I study the pullback of a map $f$, in special the pullback metric from manifold $\mathcal{Y}$ on a base smooth manifold $\mathcal{X}$. I understand that:

  1. the push forward operator $f_*: T_x \mathcal{X} \to T_y \mathcal{Y}$ maps every vector $a \in T_x \mathcal{x}$ into at least a vector $b \in T_{f(x)} \mathcal{Y}$ throughout the equality $b = F_* a$ at point $x$.
  2. The pullback $f^*\xi$ operates on space generated by push forward $f_* T_x \mathcal{X}$, for bundle $\xi$ with its fiber $\xi_x$ and base manifold $E$.
  3. The section $\Gamma(f^* T \mathcal{X})$ corresponds to every $\mathcal{C}^\infty$-map $s$ defined in an open set $\mathbb{D}_s$ from manifold $\mathcal{X}$ to $E_\xi$ with $s_x \in \xi_{f(p)}$

Given countable coordinate patches $(U, \varphi)$ and $(V, \phi)$ that covers respectively $\mathcal{X}$ and $\mathcal{Y}$ and we have the following embedding $\mathcal{X} \subseteq \mathcal{Y}$, this Wikipedia article provides the following formula in Einstein's notation: $g_{ab} = \partial_a X^\mu \, \partial_b X^\nu g_{\mu\nu}$.

How do they come to this formulation? Even in the pullback metric to coordinate patch, I fail to reproduce the result.


1 Answer 1


One defines the pullback metric as follows (and any other (0,s) tensor). If the exterior manifold is $N$ with metric $g$ and the embedded manifold is $M$, then the pullback metric is $$g_M (X,Y) := g(\phi_*(X),\phi_*(Y))$$ for all $X,Y \in \Gamma(TM)$ and where $\phi: M \to N$ injectively. Just push forward fields on the embedded manifold to the exterior one and employ its metric. As you seek the metric components, just take $X$ and $Y$ to be coordinate fields on $M$.

We define the pushforward of a vector by its action on $f \in C^{\infty}(N)$. Namely $\phi_* (X) f := X(f \circ \phi)$. So if $(\mathcal{V},y)$ is a chart on $N$, you'll first need the components of the push forwarded vectors. Consider hence for $\frac{\partial}{\partial x^i} \in \Gamma(TM)$, $$dy^a \Big(\phi_*\Big(\Big(\frac{\partial}{\partial x^i}\Big)_p\Big) \Big) = \phi_*\Big(\Big(\frac{\partial}{\partial x^i}\Big)_p\Big) y^a = \Big(\frac{\partial}{\partial x^i}\Big)_p (y \circ \phi)^a$$ I'll call this new function $y \circ \phi$ as $\Phi$. You'll note as it is a function from $M$ to $\mathbb{R}^d$ where $d = \mathrm{dim}(N)$, it amounts to the embedding function referenced in the Wikipedia article. Now attach to these components the coordinate basis induced by some chart on $N$ and apply the definition of the pullback metric, you reproduce the result. (The $X^\mu$ you gave is my $\Phi^a$): $$g_M(\partial_i , \partial_j) = g([\partial_i \Phi^a] \partial_a, [\partial_j \Phi^b] \partial_b ) = [\partial_i \Phi^a][\partial_j \Phi^b] g_{ab}$$ sorry about the different indices.

For a concrete example, take $N = \mathbb{R}^3$ with $g_{ab} = \delta_{ab}$ and $M = S^2$. Quite clearly, $\Phi$ amounts to a parametrization of the two-sphere, and because of the diagonal metric in the exterior space, the metric on the sphere are nothing but the Euclidean dot products of the partial derivatives of the parametrization.

  • $\begingroup$ Is this brackets $[]$ some advanced notation I am unaware of? $\endgroup$ Commented Jul 31, 2022 at 13:31
  • $\begingroup$ Do you have any textbook to reference these considerations of yours? I would like to consult some canonical material. $\endgroup$ Commented Jul 31, 2022 at 14:03
  • 1
    $\begingroup$ @BrunoPeixoto I think the use of bracket is to prevent from a overproliferation of parentheses $\endgroup$
    – Didier
    Commented Jul 31, 2022 at 14:42
  • $\begingroup$ Hi, Didier. We meet again. :-P. I thank you for the clarification. $\endgroup$ Commented Jul 31, 2022 at 14:46

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