Let $V^{m+1} = \mathbb{C}^{m+1}$ and let $U(1)$ act on it by its diagonal representation, so that really, it is just like scalar multiplication by a unit modulus complex number.


I am interested in the quotient $Q^m = V^{m+1}/U(1)$, which has real dimension $2m+1$. What I am interested in is the following question. When is the $U(1)$ orbit of the origin in $V^{m+1}$, which is just $\{ \mathbf{0} \}$, non-special, in the sense that the singularity is removable and the quotient space is naturally homogeneous (I want in particular that any 2 points look locally the same)?


Let me provide some examples. If $m = 0$, then $Q^0 = \mathbb{C} / U(1)$, which is just the nonnegative part of the real axis, i.e.
$$ Q^0 = \{ x \in \mathbb{R} ; x \geq 0 \}. $$
Note that $x = 0$ in the quotient space, is kind of special, because it has some punctured neighborhoods that are connected, while any punctured neighborhood of some $x > 0$ cannot be connected. So $x = 0$ is a special point in the quotient $Q^0$, which is the opposite of what we really want.


If $m = 1$, then $Q^1 = \mathbb{C}^2 / U(1)$ may be identified with $\mathbb{R}^3$ using the smooth map (essentially the Hopf map) $h: V^2 \to \mathbb{R}^3$, given by
$$ h(u, v) = \frac{1}{2} \left(u\bar{v} + \bar{u}v, i(\bar{u}v - u\bar{v}), |u|^2 - |v|^2\right).$$
Note that $\mathbb{R}^3$ is homogeneous and the origin in $\mathbb{R}^3$ is not special, in the sense that any two points of $\mathbb{R}^3$ look locally the same. There is also a group, the abelian group of translations, which acts transitively on $\mathbb{R}^3$.


What about for higher $m$'s, namely for $m > 1$? Another way to look at the quotient is as a cone over $\mathbb{C}P^m$. I suspect that any punctured neighborhood of the "problematic" point would have a punctured neighborhood which is homotopic to $\mathbb{C}P^m$, while any other point would have a punctured neighborhood which is homotopic to $S^{2m}$, so that what I am thinking about probably only happens in the special case $m = 1$.


My question is: is this the end of the story, or can something more be said? Maybe there is some kind of desingularization of $Q^m$ which looks like $\mathbb{R}^{2m+1}$, or something like that... Please allow some flexibility while interpreting my questions.


Edit: first read Robert Bryant's answer below. I will explain the construction I had in mind, when writing this post (for the interested reader). Actually, to keep the notation simple, I will take $m = 3$ (one may similarly take $m$ to be any positive odd integer, using Prof. Bryant's answer below).


Fix a submanifold $S$ of $\mathbb{H}P^1$ such that there exists at least one smooth nowhere vanishing section of the $3$-plane bundle $F$ over $\mathbb{H}P^1$ (described in the answer). For example one may take a finite collection of points on $\mathbb{H}P^1$, or a smooth curve (which may be open or closed) in $\mathbb{H}P^1$, or for example an open connected and simply connected subset of $\mathbb{H}P^1$. One may not take all of $\mathbb{H}P^1$ though, as this would violate the condition.


Having chosen $S$, let $\xi\_i$, for $i = 1, \ldots, n$, be $n$ smooth sections of $F$ over $S$, such that the graphs of the $\xi\_i$ are all disjoint. In other words, there is no point $x \in S$ and no pair of indices $i, j$, with $1 \leq i < j \leq n$, such that $\xi\_i(p) = \xi\_j(p)$.


Given $i, j$, with $1 \leq i,j \leq n$ and $i \neq j$, we form the pointwise normalization, say $p\_{ij}$, of
$$ \xi\_j - \xi\_i $$
with respect to the natural inner product on the fibers of $F$. But each unit $2$-sphere in each fiber of $F$ is naturally diffeomorphic to a corresponding real twistor line in $\mathbb{C}P^3$. I should probably explain this last statement a bit better. There is a diffeomorphism from $\mathbb{C}P^3$ onto the $2$-sphere bundle associated to $F$ which may be thought of as a generalization of the Hopf map (fiberwise, it is the Hopf map).


Thus $p\_{ij}$ is a section of the $2$-sphere bundle associated to $F$ over $S$. Using the diffeomorphism in the previous paragraph, $p\_{ij}$ allows us to define a smooth section over $S$ of the natural projection from $\mathbb{C}P^3$ onto $\mathbb{H}P^1$, mapping a complex line in $\mathbb{C}^4$ to its "quaternionification", which is a quaternionic line in $\mathbb{H}^2$. By abuse of notation, we will also denote this section by $p\_{ij}$.


Given $i$, with $1 \leq i \leq n$, we form the symmetric product
$$p\_i = \bigodot\_{j \neq i} p\_{ij}.$$
Thus, at a point $x \in S$, $p\_i(x)$ is a point in the symmetric product of $n-1$ copies of the fiber of $x$, with respect to the map $\mathbb{C}P^3 \to \mathbb{H}P^1$. But the symmetric product of $n-1$ copies of $\mathbb{C}P^1$ can be thought of as the projectivization of the polynomial space in $1$ complex variable of degree at most $n-1$.


At each fixed point $x \in S$, it is conjectured by Atiyah and Sutcliffe that $p\_1(x), \ldots, p\_n(x)$ are linearly independent over $\mathbb{C}$. I have actually given a more complicated description than their original one. However, if one things of $S$ as a parameter space, then effectively what we have done is construct a deformation of the Atiyah and Sutcliffe problem on configurations of points.


Having an explicit way of deforming that problem may help at some point in the future, so I am recording my idea here.



Is this quotient of $\mathbb{C}^{m+1}$ by $U(1)$ only "nice" for $m=1$?
The quotient is a cone on $\mathbb{CP}^m$.


When $m$ is even, $\mathbb{CP}^m$ is not the boundary of any compact smooth $(2m{+}1)$-manifold, so you can't smooth the singularity at the tip of the cone by blowing-up or any similiar local modification.


When $m$ is odd, $\mathbb{CP}^m$ is the boundary of a $3$-disk bundle over $\mathbb{HP}^{(m-1)/2}$, the set of quaternion lines through the origin in $\mathbb{H}^{(m+1)/2}$, so it's possible to `smooth' the singularity at the tip of the cone as follows: Regard $V = \mathbb{C}^{m+1}$ as $\mathbb{H}^{(m+1)/2}$ and quaternionically blow up the origin by letting $Y$ be the set of pairs $(v,L)$ in $\mathbb{H}^{(m+1)/2}\times \mathbb{HP}^{(m-1)/2}$ with $v\in L$. The map of $Y$ to $V=\mathbb{H}^{(m+1)/2}$ given by $(v,L)\mapsto v$ is a smooth diffeomorphism away from the fiber of $Y$ over $0\in V$, which is a copy of $\mathbb{HP}^{(m-1)/2}$.


Now, let $\mathrm{U}(1)\subset\mathbb{C}$ act on $Y$ by $e^{it}\cdot(v,L)= (e^{it}v,L)$. If you divide $Y$ by this circle action, you will get a smooth $3$-plane bundle over $\mathbb{HP}^{(m-1)/2}$, and the locus of the orbits of those $(v,L)$ with $|v|^2=1$ will, in the quotient, be a copy of $\mathbb{CP}^m$, as desired.


