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Is there a coordinate independent Dirac delta function? 
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#1
Dec2612, 09:02 PM

P: 978

I have been wondering exactly how one would express the Dirac delta in arbitrary spaces with curvature. And that leads me to ask if the Dirac delta function has a coordinate independent expression. Is there an intrinsic definition of a Dirac delta function free of coordinates and metrics? Or as a distribution does it inherently need a coordinized space on which to distribute its values? Thanks.



#2
Dec2612, 09:51 PM

Sci Advisor
P: 827

The only possibility I can think of would be "Dirac is the identity element for convolution".



#3
Dec2812, 12:01 PM

P: 978

What about the Dirac measure? This is called a "measure", and any measure can be integrated, right? This Dirac delta measure is defined in terms of elements and sets and not on coordinates. So does this qualify as a coordinate independent form of the Dirac delta function? Or would the coordinate free expression have to be in terms of elements/points only and not sets?



#4
Dec2812, 10:17 PM

Sci Advisor
P: 827

Is there a coordinate independent Dirac delta function?
##\int_X f(t) \delta(ta) dt = f(a)## or something similar. The problem is we use coordinate systems to express "a" itself, so if you change the coordinate system you change how you write "a". So my interpretation of "coordinate free" would be "changing coordinates doesn't change the expression", at least in the context this question. If you meant something else well you'll need to define it. Secondly, you wrote "any measure can be integrated". Well you mean "integrate with respect to the measure" not "integrate the measure itself". And yes there is a concept of "derivative of a measure" but it's very different. Now does the Dirac measure satisfy the problem? Well that depends on whether you can pick an isolated point from a set without reference to a coordinate system (remember coordinate system is really just a systematic way of writing elements with respect to each other). Can you pick 0 without knowing where it is? And if you can do this, how is this any different to "normal" way we write Dirac? 


#5
Jan313, 12:35 PM

P: 978




#6
Jan413, 03:38 AM

Sci Advisor
P: 827

As to your specific question, I guess so? Same thing here. The Dirac as a measure can be integrated against f(x)=1, but the Dirac as a Schwartz distribution can't. Nonetheless, you can take the Dirac measure and restrict is to become a Schwartz measure. Edit: Oh wait you're summing over ##p \in M##? Then that means M is countable. 


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