Visualization of $ dtdx$ and $dxdy$ term in metric tensor For the sake of simplicity, lets take a 2+1 dimensional spacetime.
Lets take the metric 
$$ds^2 = g_{tt}dt^2 + g_{xx}dx^2 + g_{yy}dy^2 + g_{tx}dtdx + g_{xy}dxdy$$ 
What is the visualization or physical interpretation of the $g_{tx}$ and $g_{xy}$ terms of the metric? Does $g_{tx}$ mean motion of space i.e. on object in this spacetime point will be moving w.r.t an observer at infinity? What would $g_{xy}$ mean? 
 A: I am thinking of this possible answer motivated by the comments of Solenodon Paradoxus.
Solenodon had mentioned that locally its possible to diagonalize the metric. I am wondering if this can be generalized, leading to an interpretation of the off-diagonal elements.
Since, the metric is a symmetric square matrix, it should be always be possible to diagonalize it (even globally) by determining the eigenvalues and eigenvectors. Let $\Lambda$ be the eigenvalue diagonal matrix and $V$ be the eigenvector matrix. Then
$g_{\mu\nu} = V^{\dagger}\Lambda V$,
where, V and $\Lambda$ may in general be complicated non-linear functions of $x$ and $t$ (depending upon the original metric $g_{\mu\nu}$).
Now, the above can be interpreted as $\Lambda$ being the diagonal metric tensor of the modified coordinate system where 
$\begin{bmatrix} dt'\\dx' \end{bmatrix} = [V] \begin{bmatrix} dt\\dx \end{bmatrix}$ 
Since $V$ can be any function of $x$ and $t$, $dx'$ and $dt'$ and hence, $x'$ and $t'$ can also be in general non-linear functions of $t$ and $x$. i.e. $t' = f(x,t)$ and $x' = h(x,t)$.
${\bf Interpretion~ of~ the~ non-diagonal~elements~of~the~metric~tensor~g_{\mu\nu}}$:
We now come to the interpretation part.
$x' = h(x,t)$, means that the co-ordinate axis $x'$ is moving/changing with time $t$. The co-ordinate axis are dynamic. A person sitting in a car at point $x'$ with his engine off, would still be moving w.r.t. and observer at infinity, based on the $t$ dependence of $x'$.
If however, $g_{xt}$ were to be zero, $x'$ would not be a function of time $t$. Meaning if $g_{xt}$ were zero, $x'$ would be curved but stationary/static.
Thus $g_{xt}$ leads to a co-ordinate axis which is changing dynamically.
Please let me know if anyone sees an issue with this solution, of if this solution is incorrect.
A: First of all, your guess seems well-founded -- the metric tensor is closely related to energy and momentum. In fact, the relation may be stated as
\begin{equation}
T^{\mu\nu}={\delta S\over \delta g_{\mu\nu}},
\end{equation}
where $S$ is some action, and $T^{\mu\nu}$ is the stress-energy tensor. For brevity, let us consider a perfect fluid that permeates all of spacetime -- this describes some mass/energy density. This is tied to the amount of matter and (and maybe photons) you have in this physical system.
Component-wise, we may write the stress-energy tensor as
\begin{equation}
\left[T^{\mu\nu}\right]=
\begin{pmatrix}
E&p^1&p^2\\
p^1&P^1&\tau\\
p^2&\tau&P^2
\end{pmatrix},
\end{equation}
where $E$ and $p^i$ are the energy and momentum densities respectively. $P^i$ describes stress along the $i$-th direction, and $\tau$ is the shear stress; these quantities are defined in analogy to components of the usual stress tensor seen in fluid mechanics. 
So $g_{tx}=g_{01}$ and $g_{xt}=g_{10}$ are related to momentum in the $x$-direction in this way. Likewise, $g_{tt}=g_{00}$ is related to energy. $g_{xy}=g_{12}$ would then be tied to the shear stress of the perfect fluid we considered earlier. 
For more information on the stress-energy tensor, you may look at this post.
I don't know enough GR to comment on how frames of reference come in though.
A: I dont know much physics but mathematically off diagonals elements on the metric tensor implies that your local coordinate system is not orthonormal. So i guess that can be interpretted as how much the spacetime looks squished from your perspective
A: Your intiuition isn't especially crazy, but one might be careful about taking it too literally.  Despite this, here is a coordinate transform that gives you a nice version of this:
Take the flat Robertson-Walker metric in standard form
$$ds^{2} = -dt^{2} + a(t)^{2}\left(dr^{2} + r^{2} d\Omega^{2}\right)$$
Now, make the coordinate transformation, $R = a\,r$, which gives $dr = \frac{dR}{a} - \frac{R{\dot a}\,dt}{a^{2}}$ 
This turns the metric into the form:
$$ds^{2} = -\left(1 - H^{2}R^{2}\right)dt^{2} - R\,H\left(2dt\,dr\right) + dR^{2} + R^{2}d\Omega^{2}$$
Where $H = \frac{\dot a}{a}$ is the Hubble "constant".  If you remember Hubble's law, it gives precisely that the recession velocity of a distant galaxy is given by $HR$, which is what $g_{tr}$ is in these coordinates (which also have the nice property of having constant-time spatial sections, so this is the coordinate system where your local rulers don't change with respect to time)
