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As far as I know, to make a precise connection, one has to invoke Hodge theory. Suppose that $X$ is a compact smooth projective variety of dimension $d$. Then Poincare duality pairs $H^n(X,\mathbb C)$ with $H^{2d-n}(X,\mathbb C),$ for any $n$. Now the Hodge decomposition gives $$H^n(X,\mathbb C) = \oplus_{p+q = n} H^q(X,\Omega^p)$$ and ...

19

One asks that $$\max\limits_x\left(\min\limits_sf(x,s)\right)\leqslant\min\limits_y\left(\max\limits_tf(t,y)\right).$$ The assertion is equivalent to the fact that, for every $x$ and $y$, $$\min\limits_sf(x,s)\leqslant\max\limits_tf(t,y).$$ Since $\min\limits_sf(x,s)\leqslant f(x,y)\leqslant\max\limits_tf(t,y)$ by definition, this holds.

10

It's the product and coproduct in an arbitrary category $C$ that are dual to each other, and this is because their definitions are categorically dual: in the dual category $C^{op}$, the product is the coproduct in $C$ and vice versa. In a particular category $C$, though, the product and coproduct may not have dual properties, and this should merely be taken ...

8

Perhaps a simple example will help. Let $f(x,y) = \sin(x+y)$. Then $\underset{y}{\text{min}} f(x,y) = -1$ for all $x$; and $\underset{x}{\text{max}} f(x,y) = +1$ for all $y$. So $\underset{x}{\text{max}}\:\underset{y}{\text{min}} f(x,y) = \underset{x}{\text{max}} (-1) = -1$; but $\underset{y}{\text{min}}\:\underset{x}{\text{max}} f(x,y) = ... 8 A duality is a pair of related concepts that display a one-to-one translation symmetry, usually (not always) as the result of some form of involution operator. In classical logic, the operators,$\vee$and$\wedge$form a dual, and negation is their involution operator. This is expressed through deMorgan's Laws:$$\neg(A\vee B) = \neg A\wedge \neg ... 7 Duality is a very general and broad concept, without a strict definition that captures all those uses. When applied to specific concepts, there usually is a precise definition for just that context. The common idea is that there are two things which basically are just two sides of the same coin. Common themes in this topic include: Two different ... 6 First, you can assume Q is symmetric, as otherwise you can convert the problem to one that does contain a symmetric matrix via P = \frac{1}{2}(Q + Q^T). It's not hard to show that P is symmetric and satisfies x^T P x = x^T Q x for all x. Vanderbei's Linear Programming: Foundations and Extensions proves that Q being positive semidefinite is a ... 5 Let \hat x,\hat y be the arguments responsible for the value \underset{x}{\text{max}}\:\underset{y}{\text{min}} f(x,y). Then f(\hat x,y)\ge f(\hat x,\hat y) for all y. For every y, the maximization \underset{x}{\text{max}}f(x,y) extends over one of these values, and thus \underset{x}{\text{max}}f(x,y)\ge f(\hat x,\hat y) for all y, and thus ... 5 I'd like to point out that there is a generalization of Poincare duality which lives purely in the land of smooth manifolds and looks like Serre duality. Let M be a compact connected smooth n-manifold. Let E be a vector bundle on M equipped with a flat (some people say integrable) connection \nabla. Let T^{\ast} be the cotangent bundle to M. ... 5 I think the problem is with the constraints in the dual (added: there's also a problem with the objective in the dual, which I just corrected), and I think it would help spot the mistake if we rewrite the primal problem to emphasize the primal variables. Doing that, the primal looks like this: max \sum_{v\in A}\left(\sum_{i\in N}\sum_{j\in ... 5 The construction you describe can be carried out in any closed monoidal category. The ones relevant to propositional logic are the ones where \otimes denotes "and" and \Rightarrow denotes "implies." See also compact closed category, Heyting algebra, and linear logic. A good general introduction to these ideas can be found in Baez's Physics, Topology, ... 5 I believe that you should look up the Fritz John conditions. My opinion is that they are superior to the KKT conditions, in that they incorporate the rather ugly issue of the "constraint qualification" into the Lagrangean by the use of an additional multiplier -and they are able to uncover solutions to an optimization problem that under KKT may pass ... 5 The case of L^p spaces for 0\lt p\lt 1 endowed with the distance$$d(f,g):=\int |f(x)-g(x)|^p\mathrm dx$$gives a normed space whose unique continuous linear functional is the null one. Indeed, each f can be written as \frac 1n\sum_{i=1}^nf_i, where d(f_i,0) is small enough (at least in the case where the measure space is the unit interval). 5 You can't map a torsion module into a torsion-free module, so it's isomorphic to 0. For any map \phi:\mathbb Z/ n\mathbb Z\to\mathbb Z,$$ \phi(1)=m\implies \phi(0)=n\cdot \phi(1)=nm=0\quad\forall n\in\mathbb Z\implies m=0. $$For the set \operatorname{Hom}_\mathbb Z(\mathbb Z,\mathbb Z/n\mathbb Z), maps are completely determined by where 1 is sent. ... 5 Nobody claims that A \times B = A + B holds in \mathcal{C}. But it is true that A \times^{\mathcal{C}} B = A +^{\mathcal{C}^{op}} B, where the supscript denotes the category in which we take the (co)product. More generally, \lim^{\mathcal{C}}=\mathrm{colim}^{\mathcal{C}^{op}}. If you want to rename your objects in the dual category - that's ok. In ... 5 One need not to know the whole dual space of C[0,1] to see the lack of reflexivity. Simply note that C[0,1]^*, opposed to C[0,1], is non-separable because for each t\in [0,1] the map$$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$is a norm-one linear functional on C[0,1] and \|\delta_t - \delta_s\| =2 for distinct s,t. To see this, ... 4 The cases you cite in the question all appear to descend, via more or less twisted paths, from Boolean algebra, where the OR operation (often written additively) and the AND operation (often written multiplicatively) are indeed duals of each other and each distribute over the other. For example, via the Curry-Howard isomorphism, these concepts find their ... 4 I personally found this question in ML class of mine and went down to solve it myself at home as the teacher did not give any proof. Here it is: Let f(x_{0}, y_{0}) = \max_x\min_y f(x, y) and f(x_{1}, y_{1}) = \min_y\max_x f(x, y). By this definition the problem is to prove that f(x_{0}, y_{0}) \leq f(x_{1}, y_{1}) provided that they exist. By ... 4 The exactness of the first sequence means that S is injective, T surjective, and the range of S meets the kernel of T just the right way in V. Okay, so to show that the second sequence is exact, we'll start by showing \circ T is injective. Let g,g' be elements of W^{*}. Suppose that g(T) = g'(T). Since T is surjective, for any w \in W ... 4 If h \in X''' is given, define the functional \tilde h \in X' by \tilde h(f) = h(J(f)) for all f \in X. For all g \in X'', you have$$J'(\tilde h)(g) = g(\tilde h) = \tilde h(J^{-1}(g)) = h(g).$$This shows that$J'(\tilde h) = h$. 4$H^0$is just global sections, so he is askign you to describe the module of globally defined differential forms on your curve$X$. Find an open covering by affine open sets of the curve, and find the coordinate ring of those sets On each of them, find a presentation of the module of Kähler differentials. Finally, see which differentials on those open sets ... 4 1) Forget about cohomology:$H^0(X,\Omega^1_{X/k})$just means$\Omega^1_{X/k}(X)$, the vector space of global sections of the sheaf$\Omega^1_{X/k}$. 2) Forget about duality, which has nothing to do with the exercise. The exercise is probably attached to section 6.4.2, devoted to the canonical sheaf. 3) The curve$X$is smooth since$n$is not ... 4 If you are satisfied with the Hilbert space situation, the natural thing to do is to generalize it. First, suppose that the norm of$X$is uniformly convex and uniformly smooth. Then for every$x\in X$there exists a unique$x^*\in X^*$such that$\|x^*\|=\|x\|$and$x^*(x)=\|x\|^2$. (Exists by Hahn-Banach, is unique by strict convexity of$X^*$.) This ... 4 In the derived category of coherent sheaves on a smooth projective scheme$X$of dimension$n$, Serre duality in the general form$\mathrm{Ext}^i(F,G \otimes \omega) \cong \mathrm{Ext}^{n-i}(G,F)^*$becomes$\hom(F,G \otimes \omega[i]) \cong \hom(G,F[n-i])^*$, or simply$\hom(F,G \otimes \omega[n]) \cong \hom(G,F)^*$. This means that tensoring with ... 4 Your first statement is too vague. Let's make it more specific: in category theory, we can dualize categorical statements by reversing the directions of all of the arrows. This always produces a second true statement, as long as you're careful to take the dual correctly. Here's an example of taking the dual incorrectly: it's true that finite limits commute ... 4 1) There is no definition of morphism. There is a definition of category, and a category consists of objects and morphisms. That is what morhpisms are. Just like there is no definition of what a vector is. There is a definition of vector space. So a vector is (precisely) anything which is an element of some vector space. Similarly, a morphism is (precisely) ... 4 My first approach with duality was in projective geometry, where it it is a very powerful tool: we can proof a theorem for points and we have a dual theorem for straight lines ! But duality is a powerful tool in many fields of math applications. Also in chemistry I suppose are used involutions that are an example of duality. There are really many ... 4$\newcommand{\dd}{\partial}$In the sense that you're asking, velocity of a point particle is a vector, and any object that "measures a velocity and returns a scalar" is a covector. Though there's a meaningful mathematical distinction between a vector space$V$and its dual space$V^{*}$, it's probably easier to see the distinction between a vector-valued ... 4 Hint: If$f(v) = 0$for all$v$, there is nothing to prove. Otherwise, pick$v_0$such that$f(v_{0}) \neq 0$, and put$c = f(v_{0})/g(v_{0})$. (Why is$g(v_{0}) \neq 0$?) Now look at the kernel of$f - cg$. 4 If$R$is noncommutative and$P$is a right$R$-module then$\text{Hom}_R(P, R)$naturally has the structure of a left$R$-module (and vice versa if$P$is a left$R$-module), so you can't even ask for this isomorphism because the two objects belong to different categories. But this is still false if$R$is commutative. Take$R = \mathcal{O}_K\$ to be the ...

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