# Pullback metric on embedded manifolds

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.

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.
• Is this brackets $[]$ some advanced notation I am unaware of? Commented Jul 31, 2022 at 13:31