unknotted $n$-dimensional knot

Question

Let $n$ be any integer. An $n$-dimensional knot is an $n$-dimensional manifold embedded smoothly into $\mathbb{R}^{n+2}$. If it is homeomorphic to a disjoint union of $n$-spheres, then it is denoted by $S^n$-knot. The $2$-dimensional knot is called surface-knot. I am asking how to define trivial or unknotted $n$-dimensional knot in general. I know how to define for special cases. For example, $S^n$-knot is called unknotted if it is obtained from some standard $n$-spheres in $(n+2)$-space. In addition a surface-knot is trivial if it is obtained from some disjoint standard surfaces in 4-space by taking a connected sum. Suppose the case where we have n-dim knot which is not homeomorphich to n-sphere and $n \neq 2$ then how to define unknotted type in this case?

Answer 1

An oriented $n-$dimensional manifold embedded in $S^{n+2}$ is unknotted if it is homeomorphic to an embedding of that manifold in $S^{n+1}\times \{0\}$.

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