Let
$$L_{e,m}^2 = L_{\text{loc}}^2([0,\infty),\mathbb{R}^m) = \left\{f:[0,\infty) \to \mathbb{R}^m\mid \int_0^T \|f(t)\|^2 dt < \infty \text{ for all } T > 0 \right\}$$
and suppose that $G:L_{e,m}^2 \to L_{e,l}^2$ is a linear operator that is time-invariant, i.e., $GS_\tau = S_\tau G$ for all $\tau > 0$, and causal i.e. $GP_Tf = P_TGP_Tf$ for all $T>0$, where $(S_\tau f)(t) = f(t+\tau)$ is the shift and $(P_T f)(t) = \mathbb{1}_{[0,T]}(t) f(t)$ is the cutoff projection.
It is well-known that if one has matrices $A \in \mathbb{R}^{n \times n}$, $B \in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{l \times n}$ and $D \in \mathbb{R}^{l \times m}$, then the mapping $u \mapsto y$ defined by $$ \dot{x} = Ax + Bu \\ y = Cx + Du \\ x(0) = 0 $$ defines such an operator $G$ (the state trajectory $x$ being absolutely continuous). In this case $G$ is said to have a finite-dimensional state space representation.
Question: Are there some explicit sufficient (and/or necessary) conditions on the operator $G$ to have a finite-dimensional spate space representation?
The delay $Gf = f(t-\tau)$ for some $\tau > 0$ is also causal and LTI, but does not a admit a representation as above. (It does however have an infinite-dimensional Hilbert state space representation).
Here is a beginning of an answer which I will update based on the comments.
A necessary and sufficient conditions for the operator $G$ to have a finite-dimensional state-space representation is that is a convolution operator with kernel $h(t)$ that can be expressed as $$h(t)=Ce^{At}B+D\delta(t)$$ for some real-valued matrices $A,B,C$ and $D$ of appropriate dimensions. This is equivalent to say that the kernel can be expressed as
$$h(t)=\sum_{i=1}^nP_i(t)e^{\lambda_it}+D\delta(t)$$
for some matrix polynomials $P_i(t)$ and complex scalars $\lambda_i$.
In the case of stable convolution operators with kernel in $L_1$ (with no impulse), the set of functions of the form $Ce^{At}B$ is dense in $L_1$.