 Jun 2021

math.stackexchange.com math.stackexchange.com

If you have a vector space, any vector space, you can define linear functions on that space. The set of all those functions is the dual space of the vector space. The important point here is that it doesn't matter what this original vector space is. You have a vector space V
One of the better "simple" discussions of dual spaces I've seen:
If you have a vector space, any vector space, you can define linear functions on that space. The set of all those functions is the dual space of the vector space. The important point here is that it doesn't matter what this original vector space is. You have a vector space V, you have a corresponding dual V∗.
OK, now you have linear functions. Now if you add two linear functions, you get again a linear function. Also if you multiply a linear function with a factor, you get again a linear function. Indeed, you can check that linear functions fulfill all the vector space axioms this way. Or in short, the dual space is a vector space in its own right.
But if V∗ is a vector space, then it comes with everything a vector space comes with. But as we have seen in the beginning, one thing every vector space comes with is a dual space, the space of all linear functions on it. Therefore also the dual space V∗ has a corresponding dual space, V∗∗, which is called double dual space (because "dual space of the dual space" is a bit long).
So we have the dual space, but we also want to know what sort of functions are in that double dual space. Well, such a function takes a vector from V∗, that is, a linear function on V, and maps that to a scalar (that is, to a member of the field the vector space is based on). Now, if you have a linear function on V, you already know a way to get a scalar from that: Just apply it to a vector from V. Indeed, it is not hard to show that if you just choose an arbitrary fixed element v∈V, then the function Fv:ϕ↦ϕ(v) indeed is a linear function on V∗, and thus a member of the double dual V∗∗. That way we have not only identified certain members of V∗∗ but in addition a natural mapping from V to V∗∗, namely F:v↦Fv. It is not hard to prove that this mapping is linear and injective, so that the functions in V∗∗ corresponding to vectors in V form a subspace of V∗∗. Indeed, if V is finite dimensional, it's even all of V∗∗. That's easy to see if you know that dim(V∗)=dimV and therefore dim(V∗∗)=dimV∗=dimV. On the other hand, since F is injective, dim(F(V))=dim(V). However for finite dimensional vector spaces, the only subspace of the same dimension as the full space is the full space itself. However if V is infinite dimensional, V∗∗ is larger than V. In other words, there are functions in V∗∗ which are not of the form Fv with v∈V.
Note that since V∗∗again is a vector space, it also has a dual space, which again has a dual space, and so on. So in principle you have an infinite series of duals (although only for infinite vector spaces they are all different).


math.stackexchange.com math.stackexchange.com

There are some very beautiful and easily accessible applications of duality, adjointness, etc. in Rota's modern reformulation of the Umbral Calculus. You'll quickly gain an appreciation for the power of such duality once you see how easily this approach unifies hundreds of diverse specialfunction identities, and makes their derivation essentially trivial. For a nice introduction see Steven Roman's book "The Umbral Calculus".
Note to self: Look at [[Steven Roman]]'s book [[The Umbral Calculus]] to follow up on having a more intuitive idea of what a dual space is and how it's useful

Dual spaces also appear in geometry as the natural setting for certain objects. For example, a differentiable function f:M→R
Dual spaces also appear in geometry as the natural setting for certain objects. For example, a differentiable function f:M→R where M is a smooth manifold is an object that produces, for any point p∈M and tangent vector v∈TpM, a number, the directional derivative, in a linear way. In other words, ==a differentiable function defines an element of the dual to the tangent space (the cotangent space) at each point of the manifold.==

The dual is intuitively the space of "rulers" (or measurementinstruments) of our vector space. Its elements measure vectors. This is what makes the dual space and its relatives so important in Differential Geometry, for instance.
A more intuitive description of why dual spaces are useful or interesting.


en.wikipedia.org en.wikipedia.org

To put it succinctly, differential topology studies structures on manifolds that, in a sense, have no interesting local structure. Differential geometry studies structures on manifolds that do have an interesting local (or sometimes even infinitesimal) structure.
Differential topology take a more global view and studies structures on manifolds that have no interesting local structure while differential geometry studies structures on manifolds that have interesting local structures.

 Mar 2016


Letβ:V×V→Wbe a symmetric bilinear form whereVand (W,h,i) arereal vector spaces of finite dimensionnandp, respectively, equipped withinner products.Thesnullityνsofβfor any integer 1≤s≤pis defined byνs= maxUs⊂Wdim{x∈V:βUs(x, y) = 0 for ally∈V}.HereβUs=πUs◦βwhereUsis anysdimensional subspace ofWandπUs:W→Usdenotes the orthogonal projection.LetR:V×V×V×V→Rbe the multilinear map with the algebraicproperties of the curvature tensor defined byR(x, y, z, w) =hβ(x, w), β(y, z)i − hβ(x, z), β(y, w)i.Lemma 4.Assume that2p < nandνs< n−2sfor all1≤s≤p. LetV=V1⊕V2be an orthogonal splitting such thatR(x, y, z, u) =R(x, y, u, v) =R(x, u, v, w) = 0for anyx, y, z∈V1andu, v, w∈V2. Then,S=span{β(x, y) :x∈V1andy∈V2}= 0.


ac.elscdn.com ac.elscdn.com

second fundamental_form h satisfies h(TpL,xTpLj =0 forallp E M
Para o nosso caso, assumir essa hipótese com respeito a decomposição do espaço tangente ao longo do bordo.

 Dec 2015

www.mathunion.org www.mathunion.org

Let M be an rcdimensional manifold of class C°° and g any given Riemannian metric on M. We will consider the following classical problem motivated by differential geometry. Does there exist an embedding u = (w1,..., uq) : M > R9 such that the usual euclidian metric of R9 induces on the submanifold u(M) the given metric gl In other words, w must satisfy E(w) := dudu = g, (1) or in local coordinates 9 du1 du1 _ ,tîâ?â?"Qij' The dot in (1) denotes the usual scalar product of R9. The notion embedding means, that w is locally an immersion and globally a homeomorphism of M onto the subspace u(M) of R*. If an embedding w : M • R9 satisfies (1) on the whole M, we speak of an isometric embedding. If w is an immersion and a solution of (1) in a (possibly small) neighbourhood of any point of M, we speak of a local isometric embedding.

 Mar 2015

arxiv.org arxiv.org

θ dμ ≥ p 16 π  Σ 
Qual a relação dessa desigualdade com a dita desigualdade de Penrose Riemanniana provada por HuiskenIlmanen e Bray?

GIBBONSPENROSE INEQUALITY
Qual a relação dessa desigualdade com a dita desigualdade de Penrose Riemanniana provada por HuiskenIlmanen e Bray?
