# Is the set of all conformal structures on $\mathbb{R}^n$ a manifold? Does it have a name?

Question: Is the set of all conformal structures on $\mathbb{R}^n$ a manifold? Does it have a name?

A pointer to a reference will suffice.

Definition: A conformal structure on $\mathbb{R}^n$ is an equivalence class of inner products, with two inner products $f,g$ equivalent, $f \sim g$, if and only $f = \lambda g$ for some $\lambda >0$. In other words, it is an inner product “up to positive scaling”.

We need to specify an inner product to determine whether a set of vectors is orthonormal, but we only need to specify a conformal structure to determine whether a set of vectors is orthogonal. An inner product determines a notion of both length and angle, but a conformal structure only determines a notion of angle, not of length. This is analogous to how a norm only determines a notion of length, but not of angle — in fact, it makes sense to think of a conformal structure as “an inner product minus a choice of norm”.

Attempt: Consider the following fact (cf. p. 201, Linear Algebra via Exterior Products):

If $\{ e_1, \dots, e_n\}$ is an arbitrary basis in $V$, then there exists an inner product with respect to which $\{e_1, \dots, e_n \}$ is an orthonormal basis.

Thus, start with the non-compact $n$-Stiefel manifold, the set of all bases on $\mathbb{R}^n$.

By the above fact, each element corresponds to an inner product (the inner product which makes the basis orthonormal). So we can impose an equivalence relation on the non-compact $n$-Stiefel manifold, saying that two bases are equivalent if and only if they are orthonormal with respect to the same inner product. Since the equivalence classes of this equivalence relation are (I imagine) diffeomorphic to the compact $n$-Stiefel manifold, this might be a homogeneous space.

Now to get from this space (the space of all inner products) to the space of all conformal structures on $\mathbb{R}^n$ is simple — we just quotient by the equivalence relation which says that two inner products are equivalent if and only if they are the same up to positive scaling, i.e. if and only if they belong to the same conformal structure. Since this is the same thing (I think) as quotienting by the action of $\mathbb{R}^{+}$, which is a Lie group, then, if the space of all inner products was a homogeneous space, the final space, the space of all conformal structures on $\mathbb{R}^n$, will also be a homogeneous space.

Since homogeneous spaces are manifolds, and the space of all conformal structures on $\mathbb{R}^n$ might be a homogeneous space, then it might be a manifold.

Obviously this says nothing about what the name of such a structure might be, except perhaps that is related to Stiefel manifolds via the actions of multiple Lie groups.

#### Solutions Collecting From Web of "Is the set of all conformal structures on $\mathbb{R}^n$ a manifold? Does it have a name?"

Inner products on $\Bbb R^n$ are defined by positive definite symmetric matrices. This is an open convex cone (open, any two points are connected by a path, $\lambda > 0, x \in C \implies \lambda x \in C$) in $\text{Sym}_n \cong \Bbb R^{n(n+1)/2}$. Scalar multiplication gives a proper free action of $\Bbb R^\times$ on this space, and indeed $\text{Inn}_n \cong \Bbb R^{n(n+1)/2-1} \times \Bbb R^+$, identifying each positive definite symmetric matrix with one of unit operator norm. So conformal structures do indeed form a manifold, one diffeomorphic to $\Bbb R^{n(n+1)/2-1}$.

More along the lines of what you were saying, inner products are homogeneous for the action of $GL(n,\Bbb R)$ (with stabilizer $O(n)$; indeed this is one way of proving that $O(n) \to GL(n,\Bbb R)$ is a homotopy equivalence: the quotient is contractible). This descends to an action of $PGL(n,\Bbb R)$ on the space of conformal structures, which is again transitive, with stabilizer $PO(n)$.