Riemannian Fisher-Rao metric and orthogonal parameter space.

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Let $\mathcal{S}$ be a family of probability distributions $\mathcal{P}$ of random variable $\beta$ which is smoothly parametrized by a finite number of real parameters, i.e., $\mathcal{S}=\left\{\mathcal{P}_{\theta}=w(\beta;\theta);\theta \in \mathbb{R}^{n}, \theta=(\theta^{i})\right\}$. The statistical model $\mathcal{S}$ carries the structure of smooth Riemannian manifold $\mathcal{M}$, with respect to which $\theta=(\theta^{i})$ play the role of coordinates of a point $\mathcal{P}_{\theta}\in \mathcal{S}$, and whose metric is defined by the Fisher's information matrix $\mbox{H}=(g_{ij}(\theta))$, where the coefficients of this matrix, which yields a positive definite metric, are calculated as the expectation of a product involving partial derivatives of the logarithm of the probability density function's (PDF)

\begin{equation*} g_{ij}(\theta)=\int^{+\infty}_{-\infty} \displaystyle \frac{\partial^{2}ln \left( w(\beta;\theta)\right)}{\partial \theta^{i} \partial \theta^{j}}w(\beta;\theta)d\beta. \end{equation*}

How do we neglect the off-diagonal terms $g_{12}$=$g_{21}$?

In other words, is there a mathematical argument, wherein it is possible to consider $g_{12}=g_{21}$=0?

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You can multiply this whole integral by the Kronecker delta $\delta^{i}_{j}$. Then, in this way, you will have zero for elements with different indices (off-diagonal elements).