it follows that $\widehat{L}_{\varepsilon}(f)$ is odd for any embedding of $K_7$.

\[k7 chiral\] $K_7$ is intrinsically chiral.

For the sake of contradiction, suppose that for some embedding $f$ of $K_7$ there is an orientation reversing homeomorphism $h$ of the pair (${S}^3$, $f(K_7)$). Let $\alpha$ denote the automorphism of $K_7$ that is induced by $h$.

Let $J$ denote the set of Hamiltonian cycles in $f(K_7)$ with non-zero Arf invariant. Since any homeomorphism of ${S}^3$ preserves the Arf invariant of a knot, the homeomorphism $h$ permutes the elements of $J$. It follows from Conway and Gordon [@Conway; @1983] that $|J|$ must be odd, and hence there is an orbit $O$ in $J$ such that $|O| = n$ for some odd number $n$. Consequently, $h^n$ setwise fixes an element of $O$. Hence some Hamiltonian cycle $C$ with non-zero Arf invariant is setwise fixed by $h^n$. We now label and orient the edges of $K_7$ as in Figure \[FigureK7\_labeled\] so that $f$ takes the 0-star of $K_7$ to $C$. Since $h^n$ leaves $C$ setwise invariant, the automorphism $\alpha^ n$ (induced on $K_7$ by $h^n$) leaves the 0-star, 1-star, and 2-star all setwise invariant.

Fix a sphere of projection $P$ in $S^3$. Since $f\circ \alpha^n(K_7)$ 