-1

Consider the mapping $f:\mathbb R^3\to\mathbb R^4$ between two Euclidean spaces: $$f(\mathbf x)=(x_1,x_2,x_3,1/\|\mathbf x\|)^T$$ How to calculate the induced metric by $f$ on $\mathbb R^3$ (i.e. the pullback metric)?

Parcly Taxel
  • 103,344
user868832
  • 9
  • 3
    Your function does not make sense. It looks like you are defining $f(x)$ to be a set (rather than a point in $\Bbb{R}^4$). Did you mean $f(x) = \left( \frac{x_1}{|x|}, , \frac{x_2}{|x|}, , \frac{x_3}{|x|}, , \frac{1}{|x|} \right)$, where $|x| = \sqrt{x_1^2+x_2^2+x_3^2}$? – Nick Jan 01 '21 at 16:48
  • 1
    @AK95 Here is MathJax help page by which you can format your question. – Infinity_hunter Jan 01 '21 at 17:10

1 Answers1

1

Assuming $f$ is defined by

$$f(\mathbf x) = f(x_1,x_2,x_3) = (x_1,x_2,x_3, \frac{1}{\sqrt{x_1^2+x_2^2+x_3^2}})$$

and the metric on $\mathbb R^4$ is defined by the Euclidean norm, it looks like the distance between $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ according to the pullback metric is

$\sqrt{ (x_1 - y_1)^2 + (x_2 - y_2)^2 + (x_3 - y_3)^2 + ( \frac{1}{\sqrt{x_1^2+x_2^2+x_3^2} }-\frac{1}{\sqrt{y_1^2+y_2^2+y_3^2}})^2}.$

D_S
  • 33,891