ##a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n \le b_2##
    ##\vdots##
    ##a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n \le b_n##
    and nonnegativity constraints
    ##x_1, x_2, \dots, x_n \ge 0##

    In some applications, the goal is to minimize an objective function, so that the primal problem does not match the standard form shown above. It turns out that a dual problem can be devised that is in standard form, thereby providing a solution to the primal problem that's not in standard form.
     
  10. Dec 3, 2015 #9

    Krylov

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    Given the varied responses in this topic, it might be nice to once produce an "insight" together, demonstrating, mainly by means of examples, how duality (I'm thinking particularly about duality of vector spaces and normed spaces, finite and infinite dimensional) can be exploited to solve concrete problems in physics and applied mathematics.

    What do the other respondents think about this?
     
  11. Dec 3, 2015 #10

    PeroK

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    Well volunteered!
     
  12. Dec 3, 2015 #11

    mathwonk

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    to give another example, note that any function of two variables sets up a certain duality between the two different variables, since we can fix either one and view the result as function of the other, so each variable can be viewed as a function on the other. This is often used nowadays to define tangent vectors to a manifold in a way that may seem somewhat unintuitive or at least non geometric, namely as derivations on functions.

    One observes that given a point p, a function f and a tangent vector v we get a number, the derivative of the function f at p in the direction of the given vector, Dpf(v). Fixing p, this can be viewed as a function of the two variables f and v. With fixed f it is linear in v, but with fixed v, it obeys the leibniz rule in f, i.e., it is a “derivation” on functions f.

    Thus we can regard a tangent vector v as a derivation on functions, (provided we prove all derivations arise this way from vectors), and they are often defined as just that, e.g. in John Lee's book on manifolds, although that is surely not what they mean intuitively to a geometer like me.

    An equivalent definition uses Taylor's theorem to show that every smooth function near a given point, can be written as a constant plus a linear function plus a linear combination of products of pairs of functions vanishing at the point. Thus the differential of a function, i.e. the linear part of this expansion, is an element of I/I^2, where I is the ideal of functions vanishing at the point, and I^2 is the ideal of functions “vanishing twice”, i.e. linear combinations of pairwise products in I.

    We know the linear part of a smooth function yields a linear function on tangent vectors as above, i.e. a dual vector, and in fact all such dual vectors arise this way. Thus the cotangent or dual tangent space can be defined first as I/I^2, and then the tangent space defined as the dual of this space, i.e. as (I/I^2)*. This is the “Zariski” tangent space used in algebraic geometry, but it is equivalent to the derivations definition of differential geometry, since derivations on functions are equivalent to linear functions on I/I^2. E.g. a derivation on functions restricts to a linear map on I. which equals zero on I^2. And a linear map on I/I^2 induces a derivation on functions f taking f to the equivalence class of f - f(p) in I/I^2. Note the resemblance to the numerator of the definition of a derivative. I.e. the definition of derivative is merely the second order value at p of the function f - f(p), which has been made to vanish once at p. Thus the derivations definition is really the double dual of the space of naive tangent vectors.
     
    Last edited: Dec 5, 2015
  13. Dec 4, 2015 #12
    That's true, but duals don't include the logarithms. I'm not sure why they are excluded. Duals need to be more geometric, I guess.
     
  14. Dec 4, 2015 #13

    Fredrik

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    Would you feel that the term "dual" is appropriate if V can be defined from V* the same way that V* can be defined from V?

    This is actually the case, sort of. [Edit: I corrected a mistake in the following sentence after reading Krylov's post below]. If we define V** as the set of linear functions on V*, and then define a map ##\phi:V\to V^{**}## by saying that for each x in V, ##\phi(x)## is the element of V** defined by
    $$\phi(x)(\omega)=\omega(x)$$ for all ##\omega\in V^*##, then ##\phi## is an isomorphism. Note that the definition of ##\phi## doesn't involve any particular basis, an inner product, or anything like that.
     
    Last edited: Dec 5, 2015
  15. Dec 4, 2015 #14

    Krylov

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    Note that ##\phi(x)## is in ##V^{**}##, not in ##V^*##.
    It depends a bit on what you assumed the topological structure on ##V## to be. When you assume ##V## is Hilbert, then yes. If you merely assume that ##V## is Banach, then not necessarily, as an isomorphism of Banach spaces is by definition surjective. (For Hilbert spaces, surjectivity is always fulfilled.) It's an isometric embedding, though. A Banach space with the property that this embedding is surjective is called reflexive. For such spaces, one can indeed identify ##V^{**}## with ##V## per ##\phi##.
     
    Last edited: Dec 4, 2015
  16. Dec 4, 2015 #15

    Krylov

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    Maybe because, among other things, logarithms are not linear?
     
  17. Dec 5, 2015 #16

    Fredrik

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    Thanks for catching the mistake. I guess I typed that part too quickly. I have corrected it now.

    Good points. Since the OP is mainly interested in differential geometry, I assumed that V is a finite-dimensional vector space over ##\mathbb R##.
     
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