# Understanding the hyperbolic metric defined by $\langle\mathbf{x}_{u}, \mathbf{x}_{u}\rangle=\frac{1}{v^{2}}$, etc.

The problem statement of this HW problem is quoted below:

Problem $$7 .$$ Let $$\mathbb{H}=\{(u, v): v>0\}$$ be the hyperbolic plane and use $$\mathbf{x}(u, v)=(u, v)$$ as the trivial coordinates. Consider the hyperbolic metric $$\left\langle\mathbf{x}_{u}, \mathbf{x}_{u}\right\rangle=\frac{1}{v^{2}}, \quad\left\langle\mathbf{x}_{u}, \mathbf{x}_{v}\right\rangle= 0, \quad\left\langle\mathbf{x}_{v}, \mathbf{x}_{v}\right\rangle=\frac{1}{v^{2}}$$ (1) Compute the length of the curve $$\alpha:[0, a] \rightarrow$$ H given by $$\alpha(t)=(0,1-t) .$$ What happens with its length when $$t \rightarrow 1^{-} ?$$

The problem was that I could not understand how to use the metric as defined above. From my understanding $$\mathbf{x}_u$$ is simply $$(1, 0)$$. Therefore $$\left\langle\mathbf{x}_{u}, \mathbf{x}_{u}\right\rangle = \left\langle(1, 0), (1, 0)\right\rangle$$ but this is $$\frac{1}{v^2}$$ (assuming $$v$$ is the common $$v$$ coordinate the two tangent vectors share) which is not defined.

I guess what I want is $$\Vert\alpha'(t)\Vert$$ (for the line length), and this expands to $$\langle (-\sin (t),\cos (t)), (-\sin (t),\cos (t))\rangle$$ and I do not see how to expand this using the definition above.

Edit: I just realized what I wrote above for $$\alpha'(t)$$ makes no sense whatsoever. I have no idea how I got that :(.

Although $$\mathbf{x}_{u}=(1,0)$$ and $$\mathbf{x}_{v}=(0,1)$$ are the "constant" vectors, we think of both $$\mathbf{x}_{u}$$ and $$\mathbf{x}_{v}$$ as vector fields on the upper half plane. So for each $$(u, v)$$ on the upper half plane, there are the tangent vectors $$\mathbf{x}_{u}$$ and $$\mathbf{x}_{v}$$ at $$(u, v)$$.
In particular, you can define the length of $$\mathbf{x}_{u}$$, $$\mathbf{x}_{v}$$ depending on the point $$(u, v)$$. The equation
$$\langle \mathbf{x}_{u}, \mathbf{x}_{u}\rangle = \frac{1}{v^2}$$
means the following: at the point $$(u, v)$$, we define the length (square) of $$\mathbf{x}_{u}$$ at that point $$(u, v)$$ to be $$1/v^2$$. For example, at the point $$(2, 3)$$, the length (square) of $$\mathbf{x}_{u}$$ is
$$\langle \mathbf{x}_{u}, \mathbf{x}_{u}\rangle = \frac{1}{3^2} = \frac 19.$$
Back to your question, $$\alpha (t) = (0,1-t)$$. So $$\alpha'(t) = (0,-1) = -\mathbf{x}_{v}$$ (again we do not think of $$\alpha'(t)$$ as a constant vector, but a vector fields along $$\alpha$$). Thus for each $$t$$, we have $$\|\alpha'(t)\|^2 = \| -\mathbf{x}_{v}\|^2_{\alpha(t)} = \frac{1}{(1-t)^2}.$$