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The metric on the sphere in the $(\theta,\phi)$ is of the form: $$ g_{\theta\theta}=r^2,g_{\theta\phi}=g_{\phi\theta}=0,g_{\phi\phi}=r^2\sin^2\theta $$

When transforming it to the $(x,y)$ coordinate on the plane using stereographic projection, I got $$ g_{xx}=g_{yy}=\frac{4}{(1+(x^2+y^2))^2},g_{xy}=0 $$

However, in the note I'm reading, it is been written in the form $$ g=\frac{4}{(1+(x^2+y^2))^2}[dx\otimes dx+dy\otimes dy] $$

Can anyone tell me how the tensor product comes here?

hxhxhx88
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  • It's just another way to denote the metric. – Michael Hoppe Oct 16 '13 at 16:28
  • @MichaelHoppe, I'm not familiar with the tensor product, could you please tell me more, or give me some reference? – hxhxhx88 Oct 16 '13 at 17:00
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    Even though nobody has told you yet, metric is a tensor field :) What you are used to is just writing its components in some basis (be it cartesian or polar). More precisely, it is a symmetric non-degenerate tensor field of rank two, meaning that it takes two arguments (two vectors at every point) and outputs a number, the inner product of those vectors, at every point. Finally, any tensor field can be written using just tensor products and tensor fields of rank 1 (vector fields and differential forms). – Marek Oct 16 '13 at 17:01
  • @Marek, I got it, thank you. – hxhxhx88 Oct 16 '13 at 18:38

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