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I am looking for a free resolution of the ideal generated by $2\times 2$-minors of a $3\times 3$ -matrix. More precisely let $M$ be a matrix (sorry but I cannot write a matrix for some TeX technical reason) $$ M=\begin{bmatrix} x_{1}& x_{2}& x_{3} \\\ x_{4}& x_{5}& x_{6} \\\ x_{7}& x_{8}& x_{9} \end{bmatrix} $$ whose entries are indeterminates. I would like to find a free resolution of the ideal generated by $2\times 2$-minors of $M$ in the ring $\mathbb{C}[x_{1},\dots,x_{9}]$.

Does anyone know the reference for this resolution? Or make one for me.

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  • $\begingroup$ The issue isn't TeX, it's Markdown, which the software processes before processing TeX and which interprets backslashes a certain way. This can be fixed by using three backslashes instead of two. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 30, 2012 at 1:24
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    $\begingroup$ For this specific case, see the answer below. For the general case, in characteristic 0, Lascoux wrote a paper in the late 70s which gave an very explicit answer up to unknown scalar constants. (He does not even prove these constants exist). Weyman and Pragacz wrote a 100 page paper in the late 80s which is explicit enough for someone who really understands it to write down matrices. Weyman also wrote a book which explains what is going on conceptually with these free resolutions but is a bit less explicit in terms of getting matrices. Much is still unknown in positive characteristic. $\endgroup$
    – Alexander Woo
    Commented Sep 30, 2012 at 5:51
  • $\begingroup$ @Qiaochu Thank you for fixing the question and clarifying what's going on. Do you know where I can read about Markdown for MO? I would like to know how to use TeX on MO for next time (or it may be written somewhere in the instruction on MO). $\endgroup$
    – Ronagh
    Commented Sep 30, 2012 at 6:46
  • $\begingroup$ @Alexander Thanks you for the information. I will take a look at it. $\endgroup$
    – Ronagh
    Commented Sep 30, 2012 at 6:47

2 Answers 2

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The standard way to construct a resolution is the following. Let $X = A^9$ and let $E$ be a 3-dim (trivial) vector bundle on $X$ such that $M \in End(E)$. Consider the relative Grassmannian $p:Gr(2,E) \to X$ and let $U$ be its tautological rank 2 bundle. Let $Z$ be the zero locus of the morphism $U \to p^*E \stackrel{\ M\ }\to p^*E$ on $Gr(2,E)$. Then it is clear that you ask for a resolution of $p(Z)$. What is good is that it is easy to write a resolution for $Z$ on $Gr(2,E)$, and then you can ush it forward to $X$.

Indeed, $Z$ is the zero locus of a regular section of the vector bundle $U^*\otimes p^*E$. Consequently, $O_Z$ has a resolution by the Koszul complex $$ \dots \to \Lambda^2(U\otimes p^*E^*) \to U\otimes p^*E^* \to O_{Gr(2,E)} \to O_Z\to 0. $$ It remains to note that $\Lambda^k(U\otimes p^*E) = \oplus \Sigma^\alpha U\otimes p^*\Sigma^{\alpha^T}E^*$, the sum is over all Young diagrams in a $2\times 3$ rectangle, $\alpha^T$ is the transposed diagram, and $\Sigma^\alpha$ is the Schur functor. To push forward to $X$ one can use the projection formula and Borel--Bott--Weil to compute the cohomology on $Gr(2,E)$. As a result you will get $$ 0 \to O_X \to E\otimes E^* \to sl(E) \oplus sl(E) \to E^*\otimes E \to O_X \to O_{p(Z)} \to 0. $$

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  • $\begingroup$ After pushing-forward a resolution, why do you still get a resolution? Do you need to play around with some double complex? $\endgroup$
    – Jiarui Fei
    Commented May 11, 2014 at 4:12
  • $\begingroup$ I'm aware this is an old answer, but do you have a reference/introduction for these kind of computations? $\endgroup$
    – Fredrik Meyer
    Commented Mar 14, 2017 at 10:57
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Macaulay2, version 1.3.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, SchurRings,
               TangentCone

i1 :  S = QQ[x_(1,1)..x_(3,3)]

o1 = S

o1 : PolynomialRing

i2 : M = transpose genericMatrix(S,x_(1,1),3,3)

o2 = {-1} | x_(1,1) x_(1,2) x_(1,3) |
     {-1} | x_(2,1) x_(2,2) x_(2,3) |
     {-1} | x_(3,1) x_(3,2) x_(3,3) |

             3       3
o2 : Matrix S  <--- S

i3 : I = minors(2,M)

o3 = ideal (- x   x    + x   x   , - x   x    + x   x   , - x   x    + x   x   , - x   x    + x   x   , - x   x    + x   x   , -
               1,2 2,1    1,1 2,2     1,2 3,1    1,1 3,2     2,2 3,1    2,1 3,2     1,3 2,1    1,1 2,3     1,3 3,1    1,1 3,3   
     ----------------------------------------------------------------------------------------------------------------------------
     x   x    + x   x   , - x   x    + x   x   , - x   x    + x   x   , - x   x    + x   x   )
      2,3 3,1    2,1 3,3     1,3 2,2    1,2 2,3     1,3 3,2    1,2 3,3     2,3 3,2    2,2 3,3

o3 : Ideal of S

i4 : (res I).dd

          1                                                                                                                                                                                                                                                                                         9
o4 = 0 : S  <------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- S  : 1
               | x_(1,2)x_(2,1)-x_(1,1)x_(2,2) x_(1,3)x_(2,1)-x_(1,1)x_(2,3) x_(1,3)x_(2,2)-x_(1,2)x_(2,3) x_(1,2)x_(3,1)-x_(1,1)x_(3,2) x_(1,3)x_(3,1)-x_(1,1)x_(3,3) x_(2,2)x_(3,1)-x_(2,1)x_(3,2) x_(2,3)x_(3,1)-x_(2,1)x_(3,3) x_(1,3)x_(3,2)-x_(1,2)x_(3,3) x_(2,3)x_(3,2)-x_(2,2)x_(3,3) |

          9                                                                                                                                                               16
     1 : S  <----------------------------------------------------------------------------------------------------------------------------------------------------------- S   : 2
               {2} | -x_(1,3) x_(2,3)  0        -x_(3,1) 0        x_(3,2)  0        0        x_(3,3)  0        0        x_(3,3)  0        0        0        0        |
               {2} | x_(1,2)  -x_(2,2) 0        0        -x_(3,1) 0        0        x_(3,2)  0        x_(3,3)  0        -x_(3,2) 0        0        0        0        |
               {2} | -x_(1,1) x_(2,1)  0        0        0        0        -x_(3,1) 0        0        0        0        0        -x_(3,2) x_(3,3)  0        0        |
               {2} | 0        0        -x_(1,3) x_(2,1)  0        -x_(2,2) -x_(2,3) 0        -x_(2,3) 0        0        0        0        0        x_(3,3)  0        |
               {2} | 0        0        x_(1,2)  0        x_(2,1)  0        x_(2,2)  -x_(2,2) 0        -x_(2,3) 0        0        0        0        -x_(3,2) 0        |
               {2} | 0        0        0        -x_(1,1) 0        x_(1,2)  0        x_(1,3)  0        0        -x_(2,3) 0        0        0        0        x_(3,3)  |
               {2} | 0        0        0        0        -x_(1,1) 0        0        0        x_(1,2)  x_(1,3)  x_(2,2)  0        0        0        0        -x_(3,2) |
               {2} | 0        0        -x_(1,1) 0        0        0        0        0        0        0        0        x_(2,1)  x_(2,2)  -x_(2,3) x_(3,1)  0        |
               {2} | 0        0        0        0        0        0        -x_(1,1) x_(1,1)  -x_(1,1) 0        -x_(2,1) -x_(1,1) -x_(1,2) x_(1,3)  0        x_(3,1)  |

          16                                                                                                9
     2 : S   <-------------------------------------------------------------------------------------------- S  : 3
                {3} | x_(3,1)  -x_(3,2) 0        -x_(3,3) 0        0        0        0        0        |
                {3} | 0        0        x_(3,1)  0        -x_(3,3) x_(3,2)  0        0        0        |
                {3} | -x_(2,1) x_(2,2)  0        x_(2,3)  0        0        0        0        0        |
                {3} | -x_(1,3) 0        x_(2,3)  0        0        0        -x_(3,3) 0        0        |
                {3} | x_(1,2)  0        -x_(2,2) 0        0        0        x_(3,2)  0        0        |
                {3} | 0        -x_(1,3) 0        0        0        -x_(2,3) 0        x_(3,3)  0        |
                {3} | -x_(1,1) 0        x_(2,1)  0        0        0        0        x_(3,2)  -x_(3,3) |
                {3} | -x_(1,1) x_(1,2)  0        0        x_(2,3)  0        0        0        -x_(3,3) |
                {3} | x_(1,1)  0        0        -x_(1,3) 0        x_(2,2)  0        -x_(3,2) 0        |
                {3} | 0        0        0        x_(1,2)  -x_(2,2) 0        0        0        x_(3,2)  |
                {3} | 0        0        -x_(1,1) 0        x_(1,3)  -x_(1,2) 0        0        0        |
                {3} | -x_(1,1) 0        0        0        x_(2,3)  -x_(2,2) -x_(3,1) 0        0        |
                {3} | 0        x_(1,1)  0        0        0        x_(2,1)  0        -x_(3,1) 0        |
                {3} | 0        0        0        -x_(1,1) x_(2,1)  0        0        0        -x_(3,1) |
                {3} | 0        0        0        0        0        0        x_(2,1)  x_(2,2)  -x_(2,3) |
                {3} | 0        0        0        0        0        0        -x_(1,1) -x_(1,2) x_(1,3)  |

          9                                              1
     3 : S  <------------------------------------------ S  : 4
               {4} | -x_(2,3)x_(3,2)+x_(2,2)x_(3,3) |
               {4} | -x_(2,3)x_(3,1)+x_(2,1)x_(3,3) |
               {4} | -x_(1,3)x_(3,2)+x_(1,2)x_(3,3) |
               {4} | x_(2,2)x_(3,1)-x_(2,1)x_(3,2)  |
               {4} | x_(1,2)x_(3,1)-x_(1,1)x_(3,2)  |
               {4} | x_(1,3)x_(3,1)-x_(1,1)x_(3,3)  |
               {4} | -x_(1,3)x_(2,2)+x_(1,2)x_(2,3) |
               {4} | x_(1,3)x_(2,1)-x_(1,1)x_(2,3)  |
               {4} | x_(1,2)x_(2,1)-x_(1,1)x_(2,2)  |

          1
     4 : S  <----- 0 : 5
               0

o4 : ChainComplexMap
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  • 1
    $\begingroup$ In other words, Macaulay is your friend! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 30, 2012 at 1:40
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    $\begingroup$ IIRC Miller and Sturmfels construct resolutions for these ideals at the end of their Combinatorial Commutative Algebra. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 30, 2012 at 1:49
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    $\begingroup$ @Mariano - that is incorrect - the current known answer is much too complicated to put in a book. $\endgroup$
    – Alexander Woo
    Commented Sep 30, 2012 at 5:46
  • $\begingroup$ @Graham Thank you for the code. I totally forgot about Macaulay! $\endgroup$
    – Ronagh
    Commented Sep 30, 2012 at 6:48

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