Post new topic Reply to topic  [ 7 posts ] 
Author Message
 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)


Report this post
Top
  
Reply with quote  
 Post subject:
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]


Report this post
Top
  
Reply with quote  
 Post subject:
PostPosted: Sat Jun 21, 2008 6:58 am 
Thanks!


Report this post
Top
  
Reply with quote  
 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?


Report this post
Top
  
Reply with quote  
 Post subject:
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.


Report this post
Top
 Profile  
Reply with quote  
 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


Report this post
Top
 Profile  
Reply with quote  
 Post subject: thanks!
PostPosted: Thu Jul 10, 2008 7:25 am 
:D


Report this post
Top
  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 7 posts ] 

You can post new topics in this forum
You can reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

It is currently Fri May 13, 2022 10:56 am
cron
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group