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Consider a form of degree $r$ in $n$, that is, a homogeneous polynomial

$$f(x_1, \ldots, x_n)=\sum_{i_1+\ldots i_n=r}\alpha_{i_1 ... i_n}x_1^{i_1} ... x_n^{i_n} $$

After the linear change of variables $$\delta : x^\prime_1\rightarrow\sum_{j=1}^n\beta_{ij}x_{j}\;\;\;\;\;\;\;\;\;\;1\leq i\leq n,$$

where $\beta_{ij}$ are real or complex numbers, we obtain a new form

$$f^\prime(x^\prime_1, \ldots, x^\prime_n)=\sum_{i_1+\ldots i_n=r}\alpha^\prime_{i_1 ... i_n}{x^\prime_1}^{i_1} ... {x^\prime_n}^{i_n} $$

An invariant of the form $f$ is a polynomial function of the coefficients $\alpha_{i_1 ... i_n}$ that changes only by a factor equal to a power of the determinant of the linear transformation $\delta$ if one replaces the coefficients $\alpha_{i_1 ... i_n}$ of the given base form by the corresponding coefficients $\alpha^\prime_{i_1 ... i_n}$of the linearly transformed form.

This is the definition of invariants given by Hilbert in "Theory of Algebraic Invariants".

My question is: Consider the determinant of the matrix $A=[x_{ij}]_{n×n}$ as a polynomial in the ring $R[x_{11},…,x_{nn}]$. The determinant is a homogeneous polynomial and hence a form. What are the invariants of the determinant form?

Remark: I put this question yesterday in https://math.stackexchange.com/questions/642843/invariants-of-the-determinant-form But was not answered. Also there are no comments.

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    $\begingroup$ I believe that you cannot speak about invariants of a single form; it's invariants of forms (say, with given $n$ and $r$). $\endgroup$
    – Alex Degtyarev
    Commented Jan 19, 2014 at 13:26
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    $\begingroup$ As Alex says, the question does not make sense. Hilbert dicusses the invariants of a generic form - in modern language, of the space of forms of given degree in a given number of variables. $\endgroup$
    – abx
    Commented Jan 19, 2014 at 13:32
  • $\begingroup$ The space of skew $n$-forms on $\mathbb R^n$ is 1-dimensional. There are just 2 orbits under $GL(n)$, zero, and the rest. So no invariants. Under $SO(n)$, each elements of $\Lambda^n (\mathbb R^n)^*$ is invariant. $\endgroup$
    – Peter Michor
    Commented Jan 19, 2014 at 13:36
  • $\begingroup$ Actually, I think that there is an interpretation of this question that does make sense: If ${\mathcal{R}}_{n,r}$ is the ring of invariants (in Hilbert's sense) of polynomials of degree $r$ in $n$ variables with coefficients in a field $F$, then any given form $f$ of degree $r$ in those $n$ variables induces a map $e_f:{\mathcal{R}}_{n,r}\to F$, and the the answer to the question "What are the invariants of $f$?" could just mean "What is $e_f$?". In this particular case, it seems that the OP would like be able to recognize when a polynomial of degree $n$ in $n^2$ variables is the determinant. $\endgroup$
    – Robert Bryant
    Commented Jan 19, 2014 at 14:08

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