Suppose $M$ is a connected closed smooth $d$-dimensional manifold, and suppose $S^1 = SO(2)$ acts smoothly on $M$. Then the fixed point set $Y = M^{S^1}$ will be a submanifold of $M$ of even codimension, and we also have $\chi(Y) = \chi(M)$. 


(For instance, there are different linear circle actions on $S^d$ for which the fixed point set will be $S^{d-2j}$; here $j$ can be any number between $0$ and $d/2$.)


My question is rather vague, I simply want to know what else we can say about $Y$. I am also very interested in seeing more examples where $Y$ is an interesting manifold. **In particular I would like to know whether it can happen that $Y$ has components of different dimensions, and whether the number of components of $Y$ can be bounded in some way.** Is there any result that tells us that the homotopy type of $Y$ cannot be much more complicated than that of $M$?



Fixed point set of smooth circle action
Given two $n$-manifolds $M\_i$ with a circle action and a choice of free orbit $\gamma\_i$ we can define the fiber connected sum $(M\_1, \gamma\_1) \# (M\_2, \gamma\_2) = (M\_1 \# M\_2, \gamma)$ by taking a tubular neighborhood of the circle, equivariantly identifying it with $S^1 \times D^n$, deleting interiors and equivariantly gluing what's left. So the moment we have disconnected manifolds with different-dimensional fixed point sets (and a free orbit) we have connected such. So in 4 dimensions already you have $(S^4, S^0)$ and $(S^4, S^2)$ from linear circle actions; the fiber connected sum of any choice of two of these is diffeomorphic to $S^2 \times S^2$. 


Whenever $X$ is a finite $S^1$-complex, you get the bound $\text{rk} H^\*(X^{S^1};\Bbb Q) \leq \text{rk} H^\*(X;\Bbb Q)$ using either Borel or Bredon equivariant cohomology, more or less the same way you do would do with the group $\Bbb Z/p$. So you certainly get a bound $b\_0(X^{S^1}) \leq \sum b\_i(X)$. I don't know if you find this satisfying.


