Number of Plücker relations for a Grassmannian 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 $Gr(2,5)$?
 A: "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 $1 \leq a < b < c < d \leq n$.
