# Tag Info

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 ...

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 ...

7

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.

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 $\underset{y}{\text{min}}\:\underset{x}{\text{max}} f(x,y) = ... 5 I think the problem is with the constraints in the dual (added: there's also a problem with the objective in the dual, which I just corrected), and I think it would help spot the mistake if we rewrite the primal problem to emphasize the primal variables. Doing that, the primal looks like this: max$\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).

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

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

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

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, ...

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

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

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

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

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

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 ...

2

(i) is true indeed, and is a special case of $\ell_p^* = \ell_{p/(p-1)}$ for $1\le p<\infty$. (I prefer $p$ as a subscript, where it does not get in the way of asterisks.) (ii) is a special case of a general fact: a normed space isometrically embeds into its second dual. In a formula, $\iota:X\to X^{**}$ is an isometric embedding, where $\iota(x)$ is a ...

2

If $G$ is a locally compact abelian group, let $G^{\vee}$ denote its Pontrjagin dual. The dual pairing $G \times G^{\vee} \to \mathbb{C}$ realizes each element of $G$ as a character of $G^{\vee}$, hence gives a canonical map $$G \ni x \mapsto (x^{\ast} \mapsto x(x^{\ast}) = x^{\ast}(x)) \in G^{\vee \vee}.$$ Pontrjagin duality asserts that this map is an ...

2

As written,the statement is false. For example, the statement $\neg (\exists x)(x\in \varnothing)$ won't flip around that way. The actual point, however, is that the set of all subsets of a set $U$ forms a lattice with join $\cup$, meet $\cap$, bottom $\varnothing$, top $U$, and ordering $\subseteq$. Thus the usual lattice dualities apply.

2

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 ...

2

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 ...

2

Try this: $\omega \in F^\bot \cap G^\bot$ implies $\omega \in F^\bot$ and $\omega \in G^\bot$. That is, $\omega (x) = 0$ for every $x \in F$ and $\omega (y) = 0$ for every $y\in G$. Hence, $\omega (x+y) = 0$ for all $x+y \in F + G$. So $\omega \in (F+G)^\bot$. So far we have proved one inclusion: $F^\bot \cap G^\bot \subset (F+G)^\bot$. For the other ...

2

The Langlands dual group is defined for reductive groups, not only for semisimple Lie groups. For example, $SL(n)$ is dual to $PGL(n)$, $SO(2n+1)$ is dual to $Sp(n)$ and $SO(2n)$ is self-dual. The group $GL(n)$ is self-dual, too. Passing to the level of Lie algebras, the Langlands duality changes the types of simple factors of the Lie algebra by taking the ...

2

Adding to @frabala 's answer: A pullback in category $C$ corresponds to a pushout in the category $C^{OP}$, so your pullback in $Set$ corresponds to a pushout in $Set^{OP}$. However the morphisms in $Set^{OP}$ are not functions. They are just formal arrows, so working in $Set^{OP}$ is not very intuitive. Now, I warn you that some working mathematician ...

2

Let $C=R[t]$ with the usual coalgebra structure $(\varepsilon(t)=1$, $\Delta(t)=t \otimes t$). Then a $C$-comodule is the same as an $\mathbb{N}$-graded $R$-module. In fact, given a coaction $\alpha : M \to M[t]$, let $M_n = \{m \in M : \alpha(m)=m \cdot t^n\}$ and show $M = \oplus_n M_n$. Thus, in this case your question is: Does every graded $R$-module $M$ ...

2

Hint. First construct a sequence of compact sets $K_n\subset\subset K_{n+1}$, such that $$\lim_{n\to\infty}\mu(K_n)=\infty.$$ Define $f_n\ge 0$ continuous to but equal to $1$ in $K_n$ and $0$ in $K_{n+1}^c$. Then look for $f=\sum a_nf_n$, for suitable $a_n>0$, $\sum a_n<\infty$, so that $\int_X f\,d\mu=\infty$ and $f\in C_0$.

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