I am trying to grasp the idea of isometry.

Under a general coordinate transformation (GCT), the metric tensor $g_{ij}$ changes as $$g^\prime_{ij}(x^\prime)=\frac{\partial x^k}{\partial x^{\prime i}}\frac{\partial x^l}{\partial {x}^{\prime j}}g_{kl}(x)\tag{1}$$ which can be seen as the prescription for how the metric at a given point P on the manifold changes as we go from the unprimed to a primed coordinate system.

Now, the isometry is defined as $$g^\prime_{ij}(x')=g_{ij}(x')\tag{2}$$

What does this mean? Here is my understanding.

If we make a coordinate transformation $$x\to x'(x),$$ in general, the metric at the point P will change by the rule Eq.$(1)$. However, if it turns out that the transformed metric at P has the same functional form as in the old coordinate system at P, then the corresponding transformation is called an isometry. So in short, my current understanding of isometry is based on doing a coordinate transformation keeping the attention fixed at a given point P.

I am not sure if my idea of isometry, as stated above, is correct! Some textbooks define isometry of two points P and Q in a manifold unlike what I have understood.

  • 1
    $\begingroup$ Some simple examples will help, e.g. transforming $ds^2 = g_{ij}(x)dx^i dx^j = dx^2 + dy^2$ to $ds^2 = g_{ij}(x')dx'^i dx'^j = dr^2 + r^2 d\theta^2$ with $g'_{ij}(x') = (r,\theta)$ i.e. $x'^1 = r, x'^2 = \theta$ is such that $g'_{22}(r,\theta) = r \neq 1 = g_{22}(r,\theta)$. Clearly eq. (2) is wrong as both sides have to be a function of the same variable. For a simple rotation of the Cartesian metric obviously the metric before and after are equal. Zee's GR book gives a good explanation of this stuff. $\endgroup$
    – bolbteppa
    Jan 2, 2020 at 20:51
  • $\begingroup$ Dear @bolbteppa, I looked at Zee's book to get an idea about the notion of isometry. I think, he treats $x=x(x^\prime)$ as an active transformation. I've narrowed down the question to first to get clarification about a very specific query. $\endgroup$
    – SRS
    Jan 4, 2020 at 7:38
  • $\begingroup$ Check Tristan Needham's Visual differential geometry $\endgroup$ Jun 20 at 20:22

2 Answers 2


You do require at least two points on the manifold to get a proper understanding of isometry. Isometry is a mapping that should preserve the proper distance between these two points. Since we define the infinitesimal proper distance element on the manifold as:

\begin{equation} ds^{2} = g_{\mu \nu} dx^{\mu} dx^{\nu} \end{equation}

then the condition $g'_{\mu \nu}(x') = g_{\mu \nu}(x')$ implies:

\begin{equation} ds'^{2} = g'_{\mu \nu}(x') dx'^{\mu} dx'^{\nu} = g_{\mu \nu}(x') dx'^{\mu} dx'^{\nu} \end{equation}

But since the metric and the infinitesimal elements $dx'^{\mu}$ are now all in the same coordinates, this is simply the original proper distance element $ds^{2}$. Therefore:

\begin{equation} g'_{\mu \nu}(x') = g_{\mu \nu}(x') \iff ds'^{2} = ds^{2} \end{equation}

The most common example of an isometry is a diffeomorphism. Assuming you have a smooth manifold, then local infinitesimally small portions of it can be approximated as Euclidean planes of the same dimensionality. We then define a diffeomorphism as:

\begin{equation} x'^{\mu} = x^{\mu} + \varepsilon \, h^{\mu} \end{equation}

where $\varepsilon \rightarrow 0$ and $h^{\mu}$ a constant vector. It's easy to see that locally $dx'^{\mu} = dx^{\mu}$ which thus preserves the proper distance between two such approximate points.


An isometry is a way of moving the manifold rigidly within itself, and is inherently an active transformation (a diffeomorphism).

From Carroll's Lecture Notes on General Relativity:

diffeomorphisms are just active coordinate transformations... We say that a diffeomorphism $\phi$ is a symmetry of some tensor $T$ if the tensor is invariant after being pulled back under $\phi$... The most important symmetries are those of the metric... A diffeomorphism of this type is called an isometry.

In the formula $$g^\prime_{ij}(x')=g_{ij}(x'),$$ it is important to note that the argument $x'$ is the same on both sides -- i.e., the original and transformed metrics are evaluated at the same coordinate values and hence at different physical locations. Of course, there is technically always the trivial isometry $x'(x) = x$. But in general, each point $P$ is mapped by the isometry to a different point $Q$.

(A passive version with $g^\prime_{ij}(x')=g_{ij}(x)$ -- note no prime on right-hand side -- would be different and less interesting. This says that the metric looks the same in the old and new coordinates at the same point $P$. Such nontrivial coordinate transformations exist for any manifold -- e.g., adding a constant to each coordinate. Nontrivial isometries, on the other hand, exist only for special manifolds.)

We can describe an isometry without even using coordinates. Given any points, curves, and other shapes of interest in a manifold, all their metrical properties and relations (such as lengths and angles) are unchanged when they are transformed into new points, curves, and other shapes by the isometry.

We can see that an isometry is very special, since a general manifold with metric will have lots of irregular (but smooth) "bumps and wrinkles" as measured by curvature, such that any point can be uniquely identified by the pattern of bumps and wrinkles around it. There will be no way to transform the manifold nontrivially while keeping these patterns the same -- i.e., no possible rigid motion of the manifold within itself.

A continuous family of isometries is particularly interesting, and leads to the notion of a Killing field.

  • $\begingroup$ Isn't the Eq. (1) in the question a coordinate transformation where we focus on how the metric changes at a given point? Is there anything wrong with the sentence I wrote below Eq. 1? $\endgroup$
    – SRS
    Jun 20 at 20:59
  • $\begingroup$ @SRS Yes, your Eq. (1) correctly describes the transformation of the metric at a given point (passive). This is the basis for my observation in the parenthetical paragraph. However, crucially, the definition of an isometry involves a "pullback" (defined by Carroll), which combines this transformation with the underlying coordinate transformation itself, so as to compare the metric at two different points (active). $\endgroup$
    – nanoman
    Jun 20 at 21:06

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