# A space with “interchangeable” coordinates, $\mathbb{R}^n / S_n$

(I'll apologize in advance for the lack of rigour in this question, I'm something of an armchair mathematician at the moment, but I do try my best):

I have a space that is similar to $\mathbb R^n$ with the following property: the point $(x_1,x_2,\ldots,x_n)$ is identified with any other point that has the same coordinates in permuted order. For concreteness, if my space is based on $\mathbb R^2$, then the point $(0,1)$ would be identified with the point $(1,0)$ and would be indistinguishable from it.

From what I can glean, this space is $\mathbb R^n$ with a group action associated with it, where that group action is due to the permutation group permuting the coordinates. Is that all it is? Does it have any nice properties (for instance, is it a manifold)? How would I begin to study this space?

• It's one of the basic examples of an orbifold. Added some tags, edited title. – zyx May 4 '16 at 19:27
• One interesting example of this is the set of all unordered pairs $\{x,y\}$ of points on the circle. Topologically this turns out to be a Möbius band. The edge of the Möbius band is is the set of pairs $\{x,x\}$. $\qquad$ – Michael Hardy May 4 '16 at 20:16

It sounds like you're describing the quotient space $\mathbb{R}^n/S_n$. Some people call this the $n^{th}$ "symmetric power" (of $\mathbb{R}$), although be a little careful with that terminology because it can be used to refer to two other related but different constructions.

This quotient is not a manifold, but can be thought of as an orbifold. For example, $\mathbb{R}^2/S_2$ is given by folding the plane in half along the diagonal $x = y$, and so fails to be a manifold (without boundary) on the diagonal, where the action of $S_2$ has nontrivial stabilizer.

If you toss out every point where the action of $S_n$ has nontrivial stabilizer (so, every point where some two of the $x_i$ are the same) you get a space called the configuration space of $n$ unordered points in $\mathbb{R}$. This is not a very interesting space, but its relatives (e.g. the configuration space of $n$ points in $\mathbb{R}^k$ for $k \ge 2$) are very interesting and extensively studied.

There is a related construction in algebraic geometry that is somewhat better behaved and also called the symmetric power.

• Hm, this is very interesting and is definitely the next direction for study for me. Now, I have another question that is the actual motivation for this one. In reality, I have a set of objects C that are identified by a point in $R^3$, a vector in $R^3$, and a vector in $R^6$. These objects are the actual coordinates of my space, and it is these objects as wholes that must be permuted, so that if $c_1 \in C$ and $c_2 \in C$ the order pair $(c_1,c_2)$ is identified with $(c_2,c_1)$. So...what did I make? Am I even allowed to treat the objects in C as separate coordinates in a space? – Michael Stachowsky May 4 '16 at 21:10
• @Michael: I'm not sure I understand your description. Are you saying that $C$ is $\mathbb{R}^{12}$ ($12 = 3 + 3 + 6$), and you want to understand the symmetric powers of $\mathbb{R}^{12}$? – Qiaochu Yuan May 4 '16 at 21:18
• To be a bit more formal: I have a construction called a "contact". It has a location $p = (x,y,z)$, which is a point in $R^3$. It also has a normal vector, $\vec{n} \in S^3$, which is a vector in $S^3$ that orients the contact, and a "wrench", which is a vector $\vec{w} \in R^6$. So if I am correct, each contact $c$ is a product manifold: $c = R^3 \times S^3 \times R^6$. I then make yet another product manifold $G = c_1 \times c_2 \times ... \times c_n$. What is G, and is it correct that the individual $c_i$ are points in the space G? Can I create the symmetric power of G? – Michael Stachowsky May 4 '16 at 21:23
• @Michael: depending on how seriously you mean the term "normal" in the phrase "normal vector," I'd be a little careful about your description of $C$; it may only be true locally rather than globally. But yes, it sounds like you want to consider the symmetric powers of $C = \mathbb{R}^3 \times S^3 \times \mathbb{R}^6$, namely the quotients $C^n/S_n$. – Qiaochu Yuan May 4 '16 at 21:26
• Thanks for the help. I do in fact mean normal as in "has length of 1". But what about G itself? I'll give a bit more insight into what I'm trying to do In essence, G is a grasp (as in, a hand touching an object and lifting it). The hand contacts the object at $n$ contacts. Let's say now that I am grasping with exactly two contacts (a "pinch"). As long as the contact positions, normals, and wrenches are the same, it doesn't matter which contact I label as 1 and which I label as 2. In essence, I have the exact same grasp whether I rotate my wrist by 180 degrees or not. – Michael Stachowsky May 4 '16 at 21:28