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I want to show that Grassmannian is a manifold in a specific case the $2$-planes in $\mathbb{R}^4$.

I'm in the following context:

$G(2,4)$ are the $2$-planes in $\mathbb{R}^4$ that we can identify with an matrix of two vectors that generate the plane (they are not unique). Considering $L(2,4)=\{A \in M_{4 \times 2 } : \text{rank} A =2\}$ and equivalence relation $A \sim B $ iff exist $g \in Gl_2(\mathbb{R})$ such that $B=Ag$.

$$G(2,4) \cong L(2,4)/\sim$$

Given $A \in M_{4 \times2}$ define $A_{ij}$ submatrix of $A$ eliminating rows $i$ and $j$ of $A$ and $V_{ij} = \{ A \in L(2,4)| A_{ij} \text{ is invertible}\}$. We can see that $V_{ij}$ is open in $L(2,4)$. Therefore $U_{ij}=\pi(V_{ij})$ is open in $G(2,4)$, where $\pi$ is a natural projection in quotient.

In this context we define the possible charts $(U_{ij}, \phi_{ij})$, $\phi_{ij}: U_{ij} \to \mathbb{R}^4$ $$\phi_{ij}([A])=A_{kl}A_{ij}^{-1}$$ where $\{1,2,3,4\}=\{i,j,k,l\}$. I'm having a hard time showing that it's a homeomorphism

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  • $\begingroup$ Can you at least show that it is bijective? $\endgroup$ – Arctic Char Sep 26 at 16:29
  • $\begingroup$ What are you $U_ij$ and what are the bracket $[] $ $\endgroup$ – EDX Sep 26 at 16:35
  • $\begingroup$ bracket just to indicate that you are in quotient $\endgroup$ – Lucas Sep 26 at 16:43
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    $\begingroup$ It's much cleaner if a priori you do away with matrices and deal with ${\rm Gr}_kV$ directly, where $V$ is any vector space. Given a surjective linear map $f: V \to \Bbb R^k$, take $$U_f = \{W \in {\rm Gr}_kV \mid f|_W \mbox{ is an isomorphism} \}$$and define $\varphi_f\colon U_f \to \prod_{i=1}^k f^{-1}(e_i)$ by sending $W$ to the unique ordered basis $\varphi_f(W)$ of $W$ which gets sent to the standard basis $(e_1,\ldots, e_k)$ of $\Bbb R^k$. Then the $(U_f,\varphi_f)$ form an atlas, and the transition maps are smooth because taking inverses is a smooth process. $\endgroup$ – Ivo Terek Sep 27 at 6:16
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    $\begingroup$ (coordinate argument for last statement: taking inverses of non-singular matrices is a smooth process because the entries of $A^{-1}$ are rational --- hence smooth --- functions of the entries of $A$) $\endgroup$ – Ivo Terek Sep 27 at 6:18

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