My question intends to take up a detail in the construction of the Hilbert variety in Harris' book Algebraic Geometry; A First Course (p 274).
Let $S=K[X_0,X_1,..., X_n]$ be the ring of polynomials in $n+1$ indeterminants and denote by $S_m$ the $K$-vector space of homogeneous polynomials of degree $m$ in $S$. It is known that $\dim_K S_m = \binom{m+n}{n}$. Let $\Lambda \subset S_m$ any $K$-subspace of dimension $d$. For any positive integer $k$ we have a natural multiplication map
$$ \Psi_{\Lambda, k}: \Lambda \otimes S_k \to S_{k+m} $$
multiplying the two homogeneous polynomials in $\Lambda \subset S_m$ ans $S_k$. Clearly it's well defined and $K$-linear.
I think that the rank of $\Psi_{\Lambda, k}$ should depend only on $d= \dim \ \Lambda$ and
$k$, but not $\Lambda$ itself; see my arguments below. But going through the conttruction of Hilbert variety in Harris' book would give strange results.
At first why I think it should depend only on $d$ (and $k$), but not on the choice of $\Lambda$ itself? If $ \overline{\Lambda}$ is another $d$-dimensional
subspace of $S_m$ then $\Lambda$ and $\overline{\Lambda}$ are isomorphic /projectively
equivalent since $PSL_n(K)$ induces an action on every homogeneous part $S_m$ and
it acts transitively on every
$d$-Grassmannian $G_d(S_m)=\{ \Lambda \ \ \Lambda \subset S_m, \dim_K \ \Lambda=d \}$.
Therefore it seems that the rank $\operatorname{rank} (\Psi_{\Lambda, k})$ depends only on $d$ and $k$, not $\Lambda$ since $\Psi_{\Lambda, k}$ and $\Psi_{\overline{\Lambda}, k}$ differ by a precomposition with an isomorphism induced by $PSL_n(K)$ that maps $\Lambda$ to $\overline{\Lambda}$.
Let now consider the sketch of the construction Hilbert variety in Harris book (p. 274):
The naively constructed set-theoretic map
$$H: \{ \text{ subvarieties } X \subset \mathbb{P}^n \text{ with Hilbert polynomial } p =p_X \} \to G= G(q(m), N(m))$$
(where $N(m)=\binom{m+n}{n}$ and $q(m)= N(m)-p(m)$) in called Hilbert map.
In fact, the open Hilbert variety is a quasi-projective subset of the Grassmannian $G= G(q(m), N(m))$. To see this, observe that for any vector subspace A c Sm and any positive integer k, we have a multiplication map
$$ \Psi_{\Lambda, k}: \Lambda \otimes S_k \to S_{k+m} $$
if we have a subspace $\Lambda \subset S_m$ of codimension $p(m)$ (so dimension $q(m)$) it will be the $m$th graded piece of the ideal $I(X)$ of a variety $X \subset \mathbb{P}^n$ with Hilbert polynomial $p$ if and only if for every $k$ the rank of $\Psi_{\Lambda, k}$ is what it should be, that is, $q(m + k)$.
But recalling my considerations above the rank of $\Psi_{\Lambda, k}$ seemingly depends only on $q(m)$ and $k$ but not subspace $\Lambda$ itself. So the rank of $\Psi_{\Lambda, k}$ is the same for all $q(m)$ dimensional subspaces $\Lambda \subset S_m$?
But in this case the definition of Hilbert scheme via ranishing rank condition make no sense since then every subspace of $S_m$ of dimension $q(m)$ would satisfy this rank condition. But that's not really what one wants, this would imply that the set of points of Hilbert variety is the complete Grassmannian $G= G(q(m), N(m))$ or it is empty! Can somebody resolve my confusion around the rank of $ \Psi_{\Lambda, k}$ and it's dependence on concrete space $\Lambda$.