In Chapter 9 of Polyakov's book Gauge Fields and Strings, 1987 , he studies measures in the space of metrics and diffeomorphisms. My question is - how does one come up with Eq. 9.23?

For some context: Consider parametrized curves with fixed endpoints, i.e. $x^{\mu}(\tau)$ for $\tau \in [0,1] $ such that $x^{\mu}(0)$ and $x^{\mu}(1)$ are fixed. Polyakov initially defines the metric for this choice of parametrization to be


Now, he wants to define a "distance" $||\delta h||^2$ between two metrics $h(\tau)$ and $h(\tau)+\delta h(\tau)$. He says that the only local expression for this distance should take the form of Eq. (9.23), which is

$$||\delta h||^2=\int_{0}^1 d \tau \left(\delta h (\tau)\right)^2 h^{-3/2}(\tau).\tag{9.23}$$

Why is this form special? For example why not something like $\int_{0}^1 d \tau \left|\delta h (\tau)\right| h^{-1/2}(\tau)$ or $\int_{0}^1 d \tau \left(\delta h (\tau)\right)^3 h^{-5/2}(\tau)$ which also have the same dimensions as Polyakov's choice?

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    $\begingroup$ How about compatibility with scalar multiplication? $|\lambda x| = |\lambda||x|$ $\endgroup$ Commented Oct 6, 2023 at 11:18
  • $\begingroup$ Thanks! Yes, that may work. But then why not something like $\left[\int_{0}^1 d \tau \left|\delta h (\tau)\right| h^{-1/2}(\tau)\right]^2$ or $\left[\int_{0}^1 d \tau \left(\delta h (\tau)\right)^3 h^{-5/2}(\tau)\right]^{2/3}$? I guess these expressions are not "local", i.e., not a sum of local terms? $\endgroup$ Commented Oct 6, 2023 at 12:59
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    $\begingroup$ Possibly. These are essentially continuous versions of the (square of) $p$-norm in $L^p$ spaces, so are all valid norms. I think you could indeed in principle work with them, but the point is that at some point you will want a path integral $\int \mathcal{D} \delta h e^{-||\delta h||^2} = 1$ to establish the normalisation for the $\mathcal{D} h$ path integral. You are going to find it difficult to do this path integral with these 'non-local' norms. Another reference aside from Polyakov is the book by Nakahara by the way. $\endgroup$ Commented Oct 7, 2023 at 12:07
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    $\begingroup$ Also note that this is all quite dubious anyway from a formal perspective. What we are really after is a metric (or volume element) on the space of (Riemannian) metrics but this is some potentially weird topological space. Instead it is much easier to work 'infinitesimally' and instead use the path integral over its tangent space to establish the path integral normalization. $\endgroup$ Commented Oct 7, 2023 at 12:07

1 Answer 1


The metric tensor $h(\tau)(\mathrm{d}\tau)^2$ and infinitesimal deviation $\delta h(\tau)(\mathrm{d}\tau)^2$ are invariant under reparametrization $$\tau\quad\longrightarrow \tau^{\prime}~=~f(\tau)\tag{9.9}$$ of the world-line (WL). The powers in the square norm$^1$ (9.23) are needed to make it

  • quadratic in $\delta h$, and

  • invariant under WL reparametrizations (9.9).


  1. A.M. Polyakov, Gauge Fields and Strings, 1987; Sections 9.1-9.2.


$^1$The square norm is related to the inner product via polarization.

  • $\begingroup$ Thanks! I was missing the quadratic in $\delta h$ condition. But I guess that there is an additional condition that the norm in (9.23) is a sum (integral) of local terms, and not a polynomial/power of an integral of local terms. Because otherwise the candidates I wrote in reply to @SvenForkbeard 's comment would also work. $\endgroup$ Commented Oct 6, 2023 at 13:27

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