In answer to a question of Long, Flapan constructed an example of a prime strongly positive amphicheiral knot that is not slice. Long had proved that all such knots are algebraically slice. Here we show that the concordance group of algebraically slice knots contains an infinitely generated free subgroup that is generated by prime strongly positive amphicheiral knots. A simple corollary of this result is the existence of positive amphicheiral knots that are not of order two in concordance.
We define a nontrivial mod 2 valued additive concordance invariant defined on the torsion subgroup of the knot concordance group using involutive knot Floer package. For knots not contained in its kernel, we prove that their iterated $(\text{odd},1)$-cables have infinite order in the concordance group and, among them, infinitely many are linearly independent. Furthermore, by taking $(2,1)$-cables of the aforementioned knots, we present an infinite family of knots which are strongly rationally slice but not slice.
A knot in $S^3$ is rationally slice if it bounds a disk in a rational homology ball. We give an infinite family of rationally slice knots that are linearly independent in the knot concordance group. In particular, our examples are all infinite order. All previously known examples of rationally slice knots were order two.
Obstructing 4-torsion in the classical knot concordance group
本文证明了：如果3-球面中纽结K的2-重分支覆盖的第一同调的阶由pm给出，其中p是3 mod 4的素同余且gcd（p，m）=1，则K在纽结和谐群中是无限阶的.这为4阶的经典纽结提供了障碍。特别是，有11个素数结与10个或更少的交叉是阶4的代数和谐组;所有的和谐是无限阶。另一个推论是，任何亚历山大多项式为5t^2 - 11 t + 5的纽结都是和谐的无穷阶;莱文证明了在高维中所有这样的纽结都是4阶的。
设K是具有2重分支覆盖空间M的3-球面中的纽结。若对某个同余于3 mod 4的素数p，M的第一同调中的p-挠是奇指数循环的，则K在纽结和谐群中是无限阶的.作为一个应用，回想一下，一个任意纽结的n-扭曲的双重在Levine的代数和谐群中有4阶当且仅当n是正的，并且一些素同余到3 mod 4在4 n +1中有奇指数;我们证明了所有这样的纽结在纽结和谐群中都是无限阶的。作为第二个应用，如果某个3 mod 4的素同余在r中有奇指数，则2-桥纽结K（r，s）在纽结和谐群中有无穷阶。
We give an infinite family of knots that are not rationally concordant to their reverses. More precisely, if R denotes the involution of the rational knot concordance group QC induced by string reversal and Fix(R) denotes the subgroup of knots fixed under R in QC, then QC/Fix(R) contains an infinite rank subgroup. As a corollary, we show that there exists a knot K such that for every pair of coprime integers p and q, the (p,q)-cable of K is not concordant to the reverse of the (p,q)-cable of K.
We show that the smooth homotopy 4-sphere obtained by Gluck twisting the m-twist n-roll spin of any unknotting number one knot is diffeomorphic to the standard 4-sphere, for any pair of integers (m,n). It follows as a corollary that an infinite collection of twisted doubles of Gompf's infinite order corks are standard.
For an L-space knot, the formal semigroup is defined from its Alexander polynomial. It is not necessarily a semigroup. That is, it may not be closed under addition. There exists an infinite family of hyperbolic L-space knots whose formal semigroups are semigroups generated by three elements. In this paper, we give the first infinite family of hyperbolic L-space knots whose formal semigroups are semigroups generated by five elements.
Two knots are homology concordant if they are smoothly concordant in a homology cobordism. The group $\hat{\mathcal{C}}_{\mathbb{Z}}$ (resp. $\mathcal{C}_{\mathbb{Z}}$) was previously defined as the set of knots in homology spheres that bounds homology balls (resp. in $S^3$), modulo homology concordance. We prove $\hat{\mathcal{C}}_{\mathbb{Z}} / \mathcal{C}_{\mathbb{Z}}$ contains a $\mathbb{Z}^{\infty}$ subgroup. We construct our family of examples by applying the filtered mapping cone formula to $L$--space knots, and prove linear independence with the help of the connected knot complex.
