-1
$\begingroup$

consider such map $P:C^2\setminus{(0,0)} \to U =\{{B \in M_2(C)|B = B^*, B^2 = B, tr(B)=1 \}}$ by $P(z_0,z_1):= \frac{1}{\|z_0\|^2+\|z_1\|^2} \begin{pmatrix} \|z_0\|^2 \ \bar{z_0}z_1 \\ \bar{z_1}z_0\ \|z_1\|^2\end{pmatrix}$

We wish to investigate whether this map is a quotient map. Now, it is possible to give a proof in the following manner:

We descend $P:C^2\setminus{(0,0)}$ to $CP^1$ and applied the result $CP^1 \cong M \cong S^2$, which will force $P$ be a quotient map.

However it is possible to directly prove it? (for example, by proving the map is an open map)

I suspect that $U$ belong to some special group and we can exploit such property, but I don't really know how to proceed.

*any comment/idea will be welcome, I lack basic knowledge in Lie Algebra and Algebraic Geometry, and I only require some inspiration to work out the full problem.

$\endgroup$
3
  • $\begingroup$ Your first method seems to be using the false premisse that if two spaces are homeomorphic then any map between them is a homeomorphism. $\endgroup$
    – Ruy
    May 14 at 3:55
  • $\begingroup$ Do I understand correctly that you have a proof, but want to have a more direct proof? $\endgroup$
    – Paul Frost
    May 14 at 23:49
  • $\begingroup$ Have you proved that $P$ is a surjection? $\endgroup$
    – Paul Frost
    May 15 at 16:27

1 Answer 1

Reset to default
-1
$\begingroup$

Consider the equivalence relation on $\mathbb C^2\setminus \{0\}$ given by $x\sim y$ iff $P(x)=P(y)$, so you have a continuous, bijective map $$ \pi:\frac{\mathbb C^2\setminus \{0\}}\sim\to U. $$

Next you should prove that the LHS is compact, and that $U$ is Hausdorff, which would imply that $\pi$ is a homeomorphism, as desired.

$\endgroup$
1
  • $\begingroup$ When downvoters act like hit and run one is sometimes left with the impression they didn't understand what's been said. $\endgroup$
    – Ruy
    May 16 at 21:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.