# Grassmannian is a manifold in a specific case the $2$-planes in $\mathbb{R}^4$

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

• Can you at least show that it is bijective? – Arctic Char Sep 26 at 16:29
• What are you $U_ij$ and what are the bracket $[]$ – EDX Sep 26 at 16:35
• bracket just to indicate that you are in quotient – Lucas Sep 26 at 16:43
• 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. – Ivo Terek Sep 27 at 6:16
• (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 $A$) – Ivo Terek Sep 27 at 6:18