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For any real $2\times 2$ matrix A, $$ \begin{pmatrix} a & b\\ c & d \end{pmatrix}\in \text{Mat}_{\mathbb{R}}(2) $$

Let $S$ be the set $$ S=\{ A\in \text{Mat}_{\mathbb{R}}(2) \mid a^2+b^2+c^2+d^2=1,\det(A)=0\}. $$ Using regular value theorem, I have proved that this is a two dimensional submanifold of $\text{Mat}_{\mathbb{R}}(2)$. And now I want to show that the map $$ \pi:S\to \mathbb{R}P(1) $$ Taking $A\in S$ to its range $ran(A)\subseteq \mathbb{R}^2$, is smooth. Since we know $A$ has rank $1$, so at least one of its one of its row or column is not a zero vector. I think it is possible to use the two charts where $U_0$ is the chart for which it has no zero rows and $U_1$ be the chart for which it has no zero columns. However, I have trouble think of the map associated with the chart. Is there any way such that we can avoid talking about the charts and prove the smoothness?

Intuitively speaking, the map $\pi$ just takes each vector to its one dimensional span, thus we can identify it with points in $\mathbb{R}P(1)$, but how can we be more a little bit more rigorous?

Also, is there a quick way of saying $S$ is a 2-torus?

Any hint would be appreciated, thanks!

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  • $\begingroup$ Try to parametrize using the columns, like in my question here. $\endgroup$ – Ivo Terek Feb 16 '17 at 23:04
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HINT: Start by writing $(a,b) = \mathbf x$, $(d,-c)=\mathbf y$, and note that your equations can be rewritten as $\|\mathbf x\|^2 + \|\mathbf y\|^2 = 1$ and $\mathbf x\cdot\mathbf y = 0$. Now you should be able to see that the submanifold is explicitly $S^1\times S^1$.

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  • $\begingroup$ Sorry, could you please elaborate on that? How can we use it to show that $\pi$ is a smooth map? $\endgroup$ – jack Feb 17 '17 at 1:24
  • $\begingroup$ Sorry about the ambiguity. I was addressing how to see the manifold is a torus. Smoothness is easy from the fact that every rank-$1$ matrix is of the form $\mathbf u\mathbf v^\top$, and you can make this locally unique and smooth if you require $\|\mathbf u\| = 1$. (Ivo's linked answer addresses this explicitly.) $\endgroup$ – Ted Shifrin Feb 17 '17 at 5:16

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