TITLE: All cohomology of quadrics comes from algebraic cycles
QUESTION [5 upvotes]: I've read in a number of place now the statement that all cohomology of quadrics (complex projective ones) comes from algebraic cycles, but I cannot find any source for this. So I hope someone here can point me to a book/paper where this is explained.

REPLY [9 votes]: Let me try to outline the proof, and you can fill in the details. We take an $n$-dimensional smooth quadric hypersurface $Q$ sitting in $\mathbf P^{n+1}$. 

By the Lefschetz hyperplane theorem, for $k \neq n$, the restriction maps

$$H^k(\mathbf P^n, \mathbf Z) \rightarrow H^k(Q,\mathbf Z)$$ 
are isomorphisms, and so the cohomology groups of $Q$ in these degrees certainly come from algebraic cycles. Moreover, by the universal coefficient theorem this also shows that $H^k(X,\mathbf Z)$ is torsion-free for all $k$. 

Using the normal bundle sequence 

$$ T_Q \rightarrow T_{\mathbf P^n} \rightarrow N_Q $$
one computes the Euler characteristic of $Q$; together with the information from Step 1 above this shows that 
$$ H^n(Q,\mathbf Z) = \begin{cases} \mathbf Z \oplus \mathbf Z \quad &\mbox{if $n$ is even} \\0 \quad &\mbox{if $n$ is odd}\end{cases}$$

Now it just remains to find two non-homologous cycles of middle dimension on an even-dimensional quadric. Thinking about the case $\operatorname{dim} Q=2$, where $Q \cong \mathbf P^1 \times \mathbf P^1$, gives you a big hint about what these should be!