Isotopic tori in $\mathbb{R}^4$ Intuitively it seems to me that two tori in $\mathbb{R}^4$ are isotopic to each other. By isotopic, I mean a smooth family of deformations beginning in one and ending in the other, and each member of the family being diffeomorphic. But I cannot write down a rigorous proof of the above "fact". Any help will be highly appreciated. Thanks in advance!
 A: This is not true. You can have knotted $T^2 \subset \mathbb{R}^4$ and you can even have knotted $S^2 \subset \mathbb{R}^4$. What is true, and perhaps is how you were visualizing this, is that any $S^1 \subset \mathbb{R}^4$ is unknotted (i.e. any two are isotopic).
Knotted surfaces in $\mathbb{R}^4$:
Given any knot $K \subset \mathbb{R}^3$ we can make a knotted $2$-torus $T_K \subset \mathbb{R}^4$ whose complement has the same fundamental group as that of the knot, i.e. with $\pi_1(\mathbb{R}^3-K) \cong \pi_1(\mathbb{R}^4-T_K)$. To do this, simply spin the knot $K$ around an axis in $\mathbb{R}^4$. More precisely, put the knot $K$ in a half-space $\mathbb{R_+}\times \mathbb{R}^2$, embed that in $\mathbb{R}^4$ as $\{0\}\times\mathbb{R_+}\times \mathbb{R}^2$, and revolve $K$ around the axis $\{0\}\times\{0\}\times \mathbb{R}^2$. We get the equality of fundamental groups from the Seifert-van-Kampen theorem.
There's a similar construction to make knotted $2$-spheres in $\mathbb{R}^4$ in which we put the knot $K$ in $\mathbb{R}_{\geq 0} \times \mathbb{R}^2$, open up the knot and place the two ends on the boundary $\{0\}\times\mathbb{R}^2$ before revolving in $\mathbb{R}^4$ to get a $2$-sphere in $\mathbb{R}^4$ we'll call $S_K$. This gives what's called a spun knot, and we again have $\pi_1(\mathbb{R}^4-S_K) \cong \pi_1(\mathbb{R}^3-K)$ by Seifert-van-Kampen.
This gives many knotted tori and spheres in $\mathbb{R}^4$, as any isotopic pair of submanifolds would have their complements have the same fundamental group.
Circles in $\mathbb{R}^4$ are unknotted:
Generally, two homotopic embeddings of a compact $k$-manifold into an $n$-manifold are isotopic if $2k+1 < n$. The basic idea is to homotop one to the other, and if at any point in the homotopy the map isn't an embedding, you'll be able to fix it by almost any tiny perturbation. (Think of two orthogonal lines in $\mathbb{R}^3$ sliding toward each other and crossing. If they were in $\mathbb{R}^4$ we could've dipped one into the $4$-th dimension to miss the other at the crossing point.) The subject to look up for the details of this type of technique is transversality.
