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--- advanced_tools:group_theory:quotient_group [2017/12/17 12:59]
jakobadmin [Student]
+++ advanced_tools:group_theory:quotient_group [2023/07/29 01:41] (current)
38.114.114.173 [Student]
@@ Line 3: / Line 3: @@
 <tabbox Why is it interesting?>
-Quotient groups are crucial to understand, for example, [[advanced_notions:symmetry_breaking|symmetry breaking]]. When a [[advanced_tools:group_theory|group]] $G$ breaks to a [[advanced_tools:group_theory:subgroup|subgroup]] $H$ the resulting [[advanced_notions:symmetry_breaking:goldstones_theorem|Goldstone bosons]] live in the quotient space: $G/H$.
+Quotient groups are crucial to understand, for example, [[advanced_notions:symmetry_breaking|symmetry breaking]]. When a [[advanced_tools:group_theory|group]] $G$ breaks to a [[advanced_tools:group_theory:subgroup|subgroup]] $H$ the resulting [[theorems:goldstones_theorem|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.
@@ Line 19: / Line 19: @@
 <tabbox Student>
-A helpful (but slightly wrong) way to think about $G/N$ is that is 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.
+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.//
@@ Line 40: / Line 40: @@
 **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
 $$ G/N=\{gN:g\in G\} $$
@@ Line 62: / Line 62: @@
 <tabbox Examples>
---> Example1#
+--> Goldstone Bosons#
+The famous [[theorems: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
 <--

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