I agree with the problem of using Glue types for the compo= sition operation in the universe.
You could imagine an extra equation t= hat Glue [ phi |-> (A, reflEquiv) ] A =3D A, but this is somehow unnatur= al.
However, it is what I was postulating in step 1 of the initia= l message of this thread.
(You then need comp Glue to respect thi= s equation, which appears impossible.)

The way to get ar= ound it is, in my opinion, to add a separate type former Box^i [ phi |->= T ] A for phi |- A =3D T(i1), satisfying the laws phi |- Box^i [ phi |->= ; T ] A =3D T(i0) and regularity: Box^i [ phi |-> A ] A =3D A.
You then need comp Box to respect Box^i [ phi |-> A ] A =3D A, an= d this is also apparently not possible in CCHM, but you now have the extra = information that A is a line of types. (Using Box to turn an equivalence in= to a line of types in the definition of comp Box is circular.)
I've developed a notion of (regular) n-dimensional composi= tion, which I think might suffice to complete the argument.
If yo= u call a notion of Box that satisfies Box^i [ phi |-> A ] A =3D A "= regular Box's", then the problem can be reduced to saying that to = give regular 1-dimensional composition of regular 1-dimensional boxes, you = need some sort of "(doubly) regular 2-dimensional composition" fo= r A, and in CCHM such we can only get this "2-dimensional composition&= quot; which is regular in one direction or the other.
Of course (= I haven't actually checked, but my intuition says that), once you have = 2-dimensional composition you will need n-dimensional composition, and then= hopefully regular (n+m)-dimensional composition for A will suffice to defi= ne regular n-dimensional composition of regular n-dimensional Box's.

Below I'll put an explanation of my idea of regu= lar n-dimensional composition (returning an (n-1)-dimensional cube), since = it feels unfair to say "I know how to do this" and hide "how= to do this".
Unfortunately, it is quite complex; read at yo= ur own peril.

I don't yet know how to intellig= ibly=C2=A0describe n-dimensional composition for Glue and Box.
Fo= r now I'll just say that they should be essentially the same as in the = 1-dimensional case, but with extra faces in the equiv and final hcomp steps= corresponding to faces in the result.

=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

In CCHM,= the basic Kan operation is (comp^i A [ phi |-> a ] a0 : A(i1) [ phi |-&= gt; a(i1) ]) for i |- A with phi, i |- a : A and a0 : A(i0) [ phi |-> a(= i0) ].
Regularity for this operation means that if A and a ar= e independent of i, then comp^i A [ phi |-> a ] a0 =3D a0.
We = have regular composition for Glue types, but we don't have comp^i (Glue= [ phi |-> (A, equivRefl) ] A) =3D comp^i A.
Simplifying thing= s a little, we can take hcom^i U [ phi |-> T ] A =3D Box^i [ phi |-> = T(i :=3D ~ i) ] A to be a new type former, and give that a regular composit= ion=C2=A0operation.
Here the thing blocking us from getting comp^= i (Box^j [ phi |-> A ] A) =3D comp^i A is that, given=C2=A0i, j |- T wit= h phi, i, j |- t : T and t0 : T(i0)(j0)[ phi |-> t(i0)(j0) ],
= we need a Path T(i1)(j1) (comp^i T(j1) [ phi |-> t(j1) ] (comp^j T(i0) [= phi |-> t(i0) ] t0)) (comp^j T(i1) [ phi |-> t(i1) ] (comp^i T(j0) [= phi |-> t(j0) ] t0)) which is refl if T is degenerate in i OR j.
<= div>This path plays the role that pres does in composition for Glue.
<= div>(I will write (i o j) for composition over j then over i, and (i =3D j)= for composition over the diagonal (i =3D j). Then we want a path from (i o= j) to (j o i) which is degenerate if T is degenerate in i OR j.
=
We can't construct such a doubly regular path, but we ca= n postulate it as a new operation our types have to have, as a sort of regu= lar composition over two dimensions simultaneously. Write (i ?'[a] j) f= or this, where a gives the location in the path.
However, my intu= ition tells me that while this two-dimensional regular composition may be p= reserved by all the usual type formers, optimistically including Glue and B= ox, because the two-dimensional composition for Box will have to satisfy an= additional law comp2^i^j (Box [ phi |-> A ] A) =3D comp2^i^j A, we will= end up needing some sort of three-dimensional analogue.
Once we = need dimensions 1 and 2, we will almost certainly need dimension n.

But what should the type of three dimensional regular com= position be? There are six ways to compose along the edges of a cube, formi= ng a hexagon that we wan't to fill, but hexagons don't fit nicely i= nto cubical type theory.
Here, we remember that there is one more= way to compose across a square: along the diagonal (comp^k T(i,j:=3Dk) [ p= hi |-> t(i,j:=3Dk) ] t0), which agrees with composing either right then = up or up then right when T is degenerate in i or j (with regularity).
=
Allowing diagonals, for each square we can give a path from the diagon= al to going up then right, with swapping i and j giving a path to going rig= ht then up. Then if both of these are degenerate when T is degenerate in i = or j, their composition is also, so we recover a path between (i o j) and (= j o i). And this formulation can be extended to higher dimensions straightf= orwardly.

For each n-cube i_1, ..., i_n |- A, we h= ave an n-dimensional regular composition operation which is a (n-1)-dimensi= onal cube. Consider notation (i_1 ?[j_1] i_2 ?[j_2] ... ?[j_n-1] i_n) This = is supposed to represent composition in an n-cube indexed by the i's, w= here our position in the resulting (n-1)-cube is given by the j's. We h= ave to describe the faces of this cube, which we describe by saying that ?[= 0] reduces to =3D and ?[1] reduces to o.
Thus for 2-dimensional A= , we have a line (i =3D j) --- (i ?[a] j) --- (i o j), and for three dimens= ional A, we have a square with four faces (i =3D j ?[a_2] k), (i o j ?[a_2]= k), (i ?[a1] j =3D k) (i ?[a1] j o k), relating the four corners given by = coercing over i then j then k, coercing over i then j and k simultaneously,= coercing over i and j simultaneously then coercing over k, and coercing ov= er i, j, and k simultaneously.
The point is that this gives a wel= l-defined notion of cube.
We also have to define the regularity c= ondition: what happens when A, a are degenerate in i_k? We require that (i_= 1 ... i_k-1 ?[j_k-1] i_k ?[j_k] i_k+1 ... i_n) is equal to (i_1 ... i_k-1 ?= [j_k-1 v j_k] i_k+1 ... i_n), treating j_0 =3D j_n =3D 1.
Extendi= ng this to =3D and o, if A, a are degenerate in i, then (i =3D j) and (i o = j) reduce to (j).
On a two-dimensional cube, this says that (i ?[= a] j) reduces to (j) when A, a are degenerate in i, and to (i) when A, a ar= e degenerate in j.
(That this is all coherent is obvious to me, b= ut I have no idea how to explain it well)

This all= gives a notion of n-dimensional regular composition, which takes an n-cube= type A and partial term phi, i1, ..., in |- a and=C2=A0a base point a0 : A= (i_k :=3D 0)[ phi |-> a(i_k :=3D 0) ] and returns=C2=A0an (n-1)-cube in = A(i_k :=3D 1) [ phi |-> everywhere a(i_k :=3D 1) ].
There are = three classes of equations that this has to satisfy, two for the face maps = j_k =3D 0 and j_k =3D 1 in the output, and one for when A, a are=C2=A0degen= erate in i_k.
In the one-dimensional case, this is the same as th= e specification of the usual regular comp.
Considering j_k to be = formally extended with j_0 =3D j_n =3D 1, we get a "zero-dimensional&q= uot; composition operation, which should be taken to be the identity. (I ha= ve no idea what it would mean to let j_0 and j_n vary)
I am reaso= nably confident that this operation is closed under usual type formers (Pi,= Sg, Path, ...) by essentially the same argument as for 1-dimensional comp.=
I think I have worked out how to give a regular n-dimensional co= mposition operation for Glue, by essentially the same argument as in CCHM, = but with additional faces for j_k =3D 1 in the equiv and final hcomp steps.=

In this setting, we want Box to also be m-dimensi= onal: j_1 ... j_n-1 |- Box^i1...^im _{j_1}..._{j_n-1} [ phi |-> T ] A fo= r phi, i1, ..., im |- T, with phi |- A =3D T(i_k :=3D 1).
I think= I have worked out how to give a regular n-dimensional composition operatio= n for m-dimensional Box, using essentially the same argument as for Glue, b= ut replacing pres with the composition of two paths given by (m+n)-dimensio= nal regular composition.
To use Box as the regular n-dimensional = composition operation for the universe, we need that (Box, comp Box) respec= ts the equations for when j_k =3D 0, j_k =3D 1, or T is independent of j.
My hope is that, as in the (1, 1) dimensional case, using (m+n)-di= mensional regular composition for pres will make this work.

<= /div>
I have sketched on paper that composition for Glue and Box are re= gular, but that is already pushing the limits of my confidence in paper cal= culations.
Checking that comp Box respects the equations that it = needs to is another layer more=C2=A0complex, and I haven't managed to d= o so on paper.
However, by analogy with the (1, 1) dimensional ca= se, I am hopeful.

On Mon, Sep 16, 2019 at 3:01 PM Licata, Dan &= lt;dlicata@wesleyan.edu> wro= te:
Wow! I read that pretty closely and couldn=E2=80=99t find a problem. That i= s surprising to me. Will look more later.=C2=A0

What I was saying about fill for the aligning algorithm is also wrong = =E2=80=94 I wasn=E2=80=99t thinking that the type of the filling was also d= egenerate, but of course it is here. So that might work too. =C2=A0

However, even if there can be a universe of regular types that is clos= ed under glue types, there=E2=80=99s still a problem with using those glue = types to show that that universe is itself regularly fibrant, I think? If y= ou define comp U [phi -> i.T] B to be Glue [phi -> T(1),...] B then no matter how nice the equivalence is (eg the = identity) when i doesn=E2=80=99t occur in T, the type will not equal B =E2= =80=94 that would be an extra equation about the Glue type constructor itse= lf. Does that seem right?

-Dan

On Sep 16, 2019, at 1:09 PM, Jasper Hugunin <jasperh@cs.washington.edu> wrote= :

Hi Dan,

Of course. I'm thinking primarily of composition for Glue given in= the CCHM paper you linked, reproduced below.
As you know, the single potential issue is that we need pres of a dege= nerate filling problem and function to be reflexivity. I claim that this ho= lds by regularity of composition in T and A, partly as a consequence of the= fact that regularity of composition implies regularity of filling (that fill of a degenerate system is refl), = which certainly holds for fill defined by connections (and I believe also h= olds for fill as defined in ABCFHL).

(a)
Given i |- B =3D Glue [ phi |-> (T, f) ] A, with psi, i |- b : B an= d b0 : B(i0)[ psi |-> b(i0) ], we want to compute b1 =3D comp^i B [ psi = |-> b ] b0 : B(i1)[ psi |-> b(i1) ].
We set a :=3D unglue b and a0 :=3D unglue b0.
Set delta :=3D forall i. phi.
Then we take:
a1' :=3D comp^i A [ psi |-> a ] a0
delta |- t1' :=3D comp^i T [ psi |-> b ] b0
delta |- omega :=3D pres^i f [ psi |-> b ] b0
phi(i1) |- (t, alpha) :=3D equiv f(i1) [ delta |-> (t1', omega)= , psi |-> (b(i1), refl=C2=A0a1') ] a1'
a1 :=3D hcomp^j A(i1) [ phi(i1) |-> alpha j, psi |-> a(i1) ] a1&= #39; (note that in the regular setting the psi face is redundant)
b1 :=3D glue [ phi(i1) |-> t1 ] a1

With given i |- f : T -> A, with psi, i |- b : T and b0 : T(i0)[ ps= i |-> b(i0) ], we define
pres^i f [ psi |-> b ] b0 =3D <j> comp^i A [ psi |-> f b, = j =3D 1 |-> f (fill^i T [ psi |-> b ] b0) ] (f(i0) b0).

(b)
Now consider the regular case, where phi, T, f, and A are independent = of i. We want that b1 =3D b0.
We have that a is independent of i, and delta =3D phi.

First consider delta (=3D phi) |- pres^i f [ psi |-> b ] b0. (This = is the explanation of your first dash)
Note that if comp^i A is regular, then fill^i A [ psi |-> b ] b0 = =3D b
This is <j> comp^i A [ psi |-> f b, j =3D 1 |-> f (fill^i = T [ psi |-> t ] t0) ] (f t0) =3D <j> comp^i A [ psi |-> f b, j = =3D 1 |-> f t0 ] (f t0) =3D <j> f t0.
Thus pres of a degenerate filling problem and function is reflexivity.=

Going back to composition of Glue,
a1' =3D a0
phi |- t1' =3D b0
phi |- omega =3D refl=C2=A0(f b0)
phi |- (t1, alpha) =3D (t1', omega) (since delta =3D phi, so we en= d up in the delta face of equiv)
a1 =3D a1' (the only dependence on j is via (alpha j), but alpha = =3D omega =3D refl, so this filling problem is degenerate)
b1 =3D glue [ phi |-> t1 ] a1 =3D glue [ phi |-> b0 ] a0 =3D glu= e [ phi |-> b0 ] (unglue b0) =3D b0 (by eta, see Figure 4 in=C2=A0CCHM)<= /div>

Thus this algorithm for composition of Glue is regular.
Other algorithms, such as the one in ABCFHL, may not be, but I am pron= e to believe that there exist regular algorithms in other settings includin= g Orton-Pitts and Cartesian cubes.

Best regards,
- Jasper Hugunin

On Mon, Sep 16, 2019 at 12:18 PM Lica= ta, Dan <dlica= ta@wesleyan.edu> wrote:
Hi Jasper,

It would help me follow the discussion if you could say a little more about= (a) which version of composition for Glue exactly you mean (because there = is at least the one in the CCHM paper and the =E2=80=9Caligned=E2=80=9D one= from Orton-Pitts, which are intensionally different, as well as other possible variations), and (b) include some of your reason= ing for why you think things are regular, to make it easier for me and othe= rs to reconstruct.=C2=A0

My current understanding is that

- For CCHM proper https://arxiv.org/pdf/1611.02108.pdf the potential issue is with the = =E2=80=98pres=E2=80=99 path omega, which via the equiv operation ends up in= alpha, so the system in a1 may not be degenerate even if the input is.=C2= =A0 Do you think this does work out to be degenerate?=C2=A0

- For the current version of ABCFHL https://github.com/dlicata335/cart-cube/blob/master/cart-cube.pdf which= uses aligning =E2=80=9Call the way at the outside=E2=80=9D, an issue is wi= th the adjust_com operation on page 20, which is later used for aligning (i= n that case beta is (forall i phi)).=C2=A0 The potential issue is that adjust_com uses a *filler*, not just a composition from 0 to= 1, so even if t doesn=E2=80=99t depend on z, the filling does, and the out= er com won=E2=80=99t cancel.=C2=A0 In CCHM, filling is defined using connec= tions, so it=E2=80=99s a little different, but I think there still has to be a dependence on z for it to even type check =E2=80=94 it s= hould come up because of the connection term that is substituted into the t= ype of the composition problem.=C2=A0 So I=E2=80=99d guess there is still a= problem in the aligned algorithm for CCHM.=C2=A0

However, it would be great if this is wrong and something does work!=C2=A0 =

-Dan

> On Sep 15, 2019, at 10:18 PM, Jasper Hugunin <jasperh@cs.washington.edu>= wrote:
>
> This doesn't seem right; as far as I can tell, composition for Glu= e types in CCHM preserves regularity and reduces to composition in A on phi= .
>
> - Jasper Hugunin
>
> On Sun, Sep 15, 2019 at 3:28 AM Anders Mortberg <anders.mortberg@math.su.se> wrote:
> Hi Jasper,
>
> Indeed, the problem is to construct an algorithm for comp (or even
> coe/transp) for Glue that reduces to the one of A when phi is true
> while still preserving regularity. It was pointed out independently by=
> Sattler and Orton around 2016 that one can factor out this step in our=
> algorithm in a separate lemma that is now called "alignment"= . This is
> Thm 6.13 in Orton-Pitts and discussed in a paragraph in the end of
> section 2.11 of ABCFHL. Unless I'm misremembering this is exactly<= br> > where regularity for comp for Glue types break down. In this step we > do an additional comp/hcomp that inserts an additional forall i. phi > face making the comp/coe irregular.
>
> One could imagine there being a way to modify the algorithm to avoid > this, maybe by inlining the alignment step... But despite considerable=
> efforts no one has been able to figure this out and I think Swan's=
> recent paper (
https://arxiv.org/abs/1808.00920v3) shows t= hat this is
> not even possible!
>
> Another approach would be to have weak Glue types that don't stric= tly
> reduce to A when phi is true, but this causes problems for the
> composition in the universe and would be weird for cubical type
> theory...
>
> In light of Swan's negative results I think we need a completely n= ew
> approach if we ever hope to solve this problem. Luckily for you Andrew=
> Swan is starting as a postdoc over in Baker Hall in October, so he can=
> explain his counterexamples to you in person.
>
> Best,
> Anders
>
> On Sun, Sep 15, 2019 at 7:57 AM Jasper Hugunin
> <jas= perh@cs.washington.edu> wrote:
> >
> > Offline, Carlo Angiuli showed me that the difficulty was in part = 1, because of a subtlety I had been forgetting.
> >
> > Since types are *Kan* cubical sets, we need that the Kan operatio= ns agree as well as the sets.
> > So part 1 could be thought of as (Glue [ phi |-> equivRefl A ]= A, compGlue) =3D (A, compA), and getting that the Kan operations to agree = was/is difficult.
> > (Now that I know what the answer is, it is clear that this was al= ready explained in the initial discussion.)
> >
> > Humbly,
> > - Jasper Hugunin
> >
> > On Fri, Sep 13, 2019 at 2:10 AM Jasper Hugunin <jasperh@cs.washington.edu<= /a>> wrote:
> >>
> >> Hello all,
> >>
> >> I've been trying to understand better why composition for= the universe does not satisfy regularity.
> >> Since comp^i [ phi |-> E ] A is defined as (roughly) Glue = [ phi |-> equiv^i E ] A, I would expect regularity to follow from two pa= rts:
> >> 1. That Glue [ phi |-> equivRefl A ] A reduces to A (a sor= t of regularity condition for the Glue type constructor itself)
> >> 2. That equiv^i (refl A) reduces to equivRefl A
> >> I'm curious as to which (or both) of these parts was the = issue, or if regularity for the universe was supposed to follow from a diff= erent argument.
> >>
> >> Context:
> >> I've been studying and using CCHM cubical type theory rec= ently, and often finding myself wishing that J computed strictly.
> >> If I understand correctly, early implementations of ctt did h= ave strict J for Path types, and this was justified by a "regularity&q= uot; condition on the composition operation, but as discussed in this threa= d on the HoTT mailing list, the definition of composition for the universe was found to not satisfy regularity.
> >> I don't remember seeing the regularity condition defined = anywhere, but my understanding is that it requires that composition in a de= generate line of types, with the system of constraints giving the sides of = the box also degenerate in that direction, reduces to just the bottom of the box. This seems to be closed under the usual typ= e formers, plus Glue, but not the universe with computation defined as in t= he CCHM paper (for trivial reasons and non-trivial reasons; it gets stuck a= t the start with Glue [ phi |-> equiv^i refl ] A not reducing to anything).
> >>
> >> Best regards,
> >> - Jasper Hugunin
> >
> > --
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