We have then to distinguish between Natural Geometry, which is the single \emph{true} geometry in the sense understood by the physicist, and World Geometry, which is the pure geometry applicable to a conceptual graphical representation of all the quantities concerned in physics. We may perhaps go so far as to say that the World Geometry is intended to be closely descriptive \index{Geometry, Riemannian!world geometry}% \index{World geometry}% of the fundamental relation-structure which underlies the various manifestations of space, time, matter and electromagnetism; that statement, however, is rather vague when we come to analyse it. Since the graphical representation is in any case conventional we cannot say that one method rather than another is right. Thus the two geometries discussed in Parts I and~II of this chapter are not to be regarded as contradictory. My reason for introducing the second treatment is that I find it to be more illuminating and far-reaching, not that I reject the first representation as inadmissible. In the following account of Weyl's theory I have not adhered to the author's \index{Weyl's theory}% order of development, but have adapted it to the point of view here taken up, which sometimes differs (though not, I believe, fundamentally) from that which he adopts. It may be somewhat unfair to present a theory from the wrong end---as its author might consider; but I trust that my treatment has not unduly obscured the brilliance of what is unquestionably the greatest advance in the relativity theory after Einstein's work. \Section{84.}{Non-integrability of length} \index{Integrability of parallel displacement!of length and direction}% \index{Length!non-integrability of}% \index{Non-integrability of length and direction}% We have found in \SecRef{33} that the change~$\delta A_{\mu}$ of a vector taken by parallel displacement round a small circuit is \begin{align*} \delta A_{\mu} &= \tfrac{1}{2} (A_{\mu\nu\sigma} - A_{\mu\sigma\nu})\, dS^{\nu\sigma}\displaybreak[0] \\ &= \tfrac{1}{2} B_{\mu\nu\sigma}^{\epsilon} A_{\epsilon}\, dS^{\nu\sigma}\displaybreak[0] \\ &= \tfrac{1}{2} B_{\mu\nu\sigma\epsilon} A^{\epsilon}\, dS^{\nu\sigma}. \Tag{(84.1)}\displaybreak[0] \intertext{Hence} A^{\mu}\, \delta A_{\mu} &= \tfrac{1}{2} B_{\mu\nu\sigma\epsilon} A^{\mu} A^{\epsilon}\, dS^{\nu\sigma} = 0, \end{align*} since $B_{\mu\nu\sigma\epsilon}$~is antisymmetrical in $\mu$ and~$\epsilon$. Hence by~\Eq{(26.4)} $\delta A_{\mu}$~is perpendicular to~$A_{\mu}$, and the \emph{length} of the vector~$A_{\mu}$ is unaltered by its parallel displacement round the circuit. It is only the direction which changes.