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19

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 ...


15

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 ...


6

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) = ...


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

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 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

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, ...


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

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 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 ...


4

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. ...


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

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

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 ...


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

$H^0$ 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 ...


3

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 ...


3

This is a very standard and yet elegant proof. We shall create a bounded linear functional on $\ell^\infty(\mathbb N)$, which is not realized by any element of $\ell^1(\mathbb N) $, using the Hahn-Banach Theorem. Let first $c(\mathbb N)$ be the set of converging sequences. Clearly $c(\mathbb N)$ is a closed subspace of $\ell^\infty(\mathbb N)$. Let ...


3

This is a longer version of my previous comments. In topology, along with the ordinary cohomology, there is also the so called cohomology with compact support, denoted by $H_c^*(X)$. It can be defined using compactly supported differential forms using singular cochains that vanish outside a compact set using the derived functor of the compactly supported ...


3

I don't see where the comment button is, maybe because I'm a newbie? This is the Riesz representation theorem which states that the topological dual space of the space of continuous functions on a compact space $X$ is the space of Borel measure on $X$. You can see a proof in Real and complex analysis by Rudin. Your thought is not true, because the ...


3

It is usually understood that $f_0(x)=+\infty$ for $x\notin D$. This would make both definitions equivalent.


3

The Pruefer group ${\mathbb Z}_{p^{\infty}}$ is the direct limit of the sequence $${\mathbb Z}/p{\mathbb Z}\hookrightarrow {\mathbb Z}/p^2{\mathbb Z}\hookrightarrow ...$$ Applying $(-)^{\wedge} = \text{Hom}_{\text{cnt}}(-,{\mathbb S}^1)$ shows that $\left({\mathbb Z}_{p^\infty}\right)^{\wedge}$ is the inverse limit of the Pontryagin duals of ${\mathbb Z}/p^k ...


3

I'm a PhD student in operator algebraic quantum groups, which is a generalisation of topological groups effectively. The other side of quantum groups are the purely algebraic versions, that is to say Hopf algebras. So quantum groups begin with the study of functions on a group (as it's a generalisation), in the Hopf algebra case we have polynomials, for ...


3

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 $M$. ...


3

There is a nice duality between $C(K)$ and $L_1(\mu)$-spaces. The double dual of $C(K)$ is of the form $C(L)$ for some huge compact space $L$. (Actually it is also isometric to $L_\infty(\nu)$ for some huge measure $\nu$.) The second dual of $L_1(\mu)$ is also of the form $L_1(\nu)$. However people rarely think of duals/biduals of these spaces like that. ...


3

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 ...


3

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 $w \in W$ ...


3

The space $C^b(\mathbb R)$ is a commutative C*-algebra, hence by the Gelfand-Naimark theorem it is *-isomorphic to some $C(K)$ space ($K$ is the spectrum of this algebra and the *-isomorphism is just the Gelfand transform). The dual of a $C(K)$ space is described by the Riesz-Kakutani theorem as the space of all regular Borel measures on $K$. EDIT: In this ...



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