By Francesco Costantino

We determine a calculus for branched spines of 3-manifolds by way of branched Matveev-Piergallini strikes and branched bubble-moves. We in brief point out a few of its attainable functions within the examine and definition of State-Sum Quantum Invariants.

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**Extra resources for A calculus for branched spines of 3-manifolds**

**Sample text**

Let S be the circle whose center is the point of intersection of the lines o and S 2 and the length of whose radius is equal to the length of the tangent from P to S 1 ; the circle S is the preimage of S under our inversion. Since S is perpendicular to both S 1 and S 2 (see above), it follows (by property C of inversion) that S is perpendicular to both S1 and S2 . All we need do now is apply an inversion whose center O is a point of intersection of S and o. One shows, just as was done before, that this inversion takes the circles S1 and S2 to two concentric circles S10 and S20 (whose common center is the point of intersection of the line o and the line S 0 , which is the image of the circle S under the inversion; see Figure 31b).

For example, the condition stated in problem 6(a) is not quite correct: the fact that the circles circumscribed about the triangles A1 A2 B3, A1 A3 B2 , and A2 A3 B1 intersect in a single point does not imply the existence of circles circumscribed about the triangles B1 B2 A3 , B1B3 A2 , and B2 B3 A1 —it is possible that one of the point-triples B1 ; B2; A3 ; B1; B3 ; A2; and B2 ; B3; A1 (or possibly all three of them) is collinear. The 29 Reflection in a circle (inversion) FIGURE 37 following is a correct formulation of the problem: If the circles circumscribed about the triangles A1 A2 B3, A1 A3 B2, and A2 A3 B1 intersect in a single point, then either the circles circumscribed about the triangles B1 B2A3 , B1 B3 A2 , and B2B3 A1 intersect in a single point (Figure 38a)— here one or two of these circles can be circles of “infinite radius,” that is, lines (Figure 38b)—or each of the point-triples B1 ; B2; A3 ; B1 ; B3; A2; and B2 ; B3; A1 is collinear (Figure 38c).

Show that the radius r of the inscribed circle of a triangle cannot exceed half the radius R of the circumscribed circle, and that the equality r D R=2 holds if and only if the triangle is equilateral. 11. Show that the nine point circle of a triangle (see problem 17(a) in NML 21) is tangent to the inscribed circle as well as to the three escribed circles (Figure 28). See also Problem 63 in Section 5. 22 Circular Transformations FIGURE 28 12. Use the properties of inversion to deduce Pascal’s theorem (problem 46, p.