Software for computing multi-graded Hilbert series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:22:55Z http://mathoverflow.net/feeds/question/20263 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20263/software-for-computing-multi-graded-hilbert-series Software for computing multi-graded Hilbert series Richard Eager 2010-04-03T20:43:32Z 2010-05-02T17:51:15Z <p>The ring of invariants $S^T$ of $k[a,b,c,d]$ under the algebraic torus action $T = k^{*}$ with weights (1,1,-1,-1) is $S = k[ac,ad,bc,bd]$ which has multigraded Hilbert series </p> <p>$\frac{1 - abcd}{(1-ac)(1-ad)(1-bc)(1-bd)}$</p> <p>Is there a software package that can compute multigraded Hilbert series? Can it be computed using Macaulay2?</p> <p>Alternatively, is there software that can compute the multigraded Hilbert series of a toric variety, specified by its fan?</p> <p>For this example $v_1 = (0,0,1), v_2 = (1,0,1), v_3 = (1,1,1), v_4 = (0,1,1)$ specify the vertices of the toric fan. The multigraded Hiblert series is given by the index which counts points in the dual cone $S_{C^{*}}$ $\sum_{m \in S_{C^{*}}} q^m = \frac{(1 - q_1)}{ (1 - q_2)(1 - q_3)(1 - q_1 q_2^{-1}) (1 - q_1 q_3^{-1}) }$</p> <p>References:</p> <p>"Linear diophantine equations and local cohomology," R.P. Stanley, 1982.</p> <p>"Combinatorial commutative algebra," E. Miller and B. Sturmfels, 2005.</p> <p>"Sasaki-Einstein manifolds and volume minimisation," Martelli, Sparks, Yau, 2006.</p> http://mathoverflow.net/questions/20263/software-for-computing-multi-graded-hilbert-series/23235#23235 Answer by Steven Sam for Software for computing multi-graded Hilbert series Steven Sam 2010-05-02T06:39:39Z 2010-05-02T06:39:39Z <p>Macaulay 2 can do multigraded Hilbert series. Let's first assume that you have a presentation of your multigraded ring. I'll mention how to calculate this below. So for your $S = k[ac,ad,bc,bd]$, we'll write it as $S = k[x,y,z,w] / (xz - yw)$.</p> <p>Assuming that each of $a,b,c,d$ has its own degree direction (so the grading is by ${\bf Z}^4$), we input $k[x,y,z,w]$ as</p> <pre><code>S = QQ[x,y,z,w, Degrees=&gt; {{1,0,1,0}, {1,0,0,1}, {0,1,1,0}, {0,1,0,1}}] </code></pre> <p>where here QQ means the rationals. Then we want the Hilbert series of the ideal $(xz - yw)$, so we put</p> <pre><code>i8 : hilbertSeries ideal(y*z-x*w) </code></pre> <p>and the answer is:</p> <pre><code> 1 - T T T T 0 1 2 3 o8 = ---------------------------------------- (1 - T T )(1 - T T )(1 - T T )(1 - T T ) 1 3 1 2 0 3 0 2 </code></pre> <p>If you also need to get the presentation, we can do this as follows. First, download normaliz: <a href="http://www.mathematik.uni-osnabrueck.de/normaliz/" rel="nofollow">http://www.mathematik.uni-osnabrueck.de/normaliz/</a></p> <p>and the Macaulay 2 interface to normaliz (I think this is automatically there in Macaulay 2 1.3.1): <a href="http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/" rel="nofollow">http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/</a></p> <p>After loading the package with</p> <pre><code>loadPackage "Normaliz" </code></pre> <p>Set the path to normaliz in Macaulay 2 using the command</p> <pre><code>setNmzExecPath("path to the executables norm32 and norm64"); </code></pre> <p>The <a href="http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/_torus__Invariants.html" rel="nofollow">torusInvariants</a> command will give you generators for the subring of invariants. Finally, we can use the <a href="http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.3.1/share/doc/Macaulay2/Macaulay2Doc/html/___Ring__Map.html" rel="nofollow">ringmap</a> command to define a surjection from a polynomial ring onto the subring to get the desired ideal.</p> http://mathoverflow.net/questions/20263/software-for-computing-multi-graded-hilbert-series/23267#23267 Answer by Brian Jurgelewicz for Software for computing multi-graded Hilbert series Brian Jurgelewicz 2010-05-02T17:51:15Z 2010-05-02T17:51:15Z <p>Piggyback question: What if the grading is by a f.g. abelian group?</p>