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visits member for 4 years, 8 months
seen 3 mins ago

Data Scientist @ Explorer Media. Statistics, Geometry and Physics.


4m
revised Brun's algorithm
added 540 characters in body
12m
comment Brun's algorithm
@CatherinePfaff do any of these examples on these slides by Shweiger match your definition?
15h
answered Max flow, min cut on manifolds
15h
answered Brun's algorithm
16h
comment Brun's algorithm
possibly, stuff by Valerie Berthé (pdf)?
Jul
19
asked why is this result about Gaussian analytic functions equivalent to the Crofton formula
Jul
17
comment Hilbert Matrix and Approximation Theory
@DavidSpeyer Legendre polynomials also arise as the matrix elements of $SO(3)$ representations. I got interested because of Hilbert's use of Minkowski theorem and geometry of numbers, but never collected my thoughts on this.
Jul
17
comment Hilbert Matrix and Approximation Theory
@DavidSpeyer Do you think Chebyshev polynoimals are a better choice than Legendre polynomials that Hilbert picked? In Legendre basis this is a rational quadratic form so it can take arbitrarily small values as $n \to \infty$, I guess.
Jul
8
revised Degrees of maps from curves to $\mathbb P^1$
typo... cuves - i was rolling on the floor for 10 minutes
Jul
6
asked 1D TQFT in Freed-Hopkins-Lurie-Teleman
Jul
6
accepted Dijkgraaf-Witten TQFT vs. Representation Theory?
Jul
4
revised conjectures regarding a new Renyi information quantity
added 46 characters in body
Jul
4
revised conjectures regarding a new Renyi information quantity
added 218 characters in body
Jul
2
comment Angles and proportions occurring in L-system fractals
You may be intersted in this book Fractal Geometry: Mathematical Foundations and Applications by Kenneth Falconer.
Jul
2
awarded  Socratic
Jul
2
awarded  Inquisitive
Jul
2
awarded  Curious
Jun
30
comment Free Boson Correlator $ \langle X(z)X(w) \rangle =- \ln |z - w| $
@MarcelBischoff $\langle x(z)x(w) \rangle = - \ln (z-w)$ is not always positive right? In fact $$\langle x(z)x(z+1) \rangle = - \ln 1 = 0$$ And it is hard to define the norm $||x(z)|| = \langle x(z)x(z) \rangle$ which would be divergent. So we can't put $x(z)$ into our Hilbert space.
Jun
29
revised Free Boson Correlator $ \langle X(z)X(w) \rangle =- \ln |z - w| $
added 517 characters in body; edited title
Jun
29
comment Free Boson Correlator $ \langle X(z)X(w) \rangle =- \ln |z - w| $
@AndréHenriques p22: the holomorphic part $x(z)$ is not a conformal field. Under a conformal map the metric transforms like $$ ds^2 \mapsto \frac{\partial f}{\partial z} \frac{\partial \overline{f}}{\partial \overline{z}} ds^2 $$ conformal "fields" $\phi(z, \overline{z})$ transform like differential forms, so that $ \phi(z, \overline{z}) dz^h d\overline{z}^{h'}$ is invariant: $$ \phi \mapsto \big(\frac{\partial f}{\partial z}\big)^h \big( \frac{\partial \overline{f}}{\partial \overline{z}}\big)^{h'} \phi $$ Maybe $x(z)$ can have a logarithmic singularity or something.