Is the quadric $3$-fold $v^2 + w^2 + x^2 + y^2 + z^2 = 0$ isomorphic to $P^3$? The subset of projective $4$-space given by $5$-tuples $[v:w:x:y:z]$ with $v^2 + w^2 + x^2 + y^2 + z^2 = 0$ is birational to projective $3$-space.  I think it has the same cohomology as projective $3$-space.  Is it actually isomorphic to projective $3$-space$?$
In other words are there homogeneous polynomials $v(a,b,c,d), w(a,b,c,d), x(a,b,c,d), y(a,b,c,d), z(a,b,c,d)$ whose squares add to $0?$
 A: Let $Q$ be the quadric 3-fold in $\mathbb{P}^4$. By adjunction formula, $K_Q=\mathcal{O}_Q(-5+2)=\mathcal{O}_Q(-3)$, and $\text{deg}\mathcal{O}_Q(-3)=2\times(-3)=-6$. On the other hand, $K_{\mathbb{P}^3}=\mathcal{O}_{\mathbb{P}^3}(-4)$ and its degree is $-4$. So $Q$ and $\mathbb{P}^3$ are not isomorphic.
A: I don't think so. Here I'm working over $\mathbb{C}$.
Let $X = \{[v : w : x : y : z] \in \mathbb{CP}^4 \mid v^2 + w^2 + x^2 + y^2 + z^2 = 0\}$. Note that $X$ is a smooth degree two hypersurface of $\mathbb{CP}^4$. On $X$, we have a short exact sequence of vector bundles $0 \to TX \to T\mathbb{CP}^4|_X \to N \to 0$ where $N$ denotes the normal bundle to $X$ in $\mathbb{CP}^4$. As $X$ is given as the zero set of a degree two homogeneous polynomial, it can be viewed as the zero set of a section of $\mathcal{O}(2)$, and hence the normal bundle $N$, which is the restriction to $X$ of the line bundle associated to the divisor $X$, can be identified with $\mathcal{O}(2)|_X$. If $i : X \to \mathbb{CP}^4$ denotes the inclusion map, then we have
$$c_1(TX) = c_1(T\mathbb{CP}^4|_X) - c_1(\mathcal{O}(2)|_X) = i^*(c_1(T\mathbb{CP}^4) - c_1(\mathcal{O}(2))) = i^*(5a - 2a) = 3i^*a$$
where $a$ denotes the first Chern class of $\mathcal{O}(1)$, the dual of the tautological line bundle on $\mathbb{CP}^4$.
As such,
\begin{align*}
c_1^3(X) &= \langle c_1(TX)^3, [X]\rangle\\ 
&= 27\langle i^*a^3, [X]\rangle\\ 
&= 27\langle a^3, i_*[X]\rangle\\ 
&= 27\langle a^3, [\mathbb{CP}^4]\cap\operatorname{PD}(i_*[X])\rangle\\ 
&= 27\langle a^3\cup \operatorname{PD}(i_*[X]), [\mathbb{CP}^4]\rangle\\ 
&= 27\langle a^3\cup c_1(\mathcal{O}(2)), [\mathbb{CP}^4]\rangle\\ 
&= 27\langle a^3\cup 2a, [\mathbb{CP}^4]\rangle\\ 
&= 54\langle a^4, [\mathbb{CP}^4]\rangle\\ 
&= 54,
\end{align*}
where I have used the fact that the Poincaré dual of a divisor is the first Chern class of the associated line bundle which in this case reads $\operatorname{PD}(i_*[X]) = c_1(\mathcal{O}(2))$.
On the other hand, $c_1(T\mathbb{CP}^3) = 4a$ where $a$ now denotes the first Chern class of $\mathcal{O}(1) \to \mathbb{CP}^3$. So 
$$c_1^3(\mathbb{CP}^3) = \langle c_1(T\mathbb{CP}^3)^3, [\mathbb{CP}^3]\rangle = \langle 64a^3, [\mathbb{CP}^3]\rangle = 64\langle a^3, [\mathbb{CP}^3]\rangle = 64.$$
As $X$ and $\mathbb{CP}^3$ have different Chern numbers, they are not biholomorphic.
