# Grassmannian manifold

Originally published at 狗和留美者不得入内. You can comment here or there.

We all know of real projective space $\mathbb{R}P^n$. It is in fact a special space of the Grassmannian manifold, which denoted $G_{k,n}(\mathbb{R})$, is the set of $k$-dimensional subspaces of $\mathbb{R}^n$. Such can be represented via the ranges of the $k \times n$ matrices of rank $k, k \leq n$. On application of that operator we can apply any $g \in GL(k, \mathbb{R})$ and the range will stay the same. Partitioning by range, we introduce the equivalence relation $\sim$ by $\bar{A} \sim A$ if there exists $g \in GL(k, \mathbb{R})$ such that $\bar{A} = gA$. This Grassmannian can be identified with $M_{k,n}(\mathbb{R}) / GL(k, \mathbb{R})$.

Now we find the charts of it. There must be a minor $k \times k$ with nonzero determinant. We can assume without loss of generality (as swapping columns changes not the range) that the first minor made of the first $k$ columns is one of such, for the convenience of writing $A = (A_1, \tilde{A_1})$, where the $\tilde{A_1}$ is $k \times (n-k)$. We get

$A_1^{-1}A = (I_k, A_1^{-1}\tilde{A_1})$.

Thus the degrees of freedom are given by the $k \times (n-k)$ matrix on the right, so $k(n-k)$. If that submatrix is not the same between two full matrices reduced via inverting by minor, they cannot be the same as an application of any non identity element in $GL(k, \mathbb{R})$ would alter the identity matrix on the left.

I’ll leave it to the reader to run this on the real projective case, where $k = 1, n = n+1$.

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