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.

  • $\begingroup$ In response to the hyperkahler question: the fixed points are then a complex manifold with the K-complex structure, and you can just use the usual holomorphic BB stratification. $\endgroup$
    – Ben Webster
    Jul 5 '21 at 19:49
  • $\begingroup$ The usual BB stratification works just in the algebrogeometric setting ,no? While the action is not algebraic in general with respect to $K$. It is not clear (at least to me) that the hypothesis for Byalinicki Birula decomposition are respected in general I think. $\endgroup$ Jul 5 '21 at 21:01
  • $\begingroup$ Well, given that you haven't indicated what you're assuming, it's a little hard to be sure, but the fixed points will be Kahler with respect to K (with the induced metric from restricting the hyperkahler one), so you're just doing Morse theory for the moment map of the U(1) action, and the basic observation you need is just that all the Morse indices are even. $\endgroup$
    – Ben Webster
    Jul 5 '21 at 21:10
  • $\begingroup$ I edited,hope it's clearer now! Sorry but I didn't get that much of your comment: in general the action of the torus is just written in terms of the manifold with respect to $I$. I don't know what are Hodge indices also, sorry. $\endgroup$ Jul 5 '21 at 21:29
  • $\begingroup$ Sorry, been thinking too much about Hodge theory lately. I meant Morse not Hodge. $\endgroup$
    – Ben Webster
    Jul 5 '21 at 21:30

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 $n+1$.

  • $\begingroup$ Thank you for the beatiful counterexample! I'd be interested in a more specific class of varieties. I don't know whether this changes something or not. I have the impression that what really fails is that the fixed point set needs not to be orientable in general $\endgroup$ Jul 5 '21 at 15:08
  • $\begingroup$ 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 $n+1$. $\endgroup$
    – Nick L
    Jul 5 '21 at 15:45
  • $\begingroup$ You are right. I imagine that should have the orientability of the full machinery of BB decomposition (like the locally closed strata ) in order to get an analogous statement. I don't know whether there are reasonable hypothesis on $X$ in order to ensure this. $\endgroup$ Jul 5 '21 at 15:55
  • $\begingroup$ I don't know sorry. By the way, it looks like there might be something meaningful happening with respect to $\mathbb{Z}_2$ homology, I haven't checked that carefully though. $\endgroup$
    – Nick L
    Jul 5 '21 at 16:06
  • $\begingroup$ With $\mathbb{Z}/(2)$ coefficients one should be able to show that the the decomposition is perfect in the Morse sense (I hope I've been clear). There are some papers concerning this. This treats the case of moduli space of vector bundles for example : arxiv.org/pdf/1109.5164.pdf . $\endgroup$ Jul 9 '21 at 14:48

There's some simple topology behind why this will work in complex cases, and not consistently in real ones: complex affine spaces are even dimensional as real manifolds, and real affine spaces are often not. So the Poincare polynomial equation you have depends on the differentials vanishing in a spectral sequence, in a way that is just not true if you can have cells of different parities (which the $\mathbb{RP}^n$ example illustrates well).

This is why it should work in the hyperkahler case you mention. In this case, the flow-in sets of the fixed point sets are holomorphic and thus even dimensional.

EDIT: Sorry to have been vague up above; in part my memory of the technical details has gotten a little misty.

The point I was trying to make is this: BB decomposition breaks our manifold up into a finite number of locally closed submanifolds. Let $X_{\leq k}$ be the union of these pieces of dimension $\leq k$. If you have isolated fixed points, this will be the union of the cells of dimension $\leq k$. There's a spectral sequence whose $E_2$ page is given by $H^*(X_{\leq k}/X_{< k})$ (where quotient in the usual topological sense of crushing down to a point). The space $X_{\leq k}/X_{< k}$ is the wedge sum of the one point compactifications of the BB strata of dimension $k$; by the Thom isomorphism, this is the same as the sum of the cohomologies $H^*(F_i)$ of the fixed points sets giving this dimension, shifted by the dimension of the directions flowing in (I think you called these $h_i$'s).

Thus, the RHS of your expression for Poincare polynomials is thus the $E_2$ page of this spectral sequence, and you'll get the equality you want if and only if all further differentials are trivial. In the case where each $F_i$ is a point, you're just computing cellular homology for the corresponding cell decomposition.

If you're working in the complex world, all of these shifts are even, so if your fixed point set has only even cohomology, then all the differentials have to vanish for parity reasons and life is great. For example, in the cellular homology case, every other term in your complex is trivial, and so all differentials are trivial. It looks to me like this should still be OK if $F_i$ has odd cohomology, but I think you have to use that $H^*(F_i)$ has a pure Hodge structure (so all the differentials have to vanish since they are morphisms of between Hodge structures of the wrong weight).

In the real world, this is all shot to pieces; you don't have any parity or Hodge structures to stop differentials from being non-zero, and the case $\mathbb{RP}^n$ shows lots of them won't be.

In the hyperkähler situation, it could be salvageable, but you need to think carefully about exactly what hypotheses you have. The thing I worry about in the situation you've mentioned is that the flow-in sets might not be holomorphic with respect to $J$ and $K$, so the fixed points of $\sigma$ could be odd dimensional, and then you're done for.

  • $\begingroup$ Sorry y background is more algebrogeometric: I often looked at BB as a purely algebraic fact and I'm trying to understand now its relation to Morse theory. Which spectral sequences are you referring to? $\endgroup$ Jul 6 '21 at 8:14
  • $\begingroup$ In general I think there's is no reason to not get odd cohomology or odd dimensional spaces unfortunately. Honestly, I don't know what happens for the $\mathbb{C}^*$ action not even in the Higgs bundle case ! Thank you anyways for the comment ! It was superclear $\endgroup$ Jul 9 '21 at 14:52

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