Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 16 / 5

False. Take $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$ and $N=S^2\times S^2$, with their standard product and connected-sum orientations. They are closed simply connected four-manifolds. For the connected sum, removing a four-ball leaves the fundamental group unchanged, and the Seifert-van Kampen theorem across the separating three-sphere preserves simple connectivity. The Mayer–Vietoris sequence computes integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, two and four, and zero in other degrees. The Künneth theorem gives the same groups for $S^2\times S^2$. Thus their Betti numbers agree and all their integral cohomology is torsion-free.

Their cohomology rings nevertheless differ. In $M$, let $u,v$ come from the two projective summands and let $z$ generate degree four. The cohomology ring of the connected sum of two complex projective planes hasThese relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfyConsequently every degree-two class of $N$ has squarean even multiple of the top generator, whereas $u^2$ in $M$ is a primitive generator. An integral graded-ring isomorphism cannot change this property, since it sends a generator of $H^4$ to its positive or negative. A homotopy equivalence induces a graded-ring isomorphism, so none exists. Equivalently, their intersection forms are $\mathrm{diag}(1,1)$ and $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, an odd form and an even form respectively.

False. Take $M=S^4$ and $N=S^3$. Both are closed oriented simply connected manifolds and $4>3$. A cohomological obstruction rules out even a topological fiber bundle.

Suppose a fiber bundle $F\to S^4\to S^3$ existed. The long exact sequence of homotopy groups of a fibration, including its component terms, makes $F$ path-connected. A local trivialization over a small open ball in $S^3$ identifies its preimage, an open subset of $S^4$, with $F\times B^3$. Projection onto $F\times \{0\}$ gives a retraction. An open smooth four-manifold has the homotopy type of a CW complex of dimension at most four, so its rational cohomology vanishes in degrees above four. A retract's cohomology injects into that of the ambient space. Therefore

Use the multiplicative cohomological Serre spectral sequence with rational coefficients. Since $S^3$ is simply connected, there is no nontrivial coefficient action, and only columns zero and three occur on $E_2$. Thus $E_3=E_2$, and the only possibly nonzero differential is $d_3$. Let $u$ be the degree-three base generator. It cannot survive, because the sequence converges to $H^{\ast }(S^4;\mathbb{Q})$ and that ring has zero degree-three part. It has no outgoing differential and its only possible incoming differential is from $E_3^{0,2}$. Hence there is $a\in H^2(F;\mathbb{Q})$ withThe class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\ge 2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,But $E_3^{3,2r-2}=H^3(S^3;\mathbb{Q})\otimes H^{2r-2}(F;\mathbb{Q})$, in which $u{a}^{r-1}\ne 0$ by minimality of $r$; also $r$ is nonzero over $\mathbb{Q}$. This is a contradiction. Thus no such fiber bundle exists. This is the special case of a sphere cannot fibre over a lower-dimensional odd sphere required here.

True for two positive-dimensional factors. In fact degree three can be replaced by any nonzero mapping degree. Suppose $M=X^k\times Y^{m-k}$ with $0<k<m$. The factors are closed and connected. They are also orientable: an orientation-reversing loop in either factor would reverse the product orientation while the other factor stays fixed, contradicting the orientation of $M$.

Let ${\omega }_X,{\omega }_Y$ be their integral top orientation classes, and let $\alpha ,\beta$ be the corresponding projection pullbacks to $M$. The Künneth theorem and the product fundamental class giveFor $f:S^m\to M$, both $f^{\ast }\alpha$ and $f^{\ast }\beta$ vanish, since a sphere has no cohomology in those intermediate positive degrees. Naturality of the cup product therefore givesOn the other hand, the definition of degree three gives $f^{\ast }{\omega }_M=3{\omega }_{S^m}\ne 0$. This contradiction proves the nonzero degree from a sphere obstructs manifold products assertion. The product with a zero-dimensional point factor is the excluded trivial case.

True: every integer occurs. The simple-connectivity assumptions in parts 1 and 2 are local to those statements and are not imposed here. For a connected closed oriented four-manifold, Poincare duality givessince $b_3=b_1$. This expression has no fixed sign or parity.

Use $\chi ({\mathbb{CP}}^2)=3$, from its three even-dimensional cells, and $\chi (S^1\times S^3)=0$, by Euler characteristic of a product. The Euler characteristic of a connected sum in dimension four isIt follows either by Mayer–Vietoris sequence after deleting balls, or by the intermediate-degree cohomology splitting and the single top and bottom generators. Therefore a connected sum of $r$ projective planes and $s$ copies of $S^1\times S^3$ hasGiven an integer $h$, choose an integer $s\ge \max \{0,\lceil (2-h)/2\rceil \}$ and put $r=h-2+2s\ge 0$. This produces a connected closed oriented four-manifold of Euler characteristic $h$. When $r=s=0$, use $S^4$. ThusIf simple connectivity were added, the possible values would instead satisfy $\chi =2+b_2\ge 2$. This distinction is recorded in Euler characteristics of closed oriented four-manifolds.