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In Némethi's book "Normal surface singularities", Example 5.1.17, there is a formula to find the Seifert invariants of a Brieskorn complete intersection $\Sigma(a_1,...,a_n)$. I am interested in the case $n=3$. I was trying to find a counterexample of a Brieskorn intersection that is not a rational homology sphere according to Nemethi's Example 5.1.17. The conditions to be a rational homology sphere are the following:

(i) $(a_1,...,a_n) = (kp_1,kp_2,p_3,...,p_n)$ where $k\geq 1$ and the integers $\{p_j\}_{j=1}^n$ are pairwise coprime, and $\gcd(k,p_j) = 1$ for any $j\geq 3$.

(ii) $(a_1,..,a_n) = (2^c p_1,2p_2, 2p_3, p_4,...,p_n)$ where the integers $\{p_j\}_{j=1}^n$ are pairwise odd and coprime, and $c\geq 1$.

I picked $\Sigma(4,4,4)$, which should not satisfy either condition to be a rational homology sphere. However, I computed the Seifert invariants and got $(b_0;q_1(\alpha_1,\omega_1), q_2(\alpha_2,\omega_2), q_3(\alpha_3,\omega_3)) = (0;4(1,-1), 4(1,0), 4(1,0))$ where $q_k(\alpha_k,\omega_k)$ means that Seifert invariant is repeated $q_k$ times. The matrix whose cokernel is isomorphic to $H_1(M,\mathbb{Z})$ is

$$\begin{pmatrix} \alpha_1 & 0 & \cdots & 0 & \omega_1 \\ 0 & \alpha_2 & \ & 0 & \omega_2\\ \vdots & \ & \ddots & \ & \vdots\\ 0 & 0 & \ & \alpha_n & \omega_n\\ 1 & 1 & \cdots & 1 & b_0 \end{pmatrix}$$

Plugging in the Seifert invariants I found, this matrix has finite cokernel. This would mean $H_1(M,\mathbb{Z})$ is finite, hence $M$ is a rational homology sphere, which it shouldn't. What went wrong?

P.S. I tried for $\Sigma(4,8,12)$ as well and got the invariants $(0;4(1,0), 4(2,-1), 4(3,1))$, or the normalized version $(-4;4(1,0), 4(2,1), 4(3,1))$. The cokernel of the matrix above is still finite even though the tuple $(4,8,12)$ doesn't satisfy conditions (i) nor (ii).

Edit: these are the conditions to compute the Seifert invariants listed in Nemethi's book. $$A := \prod_{i=1}^n a_i,\ a:=lcm\{a_i:1\leq i\leq n\},\ \alpha_i = a/lcm\{a_j: j\neq i\}$$ $$\ q_i = \frac{A\alpha_i}{aa_i},\ \ \ \ \omega_ia/a_i \equiv -1 \mod \alpha_i$$ $$ -A/a^2 = -b_0 + \sum_j \omega_j/\alpha_j$$

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    $\begingroup$ You miscalculated: in your first example, your matrix has determinant 0: the 13th row is the sum of the first 12. Not sure what's going on in the second example. $\endgroup$
    – mme
    Commented Oct 29 at 11:03
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    $\begingroup$ I now agree there's an issue somewhere. I think the issue is either in the definition of $\omega_i$ (maybe there's some additional constraint?) or in the computation of $b_0$. Now, in your computation for $(4, 4, 4)$, I'm confused by your claims about $\omega_i$ --- the condition is $\omega_i \equiv -1 \mod 1$, which allows you to choose $\omega_i$ arbitrarily. For $(4,8,12)$, choosing $0 \le \omega_i < \alpha_i$, I find $4(1,0)$ and $4(2,1)$ and $4(3,1)$, so I'm not sure where your terms are coming from. $\endgroup$
    – mme
    Commented Oct 29 at 16:33
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    $\begingroup$ @mme sorry, you are right about the (4,8,12) case, I copied my computation wrong. I edited the post now. For the condition $\omega_i\equiv -1 \mod 1\ $, I chose $\omega_i=0$ arbitrarily (or, better, $\omega_i < \alpha_i=1$). $\endgroup$
    – user13121312
    Commented Oct 29 at 16:57
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    $\begingroup$ OK, I agree then - I was confused by the insistence that $\omega_1 = -1$ instead of taking $\omega_1 = 0$. But, this is inconsequential. It seems we agree now that either a formula is mistaken or there's some missing assumption on the choice of $\omega_i$. $\endgroup$
    – mme
    Commented Oct 29 at 17:42
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    $\begingroup$ I wonder why Nemethi’s computation doesn’t mention the genus, which is an important part of the Seifert invariant. Compare for instance, with the explanation on Wikipedia: en.m.wikipedia.org/wiki/Seifert_fiber_space . It provides a quick sanity check, since if $g>0$ it can’t be a homology sphere. $\endgroup$
    – HJRW
    Commented Oct 30 at 6:59

1 Answer 1

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The key is pointed out by HJRW in his comment: there's a missing piece in your explanation, which is the genus of the base $B$ of the Seifert fibration. Némethi writes: $$ 2g-2 = (n−2)A/a−\sum q_i, $$ which in your case gives (modulo mistakes in my computation) $2g-2 = 16 - 12 = 4$, so $g=6/2=3$, and since the matrix you provided is non-singular, $b_1(\Sigma(4,4,4)) = b_1(B) = 6$.

By the way, that the genus is 3 makes sense to me, since $\Sigma(4,4,4)$ is the 4-fold cyclic cover of $S^3$ branched over the link of the curve singularity of $C = \{x^4 + y^4 = 0\} \subset \mathbb{C}^2$ at the origin. To give an explicit description of its cover, it suffices to blow up the plane at the origin and observe that $\Sigma(4,4,4)$ is the boundary of cover of the disc bundle with Euler number -1 over the 2-sphere (morally, a compact version of the total space of the blow-up at the origin) branched over four fibres (the strict transform of $C$). This new 4-manifold is again a bundle, but now the base is the 4-fold cover of the 2-sphere branched over 4-points (which is indeed a genus-3 surface) and whose Euler number is $-4$.

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    $\begingroup$ To spell something out a little: my and OP's mistake was to suggest the given matrix has cokernel isomorphic to $H_1$. This is true in the case $g = 0$, but not in general, as the matrix doesn't know about the genus of the base -- which contributed to the homology of the total space! $\endgroup$
    – mme
    Commented Oct 31 at 13:58
  • $\begingroup$ Thank you for your answer. I am still confused since Nemethi writes that " $H_1(L_X, \mathbb{Q})=0$ iff $g=0$. This in terms of the $a_i$ reads as follows: ", and then proceeds to give the conditions (i) and (ii) that I listed above. The way I understand this is that the $(a_1,...,a_n)$ from $\Sigma(a_1,...,a_n)$ satisfy conditions (i) or (ii) iff $g=0$. What is my misunderstanding here? $\endgroup$
    – user13121312
    Commented Oct 31 at 14:52
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    $\begingroup$ @user13121312 Everything you've written in your comment is correct. You can check it using the definitions you've given. The issue is the claim that $H_1(L_X)$ is the cokernel of the given matrix, which holds if $g = 0$ but not in general. Compare Proposition 3.4.2 in Nemethi's book, which gives the correct statement: there is a contribution from the base surface of the Seifert fibration. Notice that $\Sigma(4,4,4)$ and $\Sigma(4,8,12)$ are not QHS --- $g = 3$ in both cases --- and indeed not of the given form. $\endgroup$
    – mme
    Commented Oct 31 at 17:18
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    $\begingroup$ Got it now, thanks a lot! $\endgroup$
    – user13121312
    Commented Oct 31 at 17:44

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