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Let $P^{n-1}(\mathbb{R})$ the real projective space of dimension $n-1$. Consider the set

$$B=\{(x,y)\in\mathbb{R}^n \times P^{n-1}(\mathbb{R}) / x=(x_1,..,x_N), y=[y_1;..;y_N], x_iy_j=x_jy_i \hspace{0.2cm} \forall i,j =1-n\}$$

The problem is to show that $(B,i)$, being $i$ the inclusion map, is a submanifold of the manifold $\mathbb{R}^n \times P^{n-1}(\mathbb{R}) $. I have proved that $B$ is well defined and that $di_p$ is injective $\forall p \in B$ (and it's clear that $i$ is injective). But how can I show that $B$ is a submanifold? I'm not sure how to prove that it has a differential structure inside $\mathbb{R}^n \times P^{n-1}(\mathbb{R})$.

uuuuu
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  • You should let us know what you have tried and where you need help? – DSinghvi Nov 19 '14 at 17:28
  • I thought the problem would be simpler writing $B$ as the preimage of a set by a function $f$. Such function could be $f: \mathbb{R}^n \times P^{n-1}(\mathbb{R}) \longrightarrow \mathbb{R}^(n^2)$ defined by $f_{i,j}((x_1,..,x_N),[y_1,..,y_N])=x_iy_j-x_jy_i$. Then $B=f^{-1}(0)$. It seems not difficult to prove that $f$ is differentiable, but, as far as I know, I need to prove that $df_0: T_z (\mathbb{R}^n \times P^{n-1}) \longrightarrow T_0(\mathbb{R}^nm)$ is surjective $\forall z \in F^{-1}(0)$, and I'm not sure how to see this. So I suppose this is not the best way to prove the result. – uuuuu Nov 19 '14 at 17:47

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