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An immersion $F:M\to N$ into a Riemannian manifold $(N,h)$ inherits the Riemannian metric $g=F^{\ast }h$: measure two tangent vectors after mapping them into the surrounding tangent space. This is a pullback of a Riemannian metric, withThe injective derivative of an immersion makes this inner product positive-definite. For a parametrized surface in Euclidean space, its coefficients are simply the scalar products of the coordinate tangent vectors: the first fundamental form.

For the circular cylinder $F(\theta ,z)=(R\cos \theta ,R\sin \theta ,z)$, the tangent vectors $F_{\theta }$ and $F_z$ are orthogonal, with squared norms $R^2$ and $1$. ConsequentlyThe unrolled coordinate $x=R\theta$ gives $d{s}^2=d{x}^2+d{z}^2$. This calculation explains intrinsic flatness of a circular cylinder and makes Riemannian distance on a circular cylinder a planar problem with repeated endpoint lifts.

Preserving a Riemannian metric locally preserves the arc length of every mapped curve, but it need not preserve distances measured by unrestricted paths in the surrounding space. The great-circle distance between two sphere points, for example, exceeds their straight chord length unless they coincide. ^[ref].

The two coordinate tangent vectors of the embedding areTheir Euclidean inner products give the pullback of a Riemannian metric, equivalently the induced metric:Locally set $u=\rho \phi$. The line element becomes $d{u}^2+d{\hat{z}}^2$, so the induced Riemann curvature tensor vanishes identically. This is the intrinsic flatness of a circular cylinder. Its bending in the ambient space is extrinsic curvature, not intrinsic curvature; periodicity of the angular coordinate does not change the local flatness.

The induced metric is obtained from the ambient Euclidean metric by the pullback of a Riemannian metric operation. Differentiating $x^2+a^2{\rho }^2=1$ on a branch with $x\ne 0$ gives $dx=-a^2\rho d\rho /x$. Also $d{y}^2+d{z}^2=d{\rho }^2+{\rho }^{2}d{\varphi }^2$. ConsequentlyThe assumption that $(\rho ,\varphi )$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho =0$. On such a chart, $0<\rho <1/|a|$ and $E>0$.

The conformal flattening of a surface of revolution supplies the particularly simple conformal factorSince ${\sigma }^{\prime }>0$, these are isothermal coordinates with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the induced metric. For example, $(x,\varphi )$ give $g=[1+x^2/(a^2(1-x^2))]d{x}^2+(1-x^2)d{\varphi }^2/a^2$, regular at $x=0$.

Use the numerator in the original PDF: it contains all $n+1$ squared differentials. The local TeX ends that numerator at $n-1$, which would not give the stated Riemannian metric.

For the pullback of a Riemannian metric, substitute the embedding $\pi (y)=(y_1,\dots ,y_n,1)$. Its differential sends a tangent vector $v$ to $(v_1,\dots ,v_n,0)$, so $d{X}_i=d{y}_i$ for $i\le n$ and $d{X}_{n+1}=0$. ConsequentlyThe denominator is everywhere positive. For any nonzero tangent vector,so this is indeed a smooth positive-definite Riemannian metric on $U$.