## ct.category theory – Does rigidity imply a unqiue dualizing functor?

For rigid symmetric monoidal categories, there is in fact always a duality functor, unique up to isomorphism (without symmetry you would have to specify what you mean by “dual”).

Here is a possible proof of existence:

Consider the following category $$tilde C$$, its objects are quadruples $$(x,y,eta,epsilon)$$ where $$x,y$$ are objects of $$C$$ and $$eta: 1_Cto xotimes y$$ is a morphism in $$C$$, and $$epsilon : yotimes xto 1_C$$ is another morphism, together exhibiting $$y$$ as a dual of $$x$$.

A morphism $$(x_0,y_0,eta_0,epsilon_0)to (x_1,y_1,eta_1,epsilon_1)$$ is a pair of morphisms, $$f :x_0to x_1, g : y_1to y_0$$ such that $$g$$ “is dual to $$f$$” in the following sense: the composite $$y_1cong y_1otimes 1_C to y_1otimes x_0otimes y_0to y_1otimes x_1otimes y_0 to 1_Cotimes y_0cong y_0$$
is equal to $$g$$, where I’ve used the unitors of $$C$$, the map $$eta_0$$, then $$f$$, and $$epsilon_1$$ .

Composition is the obvious one, where you have to check that the “dual” of $$fcirc f’$$ is the dual of $$f’circ$$ the dual of $$f$$. This check is essentially going to come up in any proof, and is the essential part of the proof – you should try to do that check.

I then claim the following two things: the forgetful functor $$tilde Cto C, (x,y,eta,epsilon)mapsto x$$ and $$(f,g)mapsto f$$ is an equivalence of categories.

The fact that it is essentially surjective comes from the fact that $$C$$ is rigid, and fully faithfulness comes from the fact that for any $$f$$ there is a unique $$g$$ which works (namely, the composite that I described !).

Then you get the following zigzag $$Coversetsimleftarrow tilde Cto C^{op}$$ where the second arrow is projection to $$y$$ (and $$g$$)

Now, equivalences have quasi-inverses (if you have the axiom of choice, or a canonical choice of a dual and duality data for any $$x$$ – which you don’t need to be functorial a priori), so that gives you a well defined functor $$D: Cto C^{op}$$ which is obviously an equivalence.

Unicity (up to natural isomorphism) is not hard to prove, you should try that !

It gets a little bit more complicated for higher categories. A functor $$D$$ is always definable, in a similar way as above (although the definition of $$tilde C$$ needs to be changed a bit, because you need to specify a $$2$$-morphism witnessing one of the triangle identities), but saying what it does on objects and morphisms is no longer enough to fully determine it as a functor – the obstruction is exactly the same obstruction as the unicity of a functor which is the identity on objects and on $$1$$-morphisms.

## mg.metric geometry – Rigidity for convex surfaces in elliptic/hyperbolic space

From Alexandrov’s work we know that any metric on the sphere with lower curvature bound $$kappa$$ (in the sense of Alexandrov) can be realized as a closed convex surface (i.e. boundary of a compact convex domain) in the $$3$$-space form of constant curvature $$kappa$$.

For $$kappa=0$$, the surface is unique up to isometry of the surrounding space $$mathbb{R}^3$$ due to Pogorelov. Is the same true for the elliptic and hyperbolic case? In other words, are two closed convex surfaces in these spaces congruent when they are isometric with respect to their inner metrics?

I asked this question here on Math SE, but did not get an answer.

## ag.algebraic geometry – Rigidity lemma up to cover

Let $$X,Y,Z$$ be proper varieties over $$mathbb{C}$$. Let $$Wto Xtimes Y$$ be a finite flat surjective morphism, and let $$Wto Z$$ be a morphism.
Fix $$x in X$$ and suppose that $${x}times Y$$ is contracted to a point, say $$z$$, in $$Z$$ in the following sense:

The fibre over $${x}times Y$$ along $$Wto Xtimes Y$$ is contracted to $$Z$$ via $$Wto Z$$.

If $$Wto Xtimes Y$$ is an isomorphism, then the Rigidity Lemma says that $$Wcong Xtimes Yto Z$$ factors over $$Xto Z$$. (The map is “independent of $$Y$$” in some sense.)

Is there some version of the Rigidity Lemma that we can apply in the above situation to see that $$Wto Z$$ is also “independent of $$Y$$“? More precisely, I expect that there is a finite flat morphism $$W’to W$$ such that $$W’to Z$$ factors over some covering of $$X$$.

## \$l\$-adic rigidity for Milnor \$K\$-theory

Given a local henselian ring $$A$$ with the maximal ideal $$m$$, does the quotient map $$Amapsto A/m$$ induce isomorphisms on $$l$$-adic Milnor $$K$$-theories? ($$K_n^M(R)otimes mathbb{Z}_l$$, where $$l$$ is an invertible prime.)

## ag.algebraic geometry – Global version of Gabber’s rigidity theorem

I had a number of questions regarding Gabber’s rigidity, I’m not sure whether I am understanding it correctly, so please let me know if I’m making any mistakes.

Let $$A$$ be a ring (let’s assume Noetherian) and $$I$$ be an ideal, since the pair $$(hat{A},I)$$ is a henselian pair ($$hat{A}$$ is the completion along $$I$$), Gabber’s rigidity in algebraic $$K$$-theory for henselian pairs implies that $$K_*(A/I, mathbb{Z}/lmathbb{Z})simeq K_*(hat{A}, mathbb{Z}/lmathbb{Z})$$. Here algebraic $$K$$-theory of the completion is taken in the sense that $$hat{A}$$ is just a Noetherian ring and we can construct $$BGL(hat{A})^+$$ just as we can do for any ring i.e. the $$K$$-theory of finitely generated projective modules on $$hat{A}$$. There is another way to define $$K$$-theory and that is to take the $$K$$-theory of formal vector bundles on $$hat{A}$$, which means a system of vector bundles on all finite thickenings of $$I$$ with compatibility conditions under pullbacks. I don’t think these two coincide in general. (Correct me if I am wrong)

My original question is about whether there is a global version of this rigidity in the setting of formal completions or not? The following is my attempt on this question. (I am not sure whether I am making any mistakes or not)

Now we have Noetherian schemes $$X$$ and a closed subscheme $$Z$$ and we are looking at the formal completion $$X_Z$$. Well as we are dealing with formal schemes it makes sense to consider formal vector bundles and compare its $$K$$-theory with the $$K$$-theory of $$Z$$ with coefficient in $$mathbb{Z}/lmathbb{Z}$$. But this comparison possibly fails even in the affine case according to the observation above. So I think the natural generalization would be instead of considering $$X_Z$$ as a formal scheme we can consider it as a scheme just as in the affine case. This means if on an affine chart $$X_Z$$ is the formal scheme $$hat{A}$$, we consider it just as a scheme i.e. $$text{Spec}(hat{A})$$ (Please let me know if this construction is flawed because I’ve never since a formal scheme to be considered as a scheme but I do not see any issues on this case). Since completion with respect to an ideal preserves being Noetherian and having a finite Krull dimension and because of Zariski descent of connective algebraic $$K$$-theory this implies that $$K_*(Z,mathbb{Z}/lmathbb{Z})simeq K_*(X_Z,mathbb{Z}/lmathbb{Z})$$ where $$X_Z$$ is regarded as a scheme rather than a formal scheme.

For the question when the $$K$$-theory of formal scheme $$X_Z$$ should match with its $$K$$-theory as a scheme, I think if the pair $$(X,Z)$$ satisfy the property that every formal vector bundle on $$X_Z$$ can be extended to a neighborhood of $$Z$$ they should be compatible.

## dg.differential geometry – A strong form of Mostow rigidity without geometrization?

Gabai proved that homotopy hyperbolic 3-manifolds are virtually hyperbolic, in the paper of that name:

Gabai, David, Homotopy hyperbolic 3-manifolds are virtually
hyperbolic.
J. Amer. Math. Soc. 7 (1994), no. 1, 193–198.

I suspect this is the best you can do without geometrisation.

## probability theory – Criteria for the rigidity of the sequence of distribution functions

Leave $$φ_n$$ Be a sequence of characteristic functions. We know that converge at each point of a not empty
open interval around 0
yet function $$φ$$ which is also continued at 0.

Can you say that the sequence
of the distribution functions corresponding to $$φ_n$$They are tight

I am aware of the test of a different version where convergence occurs in the entire real line. But what to do in this case?

## differential equations – Non-autonomous ODE uses NDSolve, error: Step Size is effectively zero; The singularity or rigidity of the system is suspected.

I have seen this error `NDSolve :: ndsz` many times when I use `NDSolve` To obtain the solution of a non-autonomous ODE. I try but everything has failed. Thanks, Here is the code, very simple.

``````    s = NDSolve[{x'
Y
X[0] == 1, and[0] == 1}, {x
``````