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A knot can be defined as an embedded circle in $3$-dimensional Euclidean space or in the $3$-sphere $S^3$. There is also a notion of a knot in higher dimensions: an $n$-knot is an embedding of the $n$-sphere into $m$-dimensional Euclidean space where $m>n$.

For $k>3$, is an embedded circle in $k$-dimensional Euclidean space a knot?

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I think what you meant to ask is whether knots can be nontrivial in dimension greater than $3$, and the answer is no: every knot in $\mathbb{R}^d, d \ge 4$ is trivial. – Qiaochu Yuan Aug 26 '12 at 4:49

No. Every embedded circle in $\mathbb{R}^k$ for $k\geq 4$ is equivalent to the unknot.

The way to see this intuitively is to imagine two beads on a string. We cannot move those two beads past each other as they cannot pass through each other. However, if those beads now exist on a table top. We can easily move a bead around any other bead on the table while remaining on the table. By passing from 1 dimension to 2 dimensions, we have new freedoms of movement.

A similar situation occurs with embedded circles. In three dimensions, we have knots because we have segments of the embedded circle that cannot be moved around other segments. In higher dimensions, we may now pass the segments around each other because of the freedom of movement.

Note: The analogy breaks down in higher dimensions, ie there are smoothly knotted 3-spheres in the 6-spheres, it is not enough to simply have 3 extra dimensions.

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