$\textbf{Context:}$
I'm posting in Math SE instead of Physics SE because this is really a question about differential geometry and vector spaces.
Let $\mathcal{H}$ be a Hilbert Space, then $|\psi\rangle \in \mathcal{H}$ is a vector that describes a quantum mechanical state. Let $H$ be an operator such that $H |\psi \rangle = i \hbar \partial_t |\psi \rangle$. In physics, when there is dissipation it is often useful to switch to a density matrix instead of a wavefunction representation.
$\rho = \sum_i P_i |\psi\rangle_i\langle\psi|_i$ where $P_i$ is the classical probability to find the $i$th state, and $\sum_i P_i = 1$. We'll restrict ourselves to a pure state - this doesn't matter for my question but it cleans up notation, so now $\rho = |\psi\rangle \langle\psi|$. Clearly, $\rho \in \mathcal{H}\otimes\mathcal{H^*}$ because it will act left on a vector, and right on a covector to return a complex number. We can find the time evolution as follows:
$\partial_t\rho = \partial_t(|\psi\rangle \langle\psi|) = (\partial_t|\psi\rangle ) \langle\psi|+|\psi\rangle ( \partial_t \langle\psi|)= \frac{1}{i \hbar} H |\psi\rangle \langle\psi|+|\psi\rangle \langle\psi|\frac{1}{-i \hbar} H = \frac{1}{i \hbar}[H,\rho]$
where $[A,B] = A B - B A$ is defined as the commutator.
$\textbf{Question:}$
Often times, we will define the Liouville Operator as follows:
$ \hat{\mathcal{L}} \rho=\frac{1}{i \hbar}[H,\rho]$, $ \hat{\mathcal{L}} = \frac{1}{i \hbar}[H,\cdot]$, or equivalently as the time derivative operator.
So this operator takes a (1,1) tensor, $\rho$, and gives us back a new (1,1) tensor. $\textbf{Does that mean $\hat{\mathcal{L}}$ could be understood as a (2,2) tensor?}$ It does not seem like it is also a (1,1) tensor in $\mathcal{H}\otimes\mathcal{H^*}$. $\textbf{Is it another object entirely that can't be understood with just a vector space structure?}$