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 Post subject: modality of singularities
PostPosted: Wed Jun 18, 2008 2:20 pm 
I haven't found a command to calculate the modality of a singular point.
In some simple cases this can be done by running classify(f);

Guess this works for some lowest cases only.
But it does not recognize (??) e.g. the case (x^5+y^6)


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PostPosted: Fri Jun 20, 2008 5:59 pm 
Computing the modality is a non-trivial task and there is no general algorithm available besides the classification, except for plane curve singularities. Moreover, we have to distinguish the modality for right- and contact-equivalence (the latter is even more complicated than the first).

For curve singularities we can compute the right-modality with the help of an embedded resolution, cf.
G.-M. Greuel, C. Lossen, E. Shustin: Introduction to Singularities and Deformations. Springer Verlag, Berlin, Heidelberg, New York (2006), p. 373, Remarks and Exercises.
Or by the algorithm developed in
A. Campillo, G.-M. Greuel, C. Lossen: Equisingular Calculations for Plane Curve Singularities. J. Symb. Comput. 42 (2007), 89-114.
The latter is implemented in Singular in equising.lib.

If mod denotes the right modality of an isolated hypersurface singularity f then mod = dimension of the mu-constant stratum of f in the base of the semiuniversal deformation of f, where mu is the Milnor number of f.
Hence, if f is semi-quasihomogeneous or Newton-nondegenerate, it can be computed with the help of the Newton diagram of f.

A discussion of this with examples about how to use the SINGULAR library equising.lib to compute the modality or, equivalently, the codimension tau_es of the mu-constant stratum for arbitrary plane curve singularities can be found at the above mentioned place too.

Example:
LIB "equising.lib";
ring r = 0,(x,y),ds;
poly f = x5+y6;
milnor(f) - tau_es(f);

// -> 3 i.e. the right-modality is 3
(printlevel = 1; shows additional information)
[/quote]


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PostPosted: Sat Jun 21, 2008 6:58 am 
Thanks!


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 Post subject: Something is strange with \tau^{es} !!
PostPosted: Sat Jun 21, 2008 8:23 am 
The command
poly f = (x2+y2)*(y2-(x-y)^3)*(x^3+y9);
tau_es(f);

returns: tau_es(f)=37

while the command
poly f = ((x+y)^2+y2)*(y2-x3)*((x+y)^3+y9);
tau_es(f);

returns: tau_es(f)=23

note, however, that the polynomials differ by just a linear change of variables x->x+y


A bug in Singular?


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PostPosted: Mon Jun 23, 2008 2:00 am 

Joined: Mon Aug 29, 2005 9:22 am
Posts: 41
Location: Kaiserslautern, Germany
Indeed, this is a bug. :(
It will (hopefully) be fixed in the next version of SINGULAR.
So far tau_es works only correct in the Newton nondegenerate case.


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 Post subject: bug fix
PostPosted: Tue Jul 08, 2008 4:00 pm 

Joined: Wed May 25, 2005 4:16 pm
Posts: 275
get a new version of the library hnoether.lib
from
ftp://www.mathematik.uni-kl.de/pub/Math ... oether.lib
This should solve this issue.

Hans Schoenemann


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 Post subject: thanks!
PostPosted: Thu Jul 10, 2008 7:25 am 
:D


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