Why are physicists so interested in irreps if in their non-block-diagonal form they mix all components of a vector?Identity as a trivial reducible representationHilbert space decomposition into irrepsWhy is the $(frac12,frac12)$ representation of the Lorentz group realized as the vector space of complex $2times 2$ matrices?Problem with Bogoliubov transformations of an operatorConstruct an SO(3) rotation inside the two SU(2) fundamental rotations
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Why are physicists so interested in irreps if in their non-block-diagonal form they mix all components of a vector?
Identity as a trivial reducible representationHilbert space decomposition into irrepsWhy is the $(frac12,frac12)$ representation of the Lorentz group realized as the vector space of complex $2times 2$ matrices?Problem with Bogoliubov transformations of an operatorConstruct an SO(3) rotation inside the two SU(2) fundamental rotations
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
Consider a group $G,circ$, with elements $e,g_1,g_2,...$, represented by the matrices $D(e), D(g_1), D(g_2)...$. If all the matrices can be brought to block diagonal forms by a similarity transformation, such a representation is reducible.
In this block-diagonalized form, any vector on which the representative matrices act, do not mix all components of the vector. So we conclude that there exists more than one nontrivial (except the null space and whole space) invariant subspaces.
I cannot fully understand the importance of this. Irreps are not always block-diagonal to start with. But can be made block-diagonal. If they are not block-diagonal, they would mix all components of a vector. Is this not a problem?
Let me give a simple example. The 2-dimensional representation of SO(2): $$D=beginpmatrixcostheta & -sintheta\ sintheta & costhetaendpmatrix$$ is reducible to $$Dto D^prime= SDS^-1=beginpmatrixe^itheta & 0\ 0 & e^-ithetaendpmatrix$$
In this first case, the basis vectors are $(1~0)^T$ and $(0~1)^T$. In this case, $A_x$ and $A_y$ mix with each other.
In the second case, the basis vectors are $2^-1/2(1~~i)^T$ and $2^-1/2(1-i)^T$ and S is given by $$S=2^-1/2beginpmatrix1 & i\ 1 & -iendpmatrix.$$ It is easy to see that $A_pm=A_xpm iA_y$ do not mix with each other.
So if I am getting it correct, this example shows that there are two invariant 1-dimensional subspaces in this case - one spanned by $2^-1/2(1~~i)^T$ and the other by $2^-1/2(1-i)^T$. So finding irreps mean identify the basis?
group-theory group-representations linear-algebra
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
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$endgroup$
add a comment
|
$begingroup$
Consider a group $G,circ$, with elements $e,g_1,g_2,...$, represented by the matrices $D(e), D(g_1), D(g_2)...$. If all the matrices can be brought to block diagonal forms by a similarity transformation, such a representation is reducible.
In this block-diagonalized form, any vector on which the representative matrices act, do not mix all components of the vector. So we conclude that there exists more than one nontrivial (except the null space and whole space) invariant subspaces.
I cannot fully understand the importance of this. Irreps are not always block-diagonal to start with. But can be made block-diagonal. If they are not block-diagonal, they would mix all components of a vector. Is this not a problem?
Let me give a simple example. The 2-dimensional representation of SO(2): $$D=beginpmatrixcostheta & -sintheta\ sintheta & costhetaendpmatrix$$ is reducible to $$Dto D^prime= SDS^-1=beginpmatrixe^itheta & 0\ 0 & e^-ithetaendpmatrix$$
In this first case, the basis vectors are $(1~0)^T$ and $(0~1)^T$. In this case, $A_x$ and $A_y$ mix with each other.
In the second case, the basis vectors are $2^-1/2(1~~i)^T$ and $2^-1/2(1-i)^T$ and S is given by $$S=2^-1/2beginpmatrix1 & i\ 1 & -iendpmatrix.$$ It is easy to see that $A_pm=A_xpm iA_y$ do not mix with each other.
So if I am getting it correct, this example shows that there are two invariant 1-dimensional subspaces in this case - one spanned by $2^-1/2(1~~i)^T$ and the other by $2^-1/2(1-i)^T$. So finding irreps mean identify the basis?
group-theory group-representations linear-algebra
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
$endgroup$
$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:11
2
$begingroup$
By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:13
$begingroup$
$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:40
$begingroup$
Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:45
add a comment
|
$begingroup$
Consider a group $G,circ$, with elements $e,g_1,g_2,...$, represented by the matrices $D(e), D(g_1), D(g_2)...$. If all the matrices can be brought to block diagonal forms by a similarity transformation, such a representation is reducible.
In this block-diagonalized form, any vector on which the representative matrices act, do not mix all components of the vector. So we conclude that there exists more than one nontrivial (except the null space and whole space) invariant subspaces.
I cannot fully understand the importance of this. Irreps are not always block-diagonal to start with. But can be made block-diagonal. If they are not block-diagonal, they would mix all components of a vector. Is this not a problem?
Let me give a simple example. The 2-dimensional representation of SO(2): $$D=beginpmatrixcostheta & -sintheta\ sintheta & costhetaendpmatrix$$ is reducible to $$Dto D^prime= SDS^-1=beginpmatrixe^itheta & 0\ 0 & e^-ithetaendpmatrix$$
In this first case, the basis vectors are $(1~0)^T$ and $(0~1)^T$. In this case, $A_x$ and $A_y$ mix with each other.
In the second case, the basis vectors are $2^-1/2(1~~i)^T$ and $2^-1/2(1-i)^T$ and S is given by $$S=2^-1/2beginpmatrix1 & i\ 1 & -iendpmatrix.$$ It is easy to see that $A_pm=A_xpm iA_y$ do not mix with each other.
So if I am getting it correct, this example shows that there are two invariant 1-dimensional subspaces in this case - one spanned by $2^-1/2(1~~i)^T$ and the other by $2^-1/2(1-i)^T$. So finding irreps mean identify the basis?
group-theory group-representations linear-algebra
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
$endgroup$
Consider a group $G,circ$, with elements $e,g_1,g_2,...$, represented by the matrices $D(e), D(g_1), D(g_2)...$. If all the matrices can be brought to block diagonal forms by a similarity transformation, such a representation is reducible.
In this block-diagonalized form, any vector on which the representative matrices act, do not mix all components of the vector. So we conclude that there exists more than one nontrivial (except the null space and whole space) invariant subspaces.
I cannot fully understand the importance of this. Irreps are not always block-diagonal to start with. But can be made block-diagonal. If they are not block-diagonal, they would mix all components of a vector. Is this not a problem?
Let me give a simple example. The 2-dimensional representation of SO(2): $$D=beginpmatrixcostheta & -sintheta\ sintheta & costhetaendpmatrix$$ is reducible to $$Dto D^prime= SDS^-1=beginpmatrixe^itheta & 0\ 0 & e^-ithetaendpmatrix$$
In this first case, the basis vectors are $(1~0)^T$ and $(0~1)^T$. In this case, $A_x$ and $A_y$ mix with each other.
In the second case, the basis vectors are $2^-1/2(1~~i)^T$ and $2^-1/2(1-i)^T$ and S is given by $$S=2^-1/2beginpmatrix1 & i\ 1 & -iendpmatrix.$$ It is easy to see that $A_pm=A_xpm iA_y$ do not mix with each other.
So if I am getting it correct, this example shows that there are two invariant 1-dimensional subspaces in this case - one spanned by $2^-1/2(1~~i)^T$ and the other by $2^-1/2(1-i)^T$. So finding irreps mean identify the basis?
group-theory group-representations linear-algebra
group-theory group-representations linear-algebra
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
edited Aug 9 at 14:32
Qmechanic♦
116k14 gold badges231 silver badges1390 bronze badges
116k14 gold badges231 silver badges1390 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
asked Aug 9 at 13:09
mithusengupta123mithusengupta123
1,0801 gold badge18 silver badges45 bronze badges
1,0801 gold badge18 silver badges45 bronze badges
$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:11
2
$begingroup$
By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:13
$begingroup$
$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:40
$begingroup$
Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:45
add a comment
|
$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:11
2
$begingroup$
By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:13
$begingroup$
$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:40
$begingroup$
Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:45
$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:11
$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:11
2
2
$begingroup$
By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:13
$begingroup$
By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
$endgroup$
– Chiral Anomaly
Aug 9 at 23:13
$begingroup$
$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:40
$begingroup$
$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:40
$begingroup$
Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:45
$begingroup$
Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
$endgroup$
– Jyrki Lahtonen
Aug 10 at 3:45
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.
Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.
It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
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$begingroup$
"Reducible" and "irreducible" can be defined in a basis-independent way.
Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.
("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)
Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
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2 Answers
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2 Answers
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$begingroup$
The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.
Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.
It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
$endgroup$
add a comment
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$begingroup$
The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.
Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.
It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
$endgroup$
add a comment
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$begingroup$
The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.
Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.
It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
$endgroup$
The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.
Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.
It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
answered Aug 9 at 13:38
JavierJavier
16.2k8 gold badges48 silver badges86 bronze badges
16.2k8 gold badges48 silver badges86 bronze badges
add a comment
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add a comment
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$begingroup$
"Reducible" and "irreducible" can be defined in a basis-independent way.
Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.
("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)
Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
$endgroup$
add a comment
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$begingroup$
"Reducible" and "irreducible" can be defined in a basis-independent way.
Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.
("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)
Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
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$begingroup$
"Reducible" and "irreducible" can be defined in a basis-independent way.
Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.
("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)
Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
$endgroup$
"Reducible" and "irreducible" can be defined in a basis-independent way.
Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.
("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)
Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
edited Aug 9 at 13:42
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
answered Aug 9 at 13:37
Chiral AnomalyChiral Anomaly
21.4k4 gold badges29 silver badges66 bronze badges
21.4k4 gold badges29 silver badges66 bronze badges
add a comment
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add a comment
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$begingroup$
"So finding irreps mean[s] identify[ing] the basis?" Not a completely-specific basis, but a decomposition into subspaces. If the vector space $V$ on which the representation acts can be partitioned into two subspaces that are not mixed with each other by any $D(g)$, then we might as well think of them as two separate representations (intuitively). An irrep is a representation that can't be decomposed in this way any further.
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– Chiral Anomaly
Aug 9 at 23:11
2
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By the way, the example you added is reducible over $mathbbC$ but not over $mathbbR$. A representation of $G$ is a homomorphism from $G$ into a matrix algebra, and specifying whether the matrix algebra is over $mathbbC$ or $mathbbR$ is important. Your example illustrates why it's important.
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– Chiral Anomaly
Aug 9 at 23:13
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$SO(2)$ is a bit special as it is abelian, implying that all the irreps are 1-dimensional (in "sensible" cases). My mathematician's nitpick is that if you list the elements of a group $G$ like $e,g_1,g_2,ldots$ then that implies the group to be countable, which $SO(2,BbbR)$ is not :-/
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– Jyrki Lahtonen
Aug 10 at 3:40
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Possibly the source of your confusion is that for a representation to be reducible it should be the case that the same similarity transformation brings all the matrices of the group into a block diagonal form.
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– Jyrki Lahtonen
Aug 10 at 3:45