M = {\Lambda}^{n-k}\ {{ \hbox{\larger${ \hbox{\larger${ \hbox{\larger$\oplus\m@th{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$$}\m@th$\m@th$}{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\m@th$\m@th$}\m@th{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$$\m@th$}{}}\m@th$ }_{i=1}^k \frac{ \Lambda}{\left< d_i\right>} .$$It follows that $A_L= \Delta_{n-k+1}(L) \sim_{\Lambda} \prod_i d_i$; where we let $\sim_R$ mean equal up to a multiplicative factor from the units of a ring $R$. The matrix $W = (1-t)V + (1-t)V^ \textsf{T}$ presents a $\Lambda$--module, say, $N$. Since$$W = t^{-1}(1-t)(tV - V^ T),$$the same row and column operations that diagonalize $tV - V^ \textsf{T}$ also diagonalize $W$ and we see that$$ N= \Lambda^{n-1} {{ \larger${ \hbox{\larger${ \hbox{\larger$\oplus\m@th{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$$}\m@th$\m@th$}{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$\m@th$\m@th$}\m@th{ \set@fontsize {1}{\f@size}{\f@baselineskip} \hbox{$$\m@th$}{}}\m@th$ }_{i=1}^k \frac{ \Lambda}{\left< (1-t)d_i\right>}.$$\subsection{Step function} We first want to observe that, as in the case of knots, the signature function is a step function. There is a diagonalization of $W$ over the field of fractions $\mathbb{R}(t)$. That is, there is a determinant one matrix $A(t)$ with entries in $\mathbb{R}(t)$ such that $A(t) W(t) {A^ \textsf{T}}(t^{-1})$ is diagonal, with diagonal entries rational functions: $[q_1, \ldots, q_k, 0, \ldots 0]$. (Since the module $N$ described above becomes an $n-k$ dimensional vector space with $\mathbb{R}(t)$ coefficients, we can use the same value of $k$ here as above.) For all but the finite set of $\rho \in S^1$ for which $A(\rho)$ is not defined, this provides a diagonalization of $W(\rho)$. Away from this set of singular values, we see that the signature of $W$ can jump only at zeroes and poles of the diagonal entries. Thus, the signature function is a step function; in particular, it has a finite number of discontinuities. \subsection{Jumps away from $\rho = \pm 1$} Let $\rho \in S^1$ be a fixed complex number. By Lemma~\ref{diag}, we can diagonalize $W$ over the ring $\Lambda_{(f_\rho)}$. Since this matrix presents the module $N \otimes \Lambda_{(f_\rho)}$, we have that after reordering the entries, this diagonalization of $W$ has entries $[\alpha_1 (f_\rho)^{\epsilon_1}, \ldots , \alpha_k (f_\rho)^{\epsilon_k}, 0, \ldots , 0]$ where the $\alpha_i$ are units in $\Lambda_{(f_\rho)}$ and $(f_\rho)^{\epsilon_i}$ is the maximum power of $f_\rho$ dividing $d_i$. The jumps at the discontinuities, $\mathop{\mathrm{jump}}\nolimits^\pm$, arise from the diagonal terms for which $\epsilon_i >0$. It is now evident that these jumps are bounded by the multiplicity of $\rho$ in $A_L$, as was to be proved. Thus $\sigma_L(\omega)$ is a step function as claimed. \subsection{Jump at $-1$} The diagonalization lemma, Lemma~\ref{diag}, does not apply for $\rho = -1$ (see ~\S \ref{counter}). A transformation corrects for this. Let $W^*$ be the Hermitian matrix $(1-t^2)V + (1- t^{-2})V^ \textsf{T}$. The jumps of the signature function of $W$ at $-1$ correspond to the jumps of the signature function of $W^*$ at $\rho = \sqrt{-1}$. Notice that $f_\rho = t^{-1} + t$. The diagonalization of $W^*$ in $\Lambda_{({f_\rho})}$ has nonzero entries of the following types: \begin{itemize} \item $\alpha_i$, where $\alpha_i$ is a unit. \item $\beta_i (f_\rho)^{b_i}$, where $\beta_i(\rho) >0$ and $b_i$ is odd. \item $\gamma_i (f_\rho)^{c_i}$, where $\gamma_i(\rho) < 0$ and $c_i$ is odd. \item $\delta_i (f_\rho)^{d_i}$, where $\delta_i(\rho) >0$ and $d_i$ is even. \item $\eta_i (f_\rho)^{e_i}$, where $\eta_i(\rho) <0$ and $e_i$ is even. \end{itemize} Suppose that the number of elements for type $\beta$, $\gamma$, $\delta$ and $\eta$ are given by $B$, $C$, $D$, and $E$, respectively. The total jump in the signature function of $W$ at $-1$ is 0, by conjugation symmetry, so the same is true for the total jump in the signature function of $W^*$ at $\sqrt{-1}$. This implies that $B = C$. We then see that $|\mathop{\mathrm{jump}}\nolimits^\pm(W^*,\sqrt{-1})| \le |D - E |$. Since $D$ and $E$ both correspond to diagonal elements for which $f_{\rho}$ has even exponent, it is now clear that the jump is at most one half the multiplicity of $-1$ in $A_L$. It follows that $\mathop{\mathrm{mult}}\nolimits_{-1}(A_L)$ is even and $|\mathop{\mathrm{jump}}\nolimits^\pm(W, -1)| \le (1/2)\mathop{\mathrm{mult}}\nolimits_{-1} (A_L)$, as was to be proved. \subsection{Jump at $1$} Since $\sigma_L(1)=0$, $|\mathop{\mathrm{jump}}\nolimits^\pm(1)| =| \sigma_L^\pm(1)|$. We wish to show that $\mu_L-1$ is an upper bound for $| \sigma_L^\pm(1)|$. We let $t = \cos(2\theta) + \sqrt{-1}\sin(2\theta)$. Expressing $W(t) = (1-t)V + (1-t^{-1})V^ \textsf{T}$ in terms of $\theta$ and simplifying, we find$$W(t) = -\sin(2 \theta) \left( \sqrt{-1}(V - V^ T) - \tan(\theta)(V+V^ T) \right).$$The matrix $V-V^ \textsf{T}$ gives the intersection form for the Seifert surface. Thus we may take $V-V^ \textsf{T}$ to be the direct sum of $g$ copies of $[\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}]$ direct sum with a $(\mu_L-1) \times (\mu_L-1)$ zero matrix. It follows that the form $i(V - V^ \textsf{T})$ has $g$ eigenvalues $1$ and $g$ eigenvalues $-1$. After a small perturbation, the number of positive and negative eigenvalues will both continue to be at least $g$, so that the absolute value of the signature is at most $\mu_L -1$. (Alternatively, the matrix $\sqrt{-1}(V - V^ \textsf{T})$ is congruent to a diagonal matrix with its first $2g$ diagonal entries alternating between $1$ and $-1$. Thus, there are exactly $g$ sign changes in the sequence of the first leading $2g$ principle minors of this congruent matrix. By the continuity of determinants, the same is true for a small perturbation. It follows that the signature of the upper left $2g \times 2g$ block of a small perturbation is also zero. This upper left $2g \times 2g$ block is nonsingular. It follows that the absolute value of the signature of a small perturbation is at most $\mu_L-1$.) \section{Proof of Theorem \ref{bound2}} \begin{lem} If $\Delta_L(t) \ne 0$, then $\mathop{\mathrm{mult}}\nolimits_1\left(\Delta_L(t)\right) \ge \mu_L-1$. \end{lem} \begin{proof} Since $\Delta_L(t) \ne 0$, the $\Lambda$--module $M$ presented by $V - tV^ \textsf{T}$, has no free summand: $M \cong \oplus_{i=1}^n \frac{\Lambda}{\left< d_i \right>}$ where each $d_i$ divides $d_{i+1}$ and is nonzero. Tensoring with $\mathbb{R}$, viewed as a $\Lambda$--module with trivial action (formally, $\mathbb{R} \cong \frac{\Lambda}{\left< 1-t \right> }$), yields the $\mathbb{R}$--vector space presented by $V - V^ \textsf{T}$. For a link of $\mu$ components, this is isomorphic to $\mathbb{R}^{\mu -1}$ (see the description of $V-V^ \textsf{T}$ in \S\ref{subsec.jump1}). However, $\frac{\Lambda}{\left< d_i \right>} \otimes \frac{\Lambda}{\left< 1-t \right>}$ is trivial unless $d_i$ is divisible by $(1-t)$. Thus, $(\mu -1)$ of the $d_i$ are divisible by $(1-t)$. This implies the statement of the lemma. \end{proof} By Theorem~\ref{main}, $\mu_L-1 \ge |\mathop{\mathrm{jump}}\nolimits^\pm(1)|$. Using the triangle inequality, we easily obtain Theorem~\ref{bound2} from Corollary~\ref{bound}. \begin{rem} { \em For $L= 8n8(0,0,0)$ or $8n8(1,0,0)$, discussed in \S \ref{L8n8} below, we have $\Delta_L(t) = 0$, and $\mathop{\mathrm{mult}}\nolimits_1(A_L(t))= 2 <3= \mu_L-1.$ Thus Lemma \ref{estmult} cannot be modified to read $\mathop{\mathrm{mult}}\nolimits_1(A_L(t)) \ge \mu_L-1$ in the situation where $\Delta_L(t) = 0.$ By adapting the argument given above, one can show that $$\mathop{\mathrm{mult}}\nolimits_1(A_L) \ge \mu_L-h_L.$$ Note that Proposition \ref{even} says that this inequality is also a congruence modulo two. } \end{rem} \section{Proof of Theorem \ref{cong}} Let $\rho \ne \pm 1$. If $\frak{d}_i$ is an entry of the diagonal matrix $\mathbb{D}$ appearing in \S \ref{awayfpm1} and has $\mathop{\mathrm{mult}}\nolimits_\rho(\frak{d}_i)$ odd, then $\frak{d}_i$ contributes $\pm 2$ to $\mathop{\mathrm{jump}}\nolimits(\rho)$. If $\frak{d}_i$ is an entry of $\mathbb{D}$ and has $\mathop{\mathrm{mult}}\nolimits_\rho(\frak{d}_i)$ even, then $\frak{d}_i$ contributes $0$ to $\mathop{\mathrm{jump}}\nolimits(\rho)$. If $\mathop{\mathrm{mult}}\nolimits_\rho(A_L)$ is odd, then an odd number of entries $\{\frak{d}_i\}$ have $\mathop{\mathrm{mult}}\nolimits_\rho(\frak{d}_i)$ odd. If $\mathop{\mathrm{mult}}\nolimits_\rho(A_L)$ is even, then an even number of entries $\{\frak{d}_i\}$ have $\mathop{\mathrm{mult}}\nolimits_\rho(\frak{d}_i)$ odd. The result stated in the first sentence follows. If $\rho \ne \pm 1$, and $\mathop{\mathrm{mult}}\nolimits_\rho(A_L)=1$, then $\mathop{\mathrm{mult}}\nolimits_\rho(\frak{d}_i)$ is nonzero for exactly one $i$. The result stated in the second sentence follows. \section{Proof of Proposition \ref{even}} Let $W^{**}=(1-t^4)V + (1- t^{-4})V^ \textsf{T}$. The jumps of the signature function of $W$ at $1$ correspond to the jumps of the signature function of $W^{**}$ at $\rho=\sqrt{-1}$. Thus it follows that$$\mathop{\mathrm{jump}}\nolimits^+(W^{**},\sqrt{-1})= -\mathop{\mathrm{jump}}\nolimits^-(W^{**},\sqrt{-1}).$$Notice that $(1-t^4)= (1-t)(1+t)(1+t^2)$ and $f_\rho = t^{-1} + t$. Moreover, up to $\sim_\Lambda$, $f(t)=1-t$ is the only irreducible polynomial for which $f(t^4)$ has $1+t^2$ as a factor. It follows that $e_{t-1}(W)=e_{t+t^{-1}}(W^{**})$. The argument in \S \ref{subsec.jump-1} shows that $e_{t+t^{-1}}(W^{**})$ is even. Thus $e_{t-1}(W)$ is even. But$$e_{t-1}(W)=e_{t-1}((1-t)(tV-V^ T))= e_{t-1}(tV-V^ T)+ 2g +\mu_L-h_L.$$As $\mathop{\mathrm{mult}}\nolimits_1(A_L)= e_{t-1}(tV-V^ \textsf{T})$, the result follows. \section{A module approach to the signature function} The proof of Lemma~\ref{diag} and of Theorem~\ref{main} in the case of $\rho = -1$ (as given in \S \ref{subsec.jump-1}) point to a general approach to understanding the signature function. Let $V$ be an $n\times n$ Seifert matrix for a link $L$. The $\Lambda$--module $M$ presented by $V - tV^ \textsf{T}$ has a direct sum composition$$M\cong \frac{\Lambda}{\left< d_1\right>} \oplus \frac{\Lambda}{\left< d_2\right>} \cdots \oplus \frac{\Lambda}{\left< d_k\right>} \oplus \Lambda^{n-k},$$where the $d_i \in \Lambda$ are nonzero and $d_i$ divides $d_{i+1}$ for all $i < k$. We have that $A_L \sim_\Lambda \prod d_i$; also, $\Delta_L \ne 0$ if and only if $n = k$. More generally, $h_L=n-k+1$. Let $X$ denote the infinite cyclic cover of $S^3 \setminus L$ specified by the linking number. According to \citation[['Thm 6.5']]{lickorish}, the $\mathbb{Z}[t,t^{-1}]$--module $H_1(X,\mathbb{Z})$ is presented by the matrix $V - tV^ \textsf{T}$. It follows that $M$ is a description of $H_1(X,\mathbb{R})$ as a $\Lambda$--module. For any $\rho \in S^1$, let $\phi_o(L, \rho)$ and $\phi_e(L,\rho)$ to be the number of $d_i$ in which $f_\rho$ has a positive exponent that is odd or even, respectively. Bounds on the signature function are easily given in terms of these functions; the proofs follow along the same lines as our earlier work. Here is an example of the type of result that can be attained in this way. \begin{thm} If $\rho \ne \pm 1$, then $|\mathop{\mathrm{jump}}\nolimits(\rho)| \le 2 \phi_o(L, \rho)$ and $|\mathop{\mathrm{jump}}\nolimits(\rho)| = 2 \phi_o(L, \rho)$ modulo 4. Also $|\mathop{\mathrm{jump}}\nolimits_\pm(-1)| \le \phi_e(L, -1)$ and $|\mathop{\mathrm{jump}}\nolimits_\pm(-1)| = \phi_e(L, -1)$ modulo 2. Thus $$| \sigma_L - \sigma_L^+(1) | \le \phi_e(L,-1) + \sum_{\rho \in S^1 \setminus\{-1,1\}} \phi_o(L,\rho).$$ \end{thm} \section{Some examples} In this section we will present some examples that illustrate our results. Notice that all the polynomials presented are in $\mathbb{Z}[t,t^{-1}]$ and are defined up to multiplication by a unit, $\pm t^i$. When possible, we normalize polynomials $f$ so that $f \in \mathbb{Z}[t]$ and $f(0)\ne 0$. \subsection{A family of links with zero Alexander polynomial} Consider the family of 3--component links $L_n$ illustrated in Figure~\ref{figure}. \begin{figure} {\includegraphics{drawing6.pdf}} \caption{Link with $n$ denoting full twists.} \end{figure} There is an obvious disconnected Seifert surface consisting of an annulus and a punctured torus. We tube these together to obtain a connected Seifert surface with Seifert matrix$$V_n=\left( }\begin{array}{cccc} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & n & 0 \\ 0 & 0 & 0 & 0 \end{array}\hbox{ \right).$$We have that $\Delta_{L_n}=0$; $A_{L_n}=\Delta_2(L_n)= (t-1)((n+1) t^2-(n+2) t+(n+1))$;$$\sigma_{L_n}= }\begin{cases} -3 & \hbox{if} n \le -2 \\ -1 & \hbox{if} n \ge -1 . \end{cases}\hbox{$$Thus $A_{L_0}=(t-1)^3$ and $\sigma_{L_0}=-1$. As $A_{L_0}$ has no roots on $S^1\setminus \{1\}$, $\sigma_{L_0}$ is identically $-1$ on $S^1\setminus \{1\}$. Similarly, $A_{L_{-1}}=(t-1)$ and $\sigma(L_{-1})=-1$. As $A_{L_{-1}}$ has no roots on $S^1\setminus \{1\}$, $\sigma_{L_{-1}}$ is identically $-1$ on $S^1\setminus \{1\}$. For $n \notin \{0,-1\}$, $A_{L_n}$ has a single conjugate pair of roots $\alpha_n$ and $\bar{\alpha}_n$ on $S^1\setminus \{1\}$ with real part $\Re (\alpha_n)=\frac{n+2}{2(n+1)}$, where $\sigma_{L_n}$ must make a total jump of $\pm2$. The one-sided jump at $1$ must have absolute value less than or equal to $2$. It follows that if $n \le -2$, then$$\sigma_{L_n}(e^{\pi i x})= }\begin{cases} 0 & \hbox{if} x=0 \\ -1 & \hbox{if} 0

Signature Jumps and Alexander Polynomials for Links

Abstract

Signature Jumps and Alexander Polynomials for Links

Abstract

Paper Structure

Table of Contents

Key Result

Theorems & Definitions (7)