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advanced_tools:group_theory:quotient_group [2017/12/17 12:59]
jakobadmin [Student]
advanced_tools:group_theory:quotient_group [2017/12/17 13:01]
jakobadmin
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 **Definition:​** **Definition:​**
  
-For a group $G$ and a normal subgroup of it $N$, we call +For a group $G$ and a [[advanced_tools:​group_theory:​subgroup|normal subgroup]] of it $N$, we call 
  
 $$ G/​N=\{gN:​g\in G\} $$  $$ G/​N=\{gN:​g\in G\} $$ 
Line 62: Line 62:
 <tabbox Examples> ​ <tabbox Examples> ​
  
---> ​Example1#+--> ​Goldstone Bosons#
  
- +The famous [[advanced_notions:​symmetry_breaking:​goldstones_theorem|Goldstone bosons]] that appear through the process of spontaneous [[advanced_notions:​symmetry_breaking|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
 <-- <--
  
---> ​Example2:#+--> ​$\mathbb{Z}/​3\mathbb{Z}$# 
 + 
 +$\mathbb{Z}$ 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.  
 + 
 +<WRAP tip>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 $2$. </​WRAP>​ 
 + 
 +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 [[advanced_tools:​group_theory:​subgroup|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 [[https://​books.google.de/​books?​id=5Dy1hlKvmCYC&​lpg=PA590&​ots=BQWJTqWzCH&​dq=SO(3)%2FSO(2)%20equivalence%20classes&​hl=de&​pg=PA590#​v=onepage&​q&​f=false|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
 +
 <-- <--
  
advanced_tools/group_theory/quotient_group.txt · Last modified: 2023/07/29 01:41 by 38.114.114.173