Let $K$ be a field with characteristic $0$.

Let $G:=G(d,nd)$ the Grassmannian of all $d-$dimensional subspaces of $K^{nd}$ and let $H:=O_d(K)^n$ the n-fold direct product of the orthogonal group. $H$ operates on $G$ in the following way: Let $v_1, \ldots , v_d \in K^{nd}$, $U_1, \ldots , U_n \in O_d(K)$ and let $U \in K^{nd \times nd}$ be the block diagonal matrix with entries $U_1, \ldots , U_n$. We define the group operation $(U_1, \ldots , U_n) ([v_1 \wedge \ldots \wedge v_d]):= [Uv_1 \wedge \ldots Uv_d]$. I am interested in the quotient variety $G//H$. Is there already a description of it?

If not, I considered the group action of $H$ on a open affine $H-$invariant subset of $G$ and reduced it to the following question:

For $1 \leq k \leq n$ and $1 \leq i,j \leq d$ consider the variables $x_{ij}^k$ (thereby $k,i,j$ are indices). Let $R:=K[\sum_{m=1}^d x_{mi}^k x_{mj}^k : 1 \leq k \leq n, 1 \leq i,j \leq d]$ and $S:=K[\sum_{m=1}^d x_{im}^k x_{jm}^l: 1 \leq k,l \leq n, 1\leq i,j \leq d]$. In other words: $R$ is generated by all forms $([x_{1i}^k, \ldots , x_{di}^k] , [x_{1j}^k, \ldots , x_{dj}^k])$, and $S$ is generated by all forms $([x_{i1}^k, \ldots , x_{id}^k] , [x_{j1}^l, \ldots , x_{jd}^l])$. $(\cdot, \cdot)$ denote the standard scalar product. What are the generators of $R \cap S$? ($R \cap S$ is the invariant ring of the described group action on an affine subset)

  • $\begingroup$ Let $M_d$ be the space of $d\times d$ matrices, carrying an action of $GL(d)$ on the left, and $O(d) \leq GL(d)$ on the right. Then $GL(d) \backslash \backslash (M_d)^n = G(d,nd)$, so your question is about $GL(d) \backslash \backslash (M_d)^n // O(d)^n = GL(d) \backslash \backslash (M_d // O(d))^n$. I'm guessing that $M_d // O(d)$ is the space of symmetric (Gram) matrices, but I'm not sure. Anyway then the question would be about $GL(d)$ acting diagonally on $n$ copies of the symmetric $d\times d$ matrices, where each action is by $g\cdot S = g S g^T$ as usual. $\endgroup$ Commented Sep 14, 2012 at 12:55
  • $\begingroup$ Yes, that is true. Yours is a more natural formulation of my problem. I will pose a new question: mathoverflow.net/questions/107340/… $\endgroup$
    – Döni
    Commented Sep 16, 2012 at 19:21


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