Everyone: Consider the (local) change of variables from Cartesian coordinates $(x,y,z)$ to Cylindrical coordinates $( r,\theta,z)$ given by f. Does this map preserve the local geometry; does it preserve inner-product , angle, etc., i.e., given that f is a local diffeomorphism, do we have that $<v,w> _{R^3}=<f(v),f(w)>_{(r,\theta,z)}$ , where $<,>$ is the local inner-product? My reason for asking is that I have been trying to construct a orthogonal basis for the tangent space in $(r, \theta,z)$ with basis{ $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}, \frac{\partial}{\partial z}$} but in order to apply Gram-Schmidt, I need to have a notion of orthogonality, which I am trying to "ïmport" from $(\mathbb R^3, $, Cartesian), via the locally-diffeomorphic change-of-coordinate. Thanks for your comments/feedback.

  • $\begingroup$ $f$ is a map $f:\mathbb R^3 \to \mathbb R^3$ such that $f(x, y, z) = (r, \theta, z)$? $\endgroup$
    – Muphrid
    Oct 29, 2013 at 2:48
  • $\begingroup$ Hi, the map takes standard Cartesian coordinates to cylindrical coordinates $(r,\theta, z)$, where cylindrical is equal to polar in the first two coordinates and $z$ is just the height, so that $x=x(r, \theta) ; y=y(r,\theta)$ , with $x=rcos\theta, y=rsin\theta $ and $z=z$ $\endgroup$
    – user99680
    Oct 29, 2013 at 3:28
  • $\begingroup$ I'm inquiring about which direction the map goes--from cartesian or to cartesian? Writing $x = x(r, \theta)$ makes me think you're going to cartesian, but you seem to have written the question as going from cartesian. $\endgroup$
    – Muphrid
    Oct 29, 2013 at 3:36
  • $\begingroup$ The concept you're looking for is the pullback of a metric, and involves the Jacobian matrix of $f$ rather than $f$ itself. $\endgroup$ Oct 29, 2013 at 3:39
  • $\begingroup$ Well, I need to go back-and-forth in order to pullback (if possible) the standard inner-product in $\mathbb R^3$ into cylindrical coordinates, but basically, I want to know if I can go from Cartesian to Cylindrical using the local diffeo f and say $<v,w>$ (Cartesian) =$f(v),f(w)$ (Cylindrical) , so that I can transport the local geometry (inner-product) from Cartesian to Cylindrical. $\endgroup$
    – user99680
    Oct 29, 2013 at 3:42

1 Answer 1


The corresponding metric in cylindrical coordinates does not use the diffeomorphism $f$ itself, but rather the Jacobian map $J_f$ instead. It is the Jacobian that maps between the tangent spaces of the two manifolds.

Let $f(p) = p'$, for $p \in M$ and $p' \in M'$ for two manifolds $M, M'$. Let $g(u,v)$ be the metric on $M$ and $g'$ be the induced metric on $M'$. The two metrics are related by

$$g'(J_f(u),J_f(v))_{p'} = g(u, v)_{p}$$

where $u, v \in T_p M$; they are tangent vectors in the original manifold $M$ at the point $p$.

  • $\begingroup$ So, how is this concept called? Pullback? $\endgroup$
    – Evgeny
    Oct 29, 2013 at 4:36
  • $\begingroup$ My impression is that this is a pushforward, and pullbacks are usually used to bring cotangent vectors in $M'$ back to their corresponding partners in $M$--where here, a vector in $M$ is "pushed forward" to $M'$ in the same direction as the underlying transformation $f$. But this is something I'm a bit fuzzy on. It's all the Jacobian anyway, really, and I think the pullback is just the adjoint of the pushforward. $\endgroup$
    – Muphrid
    Oct 29, 2013 at 5:17
  • $\begingroup$ It's the pullback by the inverse of $f$, which you could call the pushforward by $f$ in this case of local diffeomorphism. If $f$ was an arbitrary smooth function then we could only pull back metrics, not push them forwards; so the pushforward of a metric is a rarer thing and thus a less common terminology. $\endgroup$ Oct 29, 2013 at 6:04

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