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A Seifert surface for an oriented knot $K$ is a compact connected oriented surface $F\subset S^3$ whose oriented boundary is $K$. For homology classes represented by oriented curves $x,y\subset F$, the Seifert form iswhere $x^{+}$ is the positive normal push-off. Choosing a basis of $H_1(F;\mathbb{Z})$ gives a Seifert matrix $A$.

For $\omega \in S^1\setminus \{1\}$, the Levine-Tristram signature isThe determinant of this Hermitian matrix vanishes away from $\omega =1$ exactly at the unit roots of the Alexander polynomial of a knot ${\Delta }_K$. Consequently the signature is locally constant on their complement.

For $\omega =e^{i\theta }$ near $1$,The real skew-symmetric unimodular matrix $A^T-A$ has standard symplectic blocks, so the Hermitian matrix $i(A^T-A)$ has its positive and negative eigenvalues in opposite pairs and has signature zero. Thus ${\sigma }_{\omega }(K)=0$ near $1$. If ${\Delta }_K$ has no unit roots, then $S^1\setminus \{1\}$ contains no singular point of the signature form and is connected, so local constancy gives ${\sigma }_{\omega }(K)=0$ everywhere. With the usual convention ${\sigma }_1(K)=0$, the signature vanishes identically.

Solved by gpt-5.6-sol high.

After replacing a class by a nonsingular representative, let $A$ represent a Seifert form over a field $F$ of characteristic different from two. SetA direct calculation gives $T^{T}QT=Q$, so $(F^{2g},Q,T)$ is an isometric structure. A metabolizer for $A$ corresponds to a $T$-invariant metabolizer for $Q$, and stabilization gives the canonical homomorphismConversely, for an isometric structure with $I+T$ invertible, defineThen $A+A^T=Q$ and $A^{-1}{A}^T=T$. These constructions respect orthogonal sums and metabolic structures and are inverse on Witt classes, proving ${\mathcal{A}\mathcal{C}}_F\cong {\mathcal{W}}_F$.

For an irreducible symmetric Laurent polynomial $\delta$, the primary component of an isometric structure isfor large $N$. The primary decomposition is orthogonal, so restriction of $Q$ and $T$ to $V_{\delta }$ defines the projection

Now take $F=\mathbb{R}$ and let $\delta$ have roots $\omega ,\overline{\omega }$ on the unit circle, with $\omega =e^{i\theta }$ in the upper half-plane. The isomorphism ${\mathcal{W}}_{\mathbb{R}}^{\delta }\cong 2\mathbb{Z}$ sends a class to the even signature jumpFor the class of a knot, this is precisely the jump of its Levine-Tristram signature at the root $\omega$; reversing the choice of side changes the overall sign convention.

Solved by gpt-5.6-sol high.

Over $\mathbb{R}$, the relevant part of the Alexander polynomial of a knot of $T_{2,5}$ has the two irreducible symmetric factorsTheir upper-half-plane roots are $\alpha =e^{\pi i/5}$ and ${\alpha }^3=e^{3\pi i/5}$. The supplied determinant shows that the Levine-Tristram signature can jump only at these roots and their conjugates.

For the supplied Seifert matrix, direct inertia calculations on successive arcs of the upper semicircle giveChanging the orientation convention reverses all signs but changes no conclusion. Thus the jumps at both $\alpha$ and ${\alpha }^3$ are $-2$. It follows from part a thatand in both nonzero cases the image is a generator of ${\mathcal{W}}_{\mathbb{R}}^{\delta }\cong 2\mathbb{Z}$.

Solved by gpt-5.6-sol high.

For the supplied Seifert matrix $A$,and the symmetric form $Q=A+A^T$ has signature $-2$. ThusSince the Levine-Tristram signature is an additive homomorphism on the algebraic concordance group, $K$ has infinite algebraic-concordance order.

Over ${\mathbb{Q}}_3$, reduction of the Alexander polynomial givesThe factors are coprime, nonsymmetric, and exchanged by reciprocity. Hensel lifting therefore decomposes the local isometric structure into a reciprocal pair, which is metabolic. Its class in ${\mathcal{W}}_{{\mathbb{Q}}_3}$ is zero and in particular does not have order four.

For $p=11$, diagonalization givesThe second residue at $11$ is the one-dimensional formBecause $11\equiv 3(\mathrm{mod}4)$, $W({\mathbb{F}}_{11})\cong \mathbb{Z}/4$ and this one-dimensional form is a generator. The P-adic algebraic-concordance obstruction therefore has exact order four, so the image of $K$ in ${\mathcal{W}}_{{\mathbb{Q}}_{11}}$ has order four.

Solved by gpt-5.6-sol high.

A genus-one Seifert matrix for the Stevedore knot isIts Alexander polynomial iswhose roots are $2$ and $1/2$. There are no unit roots, so Question 1(a) proves that every Levine-Tristram signature of the Stevedore knot vanishes.

On the other hand,has Smith normal form $\mathrm{diag}(1,9)$. ThereforeIf a knot is doubly slice, the linking form on the first homology of its two-fold branched cover of a knot is hyperbolic: it has two complementary metabolizers, arising from the two sides of the unknotted sphere. A cyclic group of order nine has a unique subgroup of order three, so its linking form of a branched cover cannot have two complementary metabolizers. The Stevedore knot is consequently not doubly slice, despite its identically vanishing signature function.

Solved by gpt-5.6-sol high.

For a pattern of winding number $w$,whenever the relevant signatures are defined by nonsingular forms. The usual averaged convention extends the identity through roots.