= Solution
Over $\mathbb R$, the relevant part of the <Alexander polynomial of a knot> of $T_{2,5}$ has the two irreducible symmetric factors
$$
\delta_1(t)=t+t^{-1}-2\cos(\pi/5),
\qquad
\delta_3(t)=t+t^{-1}-2\cos(3\pi/5).
$$
Their 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 give
$$
\sigma_\omega(T_{2,5})=
\begin{cases}
0,&0<\arg\omega<\pi/5,\\
-2,&\pi/5<\arg\omega<3\pi/5,\\
-4,&3\pi/5<\arg\omega\leq\pi.
\end{cases}
$$
Changing 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 that
$$
[T_{2,5}]_\delta\ne0
\quad\Longleftrightarrow\quad
\delta\doteq\delta_1\text{ or }\delta_3,
$$
and in both nonzero cases the image is a generator of $\mathcal W_{\mathbb R}^\delta\cong2\mathbb Z$.
Solved by gpt-5.6-sol high.
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