34
$\begingroup$

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 $1,...,n$.

$\endgroup$
4
  • $\begingroup$ You should probably add the answer as an actual answer. $\endgroup$ Commented Jul 3, 2011 at 22:09
  • $\begingroup$ 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? $\endgroup$
    – JeremyKun
    Commented Mar 8, 2012 at 1:22
  • $\begingroup$ @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. $\endgroup$ Commented Nov 2, 2014 at 21:39
  • $\begingroup$ 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! $\endgroup$ Commented Nov 2, 2014 at 21:53

4 Answers 4

30
$\begingroup$

Yes, the Plücker relations are written down totally explicitly in terms of the polynomials you require on page 110, equation (3.4.10), of Jacobson's book Finite-Dimensional Algebras over Fields. The proof, attributed by the author to Faulkner (a student of his?), is completely down-to-earth: no identifications, no duality,...

Edit Since Martin doesn't have access to the book, I'm adding an online presentation, with the relevant equations on page 21. It is very elementary, with concrete examples, and might appeal to readers whose interest has been whetted by Martin's question.
And the bibliography contains a reference to a masterful article by Kleiman and Laksov, which also contains the Plücker relations handled with minors of determinants and nothing else.

$\endgroup$
6
  • $\begingroup$ Thanks! It would be great if you add these equations to your post. The relevant pages are not in the google / amazon preview. $\endgroup$ Commented Jul 3, 2011 at 14:37
  • $\begingroup$ OK, I've added a document containing these equations . $\endgroup$ Commented Jul 3, 2011 at 19:13
  • 1
    $\begingroup$ The reference to the article by Kleiman-Laksov is great, thank you. $\endgroup$ Commented Jul 3, 2011 at 21:50
  • 2
    $\begingroup$ @OpalE I found an archived copy of the online presentation here: web.archive.org/web/20210507015137/https://… $\endgroup$
    – Ben West
    Commented Apr 5 at 15:21
  • 1
    $\begingroup$ Thank you, @Ben West. $\endgroup$ Commented Apr 6 at 18:00
3
$\begingroup$

I think the answer provided above could be improved a bit. (Not logically, just made a bit clearer.) My method avoids having extra variables and setting repeating sequences equal to zero.

Take a field $F$, and let $$X = \{x_H : H \subset \{1,2,...n\} \text{ and } |H| = d\}$$ So we can think of variables as being indexed by length $d$ increasing sequences. For $I,K \subset \{1,2,...,n\}$ of size $d-1$ and $d+1$, respectively, we can define the $(I,K)$ Plucker relation $Pl_{I,K}(X)$ by the formula:$$ Pl_{I,K}(X) = \sum_{k \in K - I} (-1)^{S_{I,K}(k)} x_{I \cup \{k\}} x_{K - \{k\}} $$ And the sign $S_{I,K}(k) = \#\{i \in I : k < i\} + \#\{\ell \in K: k < \ell\} $

$\endgroup$
2
$\begingroup$

Consider a vector the vector space $W=k^d$, $v_1$, $\ldots$, $v_d$ a basis of $W$. Then for every vector $v$ has a unique expression

$$v = \sum_{i=1}^d c_i v_i$$ where $$c_i = \frac{(v_1, \ldots, v,\ldots, v_d)}{(v_1, \ldots, v_d)}$$ and we denote by the bracket the mixed product (determinant) of $d$ vectors (Cramer's rule).

From the above we get the equality

$$(v_1, \ldots, v_d) v = \sum_{i=1}^d (v_1, \ldots, v, \ldots, v_d) v_i$$ This last equality is valid even if $(v_1, \ldots, v_d) = 0$ ( by continuity). This gives a general relation between $d+1$ vectors in a $d$ dimensional space.


$\bf{Added:}$ Another way to look at the above equality: the $d+1$ vectors $v_1$, $\ldots$, $v_{d+1}$ are linearly dependent, so

$$v_1 \wedge v_2 \wedge \cdots \wedge v_{d+1}=0$$ in $\wedge^{d+1}(V)$ ($0$ itself). But now consider the comultiplication map

$$\wedge^{d+1}(V) \mapsto \wedge^d (V)\otimes V$$

and see $0=v_1 \wedge \cdots v_{d+1}$ maps to

$$\sum_{k=1}^{d+1} (-1)^{d+1-k} (v_1 \wedge \hat v_k \wedge v_{d+1}) \otimes v_k = 0$$


Now, consider $d-1$ more vectors $w_1$, $\ldots$, $w_{d-1}$. From the above we get

$$ (v_1, v_2, \ldots, v_d)(v, w_1, \ldots, w_{d-1}) = \sum_{i=1}^d (v_1, \ldots, v, \ldots, v_d)( v_i, w_1, \ldots, w_{d-1})$$

Example: from $$(v_1, v_2)v_3 =(v_3, v_2) v_1 + (v_1, v_3) v_2 $$ we can get $$( v_1, v_2)(v_3, w) = (v_3, v_2)(v_1, w) + (v_1, v_3)(v_2, w) $$

Now, consider the $v_i$ and the $w_j$ as columns of a matrix $n\times d$ with entries in (a commutative ring) $k$. We get the Plücker relations.

$\bf{Added:}$ While the necessity of conditions are at the level of first year LinAlg, the sufficiency is a bit more involved.

Consider $\omega\in \wedge^d(V)$. Recall that we have a bilinear map ( internal product) $$\wedge^d(V) \times \wedge^{d-1}(V^{\star}) \to V$$

Fixing $\omega$ on the left factor, we get a map $\wedge^{d-1}(V^{\star})\to V$. Denote the image of this map by $S(\omega)$ ( a subspace of $V$ associated to $\omega$). The Plucker relations in invariant form are $$\omega \wedge w = 0$$ for all $w \in L(\omega)$.

Now, use this general lemma valid for every $\omega \in \wedge^d(V)$, $w \in V$.

$\omega \wedge w = 0$ if and only if $\omega = \eta\wedge w$ for some $\eta \in \wedge^{d-1}(W)$.

The proof is now fairly simple.

Note: for every $0\ne\omega \in \wedge^{d}(W)$ we have the space of vector $w$ such that $\omega\wedge w = 0$, denote it by $K(\omega)$ ( the nucleus of $\omega$). We also have $S(\omega)$ from above. Moreover, we also have $S'(\omega)$, the smallest subspace $V'\subset V$ such that $\wedge^d(V') \ni \omega$. Apriori, $S'(\omega) \supset S(\omega)$. It turns out that we have $$S'(\omega) = S(\omega)$$ the span of $\omega$.

Clearly, $K(\omega) \subset S(\omega)$. Moreover, $\wedge^k(K(\omega))$ is a factor of $\omega$ ( the largest possible split factor).

$\endgroup$
0
$\begingroup$

Let $B_r\in G_n^r$ be an $r$-vector in the Clifford geometric algebra $G_n$ of the Euclidean vector space $R^n$. In this setting, the Pl\"ucker relations take the elegant coordinate-free form $(A_{r-1}\cdot B_r)\wedge B_r=0$ for all $(r-1)$-vectors $A_{r-1}\in G_n^{r-1}$. When the quadratic homogeneous Pl\"ucker relations are satisfied, the $r$-vector $B_r$ can be factored into the outer product of $r$ vectors $v_1, \cdots, v_r\in G_n^1$, and an explicit formula can be given for these vectors. I have just today published the paper {\it Notes on Pl\"ucker's Relations in Geometric Algebra}. It can be found on my website: http://www.garretstar.com

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .