Does the entangibility of density operators rely on what component spaces are being specified?

Is the entangibility of density operators relied on what component spaces are being specified?

More precisely, let $$H$$ be a Hilbert space, $$\rho$$ be a density operator on $$H$$. Suppose we were not given any information about $$H$$. i.e., it is possible that $$H=H_A\otimes H_B$$, or $$H=H_C\otimes H_D$$, and perhaps that $$\dim H_A\neq\dim H_C$$ and $$\dim H_B\neq\dim H_D$$. If $$\rho$$ is, say being entangled with respect to $$H_A$$ and $$H_B$$, is $$\rho$$ also necessarily entangled with respect to $$H_C$$ and $$H_D$$?

It absolutely depends on the subdivision of the spaces. Take the 3-qubit system (qubits A, B and C) in a state $$|0\rangle_A(|00\rangle+|11\rangle)_{BC}$$ We can partition these qubits in various different ways. Clearly the $$A|BC$$ partition has no entanglement across the partition, while the $$AB|C$$ partitioning is maximally entangled. Note that if you go to, for example, 4 qubits, you can construct similar examples where the dimensions of the Hilbert spaces are all the same, so that's not the issue.
Another interesting example is a single-photon state in a superposition of two different spatial (or any other type of degree of freedom really) modes: $$\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)|0\rangle\equiv \frac{1}{\sqrt2}(|1\rangle+|2\rangle).$$ This type of states is sometimes not considered "entangled", as there is only a single particle involved. However, it definitely is entangled with respect to the bipartition of the underlying space in the Fock spaces for the two spatial modes. To see it, you can rewrite the state as $$\frac{1}{\sqrt2}(|10\rangle + |01\rangle),$$ where now $$|nm\rangle$$ denotes a state with $$n$$ photons in the first spatial mode and $$m$$ photons in the second one. This state is entangled in the bipartition $$H_1\otimes H_2$$ with $$H_i$$ the (infinite-dimensional) Fock state associated to the $$i$$-th spatial mode.