Let us have an algebraic action by a finite group G on a complex projective variety $X=\bigcup\limits\_{i=1}^N X\_i$, whose irreducible components $X\_i$ are all smooth and of the same dimension $d$, and whose singular cohomology $H^\*(X)$ is generated by algebraic cycles (hence supported in even degrees). Note that variety X itself is NOT smooth (unless N=1).  
  

Is there a fixed-points type theorem that would give us that
$$\text{rank}(H^\*(X^G))=\text{rank}(H^\*(X))?$$


NB I have EDITED the question. 



Fixed points under a finite group action on projective variety
There is actually a simple answer. Pick X to be two $\mathbb{C}P^1$'s intersecting transversally, and a $\mathbb{Z}/2$ action that swaps two spheres. Then $\text{rk } H^\*(X)=3$ whereas $\text{rk } H^\*(X^G) = 1.$


