1D topological defects in $d>3$ spatial dimensions

Question

I am trying to construct a 1D topological defect solution in 4 spatial dimensions, i.e., a solution to some PDE (likely the equations of motion of some Lagrangian) on $\mathbb{R}^{4}$ which is homotopically distinct from the vacuum solution.

In particular I want a 1D defect, but I don't know how to express the condition for this in $d>3$ spatial dimensions.

Example in $d=3$

In $d=3$ dimensions the homotopy classification of defects says that a Lagrangian associated to a continuous symmetry group $G$ which is spontaneously broken to a subgroup $H \subset G$ gives equations of motion which admit a 1D topological defect if the vacuum manifold $M = G / H$ has nontrivial first homotopy group $\pi_1(M)$.

A classic example in 3 spatial dimensions is the Lagrangian of the complex scalar field $\Phi$ given by

\begin{equation} \mathrm{L}=\left|\partial_\mu \Phi\right|^2-\frac{\lambda}{4}\left(|\Phi|^2-\eta^2\right)^2 \end{equation} which has a $U(1)$ symmetry broken down to $\{0\}$ ($\pi_1(M) = \mathbb{Z}$), and admits static 1D topological defects of the form \begin{equation} \Phi(\mathrm{x}, \mathrm{y})=\eta \,\mathrm{f}(\mathrm{m} \rho) \,\mathrm{e}^{\mathrm{in} \theta} \end{equation} with $\rho^2 = x^2 + y^2$, $n\in \mathbb{Z}$, and where $f$ is a solution to the equation of motion \begin{equation} \partial_\mu \partial^\mu \Phi=-\lambda\left(|\Phi|^2-\eta^2\right) \Phi / 2 \end{equation} with $f(0)=0$, and $f(\infty)=1$.

Intuitively this corresponds to a 1D defect given by the zero-locus of $\Phi$, the set of points $(x,y,z)$ with $x=y=0$.

Case $d=4$

My questions are thus:

1. How does the homotopy classification of defects generalize to $d>3$ spatial dimensions? Is $\pi_1(M)$ being non-trivial still the right condition?
2. Does the same Lagrangian as above, but extended to 4 spatial dimensions ($\Phi=\Phi(x,y,z,w)$), admit a 1D defect?
3. If not, is there a simple example which realizes 1D defect in 4 spatial examples?

Answer 1

1. No, if you want to have a 1D defect in $d$ spatial dimensions, you need $\pi_{d-2} (M)$ to be nontrivial.

2. The fact that your Lagrangian has nontrivial $\pi_{1} (M)$ implies, when extended to 4 spatial dimensions, that there are 2D defects.

AMENDED to address OP's follow-up question:

3. If you want a 1D defect in 4 spatial dimensions, you need to use a model that yields a nontrivial $\pi_{2} (M)$ (which would yield a 0D defect in 3 spatial dimensions, such as a magnetic monopole). Typically you'd use something with an internal $SU(2)$ symmetry, see for example this discussion.