Let $f\in k[x_0,...,x_n]_d$ be a degree $d$ homogeneous polynomial in $n+1$ variables.

Is there a way to associate to $f$ a form $g(y_1,...,y_m)$ which is symmetric in the sets of binary variables $y_i = [y_0^i:y_1^i]$ of degree $d_i$ in each set for a suitable choice of $m$ and $d_i$?

For instance we could associate to a polynomial $f\in k[x_0,...,x_3]_2$ the form $g(y_1,y_2) = \sum_{0\leq i_1,i_2\leq 3}a_{i_1,i_2}(y_0^1)^{3-i_1}(y_1^1)^{i_1}(y_0^2)^{3-i_2}(y_1^2)^{i_2}$.

  • $\begingroup$ I see the beginning of an interesting question but it would be nice to know what exactly do you mean by "associated". $\endgroup$ May 25, 2017 at 21:06

1 Answer 1


Not yet an answer but you can see $f$ as an element of the symmetric power $S^d(V^{\vee})$ for $V$ a vector space of dimension $n+1$. You can take $V=S^n(W)$ with $W$ of dimension 2. That should give something like your representation for $g$. The indices labeling the $x$ basis of $V^{\vee}$ become labels for monomials of degree $n$ in your $y$ variables.

More explicitly, if $$ f(x_0,\ldots,x_n)=\sum_{i_1,\ldots i_d=0}^n f_{i_1,\ldots,i_d}x_{1_1}\ldots x_{i_d} $$ with the (physicsy) tensor $f_{i_1,\ldots,i_d}$ being symmetric, then the associated form is $$ g(y^1,\ldots,y^d)=\sum_{i_1,\ldots i_d=0}^n f_{i_1,\ldots,i_d} (y_0^1)^{n-i_1}(y_1^1)^{i_1}\ldots (y_0^d)^{n-i_d}(y_1^d)^{i_d}\ . $$ It is of course essential to now see $f$ as a symmetric multilinear form rather than a homogeneous polynomial, i.e., to work with $n$ distinct pairs of binary variables $y$. This is the idea behind the symbolic method of classical invariant theory (see my answer to this MO question for more details).

  • $\begingroup$ So it was an answer after all. Glad to be of help. BTW may I ask about the motivation for the question? I am interested in everything that has to do with the classical symbolic method. $\endgroup$ May 26, 2017 at 13:16
  • 1
    $\begingroup$ In this book "Algebraic geometry and theta functions" Coble refers sevaral times to such a correspondence without making it explicit. I was just trying to understand what he was talking about. Thanks again. $\endgroup$
    – BlaCa
    May 26, 2017 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy