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35 votes

No canonical isomorphism

Let $X$ be a set. Permutations of $X$ are in bijection with total orderings on $X$, but (unless $\lvert X\rvert \le 1$) there is no canonical bijection. In terms of Joyal's theory of species, the ...
21 votes

No canonical isomorphism

As was mentioned in the comments, the example of a vector space and its dual can be seen as being about to "two vector spaces of the same dimension". Even in dimension one, the fact that two ...
20 votes
Accepted

Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters

Identify $\mathbb{N}$ with $\mathbb{Q}\cap[0,1]$ via a bijection, and consider the subspace $C([0,1])\subset\ell^\infty(\mathbb{N})$ of sequences which extend to a continuous function on $[0,1]$. ...
Eric Wofsey's user avatar
  • 30.3k
20 votes

No canonical isomorphism

In a fiber bundle $E \to B$ with typical fiber $F$, any two fibers $F_x$, $F_y$ over points $x,y \in B$ of the base are isomorphic (homeomorphic or diffeomorphic, depending on whether you are doing ...
20 votes

No canonical isomorphism

For a more elementary example: any two cyclic groups of order $n$ are isomorphic, but (when $n\ge3$) there is no preferred isomorphism between any two given cyclic groups of order $n$. (This is ...
20 votes

No canonical isomorphism

For $X$ a path-connected topological space, and two points $x,y\in X$, the fundamental group of $X$ based at $x$ is isomorphic to the fundamental group based at $y$, but not canonically. A choice of a ...
19 votes
Accepted

What is a module over a Boolean ring?

Theorem: Given $A$ a boolean ring/boolean algebra then there is an equivalence of categories between the category of $A$-modules and the category of sheaves of $\mathbb{F}_2$-vector spaces on Spec $A$....
Simon Henry's user avatar
  • 37.4k
18 votes
Accepted

Why is there a duality between spaces and commutative algebras?

I don't claim to have a complete answer but here are some miscellaneous comments. Note that topological spaces are already very nearly defined to be dual to certain commutative algebra-like ...
Qiaochu Yuan's user avatar
18 votes
Accepted

Are there topological versions of the idea of divisor?

Disclaimer. I am no expert at all in algebraic geometry. Therefore much of the following will be oversimplified or maybe even simply wrong. You are still invited to improve it. EDIT. There is a paper ...
17 votes
Accepted

Uniqueness of dualizing objects

If a dualizing object exists, there is a bijection between isomorphism classes of dualizing objects and isomorphism classes of $\otimes$-invertible objects (i.e. the Picard group), given by tensoring ...
Tim Campion's user avatar
  • 55.4k
16 votes

No canonical isomorphism

For examples that don't come from duality or torsors, there are cases where we have short exact sequences that split, but not naturally. For example, the universal coefficients theorem for cohomology ...
16 votes
Accepted

Why aren‘t op and co switched?

The problem is that for a long time there were only 1-categories and hence only one kind of duality, and sometimes people called it "op" and sometimes "co". Colimits and ...
Mike Shulman's user avatar
  • 62.7k
15 votes

No canonical isomorphism

Being algebraically closed fields of characteristic $0$ and transcendence degree $2^\omega$, the fields $\mathbb C$, $\widetilde{\mathbb Q}_p$, and $\mathbb C_p$ are isomorphic for any prime $p$, but ...
14 votes
Accepted

Are Spanier-Whitehead duals of general spaces expressible through some generalization of normal bundles?

Let $X$ be a finite complex. Then the functor $$\lim_X:\operatorname{Fun}(X,\operatorname{Sp})\to \operatorname{Sp}$$ sending a local system of spectra $E$ to its limit preserves all colimits. Indeed ...
Denis Nardin's user avatar
14 votes

No canonical isomorphism

A deliberately extreme example: an isomorphism of sets is a bijection, and two sets are isomorphic when they have the same cardinality. There is generally no preferred bijection between sets of the ...
14 votes
Accepted

On the Euler characteristic of a Poincaré duality space

It follows from Poincaré duality in $M\times M$. I'll suppress gradings and be vague about signs here. Let $x_i$ be a rational homology basis for $M$. Let $a_i$ be the cohomology basis that is dual to ...
Tom Goodwillie's user avatar
12 votes
Accepted

Self-dual plane curves

I believe that this is true for all plane curves parameterized by a monomial map $\mathbb{P}^1\to \mathbb{P}^2$. Denote by $[s,t]$ homogeneous coordinates on $\mathbb{P}^1$, and denote by $[u,v,w]$ ...
12 votes

Self-dual plane curves

There are also the limaçons, which are rational curves of degree 4 with one node and two cusps: There is a $1$-parameter family of these up to projective equivalence (the parameter is $a\not= \pm1,\...
Robert Bryant's user avatar
12 votes

If a $\otimes$-idempotent object has a dual, must it be self-dual?

If $C$ is not assumed to be symmetric, then the answer to questions 1 and 2 is no. Let $p^* \dashv p_* \dashv p^!$ be a fully faithful adjoint triple with $p_* : A \to B$. (For instance, $A= \mathrm{...
Mike Shulman's user avatar
  • 62.7k
12 votes
Accepted

What do you call $C$ if $[D,C] = D^\vee \otimes C$ for all $D$?

First of all, a nitpick: the condition "$[D,C] = D^\vee\otimes C$" should be stated more precisely as "the canonical map $D^\vee\otimes C \to [D,C]$ is an isomorphism". Now as you mentioned in a ...
Mike Shulman's user avatar
  • 62.7k
11 votes
Accepted

Gabriel-Ulmer duality for $\infty$-categories

I'm not aware of anyone writing the proof down, but I think we can patch it together as an easy consequence of several facts in Lurie's Higher Topos Theory (henceforth HTT). The statement, as I ...
Denis Nardin's user avatar
11 votes

No canonical isomorphism

For an example that does not (as far as I can tell) come from duality, a Drinfeld associator is an isomorphism between two operads in pro-groupoids (parenthesized braids and parenthesized chords, ...
11 votes

What are all the natural maps between iterated duals of vector spaces, and equations between these?

I think I can give a concrete description of these maps, and have an approach to proving faithfulness. Let $V^{*n}$ be the $n$'th dual of $V$. A map $V^{*n} \to V^{*m}$ is, by definition, the same ...
Will Sawin's user avatar
  • 126k
10 votes
Accepted

Applications of linear programming duality in combinatorics

How about Boosting1 and the Hardcore Lemma, as described in this paper? Trevisan, Luca, Madhur Tulsiani, and Salil Vadhan. "Regularity, boosting, and efficiently simulating every high-entropy ...
Joseph O'Rourke's user avatar
10 votes
Accepted

Are there other dualities on finite vector spaces besides the canonical one?

$FinVec$ and its opposite are enriched in finite dimensional $k$-vector spaces. Assume that $F$ is an enriched functor. Then consider the covariant functor: $$F(-)^*: FinVec \to FinVec$$ It is a $...
Chris Schommer-Pries's user avatar
10 votes
Accepted

If the closed unit ball of Banach space has at least one extreme point, must the Banach space the be a dual space?

Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $\ell_2$ and use $|...
Bill Johnson's user avatar
  • 30.7k
10 votes
Accepted

Compactness of the unit ball of a Banach space for topologies finer than the weak* topology

As pointed out by Goulifet, my previous answer was wrong. In fact almost the exact opposite is true: on any dual Banach space there is no locally convex vector space topology strictly stronger than ...
Nik Weaver's user avatar
  • 40.5k
9 votes

The concept of duality

From Quantum Field Theory III: Gauge Theory by Eberhard Zeidler (Springer, 4/2011): A) Preface (pg. XII): "It turns out that cohomology and homology have their roots in the rules for electrical ...
9 votes
Accepted

Elementary proof of a triangular grid lemma

We can prove this with a cross ratio chase, but first we need an easy lemma. Lemma: If points $A,B,C,D$ are on one line and $E,F,G,H$ are on another line, then $(A,B;C,D) = (E,F;G,H)$ if and only if ...
zeb's user avatar
  • 8,320
9 votes

Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters

Eric Wofsey's answer is very nice and simple. Nevertheless this might be interesting. Very old results of Kakutani [Concrete Representation of Abstract (M)-Spaces, Ann. of Math. 42 (1941)] show that ...
Jochen Wengenroth's user avatar

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