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Consider a (semi-)Riemannian manifold $(M,g)$ of dimension $m$, and a submanifold $(N,g|_N) \subset (M,g)$ of dimension $n$, such that the change of coordinates from $N$ to $M$ is given by the smooth map:

\begin{align*} \phi:N &\to M \\ (y^1,...,y^n) &\mapsto (\phi^1,...,\phi^n,...,\phi^m) \end{align*}

where $\{y^1,...,y^n\}$ is a local coordinatization of $N$.

If $\{x^1,...,x^m\}$ is a local coordinatization of $M$, define the line-element on M by:

$$ ds^2 := g_{\mu \nu} \ \mathrm{dx}^\mu \otimes \mathrm{dx}^\nu $$

I want to understand the following equality.

$$ \boxed{\phi^* (ds^2) = g_{\mu \nu} \ \mathrm{d\phi}^\mu \otimes \mathrm{d\phi}^\nu} $$

In particular, why is this true? An explanation in local coordinates would be helpful.

My understanding is that $\phi^* (ds^2) \equiv g|_N$.

1 Answers 1

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It follows straightforwardly from the definition of pull-backs. Note that the following properties of pull-backs hold in general.

  1. $ \phi^* (\textrm{dx}^\mu) = \textrm{d}\phi^\mu$
  2. $ \phi^* (\textrm{dx}^\mu \otimes \textrm{dx}^\nu) = \phi^* (\textrm{dx}^\mu) \otimes \phi^* (\textrm{dx}^\nu) $
  3. $ \phi^* (f\omega) = (f \circ \phi) \ \phi^*(\omega) $, where $f \in \Omega^0(M)$ and $\omega \in \Omega^k(M), 0 \le k \le n$.

From the above three identities, it follows that

$$ \phi^* (g_{\mu \nu} \ \mathrm{dx}^\mu \otimes \mathrm{dx}^\nu) = (g_{\mu \nu} \circ \phi) \ \mathrm{d\phi}^\mu \otimes \mathrm{d\phi}^\nu $$

Hope that settles your question.