Why is it interesting?

Quotient groups are crucial to understand, for example, symmetry breaking. When a group $G$ breaks to a subgroup $H$ the resulting Goldstone bosons live in the quotient space: $G/H$.

Moreover, quotient groups are a powerful way to understand geometry. Instead of a long list of axioms one can study geometry by treating the corresponding space as a homogeneous space (= coset space) and then study invariants of transformation groups of this homogeneous space.

In other words, this point of view allows us to study geometry by using the tools of group theory. This idea is known as Klein Geometry.

In addition, multi-Monopole dynamics was the first example where geodesics on a moduli space were used. (Source: page 314 in Topological Solitons by Manton, Sutcliff)

Layman

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A helpful (but slightly wrong) way to think about $G/N$ is that it consist of all elements in $G%$ that are not elements of $N$. A subgroup $N$ is defined through some special condition that its members must fulfil. Formulated differently: elements of $N$ are elements of $G$ that have some special property.

For example, the subgroup $n\mathbb{Z}$ of $\mathbb{Z}$ consist of all integers that are a multiple of $n$. (More explicitly: members of $3\mathbb{Z}\subset\mathbb{Z}$ are all integers that are divisible by $3$). Or another example: the subgroup $SO(N)$ of $O(N)$ consists of all $N \times N$ matrices with determinant equal to $1$. Now, $G/N$ consist of all elements that do not have this extra property. This isn't really correct.

We have, $N\in G/N$, but the important thing is that $N$ simply becomes the identity element of $G/N$, i.e. is trivial. This is how $N$ gets "modded out" from $G$ through the map $\varphi:G\to G/N$, which is called natural projection homomorphism. The elements are still there in $G/N$, but are mapped to the identity element. As defined above: $G/N$ is the set of all cosets and $N$ is the trivial coset. (See the examples below how this happens).

The assumption that the action of $G$ is transitive means that the structure of $M_0$ is determined by $G$ and $H=H_{\Phi_0}$ for any given $\Phi_0 \in M_0$. In fact: $$ M_0=G/H \tag{(5.37)}$$the space of right cosets of H in G. ($g_1$, $g_2$ $\in G$ are said to be in the same right coset of H in G if and only if there exists an $h \in H$ such that $g_1=g_2 h$. This defines an equivalence relation on G and the equivalence classes are the right cosets.)

Magnetic monopoles in gauge field theories by P GODDARD and DI OLIVE

Definition:

For a group $G$ and a normal subgroup of it $N$, we call

$$ G/N=\{gN:g\in G\} $$ the set of all cosets of $N$ in $G$ or equivalently the quotient group of $N$ in $G$. Thus the quotient group of $N$ in $G$ is defined as the set of all cosets of $N$ in $G$. ($G/N$ is spoken as "$G$ mod $N$").

• A great discussion of coset spaces and manifolds vs. coset spaces can be found in "Geometry of Physics" by Frankel page 456 and page 457.
• http://www.math3ma.com/mathema/2016/10/17/whats-a-quotient-group-really-part-1
• http://math.stackexchange.com/a/109114/120960
• http://www1.spms.ntu.edu.sg/~frederique/lecture9ws.pdf
• http://kias.dyndns.org/topogeo.html

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Examples

Goldstone Bosons

The famous Goldstone bosons that appear through the process of spontaneous symmetry breaking of a group $G$ to some subgroup $H$ live in the coset space $G/H$.

This can be understood by considering an explicit example:

For an $SO(N)$ gauge theory with a fundamental scalar representation $\phi$ the scalar potential reads

$$ V(\phi) = \frac{\lambda}{4}(\phi^2-v^2). $$

The fundamental representation acts on $\mathbb{R}^N$. Thus, we can divide $\mathbb{R}^N$ into orbits by using the group action. Each orbit consists of all the points that can be transformed into each other through $SO(N)$ transformations. These orbits are are $N-1$ dimensional spheres (except for the one-dimensional orbit that consist of the origin that isn't moved around at all by $SO(N)$ transformations).

We denote the radius of such a sphere by $\rho$ and with $\varphi_\alpha$, where $\alpha=1, \ldots, N$ the coordinates on the sphere (= angles).

Now, we assume the scalar representation develops a non-zero vacuum expectation value, i.e. breaks the $SO(N)$ symmetry. Scalars in the fundamental representation always break $SO(N)$ to $SO(N-1)$, because any VEV can be brought into the following form $<\phi> = (0,\ldots, v)$ through $SO(N)$ transformations.

The radius $\rho$ is then the Higgs field while the $\varphi_\alpha$ are the Goldstone bosons. This way, we can see that the Goldstone bosons are simply gauge degrees of freedom, while the Higgs field is not. (We can move around on a sphere $S^{N-1}$ through $SO(N)$ transformations).

In this sense, the Goldstone bosons live on the sphere $S^{N-1}$, which is exactly the coset space

$$ SO(N)/SO(N-1) = S^{N-1} . $$

Source: https://arxiv.org/pdf/0910.5167v1.pdf

See also: http://physics.stackexchange.com/questions/108722/in-what-sense-do-goldstone-bosons-live-in-the-coset

$\mathbb{Z}/3\mathbb{Z}$

$\mathbb{Z}$ is the set of all integers

$$ \mathbb{Z} = \{ \ldots, -3,-2,-1,0,1,2,3,\ldots \} $$

and forms a group under addition.

We can divide this set into subsets, for example, using as a criterion their remainder after a division by $3$. This way $\mathbb{Z}$ gets divided into three distinct subsets:

$$ 3\mathbb{Z} = \{ \ldots, ,-6,-3,0,3,6,\ldots \} =[0], $$ $$ 3\mathbb{Z}+1 = \{ \ldots, ,-5,-2,1,4,7,\ldots \}=[1], $$ $$ 3\mathbb{Z}+2 = \{ \ldots, ,-4,-1,2,5,8,\ldots \}=[2]. $$

The three sets $[0],[1],[2]$ form a group under addition. For example, $[1]+[2]=[0]$. This means, we can take any element of $[1]$ and add to it an element of $[2]$ and this always yields an element of $[1]$.

The set $\{[0],[1],[2] \}$ is called the quotient group $$ \mathbb{Z}/3\mathbb{Z} = \{[0],[1],[2] \} $$

The subgroup $3 \mathbb{Z}$ of $\mathbb{Z}$ becomes the identity element $[0]$ of the quotient group. The sets $[0]$, $[1]$ and $[3]$ are the cosets of $3 \mathbb{Z}$ in $\mathbb{Z}$ and $[0]$ is called the trivial coset.

This means that $ \mathbb{Z}/3\mathbb{Z}$ are all integers that aren't multiples of $3$, i.e. all integers that do not have the defining property of the set $3 \mathbb{Z}$. The cosets are defined as subsets of the integer number which are group together how much they do not satisfy the defining property of the set $3 \mathbb{Z}$. For example, the integers in the coset $[1]$ fail to be elements of $3 \mathbb{Z}$ by $1$ and $[2]$ fail to be elements of $3 \mathbb{Z}$ by $2$.

Source: https://www.quora.com/Whats-an-intuitive-explanation-of-quotient-group

$SO(3)/SO(n-1)$

$SO(n)$ is the set of all rotations in $\mathbb{R}^n$. Such rotations can be understood by investigating how points on a sphere get moved around. For example, $SO(3)$ moves points on the two-sphere $S^2$ into each other. In contrast, $SO(2)$ described rotations in $\mathbb{R}^2$ and therefore acts on the unit circle $S^1$.

Now, we are interested in the coset $g SO(2)$ for some fixed $g \in SO(3)$. If we say $SO(2)$ described rotations around some arbitrary, but fixed axis like the $z$-axis, we can say roughly that $SO(3)/SO(2)$ contains all three dimensional rotations that do not rotate around the $z$-axis.

For concreteness, we pick an arbitrary base point, say the north pole $\hat{e}_z= (0,0,1)$ on the two-sphere $S^2$. This point can be rotated into any other point $\hat u$ on $S^2$. However, it is not possible to identify points on $S^2$ with elements of $SO(3)$. The reason is that there are several rotations that rotate the north pole $\hat{e}_z$ into $\hat u$. Say an element $g_1$ of $SO(3)$ rotates $\hat{e}_z$ into $\hat u$:

$$ g_1 \hat{e}_z = \hat u . $$

We can then immediately write down another rotation that rotates $\hat{e}_z$ into $\hat u$ by combining $g_1$ with a rotation $h$ around the $z$-axis: $$ g_2 := g_1 h$$ $$ g_2 \hat{e}_z = g_1 h \hat{e}_z = \hat u ,$$ because the north pole is unchanged under rotations around the $z$-axis: $ h\hat{e}_z = \hat{e}_z$.

Rotations in a plane, like, for example, rotations in the $x-y$ plane (= rotations around the $z$-axis) are described by $SO(2)$.

In this sense, when we mod out $SO(2)$ rotations from $SO(3)$, we can identify elements of the resulting $SO(3)/SO(2)$ with elements of $S^2$. Without the $SO(2)$ rotations, we have a one-to-one correspondence between the remaining rotations ( = elements of $SO(3)/SO(2)$) and the two-sphere $S^2$. To every point on $S^2$ there is a unique element of $SO(3)/SO(2)$, namely the rotation that rotates, for example, the north pole into this point.

Take note that $S^2$ is not a Lie group, because $SO(2)$ is not a normal subgroup of $SO(3)$.

In general, the quotient space $SO(n)/SO(n-1)$ is $S^{n-1}$ (= the $n-1$-sphere).

$$S^{n} = \{ x \in \mathbb{R}^n : |x|=1\}$$

Source: Bott, R. and Tu, L. W., Differential Forms in Algebraic Topology, Springer (1982) page 195 and page 590 in Zees' Einstein Gravity in a Nutshell.

$GL(\mathbb{R},n)/SL(\mathbb{R},n)$

The group $GL(\mathbb{R},n)$ consist of all invertible $n \times n$ matrices and is called the general linear group. The subgroup $SL(\mathbb{R},n) \subset GL(\mathbb{R},n)$ consist of all invertible $n \times n$ matrices with determinant $1$. Now, a (slightly wrong but) helpful view is that $GL(\mathbb{R},n)/SL(\mathbb{R},n)$ is the set of all invertible $n \times n$ matrices with determinant not equal to $1$.

Each coset $gSL(\mathbb{R},n)$ for $g\in GL(\mathbb{R},n)$ consist of all invertible $n \times n$ matrices with a given value for the determinant. For example, all matrices with determinant $2$ live in the same coset and all matrices with determinant $\sqrt{3}$ live in another coset. (Two matrices $A$ and $B$ live in the same coset if $\det(AB^{-1})=1$, which is only true for $\det(A)= \det(B)$ because $\det(AB^{-1})=\det(A)\det(B^{-1})= \det(A)\det(B)^{-1}$).

This can be understood as follows:

We pick an element $g$ of $GL(\mathbb{R},n)$ and the corresponding coset $gSL(\mathbb{R},n)$ is the set of all matrices that we get when we act on $g$ with elements of $SL(\mathbb{R},n)$. Because $SL(\mathbb{R},n)$ is the set of invertible $n\times n$ matrices with determinant $1$ and $\det(AB) = \det(A)\det(B)$ holds it follows that all elements of a given coset have the same determinant. With a slight abuse of notation:

$$ \det(gSL(\mathbb{R},n))= \det(g) \det(SL(\mathbb{R},n)) = \det(g) 1 = \det(g)$$

Again, as in the $\mathbb{Z}/3\mathbb{Z}$ example, the cosets are elements that are grouped together depending on how far off they are from the defining condition of the subgroup that gets modded out. The cosets $gSL(\mathbb{R},n)$ are sets that are grouped depending on how far away their determinant is from $1$. Moreover, again the subgroup that gets modded out, in this case $SL(\mathbb{R},n)$ becomes the identity element of the quotient group.

Each coset is an equivalence class labelled through the value of the determinant. For the elements of $GL(\mathbb{R},n)$ the determinant can be any real number and therefore we have a one-to-one correspodence between the set of all cosets and $\mathbb{R}^n$:

$$ GL(\mathbb{R},n)/SL(\mathbb{R},n) \cong \mathbb{R}^n$$.

Source: http://www.math3ma.com/mathema/2016/12/15/a-quotient-of-the-general-linear-group-intuitively

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