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17

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 ...


11

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 ...


5

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) = ...


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

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

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 ...


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

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

$H^0$ 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 ...


3

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 ...


3

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


3

Take the algebra $\mathcal A$ generated by the closed sets in $\mathbb R$. The space of finitely-additive signed measures on $\mathcal A$, with variation norm, is the dual of $C_b(\mathbb R)$. The top reference for this and many other interesting topics: Gillman & Jerison, Rings of Continuous Functions. Note $\mathbb R$ is metrizable, so the "zero ...


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 ...


3

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 ...


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

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 ...


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

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

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

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

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 ...


3

If $V$ has an inner product $\langle \cdot,\cdot\rangle$, then for each $v\in V$ we can construct an element $f_v\in V^*$ by $$ f_v(x) = \langle x,v\rangle \quad x\in X. $$ This yields a natural embedding of $V$ in $V^*$ for inner product spaces.


3

You have such a thing in spaces with an inner product. A inner product $\langle x,y\rangle$ is an inner product on a vector space $V$ over field $K$, i.e. a map $$ \langle \cdot,\cdot \rangle \,:\, V\times V \to K $$ that is linear in it's first argument, and either linear or conjugate linear (i.e. $\langle x,\lambda y \rangle = \bar\lambda \langle ...


3

You can put Pontrjagin duality in a nice categorical framework, namely as an adjunction which turns out to be an equivalence (as explained on the nlab), but the proof that the unit of the adjunction i.e. the inclusion to the bidual $G \to \widehat{\widehat{G}}$ is an isomorphism, cannot be done by abstract nonsense. This stays a nontrivial and deep result of ...


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 ...


2

The easy answer: Yes, there is a natural definition of Hecke operators on $H_1$ that makes this work. This is explained in Stein's book, as Qiaochu has already pointed out, and in lots of other places too. But there is a more subtle phenomenon going on here. The isomorphism $H_1 \cong Hom(H^1, \mathbf{Z})$ doesn't really come from Poincare duality (it's ...



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