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For $\omega \in S^1\setminus \{1\}$, the Levine-Tristram signature is the signature of the Hermitian matrixIt is locally constant away from unit roots of the Alexander polynomial of a knot.

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.

Boundary-connected-summing minimal Seifert surfaces for $K$ and $K^{\prime }$ givesFor the reverse inequality, let $F$ be a minimal-genus Seifert surface for the connected sum of knots and let $S$ be its standard splitting sphere. A minimal-genus Seifert surface is incompressible in the knot exterior: a compression either lowers its genus or separates off a closed component that can be discarded. Put $F$ and $S$ in transverse position and minimize the number of intersection circles. An innermost circle on $S$ either gives a compression of $F$ or bounds a disk on $F$ across which it can be removed. Both alternatives contradict minimality, so $F\cap S$ consists only of the single arc joining the two points of $S\cap \partial F$.

Cutting $F$ along this arc gives Seifert surfaces $F_1,F_2$ for $K,K^{\prime }$. Their Euler characteristics satisfywhich, since all three surfaces have one boundary component, is equivalent toThis proves additivity.

The torus knot $T_{2,3}$ bounds a once-punctured torus, and its degree-two Alexander polynomial of a knot forces every Seifert surface to have genus at least one. Thus $g_s(T_{2,3})=1$. If it were a composite knot, both nontrivial summands would have positive Seifert genus, and additivity would give genus at least two. Hence $T_{2,3}$ is a prime knot.

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.