Plücker embedding – Two definitions

algebraic-geometrygrassmannianprojective-space

While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ \varphi: \mathbb{A}^{n \cdot d} \rightarrow \mathbb{P}^{\binom{n}{d}-1}, A \mapsto \text{det}(A^{(I)})$$
where $ A^{(I)}$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d \times n$. Moreover, the map is only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as

$$\psi: Gr(d,n) \rightarrow \mathbb{P}^{\binom{n}{d}-1}, U \mapsto [u_1 \wedge \dotsc \wedge u_d]$$

So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.

Thank you for your help.

Best Answer

The idea here is that that the vectors $u_i, i=1,2,\ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_{i1},a_{i2},\ldots,a_{in})$, then the coefficient of $e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_d}$ in the wedge product $u_1\wedge u_2\wedge\cdots\wedge u_d$ is equal to $\det(A^{(I)})$.

Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when $$ \begin{aligned} u_1\wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)\wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\\ &=\sum_{i=1}^4\sum_{j=1}^4a_ib_je_i\wedge e_j\\ &=(a_1b_2-a_2b_1)e_1\wedge e_2+(a_1b_3-a_3b_1)e_1\wedge e_3+(a_1b_4-a_4b_1)e_1\wedge e_4\\ &+(a_2b_3-a_3b_2)e_2\wedge e_3+(a_2b_4-a_4b_2)e_2\wedge e_4+(a_3b_4-a_4b_3)e_3\wedge e_4. \end{aligned} $$ See the six $2\times2$ minors of the matrix $$ A=\left(\begin{array}{cccc}a_1&a_2&a_3&a_4\\b_1&b_2&b_3&b_4\end{array}\right) $$ emerging?!

For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $d\times n$ matrix $A$ that has $V$ as its row space. It is essential that:

  • Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $d\times d$ matrix $M$ such that $A'=MA$.
  • When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $\det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)\to\Bbb{P}^N$ is well defined.
  • The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.

Also, proving that $\phi$ and $\psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)\times (d+1)$-determinants as linear combinations of $d\times d$-determinants.

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