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Is it true that the number of Plücker relations for a Grassmannian $Gr(k,n)$ is equal to the dimension $k(n-k)$ of said Grassmannian? So far, for $Gr(2,5)$, I get exactly five Plücker relations: $$p_{12}p_{34}+ p_{23}p_{14}- p_{13}p_{24}=0,$$ $$p_{12}p_{35}+p_{23}p_{15}- p_{13}p_{25}=0,$$ $$p_{12}p_{45}+p_{24}p_{15}- p_{14}p_{25}=0,$$ $$p_{13}p_{45}+p_{34}p_{15}- p_{14}p_{35}=0,$$ $$p_{23}p_{45}+p_{34}p_{25}- p_{24}p_{35}=0.$$ I'm not sure, but I think that three of the above relations are algebraically independent, no? Are there supposed to be six Plücker relations for $Gr(2,5)$?

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    $\begingroup$ See Chapter 14.2 of Miller and Sturmfels' Combinatorial Commutative Algebra for a detailed discussion of the minimal Plücker relations. $\endgroup$ Feb 9, 2016 at 23:10
  • $\begingroup$ The part of Chapter 14.2 concerning Gröbner bases is well beyond me. $\endgroup$
    – Libertron
    Feb 9, 2016 at 23:25

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"The number of Plucker relations" is a little ambiguous, but there is no sense in which it is $k(n-k)$.

The number of Plucker coordinates is $\binom{n}{k}$, so the number of degree $2$ monomials in Plucker coordinates is $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right)$. The vector space they span inside the homogenous coordinate ring of the Grassmannian has dimension $$\frac{1}{k+1} \binom{n}{k} \binom{n+1}{k}.$$ (Derivation available on request.) So a minimal list of relations between them would be of size $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right) - \tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k}$. Note that, if we fix $k$ and let $n$ grow, then $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right) \approx \tfrac{n^{2k}}{2 (k!)^2}$ and $\tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k} \approx \tfrac{n^{2k}}{(k+1) (k!)^2}$. So the number of relations is growing like $\tfrac{k-1}{k+1} \tfrac{n^{2k}}{(k!)^2}$, not $kn$.

Now, many people mean specific lists of relations when they say "the Plucker relations". They don't always agree on which relations they mean, and they don't always mean an irredundant list. But, adding redundant relations would just make the list longer.


In your particular case of $G(2,5)$, the $5$ relations you gave form a basis for the space of relations. In general, there is an $\binom{n}{4}$-dimensional space of relations for $G(2,n)$, and I think everyone would agree that the best basis is the relations of the form $p_{ab} p_{cd} - p_{ac} p_{bd} + p_{ad} p_{bc}$, for $1 \leq a < b < c < d \leq n$.

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  • $\begingroup$ Of the five relations I gave for $Gr(2,5)$, how do I determine which ones are algebraically independent? $\endgroup$
    – Libertron
    Feb 10, 2016 at 17:16
  • $\begingroup$ I would use the Jacobian criterion math.stackexchange.com/questions/1258530/… $\endgroup$ Feb 10, 2016 at 17:35
  • $\begingroup$ When considering index sets, are there particular criteria that lead to trivial Plücker relations? Evidently, for $k=2$ if the index sets have at most $2$ indices in common, then that should result in trivial Plücker relations I think. $\endgroup$
    – Libertron
    Feb 11, 2016 at 2:12
  • $\begingroup$ @Libertron Given we know that $G(2,5) \subset \mathbb{P}^9$ has dimension $6$ shouldn't we be able to determine there are three independent Plücker relations? I think a complete intersection of three hypersurfaces in general position $ \mathbb{P}^9$ should have dimension $6.$ $\endgroup$
    – user100272
    Aug 28, 2017 at 20:29
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    $\begingroup$ @JacobA.Gross But $G(2,5)$ is not a complete intersection! It has degree $5$, not $8$. If you intersect three of the Plucker relations for $G(2,5)$, there will be a second, degree $3$, component. $\endgroup$ Aug 28, 2017 at 22:11
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The total number of Plücker relations is not the relevant question, since they are not independent; algebraic relations hold between them. In fact on various dense (Zariski) open sets, defined by the nonvanishing of a small subset of the Plücker coordinates, all Plücker relations can be deduced from a much smaller set of 3-term Plücker ones, called "short Plücker relations", and even these are not all independent algebraically. A subset of these, however, can be chosen, which are independent, such that the number of these is: $$\frac{n!}{k!(n-k)!} − k(n−k)−1,$$ which is the codimension of the Grassmannian $Gr_k(n)$ in the projectivized exterior power space $\mathbb{P}(\bigwedge^k(\mathbb{C}^n))$. Therefore, this gives an effective representation of the image of the Plücker map embedding: $Pl : Gr_k (n) \to \mathbb{P}(\bigwedge^k(\mathbb{C}^n)),$ valid on Zariski open sets. A reference in which this is explained in detail is: Appendix C.7 of "Tau Functions and their Applications", J. Harnad and F. Balogh, Monographs on Mathematical Physics series, Cambridge University Press, Cambridge, UK (2021).

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