1

This is a question more concerned about mathematical detail involving the Polyakov path integral.

In section $3.2$ of Polchinski's 1st String Theory book it is stated the following about Polyakov path integral:

The integral runs over all Euclidean metrics and over all embeddings $X^\mu(\sigma^1,\sigma^2)$ of the world-sheet in Minkowski spacetime: $$\int [d X \, d g] \, \exp (-S). \tag{3.21}$$

That the path integral is defined in the space of embedings $X: \Sigma \to M$ is clear for instance in Polchinski's Evaluation of the one loop string path integral and Nag's Mathematics in and out of String Theory.

However, there is a reference called Quantum Fields and Strings: A Course for Mathematicians. Volume $2$, whose section $1.4$, page $818$ has the following passage:

The Polyakov action leads to well-defined transition amplitudes, obtained by integration over the space $\text{Met}(\Sigma)$ of all positive metrics on $\Sigma$ for a given topology, as well as over the space of all maps $\text{Map}(\Sigma, M)$: We define $$\tag{1.5} A = \sum_{\text{topologies}} \int_{\text{Met}(\Sigma)} Dg \, \frac{1}{\mathcal{N}(g)} \int_{\text{Map}(\Sigma, M)}Dx \, e^{-S[x,g,G]}.$$

In the above quote, $x: \Sigma \to M$ is a map, $g$ is a metric on $\Sigma$ and $G$ is a metric on the target space $M$.

My question can be summarized in the following: what is the precise space of maps over which the path integral is defined? I've heard that the Polyakov path integral, as other path integrals, must admit "non-standard" trajectories i.e. maps that are not necessarily bijective, or whose differential is not injective, etc.

I'm really confused because the Polyakov classical action requires at least the maps $X^\mu$ to be differenttiable. Shouldn't the space of embeddings be more well-suited for this integration, from a geometrical point of view?

Генивалдо
  • 1,290
  • 2
    The same issue crops up in the standard QM path integral: We write a $\dot{x}$ in the action but in fact the differentiable paths have measure zero - how can that be, when the action isn't even defined on non-differentiable paths? See this and this answer of mine – ACuriousMind Jun 25 '23 at 21:13
  • @ACuriousMind My understanding of your first answer was: The conditional Wiener measure is defined over continuous paths because you are working with Hamiltonians, so you deal with momentum $p$ instead of $\dot{x}$, so differentiability of the path is not a problem. Besides that, it seems that the equality $\mathrm{d}W_{q,q'} = \mathcal{D}q\mathrm{e}^{-S_0[q(t)]}$ is just a physicist's notation and the RHS of the equality has no formal meaning, because the action $S_0[q(t)]$ is defined only over differentiable paths. Is that right? – Генивалдо Jun 26 '23 at 14:49
  • 1
    Yes, and my second answer explains a possible formal meaning of the $\mathcal{D}q\mathrm{e}^{-S}$ - an action on some function space (e.g. the Schwartz functions) induces a measure on the dual space of these functions (e.g. the tempered distributions), so this is really supposed to be that dual measure induced by the acton. – ACuriousMind Jun 26 '23 at 15:08

0 Answers0