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I'm referring to Gel'fand, Gindikin, and Graev's Selected Topics in Integral Geometry, pages 4-5, section 1.4 (see here and here).

Now in page 5 they write that:

$$dx_1 dx_2 = d(\xi_1 x_1 +\xi_2 x_2 -p) d\mu_{\xi}$$ where $d\mu_{\xi}=\frac{dx_1}{|\xi_2|}=\frac{dx_2}{|\xi_1|}$

Now I am not sure how this is true, I mean:

$$d(\xi_1 x_1 +\xi_2 x_2 -p)d\mu_{\xi} = [\xi_1 dx_1 +\xi_2 dx_2] d\mu_{\xi} = $$

$$ = \frac{\xi_1}{|\xi_1|} dx_1 dx_2 + \frac{\xi_2}{|\xi_2|} dx_2 dx_1 $$

I am not sure how does the last line equals $dx_1dx_2$, anyone care to enlighten me?

Thanks.

Willie Wong
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MathematicalPhysicist
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1 Answers1

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The volume element $d \mu_{\xi}$ itself is a 1-form that lives on the line $\xi: \xi_1 x_1 +\xi_2 x_2 -p = 0$, not on $\mathbb{R}^2$. In the equation: $$ dx_1 dx_2 = d(\xi_1 x_1 +\xi_2 x_2 -p) d\mu_{\xi} $$ the $d \mu_{\xi}$ should be interpreted as any 1-form on $\mathbb{R}^2$ whose pullback onto $\xi$ coincides with the original $d \mu_{\xi}$. You can either use $\frac{dx_1}{|\xi_2|}$ or $\frac{dx_2}{|\xi_1|}$ for that purpose. However, $\frac{dx_1}{|\xi_2|}$ didn't equal to $\frac{dx_2}{|\xi_1|}$ on $\mathbb{R}^2$. Once you have fixed your "extension" of $d \mu_{\xi}$ onto $\mathbb{R}^2$, you cannot change that in the middle of an equation.

achille hui
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