Short answer: Yes, there are many knobs to turn to decrease the density of a metal.
Since you've precluded a transition to a foam, you're probably implying that the mass $m$ remains constant. Since $\rho=m/V$, where $\rho$ is the density and $V$ is the volume, you're really asking about scenarios in which $$\left(\frac{\partial V}{\partial X}\right)_N>0\tag{1}$$
which corresponds to an increase in volume by changing some parameter $X$ at a constant mass or number of metal molecules $N$.
Fortunately for you, this inequality is very common; that is, there are many physical parameters to tweak to obtain a volumetric change. The mathematical framework here is $$dG=-S\,dT+V\,dp+\sigma\,dA+E\,dP+B\,dM+\cdots\tag{2}$$
for a closed system, where $G$ is a flavor of energy (specifically, the Gibbs free energy), $S$ is the entropy, $T$ is the temperature, $p$ is the pressure, $\sigma$ is the surface tension, $A$ is the surface area, $E$ is the electric field, $P$ is the polarization, $B$ is the magnetic field strength, and $M$ is the magnetic moment. For different geometries and conditions, you could replace $-p\,dV$ with $f dL$ (where $f$ is a force and $L$ is the length of a bar or rod) or $s\,d(V\epsilon)$ (where $s$ is the stress and $V\epsilon$ is the volumetric strain).
What this equation says is that there are various ways to add energy to a system and that these mechanisms will generally affect the volume as well. Since $V=(\partial G/\partial p)_{X\neq p}$ by inspection of (2), inequality (1) becomes
$$\left(\frac{\partial^2 G}{\partial p\,\partial X}\right)>0\tag{1a}$$ which is an important result: Material properties are second derivatives of energies. As you've already noted, we can increase the volume (of a metal, at least) by increasing its temperature. The material property here is the volumetric coefficient of linear expansion $\alpha_V$, where $$\alpha_V=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p=\frac{1}{V}\left(\frac{\partial^2 G}{\partial p\,\partial T}\right)>0\tag{1b}$$ where the inequality holds for metals, at least. (Some materials may shrink upon heating.)
Another material property you might be interested in is the reciprocal of the bulk modulus $\kappa$, corresponding to the volumetric increase per unit stress $s$ at constant temperature: $$\frac{1}{\kappa}=\frac{1}{V}\left(\frac{\partial V}{\partial s}\right)_T=\frac{1}{V}\left(\frac{\partial^2 G}{\partial p\,\partial s}\right)>0\tag{1c}$$ The simple 1-D case is to stretch a rod or bar of metal, which increases the volume and decreases the density, as you asked about. Here, the inequality holds for all stable solids for the 1D case and all stable materials for the 3D case.
Still another case is the piezoelectric coefficient, corresponding to the volumetric change when an electric field $E$ is applied: $$\frac{1}{V}\left(\frac{\partial V}{\partial E}\right)_T=\frac{1}{V}\left(\frac{\partial^2 G}{\partial p\,\partial E}\right)\tag{1d}$$
Still another is the magnetostrictive effect that alters a material's volume in response to a magnetic field $B$: $$\frac{1}{V}\left(\frac{\partial V}{\partial B}\right)_T=\frac{1}{V}\left(\frac{\partial^2 G}{\partial p\,\partial B}\right)\tag{1e}$$
You get the idea. Essentially, there are a very large number of ways to change a metal's density because these second derivatives of energy, corresponding to material properties related to changes in volume, are generally nonzero.
Knowing this, you could take your chunk of metal to a materials scientist and say, "Please determine the coefficient of volumetric expansion of this material in response to various fields—temperature, pressure, electric, magnetic, the works." to determine how you would need to change such fields to obtain your desired decrease in density.
