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Let $\mathbf{k}$ be a commutative $\mathbb{Q}$-algebra. (We could play the same game over any commutative ring $\mathbf{k}$, but this would be a bit more technical, so let me avoid it.)

Fix a nonnegative integer $n$. A binary form shall mean a homogeneous polynomial $f=f\left( x,y\right) \in\mathbf{k}\left[ x,y\right] $. Any binary form $f\left( x,y\right) $ of degree $2n$ can be written uniquely in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n}{i}a_{i} x^{2n-i}y^{i}$ for some scalars $a_{0},a_{1},\ldots,a_{2n}\in\mathbf{k}$ (which are simply the coefficients of $f$, rescaled by binomial coefficients). Let $\mathcal{F}_{2n}$ denote the $\mathbf{k}$-module of all binary forms of degree $2n$. The matrix ring $\mathbf{k}^{2\times2}$ acts on the $\mathbf{k} $-module $\mathcal{F}_{2n}$ by the rule \begin{equation} \left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \cdot f=f\left( ax+cy,bx+dy\right) . \end{equation} (In other words, a matrix $\left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \in\mathbf{k}^{2\times2}$ acts on a binary form by substituting $ax+cy$ and $bx+dy$ for the variables $x$ and $y$. This is called a "change of variables", at least if the matrix is invertible.)

The catalecticant $\operatorname*{Cat}f$ of a binary form $f\left( x,y\right) $ of degree $2n$ is a scalar, defined as follows: Write $f\left( x,y\right) $ in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n} {i}a_{i}x^{2n-i}y^{i}$, then set \begin{equation} \operatorname*{Cat}f=\det\left( \begin{array}[c]{cccc} a_{0} & a_{1} & \cdots & a_{n}\\ a_{1} & a_{2} & \cdots & a_{n+1}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n} & a_{n+1} & \cdots & a_{2n} \end{array} \right) =\det\left( \left( a_{i+j-2}\right) _{1\leq i\leq n+1,\ 1\leq j\leq n+1}\right) . \end{equation} A classical result, going back to Sylvester (On the Principles of the Calculus of Forms) in 1852 (who, in turn, ascribes it to Cayley), then says that $\operatorname*{Cat}f$ is a $\operatorname*{GL} \nolimits_{2}\mathbf{k}$-invariant (in an appropriate sense of this word, i.e., $\operatorname*{GL}\nolimits_{2}\mathbf{k}$ transforms $\operatorname*{Cat}$ by multiplication with an appropriate power of the determinant). More precisely, any binary form $f\left( x,y\right) $ of degree $2n$ and every matrix $A\in\mathbf{k}^{2\times2}$ satisfy \begin{equation} \operatorname*{Cat}\left( A\cdot f\right) =\left( \det A\right) ^{n\left( n+1\right) }\cdot\operatorname*{Cat}f. \label{catalecticant1} \tag{1} \end{equation}

Question. Is there a purely algebraic/combinatorial proof of \eqref{catalecticant1}?

These days, \eqref{catalecticant1} is usually seen as a consequence of the fact that if $\mathbf{k}$ is an algebraically closed field of characteristic $0$, then a binary form $f\left( x,y\right) $ of degree $2n$ satisfies $\operatorname*{Cat}f=0$ if and only if $f$ can be written as $\sum_{j=1} ^{n}\left( p_{j}x+q_{j}y\right) ^{2n}$ for some $2n$ elements $p_{1} ,p_{2},\ldots,p_{n},q_{1},q_{2},\ldots,q_{n}\in\mathbf{k}$. This is proven, e.g., in J. M. Selig, Sylvester's Catalecticant. Once this fact is granted, the equality \eqref{catalecticant1} follows using some technical Nullstellensatz arguments (I believe so; the details are somewhat annoying to verify, requiring e.g. to show that $\operatorname*{Cat}f$ is irreducible as a polynomial in the $a_{0},a_{1},\ldots,a_{2n}$ for $n>0$).

I dislike this argument for the technicalities involved, including the passage to an algebraically closed field and the use of the Nullstellensatz (at least for principal ideals). It seems to me that \eqref{catalecticant1} should have a purely algebraic proof, using properties of determinants. Sylvester might have had one, but I find his paper impossible to decipher. Does anyone know such a proof?

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  • $\begingroup$ the invariance proof in paragraph 209 of An introduction to the algebra of quantics is elementary and does not seem to use the "Nullstellensatz" --- is it incomplete? $\endgroup$ – Carlo Beenakker Sep 15 '17 at 12:21
  • $\begingroup$ @CarloBeenakker: Unfortunately, the proof given there is the Nullstellensatz argument I am speaking of. It's short because some steps are missing (such as the irreducibility of $\operatorname{Cat} f$). $\endgroup$ – darij grinberg Sep 15 '17 at 13:48
  • $\begingroup$ @FrancoisZiegler: Ah, of course! I prefer the following restatement: Let $D$ be the $\mathbf{k}$-algebra of constant-coefficient differential operators on $\mathbf{k}\left[x,y\right]$. It is freely generated (as a commutative algebra) by the partial derivatives $\partial_x$ and $\partial_y$. Let $D_n$ be the $n$-th graded component of $D$; thus, $D_n$ has basis $\left(\partial_x^i \partial_y^{n-i}\right)_{0 \leq i \leq n}$. Now, for any given binary form $f$ of degree $2n$, we can ... $\endgroup$ – darij grinberg Sep 15 '17 at 14:31
  • $\begingroup$ ... define a symmetric bilinear form on $D_n$ by setting $\left< \alpha, \beta \right> := \left(\alpha \beta\right) \left(f\right)$ for all $\alpha, \beta \in D_n$ (since $\alpha \beta$ is a degree-$2n$ homogeneous differential operator, and thus $\left(\alpha \beta\right) \left(f\right)$ is a constant). The matrix representing this form with respect to the usual basis of $D_n$ is the matrix whose determinant defines $\operatorname{Cat} f$ (maybe up to some scalar factors). It remains to observe that this matrix is functorial in $f$ in an appropriate sense. $\endgroup$ – darij grinberg Sep 15 '17 at 14:37
  • $\begingroup$ Let me know if you want to turn this into an answer; elsewise I will when I get the time. $\endgroup$ – darij grinberg Sep 15 '17 at 14:41
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Dolgachev (2012, p. 57; pdf) observes that your matrix $\left( a_{i+j-2}\right) _{1\leq i\leq n+1,\ 1\leq j\leq n+1}$ (with determinant $\operatorname{Cat} f$) is the matrix of a symmetric bilinear form $\Omega_{\,f}$ on $\smash{\operatorname{Sym}^n(\mathbf k^{2})}$ in a certain basis, then states as obvious that $f\mapsto\Omega_{\,f}$ is a $\smash{\operatorname{GL}(\mathbf k^2)}$-equivariant map $$ \operatorname{Sym}^{2n}(\mathbf k^{2})^*\to\operatorname{Sym}^2(\operatorname{Sym}^n(\mathbf k^{2}))^*, $$ making the first representation a direct summand in the second. (The latter’s full decomposition is in Fulton-Harris, Exercises 6.16, 11.31, 15.45.) That should help...


Edit: Here is a detailed proof. Write $\mathcal F_n=\operatorname{Sym}^n(\mathbf k^{2})^*$. Functoriality gives us morphisms $$ \begin{array}{ccccc} \operatorname{End}(\mathbf k^2) & \overset{\operatorname{Sym}^{n*}}\longrightarrow & \operatorname{End}(\mathcal F_n) & \overset{\operatorname{Sym}^2}\longrightarrow & \operatorname{End}(\operatorname{Sym}^2(\mathcal F_n))\\ A & \longmapsto & B_n & \longmapsto & C_n \end{array} $$ which satisfy $\det(B_n)= \det(A)^{n(n+1)/2}$ and $C_n(\Omega)=\Omega({}^tB_n\,\cdot,{}^tB_n\,\cdot)=B_n\,\Omega\, {}^tB_n$ if we regard quadratic forms as symmetric matrices; ${}^t$ is transposition $\operatorname{End}(V^*)\leftrightarrow\operatorname{End}(V)$. Thus, $$ \det(C_n(\Omega))=\det(B_n)^2\det(\Omega)=\det(A)^{n(n+1)}\det(\Omega). $$ So your $(1)$ will indeed follow from the claimed equivariance of $\smash{f\mapsto\Omega_f}$. To see the latter, note that (viewing as you do members of $\mathcal{F}_{2n}$ as functions of $\smash{(\begin{smallmatrix}x\\y\end{smallmatrix})}$) Dolgachev’s definition can be written $$ \Omega_{\,f}(\xi,\eta):=\tilde f(\xi\eta), \quad\text{where}\quad \tilde f=\tfrac1{(2n)!}\mathrm D^{2n}f(0) $$ (polar form of $f$) and $\xi\eta\in\operatorname{Sym}^{2n}(\mathbf k^2)$ is the product of $\xi,\eta\in\operatorname{Sym}^n(\mathbf k^2)$. So equivariance boils down to the chain rule $\mathrm D^{2n}(\,f\circ {}^t\!A)(0) = \mathrm D^{2n}f(0)\circ{}^tB_{2n}$ (e.g. Abraham et al. (1988, Ex. 2.3E)): $$ \Omega_{\,f\circ\, {}^tA}(\xi,\eta) = (\,\tilde f\circ{}^tB_{2n})(\xi\eta) = \tilde f({}^tB_n\xi\,{}^tB_n\eta) =C_n(\Omega_{\,f})(\xi,\eta). $$ Remark: Defining coefficients $a_i$ of $f$ as you do, and writing $e_1=(\begin{smallmatrix}1\\0\end{smallmatrix})$, $e_2=(\begin{smallmatrix}0\\1\end{smallmatrix})$, the elementary relation $ \mathrm D^{2n}(x^iy^{\,j})(0)(e_1^ke_2^l)=(\partial_x^{\,k}\partial_y^{\,l})(x^iy^{\,j})=i!j!\delta_{ik}\delta_{jl} $ whenever $i+j=k+l=2n$, readily implies that the matrix of $\Omega_f$ indeed has $i,j$-th entry $ \Omega_f(e_1^{\,n-i}e_2^{\,i},e_1^{\,n-j}e_2^{\,j})=a_{i+j}$ $(i, j = 0,\dots,n). $ Note also that parts of this are already in Reznick (1992, 1.19, 1.24, 1.30, 1.31), (1996, 2.12).

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    $\begingroup$ Terminological correction: it's Toeplitz but not circulant. $\endgroup$ – Noam D. Elkies Jul 18 '18 at 22:19
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    $\begingroup$ @NoamD.Elkies I think you have a point. I was following Dolgachev who writes (p. 55 with $d=2n$): “Then $$\mathrm{Cat}_k(f)=\begin{pmatrix}a_0&a_1&\dots&a_k\\a_1&a_2&\dots&a_{k+1}\\\vdots&\vdots&\vdots&\vdots\\a_{d-k}&a_{d-k+1}&\dots&a_d\end{pmatrix}.$$A square matrix of this type is called a circulant matrix, or a Hankel matrix.” $\endgroup$ – Francois Ziegler Jul 18 '18 at 22:45
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    $\begingroup$ Hankel, yes; that and Toeplitz mean the same thing. Circulant would require the additional condition that each $a_{i+k+1} = a_i$ (and then the determinant factors into linear forms over the $k+1$st cyclotomic field). $\endgroup$ – Noam D. Elkies Jul 18 '18 at 23:34
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Let $d=2n$ be the degree of your binary form $f$. Let me introduce $n+1$ pairs of formal variables $\alpha^{(1)}=(\alpha^{(1)}_{1},\alpha^{(1)}_{2}),\ldots, \alpha^{(n+1)}=(\alpha^{(n+1)}_{1},\alpha^{(n+1)}_{2})$. Let $\mathcal{S}$ be the polynomial $$ \mathcal{S}=\prod_{1\le i<j\le n+1} (\alpha^{(i)}\alpha^{(j)})^2 $$ where I used the classical bracket notation for $2\times 2$ determinants, namely, $$ (\alpha^{(i)}\alpha^{(j)}):= \alpha^{(i)}_{1}\alpha^{(j)}_{2}-\alpha^{(i)}_{2}\alpha^{(j)}_{1}\ . $$ Let $\mathcal{D}$ be the differential operator $$ \mathcal{D}=f\left(\frac{\partial}{\partial\alpha^{(1)}_{1}}, \frac{\partial}{\partial\alpha^{(1)}_{2}}\right)\cdots f\left(\frac{\partial}{\partial\alpha^{(n+1)}_{1}}, \frac{\partial}{\partial\alpha^{(n+1)}_{2}}\right)\ . $$ Then the catalecticant is up to a numerical factor equal to the result of applying $\mathcal{D}$ to $\mathcal{S}$. The latter is the symbolic expression of the catalecticant and corresponds to a directed graph on $n+1$ vertices which is basically the complete graph with doubled edges. So each vertex is of degree $d$. As a polynomial $\mathcal{S}$ is multihomogeneous of degree $d$ in each $\alpha^{(i)}$. More generally for any graph of this kind the resulting polynomial $\mathcal{D}\mathcal{S}$ will be an $SL_2$-invariant. To prove equality with your Hankel determinant, see $\mathcal{S}$ as the square of a (homogenized $\mathbb{A}^1\rightarrow \mathbb{P}^1$) Vandermonde determinant. Take the square of the matrix and then hit its determinant with $\mathcal{D}$.

An alternate construction (essentially the same as the answer by François) is to take the determinant of the map $$ \begin{array}{ccc} {\rm Sym}^n(\mathbf{k}^2) & \longrightarrow & {\rm Sym}^n(\mathbf{k}^2)\\ g & \longmapsto & (f,g)_{n} \end{array} $$ where $(f,g)_n$ is the $n$-th transvectant, i.e., the equivariant projection $$ {\rm Sym}^{2n}(\mathbf{k}^2)\otimes {\rm Sym}^{n}(\mathbf{k}^2)\longrightarrow {\rm Sym}^n(\mathbf{k}^2) $$ You can learn more about symbolic forms and transvectants in Section 2 of my article "On the volume conjecture for classical spin networks". J. Knot Theory Ramifications 21 (2012), no. 3, 1250022. See in particular Page 14 which contains a graphical proof of the $SL_2$-invariance property. The example is that of the canonisant of the quintic, but you can remove the three $x$ legs and you get the proof for the catalecticant of the binary quartic. Essentially you need the graphical rule for the action of $g\in SL_2$ on a binary form $F$, i.e.,

enter image description here

Then you need the basic identity for determinants:

enter image description here

and finally the rest of the proof:

enter image description here

Again, remove the three $x$'s to do the catalecticant of the quartic instead of the canonisant of the quintic.

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  • $\begingroup$ I had not seen this answer to the question in its first time around! Very nice! $\endgroup$ – paul garrett Jul 18 '18 at 22:35

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