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Let $A$ consist of the columns of the $2\times (d+1)$ matrix

$$A=\begin{pmatrix} d & d-1 & \cdots & 1&0\\ 0 & 1 & \cdots & d-1 &d \end{pmatrix}$$

Then consider the map

\begin{array}{lcl}\theta_A:\mathbb{(C^*)}^2\rightarrow\mathbb{P}^d\\ (s,t)\rightarrow[s^d:s^{d-1}t:\dots:st^{d-1}:t^d]\end{array}

We know that $C_d$ is Zariski closure of image of the $\theta_A$ and $C_d$ called the rational normal curve of degree $d$.

Now I want to show that $I(C_d)=\langle x_ix_{j+1}-x_{i+1}x_j: 0\le i <j\le d-1\rangle$.

Consider $$[s^d:s^{d-1}t:\cdots:st^{d-1}:t^d]=[1:\frac{t}{s}:\frac{t^2}{s^2}:\cdots:\frac{t^{d-1}}{s^{d-1}}:\frac{t^d}{s^d}]=[1:u:u^2:\cdots:u^{d-1}:u^d]$$ Then we get $B=\begin{bmatrix}0&1&\cdots&d-1&d\end{bmatrix}$. And also it is clear that $B$ is in the row space of $A$. This gives the map \begin{array}{lcl}\theta_B:\mathbb{(C^*)}^2\rightarrow\mathbb{P}^d\\ (s,t)\rightarrow[1:t:\dots:t^{d-1}:t^d]\end{array} And also I know $C_d$ is Zariski closure of image of the $\theta_B$. But I don't know how can I show that $I(C_d)=\langle x_ix_{j+1}-x_{i+1}x_j: 0\le i <j\le d-1\rangle$. I need a hint for this.

I try to use Proposition 2.1.4 (Cox, Little, Schenk-Toric varieties) but I cant see yet.

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  • $\begingroup$ The columns of $A$ give the Laurent monomials (you see image of the map $\theta_A$) defining the rational normal curve $C_d$. $\endgroup$ Jan 30, 2020 at 18:24
  • $\begingroup$ Okay, but it is not clear to me. How can I show the ideal is equal to ideal of the rational normal curve. I need to hint. $\endgroup$ Jan 30, 2020 at 18:47
  • $\begingroup$ It can probably be done by clever tricks/brute force but to really understand why this is true one needs some representation theory of $SL_2$. If I have time I will explain. $\endgroup$ Jan 30, 2020 at 19:03
  • $\begingroup$ To "close" voters: it is true that the question as is appears as an exercise in many algebraic geometry books, but suitably enlarged (see the end of my answer) it becomes a completely open problem. $\endgroup$ Jan 30, 2020 at 23:29
  • $\begingroup$ I added the "plethysm" tag because this could have easily made it in the book by Fulton and Harris on representation theory as another "A little geometric plethysm" section. See in particular Section 11.3 of the 1991 edition. $\endgroup$ Jan 31, 2020 at 19:16

1 Answer 1

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The ideal of the rational normal curve

Let $V$ be a two-dimensional complex vector space and let $V^{\vee}$ be the dual space of linear scalar-valued functions on $V$. Although representations of $SL(V)$ are self-dual, I will keep the distinction between vectors and covectors in order to avoid confusion. Let $W={\rm Sym}^d(V^{\vee})$ be the space of homogeneous polynomial functions of degree $d$ on $V$. It's good to think of the projective space $\mathbb{P}^d$ here as $\mathbb{P}(W)$. The rational normal curve $C_d$ is the variety of projective classes of nonzero $F$'s in $W$ of the form $F=L^d$ for some linear form $L$ in $V^{\vee}$. The ideal $I$ is homogeneous and therefore admits a graded decomposition $I=\bigoplus_{N\ge 0}I_N$. Now let me take a homogeneous polynomial $P$ of degree $N$ in the ideal. Namely, $$ P\in I_N={\rm Sym}^{N}(W^{\vee})\simeq {\rm Sym}^{N}({Sym}^{d}(V))\subset V^{\otimes Nd}\ . $$ For $F\in W$, one can see the evaluation $P(F)$ as a duality pairing $$ P(F)=\langle P,F\otimes\cdots\otimes F\rangle $$ between $V^{\otimes Nd}$ on the left and $(V^{\vee})^{\otimes Nd}$ on the right, with $F$ appearing $N$ times.

Consider the $SL(V)$ module $W^{\otimes N}$ and the identity map $I_{W^{\otimes N}}$ from this module to itself. Decompose this module into irreducibles. This gives a decomposition of the identity of the form $$ I_{W^{\otimes N}}=\sum_{\alpha\in A}\iota_{\alpha}\circ\pi_{\alpha} $$ where for $\alpha\in A$, $\pi_{\alpha}:I_{W^{\otimes N}}\rightarrow {\rm Sym}^{k_{\alpha}}(V^{\vee})$ is an equivariant projection on an irreducible indicated by the nonnegative integer $k_{\alpha}$, and $\iota_{\alpha}:{\rm Sym}^{k_{\alpha}}(V^{\vee})\rightarrow I_{W^{\otimes N}}$ is a "reinjection". The complete symmetrization corresponding to $k_{\alpha}=Nd$ occurs only once, say for $\alpha=\alpha_0$.

We now have $$ P(F)= \langle P, \iota_{\alpha_0}\circ\pi_{\alpha_0}(F\otimes\cdots\otimes F)\rangle +\sum_{\alpha\in A, \alpha\neq \alpha_0} \langle P, \iota_{\alpha}\circ\pi_{\alpha}(F\otimes\cdots\otimes F)\rangle\ . $$ But the contribution of $\alpha_0$ vanishes because $P$ is in the ideal. In other words $\langle P, L\otimes\cdots\otimes L\rangle$, with $Nd$ tensor factors equal to a linear form $L$, is always zero. This is because powers linearly span the symmetric power. See, e.g., A basis of the symmetric power consisting of powers

Other terms can be analyzed as follows, but one needs to know the explicit structure of the Clebsch-Gordan decomposition. Namely, not just the list of irreducible submodules, but the decomposition of the identity with explicit intertwiners $\pi$ and $\iota$. Moreover, this Clebsch-Gordan decomposition is here iterated $N-1$ times. This involves symmetrizations as well as contractions with a nonzero $SL(V)$ invariant alternating form $\epsilon\in \wedge^2 V$ and also with its dual. Let $e_1,e_2$ be a basis of $V$ and $x_1,x_2$ be the dual basis of $V^{\vee}$. One can take $\epsilon=e_1\otimes e_2-e_2\otimes e_1$. An element $F\in W$ is thus a binary form $$ F(x_1,x_2)=\sum_{i=0}^{d}\binom{d}{i} f_i x_1^{d-i}x_2^{i} $$ with complex coefficients $f_0,\ldots,f_d$.

The key observation is: the terms with $\alpha\neq \alpha_0$ always involve at least one copy of $\epsilon$.

So such a term $\langle P, \iota_{\alpha}\circ\pi_{\alpha}(F\otimes\cdots\otimes F)\rangle$ is a linear combination of terms of the form $\langle \Gamma, F\otimes\cdots\otimes F\rangle$ where $\Gamma$ looks like $$ e_{i_1}\otimes \cdots \otimes e_{i_{a-1}}\otimes e_1\otimes e_{i_{a+1}}\otimes \cdots \otimes e_{i_{b-1}}\otimes e_2\otimes e_{i_{b+1}}\otimes \cdots \otimes e_{i_{Nd}} $$ $$ - e_{i_1}\otimes \cdots \otimes e_{i_{a-1}}\otimes e_2\otimes e_{i_{a+1}}\otimes \cdots \otimes e_{i_{b-1}}\otimes e_1\otimes e_{i_{b+1}}\otimes \cdots \otimes e_{i_{Nd}} $$ for $1\le a<b\le Nd$. Namely, $\epsilon$ has been inserted in the pair of spots given by positions $a$ and $b$ in $V^{\otimes Nd}$, while the other spots are given by basis vectors. Now with a little computation, one sees that $\langle \Gamma, F\otimes\cdots\otimes F\rangle$, if nonzero, is a monomial of degree $N-2$ in the $f$'s times a $2\times 2$ minor of the matrix $$ \begin{pmatrix} f_0 & f_1 & \cdots & f_{d-1} \\ f_1 & f_2 & \cdots & f_{d} \end{pmatrix} \ . $$ QED.

To get an idea of how the decomposition of the identity looks like when $N=2$, see this picture:enter image description here

which is taken from my article "On the volume conjecture for classical spin networks" in JKTR 2012. Depending on what one decides is $V$ or $V^{\vee}$, the lines with arrows at the bottom are $\epsilon$'s, i.e., $e_1\otimes e_2-e_2\otimes e_1$ while the ones on top are their duals $x_1\otimes x_2-x_2\otimes x_1$. The rectangular boxes correspond to symmetrizers. This "microscopic" graphical notation can be abbreviated to a "macroscopic" one as in the picture (from the same article) given by:

enter image description here

This is needed for iteration $N-1$ times.

More generally for an integer partitions $\lambda$ of $d$, one can study the defining ideals of coincident root loci, i.e., varieties $X_{\lambda}$ of forms $F$ which factor as $L_1^{\lambda_1}L_2^{\lambda_1}\cdots$ for suitable linear forms. The rational normal curve corresponds to the one part case $\lambda=(d)$. For partitions with two parts, the ideal was determined by Chipalkatti and myself in the two articles:

  1. "Brill-Gordan Loci, transvectants and an analogue of the Foulkes conjecture" in Adv. Math. 2007.
  2. "The bipartite Brill-Gordan locus and angular momentum" in Transform. Groups 2006.

For a more recent update on the ideals of such loci $X_{\lambda}$, see the article by Lee and Sturmfels, "Duality of multiple root loci", in J. Algebra 2016.


Variant: Since the explicit Clebsch-Gordan decomposition (of course known to Alfred Clebsch and Paul Gordan in the 1870's) is not well known today, one can finish the proof above in a different way. Let $m=Nd$ and consider the action of the group algebra $\mathbb{C}[\mathfrak{S}_m]$ with identity $e$, on $(V^{\vee})^{\otimes m}$. The complete symmetrization corresponds to the element $s=\frac{1}{m!}\sum_{\sigma\in\mathfrak{S}_m}\sigma$. One needs to show that $s-e$ is a linear combination of terms of the form $\rho\ (\tau-e)$ where $\rho$ is a permutation and $\tau$ is a transposition. This reduces to showing the same kind of decomposition for $\sigma-e$, for arbitrary permutations $\sigma$. Write $\sigma$ as a product of transpositions and expand as in the example $$ \tau_2\tau_1-e=\tau_2((\tau_1-e)+e)-e=\tau_2(\tau_1-e)+(\tau_2-e) $$ etc. Because of the Grassmann-Plücker relation, $\tau-e$ produces, in the big tensor product ${\rm Hom}((V^{\vee})^{\otimes m},(V^{\vee})^{\otimes m})\simeq (V^{\vee})^{\otimes m}\otimes V^{\otimes m}$, the insertion of $$ (e_1\otimes e_2-e_2\otimes e_1)\otimes (x_1\otimes x_2-x_2\otimes x_1) $$ as wanted.

Remark: The variant works for more general Veronese embedings with ${\rm dim}(V)>2$. Then, no need then to invoke Grassmann-Plücker, but the distinction of $V$ and $V^{\vee}$ is essential. Interestingly, the explicit Clebsch-Gordan decomposition is not well understood in that case, except for ${\rm dim}(V)=3$ as worked out in this article. Instead of considering the varieties $X_{\lambda}$, another direction for generalization is to look at defining ideals of secants of the Veronese. This also full of open problems which although of classical nature, have recently received much attention as central questions in Geometric Complexity Theory. See, e.g., the books "Tensors: Geometry and Applications" and "Geometry and Complexity Theory" by J.M. Landsberg. Another good reference is, the lecture notes "Introduction to geometric complexity theory" by Bläser and Ikenmeyer.

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