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Let $H$ be the upper half-plane in $\mathbb{R}^2$. How does the following expression $$ds^2= \frac{dx^2+dy^2}{y^2}$$ specify a Riemannian metric on $H$?

I don't understand what the expression means. If the $y^2$ wasn't there, then $g=dx^2+dy^2=dx\otimes dx+dy\otimes dy$ is the standard metric. This does make sense to me: at every point it associates the standard inner product. But the above expression is not even symmetric in $x,y$ so it isn't meant to be read the same way, but I don't know how to make sense of it.

Bruno Stonek
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  • Given a tangent vector based at a point, square the first component of the tangent vector, square the second component of the tangent vector, add them up and then divide by the square of the second coordinate. – Fly by Night Feb 04 '13 at 20:20
  • @FlybyNight: the square of the second coordinate -of the point-, if I understood the answer below correctly. This was my confusion; I thought I had to divide by the second coordinate -of the vector-. – Bruno Stonek Feb 04 '13 at 20:22
  • Exactly. That's why I use the words "component" and "coordinate". Components refer to vectors while coordinates refer to points. – Fly by Night Feb 04 '13 at 20:23
  • Ah! I hadn't noticed the different terminology. I wasn't aware of the distinction. – Bruno Stonek Feb 04 '13 at 20:24
  • Also, notice that the $x$ and $y$ denote the coordinates of the point $(x,y)$ while the $dx$ and $dy$ are differential forms which give the first and second components of a tangent vector as their output. – Fly by Night Feb 04 '13 at 20:27

3 Answers3

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A "fundamental form" $$ds^2=E(x,y)\>dx^2+2F(x,y)\>dx\>dy+G(x,y)\>dy^2\ ,$$ in your case $$ds^2={dx^2+dy^2\over y^2}\ ,$$ tells you how to measure the length of a curve $\gamma$ given in the form $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$ The rule is: Write formally $$dx=\dot x(t)\ dt,\quad dy=\dot y(t)\ dt$$ and compute $$L(\gamma)=\int_a^b\sqrt{E\bigl(x(t),y(t)\bigr)\dot x^2(t)+2F\bigl(x(t),y(t)\bigr)\dot x(t)\dot y(t)+G\bigl(x(t),y(t)\bigr)\dot y^2(t)}\ dt\ ,$$ in your case $$L(\gamma)=\int_a^b{\sqrt{\dot x^2(t)+\dot y^2(t)}\over y(t)}\ dt\ .$$ The intuitive idea is the following: For reasons particular to the problem at hand the length of an "infinitesimal segment" connecting the points $(x,y)$ and $(x+dx,y+dy)$ should not be the usual pythagorean $\sqrt{dx^2+dy^2}$; instead it should depend in a still "natural", but more complicated way on the vector $(dx,dy)$. In your case the length of this "infinitesimal segment" should be its euclidean length, divided by the $y$-coordinate of the point where it is located.

Christian Blatter
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As you note, $dx^2 + dy^2$ means $dx \otimes dx + dy \otimes dy$. The expression means exactly what it says: you divide this tensor by $y^2$. It may help to write it as

$$ \frac{1}{y^2} dx \otimes dx + \frac{1}{y^2} dy \otimes dy $$

While a metric needs to be symmetric, there is no reason it needs to be symmetric under swapping $x$ and $y$. Let me add color to help you through your confusion: the tensor we define is

$$ g = \frac{1}{y^2} \color{blue}{dx} \otimes \color{red}{dx} + \frac{1}{y^2} \color{blue}{dy} \otimes \color{red}{dy} $$

The requirement that $g$ be symmetric means that $g$ needs to be equal to

$$ \frac{1}{y^2} \color{red}{dx} \otimes \color{blue}{dx} + \frac{1}{y^2} \color{red}{dy} \otimes \color{blue}{dy} $$

and it is, because the color is just notation and isn't part of the metric.

An example of a tensor that isn't symmetric is $dx \otimes dy$.

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    Ok, I think I understand. The $\frac{1}{y^2}$ factor is just a constant, in the sense that if $p=(x,y)$, then in the expression for $g_p$ it will be just a constant factor. This makes sense here because we are taking the identity on $H$ as a paremetrization. Am I right? – Bruno Stonek Feb 04 '13 at 20:16
  • the term $\frac{1}{y^2}$ means that at every point $(x_0,y_0)\in H$ you have to divide the standard (Euclidean) metric by $\frac{1}{y_0^2}$. Then, excpet that for points of kind $(x_0, 1)$ or $(x_0, -1)$, you'll have another scalar product – AX.J Feb 04 '13 at 20:19
  • @BrunoStonek: at each point, yes, the $\frac{1}{y^2}$ is constant, and this just says that, infinitesimally, the hyperbolic plane is the same as the real plane. But when you actually use this Riemannian metric to measure length, or area, the $\frac{1}{y^2}$ will not be constant, but part of your integral. –  Feb 04 '13 at 20:20
  • @BrunoStonek Intuitively, the $\frac{1}{y^2}$ factor means that distances expand as you get closer and closer to the $x$-axis ($y=0$); the axis is infinitely far away from any point in the upper half-plane! – Steven Stadnicki Feb 04 '13 at 20:22
  • Why call it "$ds^2$" and not just $g$? – Bruno Stonek Feb 04 '13 at 20:29
  • @BrunoStonek Because it measures (squared) infinitesimal distances; the standard (2d) Euclidean metric $ds^2=dx^2+dy^2$ is stating an infinitesimal version of the Pythagorean theorem! – Steven Stadnicki Feb 04 '13 at 20:31
  • @StevenStadnicki: I don't see what that means; I don't understand that "infinitesimal" thing. Could you explain? Thanks. – Bruno Stonek Feb 04 '13 at 20:36
  • $ds^2$ is a symmetric bilinear form on tangent vectors; which is exactly what you're calling $g$ is. –  Feb 04 '13 at 20:38
  • Hurkyl I'm (also) kind of struggling with this stuff. I know that differential forms eat vector fields to yield scalarfields. What does $(dx \otimes dy)$ eat, and does $(dx \otimes dy)$ equal $(dy \otimes dx)$? Its very hard to find anywhere that explicit definitions are truly given. – goblin GONE May 29 '15 at 03:11
  • For instance, does $(dx \otimes dy)(\frac{\partial}{\partial x},\frac{\partial}{\partial y})$ make sense? Does $(dx \otimes dy)(\frac{\partial}{\partial x} \otimes \frac{\partial}{\partial y})$? – goblin GONE May 29 '15 at 03:15
  • @goblin: Let $u,v$ be vectors and $\omega, \xi$ be differential forms. When people speak like you do, they usually have in mind thinking of $\omega \otimes \xi$ as the bilinear function $(\omega \otimes \xi)(u, v) = \omega(u) \xi(v)$. There are, of course, lots of other things you can do with $\omega \otimes \xi$, such as define a linear functional on tensors of vectors satisfying $(\omega \otimes \xi)(u \otimes v) = \omega(u) \xi(v)$, or linearly transforming vectors into forms by $(\omega \otimes \xi)(u) = \omega(u) \xi$, or instead by $(\omega \otimes \xi)(u) = \xi(u) \omega$. –  May 29 '15 at 15:41
  • @Hurkyl, thanks. – goblin GONE May 30 '15 at 03:25
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It can help to fall back on an extrinsic point of view to get some intuition.

Let $S$ be a Riemannian manifold of dimension 2, smoothly embedded in $\mathbb R^n$. We can associate with each point in $S$ a vector in $\mathbb R^n$. Of course, $\mathbb R^n$ has an inner product, but we naturally do not expect the inner product of two points in $S$ to tell us anything about the geometry of $S$.

When you introduce a pair of coordinates to parameterize $S$, you can talk about the tangent space. Let $s$ be the position of some point in $S$, then the vectors in the tangent space taken on the form

$$e_x = \frac{\partial s}{\partial x}, \quad e_y = \frac{\partial s}{\partial y}$$

Again, $s$ is a vector in $\mathbb R^n$, so these expressions have a sensible, concrete meaning.

The metric, then, merely tells us about the vectors in the tangent space and how they are related to each other. You can write it as

$$ds^2 = (e_x \cdot e_x) dx^2 + (e_y \cdot e_y) dy^2 + 2 (e_x \cdot e_y) dx \, dy$$

So the metric tells us whether the tangent space vectors are orthogonal to one another, and about their lengths. This is the basic extrinsic viewpoint; intrinsically, one has to work a little harder, but hopefully this is enough to gain some intuition.

Muphrid
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