Plücker Relation: misunderstanding? I'm trying to understand exterior algebra better by gaining some "bare hands" understanding of the exterior powers $\Lambda^k(X)$ in more detail when $\dim(X)$ is small. I think so far I understand the cases $\dim(X) = 1,2,3$ quite well. The next case $\dim(X) =4$ is giving me more trouble. 
So, let us take $X = \mathbb{R}^4$. We have, as usual $\Lambda^1(\mathbb{R}^4) = \mathbb{R}$ and $\Lambda^1(\mathbb{R}^4) = \mathbb{R}^4$. The exterior square $\Lambda^2(\mathbb{R}^4)$ is more interesting because $4$ is the smallest dimension such that there exist elements in $\Lambda^2(\mathbb{R}^4)$ besides those of the form $u \wedge v$ for $u,v \in \mathbb{R}^4$. Notably, there are the "symplectic elements" like $e_1 \wedge e_2 + e_3 \wedge e_4$. I can see that they cannot be obtained as wedges of two vectors. Also, I think I can see that they are all congugate under the action of $GL(4)$ on $\Lambda^2(\mathbb{R}^4)$. 

Initial Question: It seems to me that there are precisely two types of elements in $\Lambda^2(\mathbb{R}^2)$, each of which constitutes an orbit of the $GL(4)$-action. The "symplectic elements" and the ones which are wedges of two vectors. Is this correct? Or is there some intermediate type of element I am missing?

In trying to confirm the answer to the above question was "yes" I spend some time browsing through wikipedia articles like these ones on Plücker coordinates. A lot of what is there is superfluous to my needs, but I think I correctly understood that there is a mapping
$$ \{ \text{planes in }\mathbb{R}^4 \} \longrightarrow \{ \text{lines in }\Lambda^2(\mathbb{R}^4)\}$$
which sends
$$ \mathrm{span}\{u,v\} \to \mathrm{span}( u \wedge v).$$
In terms of coordinates, this map sends the column space of a rank $2$ matrix 
$$ A =  \begin{bmatrix}
u_1 & v_1 \\
u_2 & v_2 \\
u_3 & v_3 \\
u_4 & v_4 \\
\end{bmatrix} $$
to 
$$ X_{12} e_1 \wedge e_2 + X_{13} e_1 \wedge e_3  + X_{14} e_1 \wedge e_4 +
X_{23} e_2 \wedge e_3  + X_{24} e_2 \wedge e_4  + X_{34} e_3 \wedge e_4 $$
where $X_{ij}$ is the determinant of the $2 \times 2$ submatrix of $A$ consisting of Rows $i$ and $j$. It's not too hard to see this map is well defined. It suffices to check performing elementary operations on the matrix can only scale the Plücker coordinates. Indeed:


*

*Switching the  columns of $A$ changes the sign of each $X_{ij}$.

*Adding a multiple of one column of $A$ to the other leaves each $X_{ij}$ invariant.

*Scaling a column of $A$ results in scaling each $X_{ij}$ by the same amount. 


Now from this section, I gathered that you are supposed to be able to detect precisely which elements of $\Lambda^2(\mathbb{R}^4)$ are wedges of two vectors by checking whether they satisfy the Plücker Relation:
$$X_{12}X_{34} − X_{13}X_{24}+ X_{23}X_{14} = 0$$
So, what I tried to do next was verify directly that, given an element $w =  \sum_{i < j} X_{ij} e_i \wedge e_j \in \Lambda^2(\mathbb{R}^4)$:


*

*If $w$ is a wedge of two vectors, then $w$  satisfies the Plücker Relation.

*If $w$ does not satisfy the Plücker Relation, then $w$ is a "symplectic element". 


However, I was confused to discover that (1) does not seem to hold

Main Question: Am I going crazy? Or does the Plücker relation
  $$X_{12}X_{34} − X_{13}X_{24} + X_{23}X_{14} = 0$$ not generally hold where
  $$
X_{ij} = 
\det \begin{bmatrix}
u_i & v_i \\
u_j & v_j \\ 
\end{bmatrix}$$
  and $u,v \in \mathbb{R^4}$ are linearly independent? I have computed this about 4 times now. Possibly I am making some mistake, but at this point I think it is more likely that I am not understanding what the Plücker relations are supposed to do properly.

Sorry for the long question. I could have been more concise, but I also wanted to record my thought process so that I could recall it later.
 A: The answer to this question turned out to be rather uninteresting. There was an error on one of the Wikipedia articles which I had consulted, and this lead to my confusion. 
By the way, having done a bit more reading, I think that using the Plücker relations the way that I did, i.e. to decide when an element $\alpha \in \Lambda^2(X)$, $X$ a finite-dimensional vector space, has the elementary form $u \wedge v$ for $u,v \in X$, is perhaps a bit strange. Looking here, one sees that, more generally, the rank $r$ of $\alpha \in \Lambda^2(X)$ is smallest number such that 
$$\underbrace{\alpha \wedge \ldots \wedge \alpha}_{r+1} =0.$$
In particular, an element $\alpha \in \Lambda^2(X)$ is elementary if and only if $\alpha \wedge \alpha =0$. Choosing a  basis $\{e_i\}_{i=1}^n$ for $X$ and writing out explicitly the coefficients of $\alpha \wedge \alpha$ with respect to the corresponding basis $\{e_i \wedge e_j\}_{1 \leq i < j \leq n}$ for $\Lambda^2(X)$, you arrive at $\binom{n}{2}$ quantities which must vanish in order for $\alpha$ to be elementary. I think these quantities are, more or less, the Plücker relations. So, for the purpose I had in mind, I think it is probably much more elegant to just say:

$\alpha \in \Lambda^2(X)$ can be written in the form $u \wedge v$ for $u,v \in X$ if and only if $\alpha \wedge \alpha = 0$. 

