dimCH2(Y,C)=22.\dim_{\mathbb C} H^2(Y,\mathbb C) \;=\; 22.dimCH2(Y,C)=22.
2. Knneth decomposition and computation of h2,2(X)h^{2,2}(X)h2,2(X)
By the Knneth decomposition for Hodge structures, the Hodge numbers of the product X=YYX=Y\times YX=YY are given by the convolution
hp,q(X)=a+c=pb+d=qha,b(Y)hc,d(Y).h^{p,q}(X) \;=\; \sum_{a+c=p}\sum_{b+d=q} h^{a,b}(Y)\cdot h^{c,d}(Y).hp,q(X)=a+c=pb+d=qha,b(Y)hc,d(Y).
For the bi-degree (2,2)(2,2)(2,2) we therefore compute
h2,2(X)=a+c=2b+d=2ha,b(Y)hc,d(Y).h^{2,2}(X) \;=\; \sum_{a+c=2}\sum_{b+d=2} h^{a,b}(Y)\,h^{c,d}(Y).h2,2(X)=a+c=2b+d=2ha,b(Y)hc,d(Y).
Using the nonzero Hodge numbers for YYY (listed above) one obtains
h2,2(X)=404.h^{2,2}(X) \;=\; 404.h2,2(X)=404.
(An explicit bookkeeping of the contributing tensor-products shows that this value equals the sum of contributions coming from H2(Y)H2(Y)H^{2}(Y)\otimes H^{2}(Y)H2(Y)H2(Y), together with the two trivial extreme terms H0H4H^0\otimes H^4H0H4 and H4H0H^4\otimes H^0H4H0; the principal bulk of the count arises from the H2H2H^2\otimes H^2H2H2 factor.)
3. Decomposition of H2(Y,Q)H^2(Y,\mathbb Q)H2(Y,Q): Nron--Severi and transcendental parts
Over Q\mathbb QQ (or Z\mathbb ZZ after tensoring with Q\mathbb QQ) the second cohomology of YYY admits an orthogonal decomposition
