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In 1960, Kervaire found the first example of a PL-manifold which does not admit a smooth structure. Since then, I understand that there are many examples of non-smoothable manifolds that can be built. My question is: Which is the "easiest" non-smoothable manifold? Easiest in the sense that among all the non-smoothable manifolds, this manifold has the easiest construction process.

Thanks everyone for your help !!

Cheers...

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Here are explicit equations for nonsmoothable manifolds (all of which admit triangulations). I do not know if these are the "easiest" but they are surely much more explicit than a description of the E8-manifolds, which is constructed as a result of some infinite, and very implicit, process (Freedman's work).

Consider the homogeneous equation $$ z_1^5 + z_2^3 + z_3^2 +z_4^2 + z_5^2 +\sum_{j=1}^5 e^{j-1} z_j^6=0 $$ in the complex projective space $CP^5$. Here instead of $e$ one can take any transcendental number. Then the solution set of this equation is a piecewise-linear complex 4-dimensional (real 8-dimensional) manifold which is not homeomorphic to a smooth manifold.

See "Algebraic equations for nonsmoothable 8-dimensional manifolds" by N.Kuiper, Math. Publ. of IHES, 1967.

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  • $\begingroup$ Thank you for your answer !! this could be what i am looking for !!! $\endgroup$ – onebengaltiger May 16 '14 at 2:49
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The $E_8$ manifold is pretty easy to construct. We may describe it by the following diagram:

E8 Lattice

The meaning of the diagram is as follows. For each dot, we take the disk bundle over $S^2$ with Euler number $2$. This gives us eight $4$-manifolds with boundary. Now we plumb together each disk bundle as indicated by each edge. The result of the plumbing is a smooth $4$-manifold with boundary $P_{E_8}$. Now the intersection form on $P_{E_8}$ is given by the matrix $$Q_{E_8} = \begin{pmatrix} 2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 \end{pmatrix}.$$ Since this matrix is unimodular, we have an isomorphism $H^2(P_{E_8}; \Bbb Z) \cong H^2(P_{E_8}, \partial P_{E_8}; \Bbb Z)$, which in turn implies that $$H_1(\partial P_{E_8}; \Bbb Z) \cong 0 \cong H_2(\partial P_{E_8}; \Bbb Z).$$ So $\partial P_{E_8}$ is an integral homology $3$-sphere (in fact, $\partial P_{E_8}$ is the Poincaré homology $3$-sphere).

Now a (nontrivial) theorem of Freedman says that every homology $3$-sphere bounds a contractible topological $4$-manifold. Let $\Delta$ be a contractible topological $4$-manifold that bounds $\overline{\partial P_{E_8}}$ (here the bar means we reverse the orientation). Glue $\Delta$ to $P_{E_8}$ along their boundary via the identity map and call the result $E_8$: $$E_8 = P_{E_8} \cup_\partial \Delta.$$ This construction gives us a simply connected closed $4$-manifold $E_8$ with intersection form $Q_{E_8}$ as above.

The easiest way to see that $E_8$ is non-smoothable is as follows. Rokhlin's theorem says that if a compact, smooth $4$-manifold $X$ has even intersection form $Q_X$, then the signature of $Q_X$ must be divisible by $16$. Now $Q_{E_8}$ is even but has signature $8$, so it follows that $E_8$ cannot possibly be smooth.

An alternative way to see that $E_8$ is non-smoothable is to use Donaldson's theorem: If $X$ is a smooth, simply connected, closed $4$-manifold with $Q_X$ positive definite, then $Q_X$ is equivalent to the $\mathrm{rk}(Q_X) \times \mathrm{rk}(Q_X)$ identity matrix. $Q_{E_8}$ is positive definite but not equivalent to $I_{8 \times 8}$, so $E_8$ cannot be smooth.

Remark: Note that $E_8$ is not even a PL manifold, since $\mathbf{PL} = \mathbf{Diff}$ in dimension $4$. Hopefully you weren't looking for a non-smoothable PL manifold.

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  • $\begingroup$ Thanks for your answer !! Just for curiosity, what would be the "easiest" non-smoothable pl-manifold ? do you know of any ? $\endgroup$ – onebengaltiger Jun 2 '13 at 3:22
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    $\begingroup$ I would not call this example easiest, since it requires Freedman's theorem which involves an infinite and nonexplicit process. $\endgroup$ – Moishe Kohan Feb 15 '14 at 20:05

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