Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 2

Orient the alpha curves and choose generators $x,y$ dual to ${\alpha }_1,{\alpha }_2$, respectively. Following ${\beta }_1$ once around the edge identifications and through the attached annulus gives, up to cyclic permutation and simultaneous inversion conventions,The dotted blue arcs pass underneath the attached annulus, so their apparent crossings with the red arc on the annulus are not intersections. The fundamental group presentation is thereforeThe exponent sums of $x,y$ in $r$ are $2,0$. Henceand abelianization sends $x$ to the order-two generator and $y$ to the infinite cyclic generator.

The curve $\mu$ crosses the alpha disk ${\alpha }_2$ once and is disjoint from ${\alpha }_1$, so in the presentation it represents $y$ up to inversion and conjugacy. The Dehn filling adds the relation $y=1$, after which the beta relator becomes $x^2$. Thus the filled genus-one Heegaard diagram has two alpha–beta intersections andCompressing the displayed genus-two diagram along $\mu$ visibly cancels the annular handle and leaves that genus-one Heegaard diagram. It is therefore the lens space

Two explicit fibrations of $L(2,1)={\mathbb{RP}}^3$ are supplied by Seifert fibrations of real projective 3-space:

Thus these are Seifert fibered descriptions of the same filled manifold with different numbers of exceptional fibers.

The free part of the abelianization is generated by $[y]$, so the fibration class sends $y\mapsto 1$ and $x\mapsto 0$. Applying Fox calculus to the relator and then substituting $x=1$, $y=t$ givesThis has breadth two. Since $K$ is assumed fibered, the breadth is twice the genus, so its fiber is a genus-one surface with one boundary component:

The monodromy can also be read directly from the relator. Put $x_i=y^{i}x{y}^{-i}$. Reidemeister rewriting givesThus the kernel is freely generated by $(x_{-1},x_0)$, and conjugation by $y$ sendsOn first homology, in the ordered basis $([x_{-1}],[x_0])$, a representative isIts characteristic polynomial is $t^2-4t+1$, agreeing up to the unit $-t$ with $4-t-t^{-1}$.

The matrix in part (d) has trace $4$ and determinant $1$. Since $|\mathrm{tr}{\varphi }_{\ast }|>2$, the once-punctured-torus case of the Nielsen–Thurston classification theorem says that $\varphi$ is pseudo-Anosov. The Hyperbolization of a pseudo-Anosov mapping torus therefore gives