### Table 4.1: Pencils of conics We remark that the base points in type are not collinear, and no 3 of the basepoints in type are collinear.

1996

Cited by 1

### Table 4.2: Nets of conics It seems to be a delicate problem to determine which nets of conics can occur as NA for a three dimensional Artin-Schelter regular algebra A. However, for generic Artin-Schelter regular algebras one can use the de ning equations (given in Section 3.2) to determine the type of NA. First, however, we remark that if A is a generic Artin-Schelter regular algebra of type B then the linear system NA is a pencil of conics (rather than a net) of type . One obtains :

1996

Cited by 1

### Table 2. Now we replace 0 p by p 2 [ 0 p] and set q = 0 q such that the di erence of the equations f2(x; x) = 0 of p and g2(x; x) = 0 of q gives exactly the equation h2(x; x) = 0 of according to (27). This yields f2(x; x) = ?4x2 ? 26y2 + 28z2 + 48z + 28 and g2(x; x) = ?8x2 ? 27y2 + 24z2 + 48z + 28 : The comparison with (22), (25) and (26) gives A1 = AT

"... In PAGE 8: ... Since for regular the (imaginary) surface is polar to the absolute conic, the pencil [ ] must be polar to the linear system of surfaces confocal to . The cylinder 0 in the rst two cases of Table2 is polar to one focal conic of with respect to .... In PAGE 9: ... cylinder inters. planes 4k4x2 + 4z2 z2 = 0 : : : two-fold plane of symmetry Table2 : Di erent cases of and 0 For a 2nd-order ex of O obeying (22) it is necessary that there is a surface p pass- ing through the sides of the skew quadrangle p0 : : : p3 and a surface q through q1 and q2 such that the pencil [ p q] spanned by p and q contains . With and p each surface 0 p in the pencil [ p] passes through the equator p0 : : : p3.... ..."

### Table 2. Now we replace 0 p by p 2 [ 0 p] and set q = 0 q such that the di erence of the equations f2(x; x) = 0 of p and g2(x; x) = 0 of q gives exactly the equation h2(x; x) = 0 of according to (27). This yields f2(x; x) = ?4x2 ? 26y2 + 28z2 + 48z + 28 and g2(x; x) = ?8x2 ? 27y2 + 24z2 + 48z + 28 : The comparison with (22), (25) and (26) gives A1 = AT

"... In PAGE 8: ... Since for regular the (imaginary) surface is polar to the absolute conic, the pencil [ ] must be polar to the linear system of surfaces confocal to . The cylinder 0 in the rst two cases of Table2 is polar to one focal conic of with respect to .... In PAGE 9: ... cylinder inters. planes 4k4x2 + 4z2 z2 = 0 : : : two-fold plane of symmetry Table2 : Di erent cases of and 0 For a 2nd-order ex of O obeying (22) it is necessary that there is a surface p pass- ing through the sides of the skew quadrangle p0 : : : p3 and a surface q through q1 and q2 such that the pencil [ p q] spanned by p and q contains . With and p each surface 0 p in the pencil [ p] passes through the equator p0 : : : p3.... ..."

### Table 161: bbding Pencils and Nibs

"... In PAGE 4: ...able 160: dingbat Pencils............................................ 50 Table161 : bbding PencilsandNibs.... ..."

### Table 161: bbding Pencils and Nibs

"... In PAGE 4: ...able 160: dingbat Pencils.......................................... 45 Table161 : bbding PencilsandNibs.... ..."

### Table 2: Eigenvalues of Hamiltonian pencils

"... In PAGE 36: ... Nevertheless we will call our form Hamilto- nian Kronecker canonical form in order to avoid confusion when generalizing these results at a later stage to singular Hamiltonian pencils. As shown in Table2 for a regular Hamiltonian pencil Mh ? Lh we have similar sym- metries in the nite spectrum. So most of the analysis in this section has to be devoted to the part of the canonical form associated with in nite eigenvalues.... ..."