Let $X$ be a smooth complex algebraic variety endowed with a $\mathbb{C}^\*$ action. We assume also to have an antiholomorphic involution $\sigma$ over $X$ such that it anticommutes with the action above i.e $$\sigma(t \cdot x)=\bar{t}\cdot \sigma(x) .$$


Let us also assume that the $\mathbb{C}^\*$-action respects the properties in order to get a well behaved Byalinicki-Birula decomposition.


For every $x\in X$ there exists $\lim\_{t \to 0} t \cdot x$ and the fixed point set is a projective variety. (I think $X$ is called semiprojective in this case).


In this case, we know that the Poincare polynomial of $X$ can be expressed as $$P(X,t)=\sum\_{i \in I}P(F\_i,t)t^{d\_i} $$ where $$X^{\mathbb{G}\_m}=\bigsqcup\_{i \in I} F\_i $$ and $F\_i$ are connected and $d\_i$ are some positive integers associated to the action.


The fixed point set $X^{\sigma}$ of the antiholomorphic involution is a smooth real manifold. The hypothesis tells us that $\sigma(X^{\mathbb{G}\_m})=X^{\mathbb{G}\_m}$. Is it still true somehow that $$P(X^{\sigma},t)=\sum\_{i \in I}P(F\_i^{\sigma},t)t^{h\_i} $$ or not ?


Is this known in the literature? I'm totally new to real algebraic geometry.


EDIT: The answer below indicates this is not true in general. However, I'd be interested in the following more specific situation. $X$ should be given the structure of an hyperkahler manifold with complex structures $I,J,K$ such that the complex algebraic variety we are looking at is the one induced by $I$.


The involution $\sigma$ should then be antiholomorphic with respect to structure $I,J$ and holomorphic with respect to $K$. The $\mathbb{C}^\*$ action should be algebraic with respect to the structure induced by $I$ while in general it is not clear what happens with respect to the other structures.


The setting to think of is that of Non Abelian Hodge theory: it is known that we have an hyperkahler manifold $M$ such that with respect to $I$ it is the moduli space of stable higgs bundle $(\mathcal{E},\phi)$ of fixed rank $n$ and degree $d$. There we have the action $$t(\mathcal{E},\phi)=(\mathcal{E},t\phi) .$$ With respect to the structure $J$ is the de Rham moduli space of connections and with respect to $K$ is the associated (twisted)character variety.


As suggested below, it is likely that hyperkahler structure should help because of parity of dimension of the cells involved somehow, but I wasn't able to prove this neither to find some references.



Bialynicki-Birula decomposition for real analytic varieties
No, consider the following $\mathbb{C}^\*$-action on $\mathbb{CP}^2$ : $$z.[z\_0:z\_1:z\_{2}] = [z\_0:z.z\_1:z^2.z\_2] ,$$ along with the antiholomorphic map $\sigma ([z\_0:z\_1:z\_2]) = [\bar{z\_{0}}:\bar{z\_{1}}:\bar{z\_2}]$. One map check that $\sigma$ anticommutes with the $\mathbb{C}^\*$-action.


This is the nicest situation that one could wish for in terms of the BB decomp, etc.


Then the fixed point set of the $\mathbb{C}^\*$-action is the three points $[1:0:0],[0:1:0],[0:0:1]$, the fixed point set of the involution is $\mathbb{RP}^2 \subset \mathbb{CP}^2$. So, the statement about the Poincare polynomials cannot hold for any choice of $h\_i$ since the total Betti numbers are different (3 and 1 respectively).


Answer to edited question:


Even when the fixed point set is orientable the statement is false. Consider the action on $\mathbb{CP}^n$ $z.[z\_{0}: z\_{1} : \ldots : z\_{n}] = [z\_{0}: z z\_{1} : \ldots : z^n z\_{n}] $ where $n>1$ is odd, and the antiholomorphic involution given by conjugating all of the co-ordinates (as above). Then the the fixed point set of the involution is $\mathbb{RP}^n$ which is an orientable manifold with total Betti number $2$, and the fixed point set of the torus action is $n+1$ points, with total Betti number $n+1$.


