I have been led to believe that all shapes, surfaces and polyhedra have Euler characteristics. Does this apply to fractals as well? If so, is this linked in any way to fractal dimensions? How would such an Euler characteristic be calculated? If fractals do not have an Euler characteristic, why is that the case?
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1Euler was a great mathematician, but.. fractals in XVII century... I know, this is not the point, but still, I don't think that Euler's characteristic can be applied to fractals at all. – ajotatxe Nov 20 '19 at 20:59
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1Possibly cross posted from Math.Overflow. – Daniele Tampieri Nov 20 '19 at 20:59
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There is a good comment by Henrik Rüping at MO. – Dietrich Burde Nov 20 '19 at 21:02
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Perhaps my previous comment sounded as 'it is strange, so throw it off'. Well, perhaps a bit, but every fractal set has infinitely many points in a 'relevant' zone. The point would be finding some asymptotic ratio between vertices and edges, and if they go to infinity at some proportional speed, then $(V+C-A)/A\to 0$ – ajotatxe Nov 20 '19 at 21:11
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The usual way to define an Euler characteristic for a topological space is $$ \chi = b_0 - b_1 + b_2 - \ldots$$ where $b_n$ is the $n$'th Betti number. For a subspace of $\mathbb R^n$, $b_k = 0$ for $k > n$. However, some of the Betti numbers could be $\infty$.
Robert Israel
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This is too long for a comment.
I am a bit confused by the question, as I imagine fractals as being more of a metric property, and the Euler characteristic as a topological one. For instance, the von Koch snowflake is homeomorphic to the circle, hence has Euler characteristic $0$, any rough curve with Hausdorff dimension $>1$ homeomorphic to a segment will have Euler characteristc $1$ (for instance the Osgood curve, of dimension $2$), the Mandelbrot set has Euler characteristic $1$ (it is path connected and all its homotopy group vanish)... I understand that some examples might be more topological in nature (for instance the Sierpinski triangle), but considering the examples above, I doubt there might be a direct link to the Hausdorff dimension.
Pierre PC
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1This presentation claims that Euler characteristic of Julia set is non-integer: https://www.maths.ed.ac.uk/~tl/swansea/bmcnotes.pdf – Anixx Apr 26 '23 at 15:50
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1@Anixx This is certainly an interesting take, but it is not the commonly accepted definition of the Euler characteristic as given e.g. on Wikipedia. I am not sure how it relates to the question however, nor to my answer specifically. – Pierre PC May 02 '23 at 12:27
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Unrelated, but I lost my reference for my claim that the Mandelbrot set is path-connected (it was likely false, I think it is unknown) so its Euler characteristic is only conjectured to be 1. It follows from a) the conjecture that it is locally path connected b) its connectedness and the Hahn-Mazurkiewicz theorem c) the fact that it is full d) the fact that full sets have trivial fundamental groups e) that subspaces of the plane are aspherical. – Pierre PC May 02 '23 at 12:38
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@Anixx is there any interpretation of a Julia set as an orbifold? Orbifolds can have rational Euler characteristic and that seems like a first guess at explaining that page of the PowerPoint you linked. – Sidharth Ghoshal Oct 08 '24 at 22:51
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