Let $a_0, \dots, a_n, b_0, \dots, b_n \in \mathbb{N}$ and consider two polynomials $f = \sum_{i = 0}^{n} z_{i}^{a_i}$, $g = \sum_{i = 0}^{n} z_{i}^{b_i}$. Given two Brieskorn manifolds $\Sigma(a_0, \dots, a_n) = f^{-1}(0) \cap S^{2n+1}$ and $\Sigma(b_0, \dots, b_n) = g^{-1}(0) \cap S^{2n+1}$ (for sufficiently small spheres), under what necessary and sufficient conditions on the exponents can one conclude a homeomorphism \begin{align} \Sigma(a_0, \dots, a_n) \cong \Sigma(b_0, \dots, b_n). \end{align} NB: I do not assume that either of the Brieskorn Manifolds are necessarily integral homology $(2n-1)$-spheres, so the exponents are not necessarily pairwise coprime.

For $n = 1$, the manifolds are torus links. Therefore, $(a_0, a_1) = (b_0, b_1)$ (ignoring sign and order).

For $n = 2$, the genera of the corresponding Seifert surfaces must be equal, \begin{align} g = \frac{1}{2}\left( \frac{d}{\tau} - l \right) + 1 = \frac{1}{2} \left( \frac{d^{\prime}}{\tau^{\prime}} - l^{\prime} \right) + 1 = g^{\prime}, \end{align} where $l = \gcd(a_0,a_1) + \gcd(a_1,a_2) + \gcd(a_2,a_1)$, $d =\gcd(a_0,a_1) \gcd(a_1,a_2) \gcd(a_2,a_1)$ and $\tau = \gcd(a_0,a_1,a_2)$ and the primed version corresponding to $(b_0, b_1, b_2)$.

Is this equality also sufficient to imply homeomorphism (for $n = 2$)?

For the general case, is it necessary and/or sufficient that the sums of the GCDs of the subsets of the same size of the exponents sets be equal?

Is there a general reference on the invariants of Brieskorn Manifolds for the general case?

  • $\begingroup$ For n=2 you get Seifert manifolds and the exponents $a_i$ are topological invariants of such manifolds. Did you read Hirzebruch's paper "Singularities and exotic spheres"? He has a thorrough discussion of topology of Brieskorn manifolds there. $\endgroup$
    – Misha
    Apr 7 '13 at 22:49
  • $\begingroup$ Hi Misha, yes, I've read the paper. As far as I can tell it doesn't answer my questions. $\endgroup$
    – user02138
    Apr 7 '13 at 23:00
  • $\begingroup$ Have checked Dimca's book on singularities of hypersurfaces? The exact title excapes me at this moment. $\endgroup$ Apr 7 '13 at 23:42
  • $\begingroup$ Hi Liviu, yes, I just thumbed through "Singularities and Topology of Hypersurfaces". $\endgroup$
    – user02138
    Apr 7 '13 at 23:46
  • 1
    $\begingroup$ The 3-dimensional Brieskorn manifolds are discussed at length by John Milnor in his paper On the 3-dimensional Brieskorn manifolds M(p,q,r), in: Knots, groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), pp. 175–225. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N. J., 1975, MR0418127. $\endgroup$
    – Alex Suciu
    Apr 8 '13 at 1:34

This is by no means a complete answer but, rather, a DIY suggestion:

Let $B$ be a $2k+1$-dimensional Brieskorn manifold. Then $B$ is $k-1$-connected. C.T.C. Wall wrote in "Classification problems in differential topology—VI. Classification of (s−1)-connected (2s+1)-manifolds" a complete list of invariants (up to diffeomorphism) for such manifolds (there are few exceptions, but one should be able to deal with them on case-by-case basis), assuming, of course, that $k\ge 2$, so the fundamental group is trivial. The invariants are (mostly) of homological nature, so you should compare them with computations done by Hirzebruch's and Milnor, to see if you get enough information from there to determine a complete list of diffeomorphism invariants for Brieskorn manifolds in terms of the parameters $a_i$.

It is quite possible that such analysis was made by Alan Durfee in his 1971 thesis "Diffeomorphism classification of isolated hypersurface singularities". You may want to ask Durfee (he is at Mount Holyoke College) for a copy. I have no idea why his thesis was never published, but people refer to it quite a bit.

Now, if $k=1$, then the situation is quite different and $B$ is a Seifert manifold. Topology of such manifolds is completely determined by "Seifert invariants" which were first computed by Milnor here if there are no singular fibers (which reduces to computing genus of the base ond Euler number of the fibration) and, in general by Neumann and Raymond here, in terms of the parameters $a_i$, following an earlier paper by Neumann which I do not have access to. So, answering your question in this case, is still a DIY project, working through the formulae in the paper by Neumann and Raymond.

Addendum: Here is the link to a scan of Neumann's thesis. Since it is a scan, it is harder to read, but, unlike the paper of Neumann and Raymond, it deals specifically with Brieskorn manifolds, not with complete intersections of such.

Here is the description (taken fron Neumann's thesis) of a complete set of topological invariants of $\Sigma=\Sigma(a_0,a_1,a_2)$ from the vector $(a_0,a_1,a_2)$ in the generic case (for nongeneric cases see Corollary 9.2 in Neumann's thesis). This is not at all pretty (to say the least), but it is what it is. Define numbers $$ d=gcd(a_0,a_1,a_2), $$ $$ a_i'= \frac{1}{a_i} lcm(a_0,a_1,a_2), $$ $$ t_i= gcd(a_j', a_k'), \{i,j,k\}=\{0,1,2\} $$ $$ s_i= \frac{1}{d} gcd(a_j, a_k), \{i,j,k\}=\{0,1,2\} $$ $$ g= \frac{1}{2}(d^2 s_0 s_1 s_2- d(s_0+s_1+s_2)) +1. $$

Genericity assumption: $t_0, t_1, t_2$ are all $\ne 1$. Now, find integers $\beta_i'$ so that $$ 0\le \beta_i'< t_i, \quad \beta_i'a_i' = 1 (mod\ t_j) $$ and set $$ b= \frac{d}{t_0t_1t_2}(1- \sum_{i=0}^2 \beta_i'a_i') $$

Then the tuple $$ (g; b; \{ds_0(t_0, \beta_0'), ds_1(t_1, \beta_1'), ds_2(t_2, \beta_2')\}) $$ is a complete topological invariant of $\Sigma$. Here $$ ds_i(t_i, \beta_i')= (ds_i t_i, ds_i \beta_i'). $$ Topological meaning of some of the quantities in this tuple:

  1. $g$ is the genus of the base-orbifold $O$ of the Seifert fibration on $\Sigma$.

  2. Under our genericity assumptions, the base-orbifold $O$ will have $3$ singular points of the orders $$ ds_i t_i, i=0, 1, 2. $$ The numbers $$ ds_i \beta_i' $$ define the second set of invariants for the Seifert fibration at the singular fibers.

  3. The number $b$ is responsible for the Euler number of the Seifert fibration (I did not bother to write a precise formula for the transition between these invariants, maybe it literally is the Euler number).

Given how complex this description is, it is very likely that the complete sets of topological/smooth invariants in higher dimensions is much messier.

Now, consider the special case where the numbers $a_0, a_1, a_2$ are pairwise coprime, and greater than $1$. Then $$ g=0, t_i=a_i, d=1, s_i=1, ds_it_i=a_i. $$ In particular, the numbers $a_i$ are orders of cone-points of the base-orbifold. In particular, for coprime numbers $a_i$, the vector $(a_0, a_1, a_2)$ is the complete topological invariant of $\Sigma$. This was the original comment made by myself and Bruno: We both missed the coprimality condition.

In this setting, your "genus" equals $2$ (I still do not know why do you call it "genus"; I would write instead: $$ \frac{1}{2}\left(\frac{d}{\tau} -l\right) + 1, $$ then, at least in the coprime case it matches the genus of the base-orbifold.) Now, it is clear that this number is insufficient to determine the topology of $\Sigma$.

  • 1
    $\begingroup$ Hi Misha, thanks for the answer. Example 2 after Theorem 7.3 states that Σ(2,9,18) and Σ(3,5,15) are diffeomorphic. Doesn't this imply a homeomorphism since they are $3$-manifolds as smooth $S^1$-bundles with equal chern number over Riemann surfaces of the same Euler characteristic (and genus)? Doesn't this refute your comment about the triples necessarily being equal? $\endgroup$
    – user02138
    Apr 8 '13 at 18:42
  • $\begingroup$ user02138: You are right, I was too hasty. I will update my answer accordingly. $\endgroup$
    – Misha
    Apr 8 '13 at 19:17
  • $\begingroup$ According to Neumann and Raymond (section 1 in "Seifert Manifolds, Plumbing...", $g$ is the genus of Seifert surface $\Sigma(a,b,c)/S^{1}$. $\endgroup$
    – user02138
    Apr 9 '13 at 2:09
  • $\begingroup$ where $g = \frac{1}{2}(\frac{d}{\tau} - l) + 1$ Thanks for catching my sign error! $\endgroup$
    – user02138
    Apr 9 '13 at 2:11
  • $\begingroup$ By the way, thank you for the detailed answer! $\endgroup$
    – user02138
    Apr 9 '13 at 2:16

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