Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 17 / 4

False. Give $E=T{S}^2\to S^2$ a fibrewise area form. This is a rank-two symplectic vector bundle. A Lagrangian subbundle would be a real line subbundle $L\subset T{S}^2$. Real line bundles are classified by their first Stiefel–Whitney class in $H^1(B;\mathbb{Z}/2)$; since this group vanishes for $S^2$, $L$ would be trivial. A nonzero section of $L$ would therefore be a nowhere-vanishing tangent vector field on $S^2$. The Hairy ball theorem rules this out: every continuous tangent vector field on an even-dimensional sphere has a zero. Hence $T{S}^2$ does not even have one Lagrangian subbundle, let alone a splitting into two. Fibrewise existence of Lagrangian subspaces does not guarantee a globally continuous choice.

True. Use the normalization in which the blowup weight $\lambda$ is the area of an exceptional projective line, equivalently the capacity $\pi r^2$ of the ball removed. The auxiliary volume formula for a point symplectic blowup in real dimension $2n$ isIt applies to each successive blowup, irrespective of where the point is chosen. Geometrically the removed standard ball has volume ${\pi }^n{r}^{2n}/n!$, while its replacement divisor has zero $2n$-dimensional volume. If $V$ is the original symplectic volume, after $k$ equal-weight blowups the volume is $V-k{\lambda }^n/n!$. A nonempty closed symplectic manifold has strictly positive volume, soThis finite upper bound suffices; it does not assert that every number below the bound is realizable. A convention taking the weight to be the radius instead changes the fixed decrement to ${\pi }^n{\lambda }^{2n}/n!$, with the same conclusion.

False even when both base and fibre are closed and oriented. Take the Hopf fibration $h:S^3\to S^2$, and define the smooth fiber bundle $p:S^3\times S^1\to S^2$ by $p(x,\theta )=h(x)$. Its local trivializations are the Hopf trivializations times $S^1$, and its fibre is $S^1\times S^1=T^2$.

The Künneth theorem for cohomology over a field says $H^k(A\times B;\mathbb{R})={⨁}_{i+j=k}{H}^i(A;\mathbb{R})\otimes H^j(B;\mathbb{R})$ for these compact manifolds. Here it gives $H^2(S^3\times S^1;\mathbb{R})=0$. If a symplectic form $\omega$ existed on this closed four-manifold, it would be exact, $\omega =d\eta$. Since $d\omega =0$,by Stokes theorem. But ${\omega }^2$ is a positive volume form in the symplectic orientation, so its integral is positive. This contradiction proves the claim fails. The mere bundle structure does not ensure that a fibrewise area class extends to the total space.

True for a smooth complex projective surface, the usual manifold convention in this question. Choose an ample divisor $H$. The auxiliary pencil-existence theorem states that, for all sufficiently large $m$, a general two-dimensional space of sections of ${\mathcal{O}}_X(mH)$ defines a Lefschetz pencil: its base points are transverse and finite, its general fibre is a smooth connected curve, and each singular fibre has exactly one ordinary node, with distinct critical values. High powers separate the required jets, so a generic pencil has precisely these singularities.

Choose such an $m$ divisible by four, and write $D=mH$. The adjunction formula for a smooth connected curve of class $D$ on a smooth projective surface is $2g-2=D\cdot (D+K_X)$. ConsequentlyWriting $m=4\ell$ gives $g=1+8{\ell }^2{H}^2+2\ell K_X\cdot H$, an odd integer. Thus a sufficiently positive four-divisible polarization gives an odd-genus Lefschetz pencil. The surface is taken to be smooth; a singular projective variety need not support a Lefschetz pencil in this smooth-manifold sense.

True. The Künneth theorem gives the Poincaré polynomialso $b_2=6$ and $b_4=2$. The proposed equality also gives $b_0=b_8=1$. Even if compactness was not stated separately, these two numbers force the putative eight-manifold to be connected and compact: the top ordinary real cohomology of a connected noncompact manifold is zero.

The auxiliary Hard Lefschetz theorem on de Rham cohomology says that, on a compact Kähler manifold of complex dimension $n$, $L^{n-k}:H^k\to H^{2n-k}$ is an isomorphism, where $L\alpha =[\omega ]⌣\alpha$. Here $n=4$, so $L^2:H^2\to H^6$ is injective. If $L\alpha =0$, then $L^2\alpha =0$ and therefore $\alpha =0$. Thus $L:H^2\to H^4$ is injective, forcing $b_2\le b_4$. The inequality $6\le 2$ is impossible. Hence these Betti numbers cannot occur for a Kähler eight-manifold.