# Plücker Relations

Let $K$ be a field, $1 \leq d \leq n$ integers and $V$ an $n$-dimensional vector space. The Plücker relations are quadratic forms on $\wedge^d V$ whose zero set is exactly the set of decomposable vectors in $\wedge^d V$ (i.e. which are of the form $v_1 \wedge ... \wedge v_d$), thus describing the ideal corresponding to the Plücker embedding $\text{Gr}_d(V) \to \mathbb{P}(\wedge^d V)$. But in every book I've read so far, these Plücker relations are constructed by means of many identifications between duals, exterior powers, etc. so that I am not able to write them down explicitely. Although I've tried it, many signs and sums confuse me.

Question. Is it possible to write down these Plücker relations explicitely as a set of polynomials in the ring $K[\{x_H\}]$, where $H$ runs through the subsets of $\{1,...,n\}$ with $d$ elements? (Of course it is possible, but I wonder how do this in general)

Edit: Following the answer below, here is the

Answer: Instead of using these subsets $H$, use indices $1 \leq i_1 < ... < i_d \leq n$, and extend the definition of $x_{i_1,...,i_d}$ to all $d$-tuples in such a way that $x_{i_1,...,i_d}=0$ if these $i_j$ are not pairwise distinct, and otherwise $x_{i_1,....,i_d} = sign(\sigma) \cdot x_{i_{\sigma(1)},...,i_{\sigma(d)}}$, where $\sigma$ is the unique permutation of $1,...,d$ which makes $i_{\sigma(1)} < ... < i_{\sigma(d)}$. Then the Plücker relations are

$\sum\limits_{j=0}^{d} (-1)^j x_{i_1,...,i_{d-1},k_j} * x_{k_0,...,\hat{k_j},...,k_d} = 0$

for integers $i_1,...,i_{d-1},k_0,...,k_d$ between $1,...,n$.

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You should probably add the answer as an actual answer. – Mariano Suárez-Alvarez Jul 3 '11 at 22:09
Does that mean there will only be one Plücker relation for some? For instance, if we look at an 8-dimensional vector space and the 4th wedge power, then there is only one sequence 1,2,3,4,5,6,7,8, so is there only one corresponding relation? – JeremyKun Mar 8 '12 at 1:22
@JeremyKun : For each pair of sequences $\{ i_1,\cdots,i_{d-1} \}, \{ k_0,\cdots,k_d \}$ you get a relation, so there are more options than you suggest. In the case $d=4$ and $n=8$, there are $\binom 83$ Plücker relations where all terms do not vanish trivially (because the sequences $\{i_1,i_2,i_3\}$ and $\{k_0,\cdots,k_4\}$ can be chosen disjoint). There is actually more Plücker relations than that, i.e. you also need to count those relations which are non-zero but for which the sets $\{i_1,i_2,i_3\}$ and $\{k_0,\cdots,k_4\}$ intersect. – Patrick Da Silva Nov 2 '14 at 21:39
You suggest that the projective Grassmannian $\mathbb G(3,7)$ of $3$-planes in $\mathbb P^7$ is a quadric in $\mathbb P(\wedge^8 K^4)$ (because the Plücker relations generate the ideal of relations of the Grassmannian and you suggest there's only one relation, a quadric), thus of projective dimension $(\binom 84 - 1)-1 = 68$. We know that the dimension of $\mathbb G(p,n)$ is $(p+1)(n-p)$, so no, in this case you miscounted : The dimension of $\mathbb G(3,7)$ is $(3+1)(7-3) = 16$. So you missed a lot of quadrics! – Patrick Da Silva Nov 2 '14 at 21:53