Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' - MathOverflow

Existence of sequence of examples of braking 'Cancellation law in homeomorphic products'

2011-04-11T21:15:05Z

I know there are manifolds (with or without boundary) $A$ and $B$ such that $A\times C$ is homeomorphic to $B\times C$ but $A$ is NOT homeomorphic to $B$.

My question is (in the Diffeomorphism Category)

Is there any infinitely many $A_i$, with same dimension of course, which are pairwise non-diffeomorphic, but $A_i\times C$ become all diffeomorphic to each other.

1) What's the answer to the question under the assumption on $A_i$, $C$: Smooth Close manifolds.

2) What if we change diffeomorphic to homeomorphic?

3) Is there any example when we require $C$ to be Torus?

Answer by Faisal for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products'

2011-04-11T21:46:37Z

The answer to 2 is yes, there is such an example. In

McMillan, D. R., Jr., Some contractible open $3$-manifolds. Trans. Amer. Math. Soc. 102 (1962), 373–382.

there is a construction of uncountably many topologically distinct, contractible (open) $3$-manifolds $M_\alpha$ such that $M_\alpha \times \mathbb R$ is homeomorphic to $\mathbb R^4$.

Take a look at this recent MO question and the Wikipedia article on the Whitehead manifold for some closely related material.

Edit. The answer to 3 is also yes, assuming by "torus" you mean $S^1$. On page 221 of

Vogt, E., Foliations of codimension $2$ with all leaves compact on closed $3$-, $4$-, and $5$-manifolds. Math. Z. 157 (1977), no. 3, 201–223.

you can find a construction of infinitely many pairwise non-homeomorphic closed Seifert 3-manifolds whose product with $S^1$ gives the same Seifert 4-manifold.

Answer by Paul for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products'

2011-04-11T23:38:55Z

Wall classified certain smooth closed 6-manifolds up to diffeomophism in his 1966 Inventiones paper. It follows that if ${M_i}$ is a collection of smooth spin, simply connected 4-manifolds that are pairwise not-diffeomorphic but all homeomorphic, then $M_i\times S^2$ are all diffeomorphic. Many collections ${M_i}$ are known, distinguished eg by Seiberg-Witten invariants, eg by performing log transforms on $K3$. So 1 is true.

In general, this is the kind of problem that surgery theory is good for. So there are examples of 2 and I think 3 also, look in Wall's book.

Answer by Igor Belegradek for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products'

2011-04-11T23:43:17Z

Take any closed $4$-manifold with infinitely many smooth structures, and multiply it by a torus. The product has only finitely many smooth structures, in fact any manifold $M$ of dimension $\ge 5$ has at most finitely many smooth structures if $H^3(M;\mathbb Z_2)$ is finite.
Take any closed manifold $X$ of dimension $\ge 5$ such that the Whitehead group of $\pi_1(X)$ is infinite (e.g. this is the case if $\pi_1(X)$ is finite cyclic of order $5$ or $\ge 7$). Then there are infinitely many closed pairwise non-diffeomorphic manifolds that are h-cobordant to $X$. On the other hand, all these manifolds become diffeomorphic after multiplying by $S^1$ (because this operation makes Whitehead torsion vanish).
By contrast, if $X$, $X'$ are closed simply-connected manifold of dimension $\ge 5$ that become diffeomorphic after taking product with $S^1$, then $X$, $X'$ are diffeomorphic (this followed from the h-cobordism theorem in the universal cover of the product).