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I try to calculate the pullback of the metric on $\mathbb R^2$. I don't know if I got this right, so I write it down for check. I've been very impetuous lately, so please let me know if I've made any mistakes.

Consider $\mathbb R^2$ with Cartesian coordinates $(x,y)$. Let $$ \phi: \mathbb R^2 \rightarrow \mathbb R^2 ~~~~~~ \phi(x,y)=(3x,2y) \tag{1} $$ Then, $\phi$ induce a tangent map, namely $$ \phi_* : T_{(x,y)}\mathbb R^2 \rightarrow T_{\phi(x,y)} \mathbb R^2 ~~~~~~ \phi_*(a\partial_x+b\partial_y) = (a, b) \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix} \tag{2} $$ Since the global Cartesian coordinates, $T_{(x,y)}\mathbb R^2$ and $T_{\phi(x,y)} \mathbb R^2$ have same basis $\partial_x, \partial_y$.

Similarly, $\phi$ induce a pullback, namely $$ \phi^*: T^*_{\phi(x,y)} \mathbb R^2 \rightarrow T^*_{(x,y)}\mathbb R^2 ~~~~~~ \phi^*(a dx + b dy) = (a,b) \begin{pmatrix} 1/3 & 0 \\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} dx \\ dy \end{pmatrix} \tag{3} $$ (3) imply $\phi^*(dx)= \frac{1}{3} dx$ and $\phi^*(dy)= \frac{1}{2} dy$. Actually, I'm not sure about (3).

Now, give the $\mathbb R^2$ a metric $$ g(x,y)= \begin{pmatrix} \frac{1}{1+x^2+y^2} & 0 \\ 0 & \frac{1}{1+x^2+y^2} \end{pmatrix} \tag{4} $$ Then, I want to calculate $\phi^*g$. First, from (4), there is $$ g_{11}(x,y) = \frac{1}{1+x^2+y^2} dx\otimes dx ~~~~ g_{22}(x,y)= \frac{1}{1+x^2+y^2} dy\otimes dy ~~~~ g_{12}(x,y)=g_{21}(x,y)=0 \tag{5} $$ Therefore, $$ (\phi^* g_{11}) (x,y) = \phi^* [g_{11}(\phi(x, y))]= \phi^*[g_{11}(3x, 2y)] =\phi^* [\frac{1}{1+9x^2+4y^2} dx\otimes dx] = \frac{1}{9(1+9x^2+4y^2)} dx\otimes dx \tag{6} $$ Similarly, $$ (\phi^* g_{22}) (x,y)= \frac{1}{4(1+9x^2+4y^2)} dy\otimes dy \tag{7} $$ So, $$ (\phi^* g)(x,y) = \begin{pmatrix} \frac{1}{9(1+9x^2+4y^2)} & 0 \\ 0 & \frac{1}{4(1+9x^2+4y^2)} \end{pmatrix} \tag{8} $$ I want to know if the process to get (8) is right ?

Enhao Lan
  • 6,682

1 Answers1

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Let $f(x,y) = \frac{1}{1+x^2+y^2}$. It is $\phi^*(dx) = d(\phi^*(x)) = d(x \circ \phi) = d(3x) = 3dx$ and similar $\phi^*(dy) = 2dy$. Simply calculate $$\phi^*(g) = \phi^*(f(dx\otimes dx + dy\otimes dy)) = f\circ\phi \cdot(\phi^*(dx)\otimes\phi^*(dx) + \phi^*(dy)\otimes\phi^*(dy)) \\ = f\circ \phi\cdot(9dx\otimes dx + 4 dy \otimes dy).$$

So, $\phi^*(g)(x,y) = \begin{pmatrix} \frac{9}{1+ 9x^2 + 4y^2} & 0 \\ 0 & \frac{4}{1+9x^2 + 4y^2} \end{pmatrix}$.

psl2Z
  • 5,772
  • Could you explain why $\phi^(dx) = d(\phi^(x)) = d(x \circ \phi) = d(3x)$ ? I can't understand it. Or tell me a reference is enough. Thanks. – Enhao Lan Oct 07 '24 at 11:06
  • Seemly, you treat $x$ as a function. – Enhao Lan Oct 07 '24 at 11:07
  • @EnhaoLan See for example here https://math.stackexchange.com/q/1586606 that pullback and exterior derivative commute. – psl2Z Oct 07 '24 at 12:08
  • Thank, I can accept it. In fact, the matrices of (3) and (2) are the same, and I remember incorrectly, I remember that they be inverse. Thanks again. – Enhao Lan Oct 07 '24 at 12:46