The following determinant has come up in my research: \begin{align} D(x,y,z)=\det\begin{pmatrix} x & 0 & 0 & \nu_{11} & \nu_{21} & \nu_{31} \\ 0 & y & 0 & \nu_{12} & \nu_{22} & \nu_{32} \\ 0 & 0 & z & \nu_{13} & \nu_{23} & \nu_{33} \\ y & x & 0 & \nu_{14} & \nu_{24} & \nu_{34} \\ 0 & z & y & \nu_{15} & \nu_{25} & \nu_{35} \\ z & 0 & x & \nu_{16} & \nu_{26} & \nu_{36} \\ \end{pmatrix} \end{align}

Here $x,y,z$ are variables and I think of the last 3 columns as defining some point $\nu$ in the Grassmannian of 3-planes in $\mathbb{R}^6$, $Gr(3,6)$. Note that $D$ is a real homogeneous cubic polynomial in $x,y,z$ so (when it is not the zero polynomial) its vanishing defines a real projective plane cubic curve $C_\nu\subset \mathbb{RP}^2$ (it is not hard to check that this is well-defined). Further note that the space of cubic plane curves and $Gr(3,6)$ are both 9 dimensional.

edit (23 March 2016): I have since found out that I should probably think of the first three columns as arising from the Jacobian of the Veronese surface (modulo factors of 2 which I could add back). Still no significant progress with the following though.

I would like to understand the map between $\nu$ and the curve $C_\nu$ better, but as I am a novice both in algebraic geometry and the study of the Grassmannian, I don't know where to begin, nor even to what extent this is reasonable.

Here are some questions that come to mind:

Can $C_\nu$ be any real cubic plane curve as $\nu$ ranges over $Gr(3,6)$? (What's the best way of finding $\nu$ given a cubic curve (e.g. what equations to solve numerically, or what geometrical construction)?)

Can the sign of the discriminant of $C_\nu$ be understood in terms of some properties of $\nu$, i.e. whether it lies in some particular Schubert cells or cells in some other kind of decomposition of $Gr(3,6)$?

I would appreciate any advice or pointers to related literature.

A few things that I tried to get some insight:

This determinant can be computed by a Laplace expansion in complementary 3 by 3 minors. The bracket $[ijk]$ denotes the determinant of the 3 by 3 matrix formed by rows $i,j,k$ of the last 3 columns, so for example, \begin{align} [235]&=\det\begin{pmatrix} \nu_{12} & \nu_{22} & \nu_{32} \\ \nu_{13} & \nu_{23} & \nu_{33} \\ \nu_{15} & \nu_{25} & \nu_{35} \end{pmatrix}. \end{align}

Then (modulo errors in transcription) we have: \begin{align} D(x,y,z)&=[235]x^3+[136]y^3-[124]z^3 +([236]-[345])x^2y+([234]-[256])x^2z \\ &+(-[135]+[346])xy^2-([134]+[156])y^2z +([125]+[246])xz^2\\ &+([126]-[145])yz^2 +(-2[123]+[456])xyz. \end{align}

It was not clear to me that staring at this or blindly plugging this into the some explicit formula for the cubic discriminant would get me anywhere.

edit: After reading the answer of dhy below, I realized that I should point out that the brackets [146], [245] and [356] do not appear in the expression above. Thus the points in $Gr(3,6)$ where these are the only nonzero Plucker coordinates will cause $D$ to vanish identically. In fact, the coefficients of $D(x,y,z)$ are all linear combinations of singletons or pairs of Plucker coordinates, and the brackets appearing in distinct coefficient are distinct. This yields another question about the map above.

How do I characterize the set of $\nu$ that cause $D$ to vanish? (Do I just have to compute some Groebner basis of an ideal generated by the Plucker relations and these coefficients?) (I asked a separate question on this here.)

Another thought was to try to exploit the interpretation of $Gr(3,6)$ as the space of rank 3 configurations of 6 vectors in $\mathbb{R}^3$ modulo $GL(3,\mathbb{R})$ but I also didn't get anywhere with that.

  • 4
    $\begingroup$ Here's a possibly silly question, but I just want to be sure before I thought about it any further: Is it possible that you have a misprint in your matrix? Based on symmetry considerations, it would make more sense (i.e., be more symmetric) if the 4-th row started out $(y,x,0)$ rather than $(x,y,0)$. (I would also have rearranged the last 3 rows, moving the 4-th row to the 6-th row and moving the 5-th and 6-th up to 4-th and 5-th respectively, which would make the putative symmetry more apparent.) $\endgroup$ Mar 7, 2016 at 12:22
  • $\begingroup$ @RobertBryant Thanks for the correction! I will leave the ordering of the last 3 rows alone for now as I don't want to have to edit my bracket expression. $\endgroup$
    – j.c.
    Mar 7, 2016 at 14:54
  • $\begingroup$ Ah, so that does make more sense. The row-permutation is minor, of course, though it shouldn't be hard to change in the bracket expressions, just cyclically permute (4,5,6). $\endgroup$ Mar 7, 2016 at 16:22

1 Answer 1


Your map (assuming that the determinant is nonzero for all elements of the grassmannian) is a projection of the Plucker embedding. This tells you that the degree of the corresponding map of complex varieties is 42. Apparently in this situation the number of real points in a fiber is congruent to the number of complex points mod $4$ (as opposed to the usual $2$); see http://arxiv.org/abs/1211.7160v3. This immediately tells you that your map is surjective on real points.

  • $\begingroup$ Thanks for this answer. It appears that the determinant does vanish for some elements of the Grassmannian. I should have checked this earlier. For instance, in the Laplace expansion above, there are several brackets which do not appear ([146], [245], and [356]). Thus the point where e.g. [146] is the only nonvanishing Plucker coordinate is one where the determinant vanishes. I may need some more details for the rest of your answer though I will study the paper you linked. $\endgroup$
    – j.c.
    Mar 7, 2016 at 6:15

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