In classic, Euler numbers associated to circle bundles over a fixed surface classify all possible such bundles. But the construction of Euler class in general requires the fact that any fiber bundle has a local trivialization which is not the case for Seifert fibration. I wondered if the same is true for Seifert fibration via some clever generalization of Euler number and it seems true according to Scott's paper The geometries of 3-manifolds. In his paper he discussed the Euler number of a circle bundle over a surface but didn't define formally although it seems to be equal to the classical notion of it. The problem is that he also didn't define formally what the Euler number of Seifert fibration is. Here's the context where he discussed invariants of a Seifert fibration:

(p.435) The first invariant of a Seifert bundle is the base orbifold $X$. Let $X'$ denote $X$ with the interior of a regular neighborhood $N$ of the singular set removed. Then we have a circle bundle over $X'$. Our given Seifert bundle over $X$ is completed by adding the fibered solid tori corresponding to the cone points of $X$ (i.e. Dehn filling). Each fibered solid torus determines a pair of coprime integers $(p,q)$ with appropriate normalization so that $0\leq q<p$, called the orbit invariants. One more invariant is needed to complete the classification of Seifert bundle in the case when the base space is a closed orbifold. this is a generalization of the invariant $b$ defined earlier (in the paper) for circle bundles over closed surfaces. As in that case, $b$ is an integer if the total space of the Seifert bundle is orientable and lies in $\Bbb Z_2$ otherwise. Seifert showed that the invariants described here determine a Seifert bundle up to bundle isomorphism. thus a Seifert bundle $\eta$ is specified by the orbifold $X$, the circle bundle $\eta|X'$ where $X'$ is $X$ with the singular points removed, the orbit invariants of the critical fibers and the value of the invariant $b$.

And he later said this invariant $b$ does not satisfy some kind of naturality properties under finite covering so Neumann and Raymond defined a new Euler class $e$ with those properties but didn't spell out what that is. He also gave a connection between $e$ and $b$ as the following: Given orbit invariant $(p,q)$ define a Seifert invariant $(\alpha,\beta)$ defined by $\alpha = p$ and $\beta$ is given by $\beta q\equiv 1\bmod p$ and $\beta$ is normalized so that $0<\beta<\alpha$. Then, $$e = -\left(b+\sum_{i=1}^r \beta_i/\alpha_i\right).$$ As far as I searched, that relation appears only on Scott's paper (sorry if not). Actually, many other literatures do not define Euler number of a Seifert bundle in that way, for example in Martelli's geometric topology textbook, if a Seifert manifold is $(S,(p_1,q_1),\ldots,(p_n,q_n))$ for $p_i\geq 1$, then the Euler number is $\sum_{i=1}^n{q_i/p_i}$. Some paper put a minus sign in front and I think I should put minus if I want to compare with Scott's definition.

Although Scott's definition is not very clear, I prefer that definition because it feels to me that with that definition, the Euler number on a Seifert bundle is a generalization of the Euler number on a circle bundle. Just $-\sum_{i=1}^n{q_i/p_i}$ is not so natural to me (or maybe it is, I just can't see). My questions are simple

  1. Why the Euler number of a Seifert bundle is a "natural" generalization of a circle bundle over a surface?
  2. How is Scott's definition equal to Martelli's definition?

I thought if I understand what $b$ is for orbifold in Scott's paper, maybe I can see the "natural generalization" but as I said, there's no other paper defining Euler class in that way. I tried to read Seifert's original paper but it's not in English so I couldn't understand. Could you explain?

  • $\begingroup$ In addition to Hatcher's notes mentioned in Golla's answer, I personally like the lecture notes by Jankins and Neumann, which are available for free in the web, and cover the relevant material. $\endgroup$ Apr 3 at 18:52

1 Answer 1


Why is the Euler number of a Seifert bundle a "natural" generalization of a circle bundle over a surface?

I can see at least three reasons:

  1. A circle bundle is a Seifert manifold with no singular fibres, and for them (according to the definition in Scott's paper) it agrees with the classical Euler number.

  2. It is an obstruction for the existence of a section. This is stated as an exercise in Hatcher's Notes on basic 3-manifold topology (available here).

  3. It is somewhat multiplicative with respect to covers. This is stated in Neumann and Raymond's Seifert manifolds, plumbing, $\mu$-invariant and orientation reversing maps (reviewed here).

How is Scott's definition equal to Martelli's definition?

I think this is fairly well-explained in a couple of places, for instance in Fomenko and Matveev's Algorithmic and computer methods for 3-manifolds or in Hatcher's notes I mentioned above. To give a quick explanation, the invariants $\alpha$ and $\beta$ associated to a singular fibre of a Seifert fibration are not uniquely defined, but depend on the choice of a trivialisation of the circle bundle comprising all singular fibres. This trivialisation is related to the invariant $b$ that appears in Scott's formula. You can change trivialisation around a singular fibre, and this simultaneously changes $b$ by a certain integer and $-\beta/\alpha$ by the same integer, so the sum is invariant under this move. Any two trivialisations differ by a sequence of these moves, so you get an invariant.

Another thing you can do is normalise the invariants so that $0 < \beta < \alpha$ (by appropriately twisting the trivialisations), and this determines $b$. The other "natural" thing you can do is twist until $b = 0$ (which does not determine all invariants for singular fibres), and then you get Martelli's formula.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.