On the upper half-plane with standard $x,y$ coordinates we have the hyperbolic metric given by $g_{11}=g_{22}=1/y^2$ and $g_{12}=0$.

In terms of the first fundamental form, we write $$ds^2=\frac{dx^2+dy^2}{y^2}= \frac{-4\,dz\,d\bar{z}}{(z-\bar{z})^2}.$$

I can interpret $dz$ and $d\bar{z}$ as (complex-valued) $1$-forms on the upper half plane.

Suppose I now want to show that a fractional linear transformation $z\mapsto z'$ where $z'=\frac{az+b}{cz+d}$ and $ad-bc=1$ gives an isometry.

By definition of an isometry, I would want to show that the pushforward of two tangent vectors in the domain yield the same result evaluated under the metric.

But when I see other people approach the problem, what they do is calculate $dz'$ and $d\bar{z'}$, and then prove that $$\frac{-4\,dz'\,d\bar{z'}}{(z'-\bar{z'})^2}=\frac{-4\,dz\,d\bar{z}}{(z-\bar{z})^2}.$$ I'm a little confused by what this calculation is actually doing. I would think that what I need to do is evaluate the same fundamental form (without the 's in the numerator) at the point $z'$ and $\bar{z'}$ on the pushforward of some tangent vectors in the domain. I am not sure why it is valid to simply substitute in $dz'$ and $\bar{dz'}$ and compute "the same thing", and along the way I seem to have lost any intuition of what $dz'$ and $d\bar{z'}$ really mean as $1$-forms.

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    $\begingroup$ Rather than pushing forward tangent vectors, it's far more efficient to pull back differential forms and covariant tensors. If $f(z)=z'$, we're computing the pullback of the $2$-tensor, i.e., $f^*(\frac{dz'\otimes d\bar z'}{(z'-\bar z')^2})$. $\endgroup$ Feb 6, 2022 at 5:18
  • $\begingroup$ @TedShifrin I'm confused about why we are computing that pullback. Shouldn't I be using $dz$ and $d\bar{z}$ for the fundamental form, since that's not changing? $\endgroup$
    – Vasting
    Feb 6, 2022 at 6:50
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    $\begingroup$ To put Ted's comment and Compacto's answer in practical (but literal, if extreme) terms, suppose you've measured a house (pair of tangent vectors) with a tape measure (metric). Now you want to measure another house, supposedly of the same blueprint, to see if it's the same size. You can either (i) Move the other house to where your tape measure is (push forward tangent vectors), or (ii) Move your tape measure to where the other house is (pull back the metric). $\endgroup$ Feb 6, 2022 at 14:51
  • $\begingroup$ @AndrewD.Hwang I guess I don't understand why the thing I'm supposed to pull back is just the original fundamental form but with $z'$ coordinates--is that like the "new" metric on the image? $\endgroup$
    – Vasting
    Feb 6, 2022 at 16:57
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    $\begingroup$ The philosophical content of compacto's answer is: A metric is a mathematical structure that "measures" tangent vectors. The expression of a metric in coordinates is merely a way of representing this structure, in a way that gives meaningful results when applied to tangent vectors in those same coordinates. One way to detect an isometry is to pair a single metric representation with vectors before and after the isometry; another is to pair a single pair of tangent vectors with the metric before and after pullback. $\endgroup$ Feb 6, 2022 at 21:41

1 Answer 1


As you know, the metric $g_p$ at each point $p$ of the space $M$ is just a bilinear form on tangent vectors at the point $p$, that is, a function that associates to each pair of vectors $X_p, Y_p$ their "generalized scalar product" $g_p(X_p, Y_p)$.

To prove that a map $T:M \rightarrow M$ is an isometry, you can do what you said and show that for each point $p$ and each pair $X_p, Y_p$ of tangent vectors at $p$, we have following identity:

$g_p(X_p, Y_p) = g_{T(p)}({T_{*,p}(X_p), T_{*,p}(Y_p)})$

where $T_{*,p}$ is the pushforward, that is, the map on tangent vectors induced by $T$. In terms of matrices of coordinates, what you are proving is that

$V^t G_p W = (AV)^t G_{T(p)} AW$

where $V, W$ are the column vectors formed with the coordinates of $X_p, Y_p$, $A$ is the matrix of the pushforward map $T_{*,p}$, and $G_q$ is the matrix of the metric at each point $q$.

Modifying this equation we find that is equivalent to

$V^t G_p W = V^tA^t G_{T(p)} AW = V^t(A^tG_{T(p)}A)W = V^tG'_p W$

And we just need to show the identity between the two matrices $G_p$ and $G_p' = A^t G_{T(p)} A$, which represent two metrics at the point $p$. One is $g_p$, our first fundamental form, and the other one is $g_p'$, defined by:

$g_p'(X_p,Y_p) = (AV)^t G_{T(p)} (AW) = g_{T(p)}({T_{*,p}(X_p), T_{*,p}(Y_p)})$

this metric is also called the pullback of $g$ by $T$, and denoted by $T^*g$. As it should be obvious by now, $T$ is an isometry if and only if $g = T^*g$ at each point, which is what you are encountering, because the pullback metric is obtained from the coordinate expression of $g$ simply by replacing each coordinate $z, \overline{z}$ (or $x, y$) with the same coordinate transformed by $T$, $z' = z\circ T, \overline{z}' = \overline{z}\circ T$ (or $x' = x\circ T, y' = y\circ T$).

If you think about this, you can discover that every multilinear form on tangent vectors is transformed by each map $T$ via some definition of pullback. For example, the area form $\Omega$, which associates to every pair of vectors the (oriented) area of the parallelogram which they span, can be pulled back the same way:

$(T^*\Omega)_p(X_p, Y_p) = \Omega_{T(p)}(T_{*,p}(X_p), T_{*,p}(Y_p))$

And the same is true for linear functions of more arguments (also called covariant tensors), like the (purely covariant) curvature tensor. The name "covariant" comes precisely from the way they "vary", that is, the way they transform under smooth maps. In general, tensors represent geometric concepts, and smooth maps such as $T$ don't transform points and vectors only: they transform angles, distances, volumes, curvatures and everything tensorial. Isometries are just those maps that leave the metric tensor invariant.


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