Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be $k^{n}$, where $k$ is an algebraically closed field. Then we can compute the homogeneous coordinates of $\mathbb{P}V$ as follows: for a point $p\in \mathbb{P}V$ ($p$ is a line in $V$), take the coordinate (coordinate in $V$) of a non-zero point of $p$.

If $V$ is replaced by $\Lambda^{s}V$, how to compute the homogeneous coordinates of points in $\mathbb{P}(\Lambda^{s}V)$? Thank you very much.

Edit: Let $G(s, V)$ be the set of subspaces of $V$ of dimension $s$. There is a map: $\phi: G(s, V) \to \mathbb{P}(\Lambda^{s}V)$ defined by sending $M\in G(s, V)$ to the homogeneous coordinate of $M$ in $\mathbb{P}(\Lambda^{s}V)$. By identify $V$ with $k^{n}$, $M$ can be represented by a $k\times n$ matrix. Why the homogeneous coordinate of $M$ are minors of the matrix.

Edit: I know the answer now.

share|cite|improve this question
I don't really see the difference. $V$ is some arbitrary vector space of dimension $n$ (and you've chosen a basis so that coordinates make sense). Can't you just think of $\wedge^sV$ as a vector space of dimension $n$ choose $s$. You have coordinates on it inherited from $V$, but you can now safely forget that it is $\wedge^sV$ and instead a new vector space $W$ of some new dimension. Now the same type of coordinates should work. – Matt Aug 23 '11 at 17:16
@Matt, thanks. But why the homogeneous coordinates are minors of a matrix? – LJR Aug 23 '11 at 17:58
@user9791: If you know the answer, you should either provide the answer yourself and accept it, or delete the question; otherwise it will hang around as an unanswered question. – joriki Aug 24 '11 at 4:29
up vote 0 down vote accepted

Let $W=\langle w_1, \ldots, w_s \rangle \in G(s, V)$. Then $w_i = a_{i1}v_1+\cdots + a_{in}v_n$ for some $a_{ij}$. Therefore $W$ can be represented by a matrix $M$ whose i-th row are $a_{i1}, \ldots, a_{in}$. The map from $G(s, V)$ to $\mathbb{P}(\Lambda^{s}V)$ is given by $W \mapsto w_1 \wedge \cdots \wedge w_s$. The basis of $\Lambda^{s}V$ is $B = \{v_{i_1}\wedge \cdots \wedge v_{i_s} \mid i_1, \ldots, i_s \in \{1, \ldots, n\} \text{ and } i_k \neq i_l \text{ for } k \neq l \}$. If we compute the coordinates of $w_1 \wedge \cdots \wedge w_s$ respect to the basis $B$, then the coordinates are just the $s\times s$ minors of $M$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.