# Why is the dimension of this Grassmann manifold $G_{d, n}$ equal to $(d+1)(n-d)$ formed by the Plucker coordinates of a $d$-plane?

A $d$-plane $L \subset \mathbb{P}^{n}$ is defined as the set of points $P=(p(0), p(1), \ldots, p(n)) \in \mathbb{P}^{n}$ that satisfy equations $\sum_{j=0}^{n} b_{\alpha j}p(j) = 0$, where $\alpha = 1, \ldots, (n-d)$. If these equations are independent, then we say that the dimension of $L$ is $d$.

Form the $(d+1)\times(n+1)$ matrix $[p_{i}(j)]$ formed from the spanning vectors $P_{i} = (p_{i}(0), p_{i}(1), \ldots, p_{i}(n))$ of $L$ and consider all possible determinants of the $(d+1)\times(d+1)$ minors. There are $\binom{n+1}{d+1}$ such determinants, and we form a point $(d_{1}, \ldots, d_{N+1}) \in \mathbb{P}^{N}$, where $N = \binom{n+1}{d+1} - 1$.

There is a bijective correspondence between such points in $\mathbb{P}^{N}$ that satisfy certain quadratic relations and the $d$-planes $L \subset \mathbb{P}^{n}$.

This is where I'm confused:

There is a proposition stating that there is a bijective correspondence between the set of points of $\mathbb{P}^{N}$ (as defined above) and the affine $(d+1)(n-d)$-space of $(d+1)(n+1)$ matrices $[p_{i}(j)]$ with $i=0, \ldots, d$ and $j=0, \ldots, n$ such that the $(d+1)\times(d+1)$-submatrix $[p_{i}(k_{\gamma})]$ with $i, \gamma = 0, \ldots, d$ is the identity.

This set of points in $\mathbb{P}^{N}$ is covered by $(N+1)$ copies of affine $(d+1)(n-d)$-space, called the Grassmann Manifold (of $d$-planes in $n$-space).

Why is the dimension of $G_{d, n}$ equal to $(d+1)(n-d)$? I would have thought it's just $(d+1)(n+1)$. [I hope this is the appropriate place to post this question!]

• The grassmannian is only in bijection with a subset of $\mathbb{P}^N$, not the whole thing. As you pointed out, an open dense subset $U \subset G_{d,n}$ is given by size $(d+1) \times (n+1)$ matrices in which a specified $(d+1) \times (d+1)$ square contains an identity matrix, and this subset is an affine space of dimension $(d+1)(n-d)$. The dimension of the closure of $U$ is the same as the dimension of $U$. – Jake Levinson Oct 14 '15 at 23:03