Algebraically independent Plücker relations

Question

I'm trying to find algebraically independent Plücker relations for $Gr(2,\mathbb{C}^5)$ which generate the ideal. How do I find them and how do we prove if a Plücker relation is algebraically independent?

Answer 1

Question: "I'm trying to find algebraically independent Plücker relations for $Gr(2,C^5)$ which generate the ideal. How do I find them and how do we prove if a Plücker relation is algebraically independent?"

Answer: There is a book (a french and english version) By Manivel ("Symmetric functions and degenrazy loci") where the grassmannian (and flag variety) is studied. In Thm 3.1.6 in the book he gives a set of generators of the ideal $I(\mathbb{G}_{m,n})$ of the grassmannian $\mathbb{G}(m,n)$ (called the "plucker relations"). He also studies in Thm 3.3.4 the graded coordinate ring $R(m,n):=\mathbb{C}[x_i]/I(\mathbb{G}_{m,n})$ and gives a basis for $R(m,n)$ as complex vector space using "standard monomials". There is no mention of "plucker relations that are "algebraically independent" in the sense I describe above.

Note: For $\mathbb{G}(2,4)$ (the grassmannian of $2$-dimensional subspaces of $\mathbb{C}^4$) which is a hypersurface, it follows a generator for this ideal is "algebraically independent". The polynomial ring $A:=\mathbb{C}[x_i]$ is an integral domain and for any non-constant polynomial $f\in A$, it follows $f$ is trivially "algebraically independent" over $\mathbb{C}$. In your case with $n=5$ it follows by a dimension calculation that $\mathbb{G}(2,5)\subseteq \mathbb{P}(\wedge^2 \mathbb{C}^5)\cong \mathbb{P}^9$ and $dim(\mathbb{G}(2,5))=6$, hence $\mathbb{G}(2,5)$ is not a hypersurface.

Answer: In the case of $\mathbb{G}(2,4)$ it follows the ideal $I(\mathbb{G}(2,4))=(H)$ is generated by a polynomial $H$ which is "algebraically independent over $\mathbb{C}$" (in the above sense), and in the above mentioned book you will find a proof that this hypersurface generates the ideal of your grassmannian. In your more general case you must look for a set of generators of $I(\mathbb{G}(2,5))$ that are algebraically independent in the above sense. Maybe the above mentioned book gives references.

Note: Since $X:=\mathbb{G}(m,n) \subseteq \mathbb{P}^d$ is smooth it follows $X$ is a local complete intersection (Hartshorne Thm.II.8.17) and for any closed point $x\in X$ it follows there is an isomorphism

$$\mathcal{O}_{\mathbb{P}^d,x}/I\cong \mathcal{O}_{X,x},$$

and the ideal $I$ can be generated by a regular sequence (HH, Thm.II.8.12A).

A regular sequence $x_1,..,x_n$ in a maximal ideal $\mathfrak{m} \subseteq A$ where $A$ is a finitely generated $\mathbb{C}$-algebra has the property that

$$Gr(A,\mathfrak{m}):=\oplus_n \mathfrak{m}^n/\mathfrak{m}^{n+1} \cong A/\mathfrak{m}[t_1,..,t_n] \cong \mathbb{C}[t_1,..,t_n]$$

(HH.Thm.II.8.21A.e) hence $x_i$ have properties similar to a set of algebraically independent variables.

A projective scheme $X \subseteq \mathbb{P}^d_k$ is a (strict) complete intersection iff the ideal $I(X)$ of $X$ can be defined by $d-dim(X)$ equations. The twisted cubic $i:\mathbb{P}^1 \rightarrow C \subseteq \mathbb{P}^3$ is a smooth curve whose ideal $I(C)$ cannot be generated by two elements, hence $C$ is not a strict complete intersection in $i$. Since $C$ is smooth it is a local complete intersection. I believe $\mathbb{G}(m,n)$ is not a strict complete intersection in the Plucker embedding in general.

The flag variety is projectively normal in the Plucker embedding:

Ramanan, S.; Ramanathan, A. "Projective normality of flag varieties and Schubert varieties". (English) Zbl 0553.14023 Invent. Math. 79, 217-224 (1985).