8
$\begingroup$

Setting: There are two objects in knot theory that are commonly referred to as the Casson-Gordon invariants: the invariant $\sigma$, and the invariant $\tau$ (see for example A. Conway’s notes Algebraic Concordance and Casson-Gordon Invariants [3] for an introduction to these invariants). When it comes to the $\sigma$-invariant, usual references in the literature include Casson and Gordon’s original papers Cobordism of Classical Knots [1], where the invariant appears as $\sigma(M, \chi)$, and On Slice Knots in Dimension Three [2], where the invariant appears as $\sigma_r(M, \chi)$. I am currently trying to understand the relation between these two different formulations of the $\sigma$-invariant, but I am having some difficulties with it.

To be more precise, let me quickly outline the construction of the invariant $\sigma(M, \chi)$, as found in [3] on p.14, (resp. [1] on p. 183), and of $\sigma_r(M, \chi)$, as found in [2] on p. 41/42 (readers that are already familiar with the definitions and/or cited papers might want to skip the next two paragraphs).

Definition of $\sigma(K, \chi)$: Given a compact $4$-manifold $W$ and a morphism $\psi: \pi_1(W) \rightarrow \mathbb{Z}_m$, let $\widetilde W \rightarrow W$ be the associated $\mathbb{Z}_m$-covering. Then $H_2(\widetilde W; \mathbb{Z})$ is a $\mathbb{Z}[\mathbb{Z}_m]$-module. By mapping the generator of $\mathbb{Z}_m$ to $\omega := e^{\frac{2\pi i}{m}}$, we get a map $\mathbb{Z}_m \rightarrow \mathbb{Q}(\omega)$, which endows $\mathbb{Q}(\omega)$ with a $(\mathbb{Q}(\omega), \mathbb{Z}[\mathbb{Z}_m])$-bimodule structure. Set

\begin{equation} H_*(W; \mathbb{Q}(\omega)) = \mathbb{Q}(\omega) \otimes_{\mathbb{Z}[\mathbb{z}_m]} H_*(\widetilde W; \mathbb{Z}). \end{equation}

Then the $\mathbb{Q}(\omega)$-vector space $H_2(W; \mathbb{Q}(\omega))$ is endowed with a $\mathbb{Q}(\omega)$-valued Hermitian form $\lambda_{\mathbb{Q}(\omega)}$ (whose definition is analogous to the ordinary intersection form on $H_2(W; \mathbb{Z})$). Define the signature of $W$ twisted by $\psi$ as

\begin{equation} \textrm{sign}^\psi(W) := \textrm{sign}(\lambda_{\mathbb{Q}(\omega)}). \end{equation}

Now, given a closed $3$-manifold $M$ and a homomorphism $\chi: \pi_1(M) \rightarrow \mathbb{Z}_m$, bordism theory implies that there exists a non-negative integer $k$, a $4$-maifold $W$ and a homomorphism $\psi: \pi_1(W) \rightarrow \mathbb{Z}_m$ such that $\partial(W, \psi) = k(M, \chi)$. Define the invariant $\sigma(M, \chi)$ as

\begin{equation} \sigma(M, \chi) := \frac{1}{k}(\textrm{sign}^\psi(W) - \textrm{sign}(W)) \in \mathbb{Q}. \end{equation}

Definition of $\sigma_r(M, \chi)$: Let $M$ be a closed $3$-manifold and $\chi: H_1(M) \rightarrow \mathbb{Z}_m$ an epimorphism. Again, $\chi$ induces an $m$-fold cyclic covering $\widetilde M \rightarrow M$ with a canonical generator of the group of covering translations, corresponding to $1 \in \mathbb{Z}_m$. Suppose that for some positive integer $k$ there exists an $mk$-fold cyclic branched covering of $4$-manifolds $\widetilde W \rightarrow W$, branched over a surface $F \subset \textrm{int }W$, such that $\partial(\widetilde W \rightarrow W) = k(\widetilde M \rightarrow M)$, and such that the covering translation of $\widetilde W$ that induces rotation through $2\pi/m$ on the fibers of the normal bundle of $\widetilde F$ restricts on each component of $\partial\widetilde W$ to the canonical covering translation of $\widetilde M$ determined by $\chi$. Let this covering translation induce $\tau$ on $H = H_2(\widetilde W) \otimes \mathbb{C}$. Further, let $\cdot_H$ denote the intersection form of $H_2(\widetilde W)$ extended to $H$ (which is Hermitian, but in general not non-singular). Then $(H, \cdot)$ decomposes as an orthogonal direct sum of eigenspaces of $\tau$, corresponding to the eigenvalues $\omega^r$, $0 \leq r < m$, where $\omega := e^{\frac{2\pi i}{m}}$. Let $\varepsilon_r(\widetilde W)$ denote the signature of $\cdot_H$ restricted to the eigenspace corresponding to the eigenvalue $\omega^r$. Define, for $0 < r < m$, the invariant $\sigma_r(M, \chi)$ as

\begin{equation} \sigma_r(M, \chi) := \frac{1}{k}\left(\textrm{sign}(W) - \varepsilon_r(\widetilde W) - \frac{2[F]^2r(m-r)}{m^2}\right) \in \mathbb{Q}, \end{equation}

where $[F]$ denotes the self-intersection number of the branching surface $F$.

Problem: On pages 185-186 in [1], Casson and Gordon explain the relation of $\sigma(M, \chi)$ to the Atiyah-Singer G-signature. In particular, they conclude on page 186 that (with the notation from above),

\begin{equation} k\sigma(M, \chi^r) + \textrm{sign}(W) = \varepsilon_r(\widetilde W), \end{equation}

which reads for $r = 1$ in particular as

\begin{equation} k\sigma(M, \chi) + \textrm{sign}(W) = \varepsilon_1(\widetilde W). \end{equation}

Further, Proposition 2.19 on p. 18 in [3] states that the formula for computing $\sigma_r(M, \chi)$ in terms of a surgery description of $M$ (Lemma 3.1 on p. 42 in [2]) is also valid for $\sigma(K, \chi)$. From this, I deduced that we (should) have a relation like $\sigma(M, \chi^r) = -\sigma_r(M, \chi)$ for $0 < r < m$. However, there is the summand $\frac{2[F]^2r(m-r)}{m^2}$ appearing in the definition of $\sigma_r(M, \chi)$, and I don't see where or in what form this summand appears in the definition of $\sigma(M, \chi)$ (I do see however why it appears in the definition of $\sigma_r(M, \chi)$, namely because of the fact that for a branched covering $\widetilde W \rightarrow W$ of closed 4-manifolds, there is the formula $\varepsilon_r(\widetilde W) = \textrm{sign}(W) - \frac{2[F]^2r(m-r)}{m^2}$ (Lemma 2.1 on p. 40 in [2])). So my question is the following:

Question: Is the relation $\sigma(M, \chi^r) = -\sigma_r(M, \chi)$ correct? If so, where does the summand $\frac{2[F]^2r(m-r)}{m^2}$ appear in $\sigma(M, \chi)$? If not, is there another relation that holds for $\sigma(M, \chi)$ and $\sigma_r(M, \chi)$?

I assume that my question has something to do with the branching set $F$ as I don't see it mentioned in [1] or [3]. However, I'm too unexperienced in the subject to come to a conclusion on my own. Thus, any elaboration would be greatly appreciated. Thanks in advance!

References:

[1] A. J. Casson; C. Mc A. Gordon, Cobordism of Classical Knots, À la Recherche de la Topologie Perdue, Progress in Mathematics, Volume 62, pp. 181 - 199, Birkhäuser Boston, 1986. With an appendix by P. M. Gilmer (available here)

[2] A. J. Casson; C. Mc A. Gordon, On Slice Knots in Dimenstion Three, Proceedings of Symposia in Pure Mathematics, Volume 32, pp. 39 - 53, American Mathematical Society, 1978 (available here).

[3] A. Conway, Algebraic Concordance and Casson-Gordon Invariants, notes of a reading group held in Geneva, Spring 2017 (available here).

$\endgroup$
  • $\begingroup$ If you believe that the definition with branched covers is independent of $F$ (see pages 41-42 of Casson-Gordon), then you can take $F=\emptyset$. Since, the signature $\epsilon_r(\widetilde{W})$ is equal to the twisted signature, I think the result follows. $\endgroup$ – Anthony Conway Nov 27 '18 at 22:14
5
$\begingroup$

I will expand on my comment. Since $\sigma_r(M,\chi)$ is independent of $F$, you can take $F=\emptyset$ and therefore $\sigma_r(M,\chi)=\frac{1}{k}(\operatorname{sign}(W)-\epsilon_r(\widehat{W}))$, where I write $\widehat{W} \to W$ for the $m$-fold cover induced by $\psi$ (I will use $\widetilde{W}$ for the universal cover). It remains to show that $\epsilon_r(\widehat{W})=\operatorname{sign}^\psi(W)$, where $\psi$ extends $\chi^r$.

Let $H_{\omega^r}$ be the $\omega^r$-eigenspace of the $\mathbb{C}$-vector space $H_2(\widehat{W};\mathbb{C})$ (untwisted coefficients). I will use $\mathbb{C}$ instead of $\mathbb{Q}(\omega)$, but everything basically works the same. It is not too difficult to show that $H_{\omega^r} \cong \mathbb{C}_{\omega} \otimes_{\mathbb{Z}[\mathbb{C}_m]} H_2(\widehat{W};\mathbb{C})$. Since $\mathbb{Z}_m$ is finite, Maschke’s theorem implies that $\mathbb{C}[\mathbb{Z}_m]$ is a semi-simple ring. Consequently all its (left)-modules are projective and in particular flat. It follows that $$H_{\omega^r} \cong \mathbb{C}_{\omega} \otimes_{\mathbb{C}[\mathbb{Z}_m]} H_2(\widehat{W};\mathbb{C})=H_2(\mathbb{C}_{\omega} \otimes_{\mathbb{Z}[\mathbb{C}_m]} C_*(\widehat{W}) ).$$ Writing $C_*(\widehat{W})=\mathbb{Z}[\mathbb{Z}_m] \otimes_{\mathbb{Z}[\pi_1(W)]} C_*(\widetilde{W})$, you can then conclude that $$H_{\omega^r} \cong H_2(W;\mathbb{C}_{\omega}).$$ I slightly abused notations: I wrote $\mathbb{C}_{\omega}$ both for the $\mathbb{C}[\mathbb{Z}_m]$-module structure and for the $\mathbb{Z}[\pi_1(W)]$-module structure induced by $\chi^r$.

You can check that this isomorphism preserves the intersection forms and the equality of signatures follows. Finally, note that on page 42 of the paper you cite, Casson and Gordon write "We shall, however, always be in a situation where either $n=1$ or $F=\emptyset$".

$\endgroup$
  • $\begingroup$ Thank you very much for your answer, knowing that $\sigma_r(M, \chi)$ is independent of $F$ already explains a lot. Maybe one more question: there is this term in the computation of $\sigma_r(M, \chi)$ using a surgery description of $M$ (Lemma 3.1 in Casson-Gordon) including the sum over the matrix of linking numbers, coming from the term $(2[F]^2r(m-r))/m^2$. When we assume that $F = \emptyset$, is it correct that this term in Lemma 3.1 vanishes (i.e. when computing $\sigma(M, \chi)$ using a surgery description of $M$)? $\endgroup$ – user331406 Nov 28 '18 at 13:07
  • $\begingroup$ No, this is not true. The linking matrix in the surgery formula really comes from (a surgery description for) the 3-manifold M. Pick a surgery description for $M$ (forget completely about what $F$ you choose) and you have this matrix; this has nothing to do with $F$! $\endgroup$ – Anthony Conway Nov 28 '18 at 13:42
  • $\begingroup$ Ok, I think I understand now (I was confusing something in the proof of Lemma 3.1 in Casson-Gordon). Thank you very much again! $\endgroup$ – user331406 Nov 28 '18 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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