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Let $M, N$ are two manifolds. Now I am interested in defining the pullback of an arbitrary tensor field of type $(r,s)$ under the diffeomorphism $\phi : M \rightarrow N$ as follows:

$\phi^* T(\eta_1,\dots, \eta_r, X_1, \dots, X_s) = T( (\phi^{-1})^*(\eta_1), \dots, (\phi^{-1})^*(\eta_r), \phi_* X_1, \dots, \phi_* X_s)$.

where $\eta_i \in T_p^*(M)$ is a covector and $X_j \in T_p(M)$ is a vector.

Actually I am interested in pullback of a metric tensor.

We know that it is a $(0, 2)$ tensor. Therefore, the above formula becomes $$\phi^*g(X,Y) = g(\phi_*X, \phi_*Y)$$.

Now consider $g_{\alpha \beta}$ is a $(0, 2)$ tensor on $N$. Now my question is can one write the formula as follows

$$(\phi^*g)_{\mu \nu} = \frac{\partial y^{\alpha} }{\partial x^{\mu}} \frac{\partial y^{\beta} }{\partial x^{\nu}} g_{\alpha \beta}$$.

Please help me. Thanking in advanced.

glS
  • 6,818

1 Answers1

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Let $x=(x^1,\dots, x^m)$ and $y=(y^1,\dots,y^n)$ be local coordinates on $M$ and $N$ respectively.

Locally, $\phi: M\rightarrow N$ is $$ \phi(x)=(\phi^1(x),\dots,\phi^n(x)) $$ Then $$ \begin{align*} \phi_\ast \frac{\partial}{\partial x^j}= \frac{\partial \phi^i}{\partial x^j}\frac{\partial}{\partial y^i} \end{align*} $$ Locally, $g=g_{kl}dy^k\otimes dy^l$. So $$ \begin{align*} (\phi^\ast g)(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})&= \frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g(\frac{\partial}{\partial y^k}\frac{\partial}{\partial y^l})\\ &=\frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g_{kl} \end{align*} $$ So $$ (\phi^\ast g)_{ij}=\frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g_{kl} $$

Gilbert Easton
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