User johannes nordström - MathOverflow

A lost lemma about periodicity in a grid of long exact sequences?

2013-01-03T19:43:31Z

This is a question about finding references and hopefully a larger context for a lemma in homological algebra I proved recently. The motivation is to understand properties of characteristic classes of $T_f$, the mapping torus of a diffeomorphism $f$ of a closed manifold, by applying the lemma to Mayer-Vietoris and a change-of-coefficients sequence for the cohomology of $T_f$.

Let $C_{ij}, 1 \leq i,j \leq 3$ be cochain complexes, and $$ \begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & C_{11} & \stackrel{g}\to & C_{21} & \stackrel{h}\to & C_{31} & \to & 0 \\ & & {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ & & \\ 0 & \to & C_{12} & \stackrel{g}\to & C_{22} & \stackrel{h}\to & C_{32} & \to & 0 \\ & & {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ & & \\ 0 & \to & C_{13} & \stackrel{g}\to & C_{23} & \stackrel{h}\to & C_{33} & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0 & & 0 & & 0 & & \end{matrix}$$

a commuting diagram where the rows and columns are short exact sequences. Let $\delta_H : H^k(C_{3j}) \to H^{k+1}(C_{1j})$ and $\delta_V : H^k(C_{i3}) \to H^{k+1}(C_{i1})$ denote the boundary homomorphisms in the associated long exact sequences. The long exact sequences can be arranged into a commuting grid

$$ \begin{matrix} H^{k-2}(C_{33}) & \stackrel{\delta_H}\to & H^{k-1}(C_{13}) & \stackrel{g}\to & H^{k-1}(C_{23}) & \stackrel{h}\to & H^{k-1}(C_{33}) & \stackrel{\delta_H}\to & H^k(C_{13}) \\ {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ \\ H^{k-1}(C_{31}) & \stackrel{\delta_H}\to & H^k(C_{11}) & \stackrel{g}\to & H^k(C_{21}) & \stackrel{h}\to & H^k(C_{31}) & \stackrel{\delta_H}\to & H^{k+1}(C_{11}) \\ {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ & & {\scriptstyle u}\downarrow\ \\ H^{k-1}(C_{32}) & \stackrel{\delta_H}\to & H^k(C_{12}) & \stackrel{g}\to & H^k(C_{22}) & \stackrel{h}\to & H^k(C_{32}) & \stackrel{\delta_H}\to & H^{k+1}(C_{12})\\ {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ & & {\scriptstyle v}\downarrow\ \\ H^{k-1}(C_{33}) & \stackrel{\delta_H}\to & H^k(C_{13}) & \stackrel{g}\to & H^k(C_{23}) & \stackrel{h}\to & H^k(C_{33}) & \stackrel{\delta_H}\to & H^{k+1}(C_{13}) \\ {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ \\ H^k(C_{31}) & \stackrel{\delta_H}\to & H^{k+1}(C_{11}) & \stackrel{g}\to & H^{k+1}(C_{21}) & \stackrel{h}\to & H^{k+1}(C_{31}) & \stackrel{\delta_H}\to & H^{k+2}(C_{11}) \\ \end{matrix}$$

The grid is symmetric under translation by 3 steps up and 3 to the right.

Lemma. If $[\alpha] \in H^k(C_{12})$ and $[\beta] \in H^k(C_{21})$ are classes such that $g[\alpha] = u[\beta] \in H^k(C_{22})$ then there is some $[\gamma] \in H^{k-1}(C_{33})$ such that both $\delta_H[\gamma] = v[\alpha] \in H^k(C_{13})$ and $\delta_V[\gamma] = -h[\beta] \in H^k(C_{31})$ .

Proof. Take $\chi \in C^{k-1}_{22}$ such that $d\chi = g\alpha - u\beta$. By the definition of the boundary homomorphisms, $d(v\chi) = g(v\alpha)$ implies that $\delta_H([h(v\chi)]) = [v\alpha]$, and $d(h\chi) = -u(h\beta)$ implies that $\delta_V([v(h\chi)]) = -[h\beta]$. Hence we can set $\gamma = vh\chi$.

Does this lemma look familiar? Do you know some place where it's written down?

Edit: Corrected subscripts in statement of lemma.

Update: Thanks for the alternative proofs. However, what I'm after is rather a bibliography reference that I can cite when writing up my application, just to emphasise that it is an instance of something that someone somewhere has already considered (as I imagine it is).

Answer by Johannes Nordström for Lines on degree 2n-3 Fermat hypersufaces

2013-01-04T10:46:26Z

Regarding question 1, any line in the Fermat cubic $C = \{X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0\}$ must meet the coordinate hyperplane $H_0 = \{X_0 = 0\}$ . So which points $x \in (C \cap H_0)$ can lie on lines? If $Y, Z$ are homogenous coordinates on $T_x(C \cap H_0) \cong \mathbb{P}^1$ , then the restriction of $X_0^3 + X_1^3 + X_2^3 + X_3^3$ to $T_x C$ is of the form $X_0^3 + F(Y,Z)$ for a homogeneous cubic $F$ . For $x$ to lie on a line, $X_0^3 + F$ must factorise, so $F$ is a cube. This means that $x$ is an inflection point of the plane cubic curve $C \cap H_0 = \{X_1^3 + X_2^3 + X_3^3 = 0\}$ . The inflection points are given by intersection with the zero set of the Hessian determinant $216X_1X_2X_3$ . Hence the intersection of any line in $C$ with any coordinate hyperplane must actually have two corrdinates equal to 0, and it follows that the lines consist of $\{X_0^3 + X_1^3 = X_2^3 + X_3^3 = 0\}$ and its two images under permutating the coordinates (9 lines in each).

P.S. Here is a related exercise I like. Once one has identified the 27 lines in the Fermat cubic $C$ , one can use the symmetries of $C$ to guess how to arrange 6 points in $\mathbb{P}^2$ so that the blow-up is isomorphic to $C$ , and then write down an explicit rational map $\mathbb{P}^2 \dashrightarrow \mathbb{P}^3$ that maps birationally onto $C$ .

Answer by Johannes Nordström for Almost parallelizable 4-manifolds

2013-01-01T11:28:56Z

You want to trivialise the restriction of the tangent bundle to the 3-skeleton of $M$. Since $\pi_0 O(4) = \pi_1 O(4) = Z/2$, there are obstructions $w_1(E) \in H^1(X; Z/2)$ and $w_2(E) \in H^2(X;Z/2)$ to trivialising a rank 4 bundle over the 1- and 2-skeleta of a cell complex $X$. Because $\pi_2 O(4)$ is trivial, there is no further obstruction to extending a trivialisation from the 2-skeleton to the 3-skeleton. This is outlined in a nice way at the beginning of chaper 3 in Hatcher's book on vector bundles.

Answer by Johannes Nordström for Isometry of K3 surface.

2012-12-30T16:03:44Z

There is a unique Ricci-flat Kähler metric in each Kähler class of $S$. Thus, for any holomorphic automorphism $\iota$ of $S$, a Ricci-flat Kähler metric $g$ is invariant under $\iota$ if and only if its Kähler class $[\omega_g] \in H^{1,1}(S)$ is.

Answer by Johannes Nordström for Chern classes of a blow-up at a point

2012-03-30T11:03:24Z

Like Georges says, 15.4 of Fulton's Intersection Theory deals with the general theory. For this special case it's not too hard to work out the Chern classes by hand though.

Let $f : \widetilde X \to X$ be the projection and $E \cong \mathbb{C}P^{n-1}$ the exceptional divisor. $H^*(\widetilde X) \cong f^*H^*(X) \oplus \langle \textrm{Poincare duals of planes } P_k \textrm{ in } E $ $\textrm{of dimension }k = 1,\ldots, n-1\rangle$. Note that $[P_{n-i}][P_{n-j}] = -[P_{n-i-j}]$, while $(f^*\alpha) [P_k] = 0$ for any $\alpha \in H^i(X)$ ($i, k > 0$).

$f^* c_i(X)$ and $c_i(\widetilde X)$ are equal outside the exceptional divisor, so their difference is Poincare dual to something in $E$. On the other hand the restriction of $f^*c_i(X)$ to $E$ is 0 (for $i > 0$), while the restriction of $c_i(E)$ is $c_i(\mathcal{O}(1)^n \oplus \mathcal{O}(-1)) = \left({n\choose i} - {n \choose i-1}\right)H^i$, where $H \in H^2(E)$ is the hyperplane class. For $0 < i < n$ we deduce that $c_i(\widetilde X) = f^*c_i(X) - \left({n\choose i} - {n \choose i-1}\right)[P_{n-i}]$ by comparing the evaluations on $P_i$.

Answer by Johannes Nordström for Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$

2012-03-29T16:44:29Z

You can compute the invariants necessary to determine the lattice structure of $H^2(S_d)$ from the characteristic classes. $(1+d\sigma)(1+c_1+c_2) = (1+4\sigma+6\sigma^2)$, so $c_1 = (4-d)\sigma$ and $c_2 = (d^2 -4d + 6)\sigma^2$ (where $\sigma$ is the hyperplane class on $S_d$). Identifying top classes with their integrals $\sigma^2 = d$, so the Euler characteristic is $\chi = c_2 = d^3 -4d^2 +6d$ and the signature is $\frac{1}{3}p_1 = \frac{1}{3}(-c_2 + c_1^2) = -\frac{d^3 - 4d}{3}$. ($H^1(S_d) = 0$, so the rank of $H^2(S_d)$ is $\chi -2$.) The Stiefel-Whitney class $w_2 = c_1$ mod 2, so vanishes if and only if $d$ is even.

So for $d = 5$ we get the odd unimodular lattice of rank $53$ and signature $-35$, i.e. $H^2(S_5) = 9\langle 1 \rangle \oplus 44\langle -1 \rangle$. For $d = 6$ we get the even unimodular lattice of rank $106$ and signature $-64$, i.e. $H^2(S_6) = 8E_8 \oplus 19H$. (Sanity check: for $d = 3$ the argument gives that $H^2(S_3)$ is odd of rank $7$ and signature $-5$, as it should be since $S_3$ is the blow-up of $\mathbb{P}^2$ at 6 points.)

There's not much to do to identify the hyperplane class $\sigma$ in the cohomology lattice; the lattice has plenty of automorphisms, so any two primitive elements of the same norm are equivalent. Similarly, if you want to know the relation between $\sigma$ and the cohomology class of $C$, then you just need to know the intersection form on the sublattice spanned by $\sigma$ and $C$; the fact that $H^2(S_d)$ is indefinite implies that any two (primitive) isometric sublattices of rank $< \frac{rk \; H^2(S_d)}{2} - 1$ are equivalent under automorphisms of $H^2(S_d)$ (see Theorem 1.4.8 of Dolgachev, "Integral quadratic forms: Applications to algebraic geometry [After V. Nikulin]" for the even case (where the inequality does not have to be strict); the odd case can be deduced from Theorem 1.16.10 in Nikulin, "Integral symmetric bilinear forms and some of their applications").

I'm not sure this is quite what you were asking for, but perhaps it's a start.

Answer by Johannes Nordström for Calculating chern numbers yields a contradiction, why?

2012-03-13T17:35:19Z

Note that $P^2 = 0$, since we blow up the self-intersection of a hyperplane.

The pull-back of the hyperplane class is $P+E$, so $c_1(\tilde Y) = 5P + 4E$, and $c_2(\tilde Y) = 10(P+E)^2 - 3EP - 4E^2 = 17EP + 6E^2$.

This still does not yield $c_1(\tilde Y)^2 c_2(\tilde Y) = 250$ (I think it's $512 - 3\cdot 96 = 224$), but I don't see why this Chern number should be preserved by the blow-up.

Answer by Johannes Nordström for (Second) Chern class of projective space, blown up in a linear subvariety

2012-03-07T23:45:49Z

My understanding of this is very unsophisticated, but perhaps that means that what I can explain is precisely what you want.

To understand $f^*[H] \in H^2(\tilde Y)$ , where $H \subset Y = \mathbb{P}^n$ is a hyperplane, it may help to think of $H$ as the zero set of section $s$ of the anticanonical bundle $\mathcal{O}_Y(1)$. Then the zero set of $f^*s$ (essentially just the preimage of $H$) is a divisor representing $f^*[H]$ . If $H$ contained the blow-up locus $X$, then the resulting divisor is the sum of the proper transform $P$ of $H$ (the blow-up of $H$ at $X$) and the exceptional set $\tilde X$. If $H$ was transverse to $X$, then the divisor is the proper transform of $H$ (which is exactly the pre-image of $H$ in this case), which is the blow-up $\tilde H$ of $H$ at $H \cap X$.

I presume $d = n-m$, the codimension of $X$ in $Y$.

If $D$ is a divisor in $\tilde X$ representing $[D] \in H^2(\tilde X)$, then $j_*[D]$ is the class $[D] \in H^4(\tilde Y)$ that you get by considering $D$ as a cycle in $\tilde Y$. $g^*\xi$ can be represented by a divisor that is the preimage in $\tilde X$ of a hyperplane in $X$. Writing that hyperplane as the intersection of $X$ with a transverse hyperplane $H \subset Y$, we find that $j_*g^*\xi = [\tilde H \cap \tilde X] \in H^4(\tilde Y)$ . $\zeta \in H^2(\tilde X)$ corresponds to the conormal bundle of $\tilde X$ in $\tilde Y$, so it is the restriction of $-[\tilde X] \in H^2(\tilde Y)$ to $\tilde X$. Therefore $j_*\zeta = -[\tilde X]^2 \in H^4(\tilde Y)$ .

To describe $j_*\zeta$ another way, note that since $\mathcal{N} = \mathcal{O}_X(1)^d$, $\tilde X$ is a trivial bundle $X \times \mathbb{P}^{d-1}$. You can get an explicit trivialisation by picking copy of $\mathbb{P}^{d-1} \cong Z \subset Y$ disjoint from $X$: given points $x \in X$ and $z \in Z$, the line from $x$ to $z$ defines an element in the projectivisation of the fibre of $\mathcal{N}$ over $x$. Let $h$ be the projection $\tilde X \to Z$. Then $\mathcal{O}_{\tilde X}(-1) = g^*\mathcal{O}_X(1) + h^*\mathcal{O}_{Z}(-1)$ . $h^*\mathcal{O}_{Z}(1)$ corresponds to a trivial $\mathbb{P}^{d-2}$ subbundle of $\tilde X$. Such a divisor is the intersection of $\tilde X$ with the proper transform $P$ of a hyperplane $H$ containing $X$ and some hyperplane in $Z$. In other words, $-\zeta = g^*\xi -[P \cap \tilde X] \in H^2(\tilde X)$ , so $j_*(\zeta + g_*\xi) = [P \cap \tilde X] \in H^4(\tilde Y)$ .

Sanity check: $-[\tilde X] + [\tilde H] = [P]$ implies $-[\tilde X]^2 + [\tilde H \cap \tilde X] = [P \cap \tilde X]$, so it adds up.

Answer by Johannes Nordström for Why do I need densities in order to integrate on a non-orientable manifold?

2012-03-07T16:38:44Z

You would expect the zero set of an $n$-form to have codimension 1 rather than being countable. Your suggestion of choosing some $n$-form on a non-orientable manifold $M^n$ and defining integrals relative to that essentially amounts to cutting $M$ into two orientable pieces along a codimension 1 submanifold, choosing an orientation on each, and adding the integrals on the two pieces. You can certainly do that, but since the answer depends on the choice of $n$-form/cutting it is not very natural or interesting (whereas the integral on an oriented manifold only depends on the orientation and not on the choice of orientation form).

Answer by Johannes Nordström for Poincare duality with boundary conditions

2011-02-20T00:14:39Z

An example where the duality fails is when $M^n$ is the closed unit ball $B^3 \subset \mathbb{R}^3$, and its boundary $S^2$ is divided into four quarters by 2 great circles. If $V = \mathbb{R}$, $V_F = V$ for 2 opposite quarters $F$ and $V_F = 0$ for the other two, then $H^1_{V, \{ V_F \}}(M) = 0$ while $H^2_{V^*, \{\text{ann} V_F\}} \cong \mathbb{R}$ (essentially, they are $H^1_c$ and $H^2_c$, respectively, of the product of an open 2-disc and a closed interval).

In a sense, the reason that the duality fails is that near the intersection of the two great circles, the set of boundary points where the forms are allowed to be non-zero is disconnected, and that no matter how small a neighbourhood we choose in $B^3$ for the intersection point, its cohomology will therefore not be entirely elementary. This can be prevented by demanding that every point in $\partial M$ has an "elementary" neighbourhood $U \cong \mathbb{H}^{n}$ such that

the subdivision of $\partial U$ into faces is diffeomorphic to a complete fan (a subdivision of $\mathbb{R}^{n-1}$ into simplicial cones),
$V$ has a basis $\{e_i\}$ such that for each face $F$ meeting $U$, $V_F$ is spanned by a subset,
for each $e_i$, the interior in $U$ of the union of the faces $F$ such that $e_i \not\in V_F$ is connected.

Essentially, 1. says that the subdivision of $\partial M$ is sensible, 3. prevents situations like in the example above, and 2. makes sure we can state 3. sensibly when $\dim V > 1$ (see example in Trial's comment below). I think that if $M^n$ is oriented with boundary and possesses such "elementary" neighbourhoods, then $$H^k_{V, \{V_F\}}(M) \cong H^{n-k}_{c, V^*, \{ \text{ann} V_F\}}(M)^*$$ where the subscript $c$ indicates the cohomology of a complex with compact supports. It should be possible to prove this using induction on a good cover (and the duality between the Mayer-Vietoris sequences for normal and compactly supported de Rham cohomology) like for standard Poincaré duality, provided that the statement is true for open subsets $U \subset M$ diffeomorphic to $\mathbb{R}^n$ and for the "elementary" neighbourhoods.

For $U \cong \mathbb{R}^n$ this is just usual Poincaré duality tensored with $V$. For an "elementary" neighbourhood $U$, $$H^k_{V, \{V_F\}}(U) = \bigoplus_i H^{k}_{V_i, \{V_F \cap V_i\}}(U) $$ $$H^k_{c, V^*, \{\text{ann} V_F\}}(U) = \bigoplus_i H^{k}_{c, V_i^*, \{\text{ann} (V_F \cap V_i) \}}(U), $$ where $V_i$ is the span of the element $e_i$ of the basis from condition 2. The terms on the right hand side all vanish, except that if $e_i \in V_F$ for all $F$ meeting $\partial U$ then $H^0_{V_i, \{V_F \cap V_i\}} \cong V_i$ and $H^{n}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) \cong V_i^*$ (3. is used to show that $H^{n-1}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) = 0$ ). So the duality holds for the "elementary" neighbourhoods.

Answer by Johannes Nordström for The vanishing of the 2nd plurigenus of a sextic threefold

2011-02-21T18:09:20Z

$h^{q,0}(X) = h^{0,q}(X) = 0$ for any Fano $X$ and $q > 0$, because $-K_X$ ample implies $$ H^q(X, \Omega^0) = H^q(X, K_X \otimes (-K_X)) = 0 $$ by Kodaira vanishing.

Comment by Johannes Nordström

2013-02-03T10:12:46Z

From the thesis, I see that one is supposed to make some choices of $v \in \Lambda^6 V^*$ and $\theta \in V^*$ . $i_X v$ and $\psi \wedge \phi$ are both in $\ker\; i_X \cong \Lambda^5 W^* \subset \Lambda^5 V^*$ , so must be proportional. $\ker \theta \to \Lambda^4 W^*, \; Y \mapsto i_Y v_0$ is an isomorphism.

Comment by Johannes Nordström

2013-01-06T22:14:50Z

I don't have any references for the specifics of my answer, but I've found the first of Reid's Chapters on Algebraic Surfaces (arxiv.org/abs/alg-geom/9602006) a useful general reference for cubic surfaces.

Comment by Johannes Nordström

2013-01-04T21:33:19Z

In the same order as your questions: Yes, I do. If $x$ lies on a line, then that line is contained in $T_xC$ , and its defining equation is a factor in $X_0^3 + F$ . $F(Y,Z)$ is indeed intended to mean an arbitrary cubic on the line $T_x(C \cap H_0)$ ; what I try to emphasise (perhaps unsuccessfully) is that the restriction of equation of $C$ to $T_xC \cong \mathbb{P}^2$ , which is a cubic in $X_0$ , $Y$ and $Z$ , does not contain any cross-terms like $X_0Y^2$ .

Comment by Johannes Nordström

2013-01-03T23:19:38Z

Thanks. The claim in the example from your paper looks slightly different to me, but this kind of reference is helpful.

Comment by Johannes Nordström

2012-12-30T23:36:47Z

Sure, that tells you that for any automorphism of $S$ (of finite order at least) there exists some invariant Kähler class, and hence an invariant Ricci-flat Kähler metric. But it does not mean that a given Ricci-flat Kähler metric is invariant, which is what your question seems to ask.

Comment by Johannes Nordström

2012-12-30T15:12:40Z

I don't understand the question. Are you implicitly assuming that $g$ is invariant under $\iota$? Why would the choice of complex structure affect whether a map is an isometry?

Comment by Johannes Nordström

2012-12-25T16:16:04Z

Could you explain what trivialisation you use when obtaining the local expression $f_1 \rightarrow i\frac{df_1}{dx}$ for the Dirac operator? The transition function $f_1(\frac{1}{x}) = f_2(x)$ suggests the trivialising sections of the spinor bundle is intended to have constant norm. But then the local expression looks to me like the Dirac operator of the Euclidean metric on $\mathbb{R}$, rather than with respect to the metric pulled back from the circle by stereographic projection.

Comment by Johannes Nordström

2012-09-20T09:33:35Z

Connection 1-forms do not transform in the same way as ordinary 1-forms, so the local expressions you have written do not patch up to a well-defined connection. Otherwise you could just set all the connection 1-forms to zero and get a flat connection on any vector bundle.

Comment by Johannes Nordström

2012-07-13T08:49:27Z

Then the zero section of F would give a canonical choice of metric on E, which does not seem reasonable.

Comment by Johannes Nordström

2012-06-21T08:03:12Z

Where do the coefficients come from? Note that the line $Z_3 = Z_4 = 0$ is contained in the intersection of any two quadrics with $q_1 = 0$.

Comment by Johannes Nordström

2012-04-23T02:15:55Z

The connection 1-form depends on the choice of local trivialisation for the line bundle. If you choose a hermitian trivialisation, in the sense that $h \equiv 1$, then the connection 1-form is imaginary.

Comment by Johannes Nordström

2012-04-13T07:26:55Z

Alternatively, use that $SO(n+1)$ acts on $\mathbb{C}P^n$ with cohomogeneity 1. The special orbits are $\mathbb{R}P^n$ and $Q$. By the theory of cohomogeneity 1 manifolds, removing one special orbit leaves a vector bundles over the other.

Comment by Johannes Nordström

2012-04-08T17:41:28Z

You can play the same game with other $M$, e.g. you can realise all subgroups of the group $\mathbb{Z}_{28}$ of exotic smooth 7-spheres as inertia groups by looking at $S^3$-bundles over $S^4$ with Euler number 0; according On the inertia group of certain manifolds by Wilkens the inertia group of $M^7$ is determined by the greatest divisor of the obstruction in $H^4(M;\mathbb{Z})$ to stable parallelisability of $M$ (half of $p_1$), which we can prescribe for the sphere bundle by choosing the right clutching function. This way you can also get $M$ without orientation-reversing diffeomorphism.

Comment by Johannes Nordström

2012-03-30T11:54:16Z

$p^*(c_2(X))\cdot E$ is the integral of $p^*(c_2(X))$ over $E$. But $p$ restricted to $E$ is the constant map.

Comment by Johannes Nordström

2012-03-16T22:00:30Z

But $[H]=[H']$ does not imply $[P] = [P']$.