Why isn't the plücker embedding surjective?

Question

This is basic algebraic geometry, there's a question at the end but please check that my reasoning all the way up to the question also holds:

Given a grassmannian $Gr(k,V)$, we can embed it to the projectiviation of a wedge space, by the map:

$$\iota : Gr(k,V) \to \mathbb{P}(\wedge^k V) \\ W \mapsto [w] $$

In the case that $k=3$ and dim $ V = 5$, we can compute the maximal $3 \times 3$ minors of the matrix consisting of the vectors spanning $W$, there are ${5\choose3} = 10$ such minors and that will provide homogeneous coordinates $w = [Z_0...Z_9] \in \mathbb{P}(\wedge^3 K^5) \sim \mathbb{P}^9$.

For the life of me, I can't figure out why this isn't surjective. Can someone give me an example of a $p \in \mathbb{P}^9 $ that isn't hit by any subspace $W \in Gr(3, K^5)$ and explain why it is so?

Answer 1

But $G(3,5)$ has dimension $3(5-3)=6$, which is far less than $9$.

The key idea you're missing is that the image of the Plücker map consists of all (projectivized) decomposable $k$-vectors, which is in general a very thin subset of $\Bbb P(\Lambda^k V)$.

Answer 2

To add a slightly more concrete perspective to Ted Shifrin's answer, why don't you try the simpler example of $G(2,4)$ instead? There you have 6 minors, giving an embedding into $\mathbf P^5$. But you can check that the 6 minors of a $2 \times 4$ matrix always satisfy a certain degree-2 equation. (I won't write the equation here, but it is easy to look up, or better, try to discover yourself.) So $G(2,4) \hookrightarrow \mathbf P^5$ is a a quadric hypersurface.

Once you are comfortable with that example, it won't be too hard to convince yourself that for any $G(k,n)$, points in the image of the Plücker embedding always satisfy a (nonempty) collection of degree-2 equations.