Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:29:58Z http://mathoverflow.net/feeds/question/61336 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61336/existence-of-sequence-of-examples-of-braking-cancellation-law-in-homeomorphic-pr Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' MG 2011-04-11T21:15:05Z 2011-04-13T08:16:19Z <p>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 $B$. </p> <p>My question is (in the Diffeomorphism Category)</p> <p>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. </p> <p>1) What's the answer to the question under the assumption on $A_i$, $C$: Smooth Close manifolds.</p> <p>2) What if we change diffeomorphic to homeomorphic?</p> <p>3) Is there any example when we require $C$ to be Torus?</p> http://mathoverflow.net/questions/61336/existence-of-sequence-of-examples-of-braking-cancellation-law-in-homeomorphic-pr/61341#61341 Answer by Faisal for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' Faisal 2011-04-11T21:46:37Z 2011-04-11T23:48:12Z <p>The answer to 2 is yes, there is such an example. In</p> <blockquote> <p>McMillan, D. R., Jr., <em>Some contractible open $3$-manifolds</em>. Trans. Amer. Math. Soc. <strong>102</strong> (1962), 373–382.</p> </blockquote> <p>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 $\mathbb R^4$.</p> <p>Take a look at <a href="http://mathoverflow.net/questions/60113/contractible-manifolds" rel="nofollow">this recent MO question</a> and the <a href="http://en.wikipedia.org/wiki/Whitehead_manifold" rel="nofollow">Wikipedia article</a> on the Whitehead manifold for some closely related material.</p> <p><strong>Edit.</strong> The answer to 3 is also yes, assuming by "torus" you mean $S^1$. On page 221 of</p> <blockquote> <p>Vogt, E., <em>Foliations of codimension $2$ with all leaves compact on closed $3$-, $4$-, and $5$-manifolds</em>. Math. Z. <strong>157</strong> (1977), no. 3, 201–223.</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/61336/existence-of-sequence-of-examples-of-braking-cancellation-law-in-homeomorphic-pr/61346#61346 Answer by Paul for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' Paul 2011-04-11T23:38:55Z 2011-04-11T23:38:55Z <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/61336/existence-of-sequence-of-examples-of-braking-cancellation-law-in-homeomorphic-pr/61347#61347 Answer by Igor Belegradek for Existence of sequence of examples of braking 'Cancellation law in homeomorphic products' Igor Belegradek 2011-04-11T23:43:17Z 2011-04-11T23:43:17Z <ol> <li><p>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. </p></li> <li><p>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).</p></li> <li><p>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).</p></li> </ol>