Homotopy type of the subspace $Q=\{[z_0,...,z_n] \in \Bbb CP^n : z_0^2+...+z_n^2=0 \}$ of the complex projective space Consider the standard projection $p:\Bbb C^{n+1}-0 \to \Bbb CP^n$ of the complex projective space $\Bbb CP^n$. Let us use the notation $[z_0,...,z_n]$ for $p(z_0,...,z_n)$. Now, let $Q$ denote the subspace $Q=\{[z_0,...,z_n] \in \Bbb CP^n : z_0^2+...+z_n^2=0 \}$ (It is obvious that $Q$ is well-defined). 
I want to show that $Q$ is homotopy equivalent to $S^0$ when $n=1$, $S^2$ when $n=2$, and $S^2 \times S^2$ when $n=3$. 
For $n=1$, it is obvious, since we can identify $\Bbb CP^1$ with $S^2=\Bbb C \cup \{\infty\}$ via the homeomorphism $[z,w] \mapsto z/w $, and then $Q$ is just two points, consisting of $+i$ and $-i$. 
However, I have no idea when $n=2$ or when $n=3$. 
Also, what I am wondering is, that is $Q$ homotopy equivalent to a familiar space when $n>3$?
Looking at the definition of $Q$, I think it would not be so surprising if the space $Q$ has its own name.
 A: Here's a cute argument for the case $n=2$ (thanks to Ted Shifrin for telling me about it a couple of years ago): Consider $S^2$ as the unit sphere in $\mathbb{R}^3$, and endow it with the induced (round) Riemannian metric. Up to $U(1)$-transformation, the tangent space $T_x S^2$ at a point $x\in S^2$ has a unique oriented, orthonormal basis $\{u_x,v_x\}$. Now consider $z_x=u_x+i v_x\in \mathbb{R}^3\otimes \mathbb{C}\cong \mathbb{C}^3$. Then $z_x$ satisfies the equation $(z_x)_1^2+(z_x)_2^2+(z_x)_3^2=0$ and thus defines a point in $Q\subset \mathbb{C}\mathrm{P}^2$. Since $z_x$ is unique up to $U(1)$-transformation---which acts trivially on the projective space---we obtain a well-defined, smooth map $f:S^2\to Q$. 
We can construct its inverse as follows: Given a point in $Q$, we pick a representative $z\in\mathbb{C}^3$ and wrote it as $z=u+iv$ for $u,v\in \mathbb{R}^3$. Then $\sum_i z_i^2=(z,z)=(u,u)-(v,v)+2i(u,v)=0$, where $(-,-)$ is the $\mathbb{C}$-bilinearly(!) extended inner product on $\mathbb{R}^3$. Since $u,v\in \mathbb{R}^3$, and $(-,-)$ restricts to the standard inner product on $\mathbb{R}^3$ on this subspace, we see that $u,v$ must be orthogonal and of equal length. Thus, after scaling them to be of unit length, we obtain an orthonormal pair of vectors $u',v'\in \mathbb{R}^3$. They span an oriented $2$-plane, which we identify as $T_x S^2$ for a unique $x\in S^2$. Indeed, every oriented $2$-plane in $\mathbb{R}^3$ occurs in this way---the space of oriented two-planes is $SO(3)/SO(2)\times SO(1)=SO(3)/SO(2)\cong S^2$. Since the inverse mapping is also smooth, we conclude that $f$ is a diffeomorphism.
There is a nice generalization, which is proven in exactly analogous fashion:
Proposition: For every $n\geq 2$, the Grassmannian $\widetilde{\mathrm{Gr}_2}(\mathbb{R^n})$ of oriented two-planes in $\mathbb{R}^n$ is diffeomorphic to $Q_n\subset \mathbb{C}\mathrm{P}^{n-1}$.
There is another interesting fact which is a little bit relevant here. In the construction of the map $f$ above, I used that each point in $S^2$ corresponds uniquely to an oriented two-plane in $\mathbb{R}^3$, namely its tangent space. For higher-dimensional spheres, this is no longer the case, but we can still say something interesting in case $S^n$ admits an almost complex structure (this happens only in dimensions $2$ and $6$). We can then consider $TS^n$ as a complex vector bundle, and form the smooth manifold $\mathbb{P}(TS^n)$, where $\mathbb{P}$ denotes the (complex) projectivization, i.e. points in this space are precisely the complex lines in the tangent spaces. 
In case $n=2$, there is only one such line (namely the entire tangent space itself), but for $n=6$ this space is a bundle with fiber $\mathbb{C}\mathrm{P}^2$ over $S^6$. It turns out that this space is again diffeomorphic to  a quadric hypersurface, namely $Q_5\subset \mathbb{C}\mathrm{P}^6$. In view of the above, it suffices to prove that it is diffeomorphic to the space of oriented two-planes in $\mathbb{R}^7$, which can be done using the algebraic structures derived from the identification $\mathbb{R}^7=\operatorname{Im}(\mathbb{O})$, in particular the cross product. There's an analogy here with the case of $S^2$, because in that case one can also work in terms of the more classical, three-dimensional cross product induced by the identification $\mathbb{R}^3=\operatorname{Im}(\mathbb{H})$.
