There are a few good posts/notes (see here, and here) giving high level geometric intuition of a complete statistic ($E_{T}[g(T); \theta] = 0 \Rightarrow P(g(T)=0; \theta) = 1 \text{ almost everywhere}$). The geometric intuition can be illustrated by using a discrete example of a sufficient statistic: $$ \begin{align*} E_{T}[g(T); \theta] = \sum_{t} g(t)P(T=t; \theta) &= 0 \Rightarrow g(t) = 0\\ \Rightarrow \begin{bmatrix}g(t_1) \\ g(t_2) \\ \vdots \\ g(t_n)\end{bmatrix}^T\begin{bmatrix}p_\theta(t_1) \\ p_\theta(t_2) \\ \vdots \\ p_\theta(t_n)\end{bmatrix} &= 0 \Rightarrow g(t) = 0 \end{align*} $$ Which is equivalent to saying that if we have a linear map $L: P_\theta \rightarrow V$, such that $L\left(\begin{bmatrix}p_\theta(t_1) \\ p_\theta(t_2) \\ \vdots \\ p_\theta(t_n)\end{bmatrix} \right) = \sum_{i=1}^n g(t_i)p_\theta(t_i)$, the null space of $L$ is the $0$ vector.
My question is, what is $V$ in my notation, intuitively? What space are we mapping to when considering this expectation? Equivalently, when $T$ is completely sufficient, what is the functional space that the class of probability distributions is spanning?
