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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 $M$?

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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.

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    $\begingroup$ @JensReinhold: This answer asserts that the fiber-sum is $S^2 \times S^2$. This is incorrect. It is $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$. Thanks to Danny Ruberman for pointing this out to me (about a year ago today). Actually, I believe there are no circle actions on $S^2 \times S^2$ whose fixed-sets have different dimensions. $\endgroup$
    – mme
    Mar 7 at 14:56
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To answer your question: Yes $Y$ can have components in different dimensions: Consider for example the following action on $\mathbb{CP}^2$, where we identified $S^1\subset \mathbb{C}$:

$$ w\cdot [z_0,z_1,z_2]=[w z_0,z_1,z_2]. $$

Then $[1,0,0]$ is an isolated fixed point, and also $[0,z_1,z_2]$ is a fixed $\mathbb{CP}^1$. To get more of such examples you should have a look at Delzant's classification of Hamiltonian Torus actions. Suitably choosing an $S^1$ sitting inside the torus will produce many more examples.

There is a large literature on $S^1$ actions. Even the case of a discrete fixed point set is very interesting. In the almost complex setting one could have a look at https://arxiv.org/pdf/1411.6458.pdf and references therein.

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Some homotopy theoretic results can be found in:

Yoshida, Tomoyoshi, On the rational homotopy of fixed point sets of circle actions, Math. J. Okayama Univ. 21, 149-153 (1979). ZBL0429.55002.

Or the more recent reference:

Hauschild, Volker, Rational homotopy of circle actions, Pac. J. Math. 191, No.2, 275-311 (1999). ZBL1012.57051.

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