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in Curved space it seems $d(w^2)=(dw)^2$ how is it possible!?

$$x^2+y^2+z^2+w^2=\kappa^{-1}R^2,$$ first: $$dw=-w^{-1}(xdx+ydy+zdz),$$

$$\kappa^{-1}R^2-(x^2+y^2+z^2)=w^2,$$ $$dl^2 = dx^2 + dy^2 + dz^2+dw^2,$$

$$dl^2 = dx^2 + dy^2 + dz^2 +\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$$

Remark: in general as i know $dx^2=2xdx$ so $dw^2$ should be:

$$d(w^2)=(2w)w^{-1}(xdx+ydy+zdz)=2(xdx+ydy+zdz),$$ but here

$$d(w^2)=(dw)^2=w^{-2}(xdx+ydy+zdz)^2,$$ how is it possible?

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    It seems your question is why in the expression $dl^2 = dx^2 + \cdots + dw^2$ we have that $dw^2 = (dw)^2$. The answer would be: it just is. Conventional notation demands that $dw^2$ be read with the parenthesis as in $(dw)^2$ and not as $d(w^2)$.2012-12-07
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    To be more precise, the expression really defines a covariant 2-tensor $$ \mathrm{d}l \otimes \mathrm{d}l = \mathrm{d}x\otimes\mathrm{d}x + \cdots + \mathrm{d}w \otimes \mathrm{d}w $$ In context it is usually clear how to interpret $dw^2$; if you are newly starting to read stuff in the subject, it may take a little bit of getting used to.2012-12-07

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