$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.

• Are you sure you want $E$ and not $C$?

In either case, the manifold $X = X_E = \SO(3)/\Stab(E)$ is called a "spherical space form" and also an "elliptical manifold". This is because $X$ is a quotient of the three-sphere by a freely acting subgroup of $\SO(4)$ -- note that $\SO(3)$ is a copy of $\RP^3$ so it is also a spherical space form. Jeff Weeks' book "The shape of space" has an elementary treatment of the geometry and topology involved. Also, here is link to Wikipedia:

https://en.wikipedia.org/wiki/Spherical_3-manifold

The manifold $X$ is called a "dihedral manifold" there. Now, you ask

Does this space admit a simple parameterization, measure, or metric?

and you ask for

[e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like?

You haven't said why you are looking for this, beyond mentioning a computational application. There are lots of embedding theorems (eg the Whitney embedding theorem) and there are ways to describe the induced metrics. These will (almost certainly) be useless for computational purposes.

Probably you will be much happier with the following much simpler approach. Encode elements of $X = \SO(3)/\Stab(E)$ as objects $[M]$ where

• $M$ is an orthogonal matrix (of determinant one)
• and we have a new equality method: $[M] = [N]$ if and only if $N^{-1} M \in \Stab(E)$.

I'll guess that these new objects (aka cosets) will do everything you want, while avoiding dangerous mucking about with wacky embeddings. For example, computing distance in $X$ reduces to three distance computations in $\SO(3)$. (Note that if you are using floats as entries in your matrices, then you will encounter the usual problem when testing for equality. This will be solved in the usual way.)

$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.

• Are you sure you want $E$ and not $C$?

In either case, the manifold $X = X_E = \SO(3)/\Stab(E)$ is called a "spherical space form" and also an "elliptical manifold". This is because $X$ is a quotient of the three-sphere by a freely acting subgroup of $\SO(4)$ -- note that $\SO(3)$ is a copy of $\RP^3$ so it is also a spherical space form. Jeff Weeks' book "The shape of space" has an elementary treatment of the geometry and topology involved. Also, here is link to Wikipedia:

https://en.wikipedia.org/wiki/Spherical_3-manifold

The manifold $X$ is called a "prism manifold" there. Now, you ask

Does this space admit a simple parameterization, measure, or metric?

and you ask for

[e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like?

You haven't said why you are looking for this, beyond mentioning a computational application. There are lots of embedding theorems (eg the Whitney embedding theorem) and there are ways to describe the induced metrics. These will (almost certainly) be useless for computational purposes.

Probably you will be much happier with the following much simpler approach. Encode elements of $X = \SO(3)/\Stab(E)$ as objects $[M]$ where

• $M$ is an orthogonal matrix (of determinant one) and
• $[M] = [N]$ if and only if $N^{-1} M \in \Stab(E)$.

I'll guess that these new objects (basically, cosets) will do everything you want, while avoiding dangerous mucking about with wacky embeddings. For example, computing distance in $X$ reduces to three distance computations in $\SO(3)$.

$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.

• Are you sure you want $E$ and not $C$?

In either case, the manifold $X = X_E = \SO(3)/\Stab(E)$ is called a "spherical space form" and also an "elliptical manifold". This is because $X$ is a quotient of the three-sphere by a freely acting subgroup of $\SO(4)$ -- note that $\SO(3)$ is a copy of $\RP^3$ so it is also a spherical space form. Jeff Weeks' book "The shape of space" has an elementary treatment of the geometry and topology involved. Also, here is link to Wikipedia:

https://en.wikipedia.org/wiki/Spherical_3-manifold

The manifold $X$ is called a "dihedral manifold" there. Now, you ask

Does this space admit a simple parameterization, measure, or metric?

and you ask for

[e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like?

You haven't said why you are looking for this, beyond mentioning a computational application. There are lots of embedding theorems (eg the Whitney embedding theorem) and there are ways to describe the induced metrics. These will (almost certainly) be useless for computational purposes.

Probably you will be much happier with the following much simpler approach. Encode elements of $X = \SO(3)/\Stab(E)$ as objects $[M]$ where

• $M$ is an orthogonal matrix (of determinant one)
• and we have a new equality method: $[M] = [N]$ if and only if $N^{-1} M \in \Stab(E)$.

I'll guess that these new objects (aka cosets) will do everything you want, while avoiding dangerous mucking about with wacky embeddings. (Note that if you are using floats as entries in your matrices, then you will encounter the usual problem when testing for equality. This will be solved in the usual way.)

$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.

• Are you sure you want $E$ and not $C$?

In either case, the manifold $X = X_E = \SO(3)/\Stab(E)$ is called a "spherical space form" and also an "elliptical manifold". This is because $X$ is a quotient of the three-sphere by a freely acting subgroup of $\SO(4)$ -- note that $\SO(3)$ is a copy of $\RP^3$ so it is also a spherical space form. Jeff Weeks' book "The shape of space" has an elementary treatment of the geometry and topology involved. Also, here is link to Wikipedia:

https://en.wikipedia.org/wiki/Spherical_3-manifold

The manifold $X$ is called a "dihedral manifold" there. Now, you ask

Does this space admit a simple parameterization, measure, or metric?

and you ask for

[e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like?

You haven't said why you are looking for this, beyond mentioning a computational application. There are lots of embedding theorems (eg the Whitney embedding theorem) and there are ways to describe the induced metrics. These will (almost certainly) be useless for computational purposes.

Probably you will be much happier with the following much simpler approach. Encode elements of $X = \SO(3)/\Stab(E)$ as objects $[M]$ where

• $M$ is an orthogonal matrix (of determinant one)
• and we have a new equality method: $[M] = [N]$ if and only if $N^{-1} M \in \Stab(E)$.

I'll guess that these new objects (aka cosets) will do everything you want, while avoiding dangerous mucking about with wacky embeddings. For example, computing distance in $X$ reduces to three distance computations in $\SO(3)$. (Note that if you are using floats as entries in your matrices, then you will encounter the usual problem when testing for equality. This will be solved in the usual way.)