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### 7 Double Groupoids

A double groupoid is a double category in which all the category structures are groupoids. For the most general type of double groupoid there is also an associated pre-crossed module. In this package we consider only basic double groupoids, which do not involve pre-crossed modules. The more general case will be discussed in the XMod package.

In a double groupoid, as well as objects and arrows, we need a set of squares. A square is bounded by four arrows, two horizantal and two vertical, and there is a horizantal groupoid structure and a vertical groupoid structure on these squares.

Double groupoids can be considered where the vertical arrows come from one groupoid, and the horizantal arrows from another. The double groupoids constructed here are special in that all four arrows come from the same groupoid. We call these edge-symmetric double groupoids.

This addition to the package is very experimental, and will be extended soon.

#### 7.1 Double groupoid squares

Let $$G$$ be a connected groupoid with object set $$\Omega$$. Let $$\Box(G)$$ be the set of squares with objects from $$\Omega$$ at each corner; plus two vertical arrows and two horizantal arrows from Arr$$(G)$$. The following picture illustrates the situation:

$\vcenter{\xymatrix @=4pc{ u_1 \ar[r]^{a_1} \ar[d]_{d_1} & u_2 \ar[d]^{e_1} \\ v_1 \ar[r]_{b_1} & v_2 }}$

We think of the square $$s_1$$ being based at the bottom, right-hand corner, $$v_2$$. The boundary of the square is the loop $$(v_2, b_1^{-1}d_1^{-1}a_1e_1,v_2) = (v_2,\delta(s_1),v_2)$$. The horizantal composition is illustrated by

$\vcenter{\xymatrix @=4pc{ u_1 \ar[r]^{a_1} \ar[d]_{d_1} & u_2 \ar[r]^{a_2} \ar[d]^{e_1} & u_3 \ar[d]^{f_1} \ar@{}[dr]|= & u_1 \ar[r]^{a_1a_2} \ar[d]_{d_1} & u_3 \ar[d]^{f_1} \\ v_1 \ar[r]_{b_1} & v_2 \ar[r]_{b_2} & v_3 & v_1 \ar[r]_{b_1b_2} & v_3 }}$

Notice that the boundary of the composite satisfies the identity:

$\delta(s_1s_2) ~=~ (b_1b_2)^{-1}d_1^{-1}(a_1a_2)f_1 ~=~ b_2^{-1}(b_1^{-1}d_1^{-1}a_1e_1)b_2(b_2^{-1}e_1^{-1}a_2f_1) ~=~ (\delta s_1)^{b_2} (\delta s_2).$

The horizantal groupoid on $$\Box(G)$$ has the vertical arrows as objects and pairs of horizantal arrows as arrows. Similarly, the vertical groupoid on $$\Box(G)$$ has the horizantal arrows of as objects and pairs of vertical arrows as arrows. Vertical composition is illustrated by

$\vcenter{\xymatrix @=2pc{ u_1 \ar[rr]^{a_1} \ar[dd]_{d_1} && u_2 \ar[dd]^{e_1} & & && \\ && & & u_1 \ar[rr]^{a_1} \ar[dd]_{d_1d_2} && u_2 \ar[dd]^{e_1e_2} \\ v_1 \ar[rr]_{b_1} \ar[dd]_{d_2} && v_2 \ar[dd]^{e_2} &=& && \\ && & & w_1 \ar[rr]_{c_1} && w_2 \\ w_1 \ar[rr]_{c_1} && w_2 }}$

This time the boundary condition satisfies the identity:

$\delta(s_1s_3) ~=~ c_1^{-1}(d_1d_2)^{-1}a_1(e_1e_2) ~=~ (c_1^{-1}d_2^{-1}b_1e_2)e_2^{-1}(b_1^{-1}d_1^{-1}a_1e_1)e_2 ~=~ (\delta s_3)(\delta s_1)^{e_2}.$

Vertical and horizantal compositions commute, so we may construct products such as:

$\vcenter{\xymatrix @=2pc{ u_1 \ar[rr]^{a_1} \ar[dd]_{d_1} && u_2 \ar[rr]^{a_2} \ar[dd]|{e_1} && u_3 \ar[dd]^{f_1} & & &&& \\ && && & & u_1 \ar[rrr]^{a_1a_2} \ar[dd]_{d_1d_2} &&& u_3 \ar[dd]^{f_1f_2} \\ v_1 \ar[rr]|{b_1} \ar[dd]_{d_2} && v_2 \ar[rr]|{b_2} \ar[dd]|{e_2} && v_3 \ar[dd]^{f_2} &=& &&& \\ && && & & w_1 \ar[rrr]_{b_3b_6} &&& w_3 \\ w_1 \ar[rr]_{c_1} && w_2 \ar[rr]_{c_2} && w_3 }}$

Now $$Box(G)$$ is the maximal double groupoid determined by $$G$$, but in general many substructures may be formed. Since the horizantal and vertical groupoid structures must contain the identity at each object, a double groupoid on $$G$$ contains all squares of the form:

$\vcenter{\xymatrix @=4pc{ u \ar[r]^{a} \ar[d]_{1} & v \ar[d]^{1} && u \ar[r]^{1} \ar[d]_{b} & u \ar[d]^{b} \\ u \ar[r]_{a} & v && v \ar[r]^{1} & v }}$

For an example we take for our groupoid Gd8 with group $$D_8 = \langle g = (1,2,3,4), h = (1,3) \rangle$$, with the complete graph on $$\{-9,-8,-7\}$$. It is then routine to check that the two ways of computing the product of four squares below agree.

$\vcenter{\xymatrix @=2pc{ -9 \ar[rr]^{()} \ar[dd]_{(1,2,3,4)} && -7 \ar[rr]^{(2,4)} \ar[dd]|{(1,3)} && -7 \ar[dd]^{(1,4,3,2)} & & &&& \\ && && & & -9 \ar[rrr]^{(2,4)} \ar[dd]^{(1,2)(3,4)} &&& -7 \ar[dd]_{(1,4)(2,3))} \\ -9 \ar[rr]|{(2,4)} \ar[dd]_{(1,3)} && -8 \ar[rr]|{()} \ar[dd]|{(1,2,3,4)} && -9 \ar[dd]^{(1,3)} &=& &&& \\ && && & & -8 \ar[rrr]_{(1,3)(2,4))} &&& -7 \\ -8 \ar[rr]_{(1,2)(3,4)} && -8 \ar[rr]_{(1,4)(2,3)} && -7 }}$


gap> g := (1,2,3,4);;  h := (1,3);;
gap> gend8 := [ g, h ];;
gap> d8 := Group( gend8 );;
gap> SetName( d8, "d8" );
gap> Gd8 := Groupoid( d8, [-9..-7] );;
gap> SetName( Gd8, "Gd8" );
gap> D1 := SinglePieceBasicDoubleGroupoid( Gd8 );;
gap> D1!.groupoid;
Gd8
gap> D1!.objects;
[ -9 .. -7 ]
gap> a1 := Arrow(Gd8,(),-9,-7);;         a2 := Arrow(Gd8,(2,4),-7,-7);;
gap> b1 := Arrow(Gd8,(2,4),-9,-8);;      b2 := Arrow(Gd8,(),-8,-9);;
gap> c1 := Arrow(Gd8,(1,2)(3,4),-8,-8);; c2 := Arrow(Gd8,(1,4)(2,3),-8,-7);;
gap> d1 := Arrow(Gd8,g,-9,-9);;          d2 := Arrow(Gd8,(1,3),-9,-8);;
gap> e1 := Arrow(Gd8,(1,3),-7,-8);;      e2 := Arrow(Gd8,g,-8,-8);;
gap> f1 := Arrow(Gd8,g^-1,-7,-9);;       f2 := Arrow(Gd8,(1,3),-9,-7);;

gap> bdy1 := b1![1]^-1 * d1![1]^-1 * a1![1] * e1![1];;
gap> sq1 := SquareOfArrows( D1, bdy1, a1, d1, e1, b1 );
[-9] ---- () ---> [-7]
|                         |
(1,2,3,4)    (1,4,3,2)    (1,3)
V                         V
[-9] ---- (2,4) ---> [-8]
gap> UpArrow( sq1 );
[() : -9 -> -7]
gap> LeftArrow( sq1 );
[(1,2,3,4) : -9 -> -9]
gap> RightArrow( sq1 );
[(1,3) : -7 -> -8]
gap> DownArrow( sq1 );
[(2,4) : -9 -> -8]

gap> bdy2 := b2![1]^-1 * e1![1]^-1 * a2![1] * f1![1];;
gap> sq2 := SquareOfArrows( D1, bdy2, a2, e1, f1, b2 );;
gap> LeftArrow( sq2 ) = RightArrow( sq1 );
true
gap> bdy3 := c1![1]^-1 * d2![1]^-1 * b1![1] * e2![1];;
gap> sq3 := SquareOfArrows( D1, bdy3, b1, d2, e2, c1 );;
gap> UpArrow( sq3 ) = DownArrow( sq1 );
true
gap> bdy4 := c2![1]^-1 * e2![1]^-1 * b2![1] * f2![1];;
gap> sq4 := SquareOfArrows( D1, bdy4, b2, e2, f2, c2 );;
gap> UpArrow(sq4)=DownArrow(sq2) and LeftArrow(sq4)=RightArrow(sq3);
true

gap> sq12 := LeftRightProduct( D1, sq1, sq2 );
[-9] ---- (2,4) ---> [-7]
|                         |
(1,2,3,4)    (1,3)(2,4)    (1,4,3,2)
V                         V
[-9] ---- (2,4) ---> [-9]

gap> sq34 := LeftRightProduct( D1, sq3, sq4 );
[-9] ---- (2,4) ---> [-9]
|                         |
(1,3)    (1,3)(2,4)    (1,3)
V                         V
[-8] ---- (1,3)(2,4) ---> [-7]

gap> sq13 := UpDownProduct( D1, sq1, sq3 );
[-9] ---- () ---> [-7]
|                         |
(1,2)(3,4)    (1,4)(2,3)    (1,4)(2,3)
V                         V
[-8] ---- (1,2)(3,4) ---> [-8]

gap> sq24 := UpDownProduct( D1, sq2, sq4 );
[-7] ---- (2,4) ---> [-7]
|                         |
(1,4)(2,3)    (1,4,3,2)    (1,4)(2,3)
V                         V
[-8] ---- (1,4)(2,3) ---> [-7]

gap> sq1324 := LeftRightProduct( D1, sq13, sq24 );
[-9] ---- (2,4) ---> [-7]
|                         |
(1,2)(3,4)    ()    (1,4)(2,3)
V                         V
[-8] ---- (1,3)(2,4) ---> [-7]

gap> sq1234 := UpDownProduct( D1, sq12, sq34 );;
gap> sq1324 = sq1234;
true



#### 7.2 Further developments

This part of the package is at a very early stage. The following are now urgently needed:

• generators of a double groupoid on $$G$$;

• the double groupoid generated by a set of squares;

• a method for Size;

• and lots more.

#### 7.3 Starting with two groupoids

In the literature on double groupoids the construction often starts with two groupoids $$G_1, G_2$$, and squares have horizantal arrows chosen from $$G_1$$ and vertical arrows chosen from $$G_2$$. When that is the case, the boundary of a square is not defined, since arrows from $$G_1$$ do not compose with those from $$G_2$$. Thids situation is modelled here by constructing the direct product groupoid $$G = G_1 \times G_2$$ and forming a double groupoid on $$G$$ in which squares have the form

$\vcenter{\xymatrix @=4pc{ [u_1,v_1] \ar[r]^{[a,1]} \ar[d]_{[1,b]} & [u_2,v_2] \ar[d]^{[1,b']} \\ [u_3,v_3] \ar[r]_{[a',1]} & [u_4,v_4] }}$

#### 7.4 Commutative squares

The square

$\vcenter{\xymatrix @=4pc{ u_1 \ar[r]^{a_1} \ar[d]_{d_1} & u_2 \ar[d]^{e_1} \\ v_1 \ar[r]_{b_1} & v_2 }}$

is commutative if $$a_1*e_1 = d_1*b_1$$, so that its boundary is the identity. The set of commutative squares in $$\Box(G)$$ forms the commutative sub-double groupoid of $$\Box(G)$$.

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