What is the easiest construction of the Casson invariant? The original construction using representation spaces (as found, for instance, in Akbulut-McCarthy) is very technical since you have to perturb things to make them transverse.

One possibility would be to do some kind of Kirby calculus construction. Such a thing can be found in Walker's book and Lescop's book, but they constructed much more general things so their work is also very technical.


Hoste has a combinatorial formula for the Casson invariant for integral homology $3$-spheres. Let $\Sigma$ be an integral homology $3$-sphere obtained via surgery on a framed link $L = K_1 \cup \dots \cup K_n$ in $S^3$ with framings $1/q_i$. Additionally, we assume $\mathrm{lk}(K_i, K_j) = 0$ for $i \neq j$ (this can always be arranged, see Lemma 12.2 in Saveliev's Lectures on the Topology of $3$-Manifolds. In fact, that lemma says that we can take $q_i = \pm 1$). We can write the Conway polynomial of $L$ in the form $$\nabla_L(z) = z^{n-1}(a_0(L) + a_1(L) z^2 + \cdots + a_m(L) z^{2m}).$$ With the above notations, Hoste's formula is $$\lambda(\Sigma) = \sum_{L' \subset L} \left( \prod_{i : K_i \subset L'} q_i\right) a_1(L').$$ The sum is taken over all sublinks $L'$ of $L$.

You can see the details of Hoste's formula in his original paper:

Hoste, Jim. A formula for Casson's invariant. Trans. Amer. Math. Soc. 297 (1986), 547-562

Hoste's original paper does not show that $\lambda$ as defined by his formula is actually an invariant of integral homology $3$-spheres, however. The missing ingredient is a Kirby move that preserves the unlinking of all components of $L$ as required for Hoste's formula to be stated. Such a move is constructed by Habiro in the following paper:

Habiro, Kazuo. Refined Kirby calculus for integral homology spheres. Geometry & Topology 10 (2006), 1285–1317.

Combining the results of the above two papers gives a nice combinatorial construction of the Casson invariant as an invariant of integral homology $3$-spheres.

  • $\begingroup$ Lovely, thanks! (ps : sorry I can't up vote this; I'm too lazy to register my account) $\endgroup$ – Bill Jan 20 '14 at 4:50

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