It is also possible to compute period sequence 46 (that should correspond to #2.20) in a similar way:

consider non-abelianization of complete intersection of line bundles with degrees

(1,0,1), (0,1,1) and thrice (1,1,0) on (we consider only involution permuting two four-spaces).

This non-abelianization is a complete intersection of three hyperplane sections and (dual?) universal bundle twisted by on .

Similarly.

Abelianization for #2.17 seem to be complete intersection in

(involution exchanges first two 3-spaces)

of line bundles O(1,0,1),O(0,1,1),O(1,1,0),O(1,1,1).

Abelianization for #2.21 seem to be complete intersection in

(involution exchanges first two 3-spaces)

of line bundles O(1,0,1) twice, O(0,1,1) twice and O(1,1,0).

Abelianization for #2.22 seem to be complete intersection in

( permutes three 4-spaces)

of line bundles O(1,0,0,1),O(0,1,0,1),O(0,0,1,1) and thrice O(1,1,1,0).

#2.14 is a relatively easy case (just the hypersurface in ), so can be reduced to easily, and it can also be done via abelianization:

is an intersection of 3 hypersurfaces in , abelianization of is ,

so abelianization of #14 is an intersection of 4 hypersurfaces in

of degrees (1,1,0), (1,1,0), (1,1,0), (1,1,1). Relative abelian/non-abelian tangent bundle is , and its “square root” (Vandermonde) is O(-1,1,0).

Here is the code for and #2.14.

v5 = period( polcoeff( sum(a=0,N,sum(b=0,N, o + t^(2*(a+b)) * gg(a+b,u*(A+B))^3 / (gg(a,u*A) * gg(b,u*B))^5 * (b-a+u*(B-A))*(-1)^(b-a) )) + O(u^2) , 1,u) /(B-A) + o )

mm214 = period( subst(polcoeff( sum(a=0,N,sum(b=0,N,sum(c=0,N, o + t^(a+b+c) * ( gg(a+b,u*(A+B))^3 * gg(a+b+c,u*(A+B)+C) ) / (gg(a,u*A) * gg(b,u*B))^5 / gg(c,C)^2 * (b-a+u*(B-A))*(-1)^(b-a) ))) + O(u^2) , 1,u) /(B-A) + o, C,0) )

```
```

`%3 = 1 + 6*t^2 + 114*t^4 + 2940*t^6 + 87570*t^8 + O(t^10)`

%4 = 1 + 16*t^2 + 90*t^3 + 1104*t^4 + 11460*t^5 + 133990*t^6 + 1588860*t^7 + 19463920*t^8 + 242996040*t^9 + O(t^10)

Note there is a sign as explained in **6**. In case of Gr(3,7) there were 3 signs that cancelled altogether.