Circle action with a single fixed point Is there a closed manifold (smooth, compact with no boundary) that admits a smooth circle action with a single fixed point?
From what I understand such a manifold must have Euler characteristic one. An example of a closed manifold with Euler characteristic one should be the connected sum $\mathbb{C}\mathrm{P}^2\#T^4$ of the 2-complex projective space with the 4-torus, since
$$\chi (\mathbb{C}\mathrm{P}^2\#T^4)= \chi (\mathbb{C}\mathrm{P}^2)+ \chi (T^4) - \chi(S^4)=3+0-2=1. $$
Does this example already admit such a circle action?
 A: Yes. In fact, every $\mathbb{R}P^{2n}$ admits a (smooth) circle action with exactly one fixed point. You can find a proof (as well as an extended discussion of related problems) in
Circle Action with a Prescribed Number of Fixed Points by Ping Li.
For completeness, here's an excerpt from Example $2.1$ in that paper:

We view $\mathbb{R}P^{2n}$ as the unit ball in $\mathbb{C}^{2n}$ modulo the
antipodal map on the boundary. It is easy to see the standard circle action
on $\mathbb{C}^{2n}$ given by
$(z_1, z_2, \ldots, z_n) \mapsto (z \cdot z_1, z \cdot z_2, \ldots, z \cdot z_n)$
is compatible with the antipodal map on the boundary and thus can be factored to
a circle action on $\mathbb{R}P^{2n}$. This circle action has precisely one
fixed point, the origin.

Note the author uses superscripts to indicate the real dimension of a space, so $\mathbb{C}^{2n}$ is the complex space with $2n$ real dimensions that we would normally call $\mathbb{C}^n$.
Your particular example $\mathbb{C}P^2 \# T^4$ is orientable, which means it cannot admit an action with exactly $1$ fixed point. In Circle Actions on Oriented Manifolds with Discrete Fixed Point Sets and Classification in Dimension $4$ by Donghoon Jang, we see (on page $2$):

First, let us begin with small numbers of fixed points. If there is one fixed
point, then the manifold must be a point.

I̶ ̶d̶o̶n̶'̶t̶ ̶h̶a̶v̶e̶ ̶a̶ ̶p̶r̶o̶o̶f̶ ̶o̶f̶ ̶t̶h̶i̶s̶ ̶c̶l̶a̶i̶m̶ ̶o̶n̶ ̶h̶a̶n̶d̶, but it means you'll need to consider nonorientable manifolds to get what you want.
Edit:
I found a proof in a different paper by Donghoon Jang, Circle Actions on Oriented Manifolds with Few Fixed Points. This is part of Lemma $2.1$, which crucially uses the Atiyah-Singer Index Theorem.

I hope this helps ^_^
