Metric Tensor Grid

Question

Let, we are in a 2d metric where $g_{xx}=1, g_{yy}=x^2$, therefore $|e_x|=1$, $|e_y|=x$. If we try to draw the metric in a grid - it looks something like the image I uploaded. Note that, along the X axis the $e_x$ basis is always 1 unit long, but $e_y$ grows as we move along the X axis. And that causes a problem. $g_{xx}=1$, therefore a constant, and should not grow as we move along the Y axis. But as we can see, $g_{xx}$, which is the squared length of $e_x$ grows, $CD >AB$. What is the reason for this inconsistency? What am I missing? Use Christoffel Symbols in answer if needed.

Answer 1

There is no inconsistency. Perhaps only not a very smart way to define a metric.

You are just saying that the coordinate basis $e_x,e_y$ in this metric becomes non orthonormal (it is only orthogonal). The Christoffel symbols are formally the same as those for 2d polar coordinates where the metric is, as we know, $ds^2=dr^2+r^2\,d\theta^2\,.$ Therefore, $$ {\Gamma^x}_{yy}=-x\,,\quad {\Gamma^y}_{xy}={\Gamma^y}_{yx}=\frac{1}{x}\,. $$ This leads to a vanishing Riemann curvature tensor, as it must, because the 2D plane stays flat when we only change the coordinate system from Cartesian to polar.

What happens however is that the vector fields $e_x,xe_y$ no longer commute: move one step to the right from $x$ to $x+\Delta$ then one step up by the length $(x+\Delta)$ of the vertical basis vector at that point, then one step to the left and then one step down by the length $x$ of the vertical basis vector at that point. This loop will not be closed.

One of your sentences I found confusing: "But as we can see, $g_{xx}\,,$ which is the squared length of $e_x$ grows, $CD>AB\,.$" Can you clarify this?