Quoting p. 142 of the trusty Ashcroft-Mermin (who write $\mathcal E_F$ for your $\mu$):
For each partially filled band there will be a surface in $k$-space separating the occupied from the unoccupied levels. The set of all such surfaces is known as the Fermi surface, and is the generalization to Bloch electrons of the free electron Fermi sphere. The parts of the Fermi surface arising from individual partially filled bands are known as the branches of the Fermi surface. We shall see (Chapter 12) that a solid has metallic properties provided that a Fermi surface exists.
Analytically, the branch of the Fermi surface in the $n$th band is that surface in $k$-space (if there is one) determined by $$\mathcal E_n(\mathbf k)=\mathcal E_F.$$
As to "what $\mathcal E_n(\mathbf k)$ is": it arises from looking for eigenfunctions of a "single-electron" hamiltonian $$ \left(-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf r)\right)\psi(\mathbf r)=E\psi(\mathbf r) $$ in the "Bloch" form: $\psi(\mathbf r)=e^{i\mathbf k\cdot\mathbf r}u(\mathbf r)$ with (Bravais lattice) periodic boundary conditions on $u$, i.e. $$ \left(\frac{\hbar^2}{2m}\left(\frac1i\nabla + \mathbf k\right)^2+U(\mathbf r)\right)u_{\mathbf k}(\mathbf r)=\mathcal E(\mathbf k)u_{\mathbf k}(\mathbf r). $$
Because of the periodic boundary condition we can regard [this] as a Hermitian eigenvalue problem restricted to a single primitive cell of the crystal. Because the eigenvalue problem is set in a fixed finite volume, we expect on general grounds to find an infinite family of solutions with discretely spaced eigenvalues, which we label with the band index $n$.