I don't think the question makes sense as stated. What is the definition of the integration of $\int_{W_i} \colon H^{2n-2k}(X,\mathbb C) \to \mathbb C$ for $W_i$ a variety of codimension $n-k$? This map is defined for $W_i$ a variety of *dimension* $n-k$, for example by integrating $2n-2k$-forms over a space of real dimension $2n-2k$.

If you ask the question with integration over varieties of dimension $n-k$, the answer is positive.

Since $Z$ is a subvariety of codimension $k$ in a variety of dimension $n-k$, $Z$ itself is a variety of dimension $n-k$. So we can take $W_i=Z$, and ask whether $\phi_Z = \int_Z$.

This is true by the definition of the cohomology class of an algebraic cycle. If we define $[Z]$ as the Poincare dual to the homology class of $Z$, then $\phi[Z]$ is the linear form on cohomology classes given by pairing with the homology class of $Z$, which for differential forms is the same as integration over $Z$.

For the hard Lefschetz theorem or Lefschetz type standard conjecture to appear, the hyperplane class would have to be involved in some way. It's possible to introduce this into the definition of $\int_{W_i}$, although, I think, rather strangely.

For $k> n/2$ you can define $\int_{W_i}$ by first cupping with the $2k-n$th power of the hyperplane class. In this case, the claim is equivalent to the existence of a cycle whose class, cupped with $[H]^{2k-n}$, is $[Z]$. This would indeed be a form of the Standard Conjecture A as described in Donu Arapura's answer.

For $k< n/2$ you can *define* $\int_{W_i}$ using the Hard Lefschetz theorem, i.e. by composing the Hard Lefschetz map $H^{2n-2k}(X, \mathbb C) \to H^{2k}(X, \mathbb C)$ with the usual integral of cohomology classes over cycles. In this case, we just need to take $W_i$ to be components of a cycle with cycle class $[Z] \cup [H]^{n-2k}$, which exists by the construction of the intersection product on cycles - we just take the intersection of $Z$ with $n-2k$ copies of $H$, or more concretely the literal intersection with $n-2k$ generic hyperplanes.

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are not obeyed outside of math mode, so, to get the effect of`Poincar\'{e}`

, you must (do something like) manually type the appropriate Unicode character. I have edited accordingly. $\endgroup$1more comment