Mixing
$mathbb{C}^2 otimes mathbb{C}^2 otimes mathbb{C}^2$ representation of $B_3$ braid group
$begingroup$
I've been trying to find a representation of the braid group $B_3$ acting on $mathbb{C}^2 otimes mathbb{C}^2 otimes mathbb{C}^2$ but I can't find it anywhere.
From what I understand I have to find two $8 times 8$ matrices $sigma_i$ satisfying $sigma_1 sigma_2 sigma_1 = sigma_2 sigma_1 sigma_2$. I tried doing it by hand but it proved to be harder than anticipated. Is there any known solution to this?
Thanks!
representation-theory knot-theory quantum-computation braid-groups
edited Mar 15 at 1:28
Juan Ospina
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
$endgroup$
|
show 2 more comments
$begingroup$
I've been trying to find a representation of the braid group $B_3$ acting on $mathbb{C}^2 otimes mathbb{C}^2 otimes mathbb{C}^2$ but I can't find it anywhere.
From what I understand I have to find two $8 times 8$ matrices $sigma_i$ satisfying $sigma_1 sigma_2 sigma_1 = sigma_2 sigma_1 sigma_2$. I tried doing it by hand but it proved to be harder than anticipated. Is there any known solution to this?
Thanks!
representation-theory knot-theory quantum-computation braid-groups
edited Mar 15 at 1:28
Juan Ospina
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
$endgroup$
$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
1
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
1
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
1
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
1
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27
|
show 2 more comments
$begingroup$
I've been trying to find a representation of the braid group $B_3$ acting on $mathbb{C}^2 otimes mathbb{C}^2 otimes mathbb{C}^2$ but I can't find it anywhere.
From what I understand I have to find two $8 times 8$ matrices $sigma_i$ satisfying $sigma_1 sigma_2 sigma_1 = sigma_2 sigma_1 sigma_2$. I tried doing it by hand but it proved to be harder than anticipated. Is there any known solution to this?
Thanks!
representation-theory knot-theory quantum-computation braid-groups
edited Mar 15 at 1:28
Juan Ospina
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
$endgroup$
I've been trying to find a representation of the braid group $B_3$ acting on $mathbb{C}^2 otimes mathbb{C}^2 otimes mathbb{C}^2$ but I can't find it anywhere.
From what I understand I have to find two $8 times 8$ matrices $sigma_i$ satisfying $sigma_1 sigma_2 sigma_1 = sigma_2 sigma_1 sigma_2$. I tried doing it by hand but it proved to be harder than anticipated. Is there any known solution to this?
Thanks!
representation-theory knot-theory quantum-computation braid-groups
representation-theory knot-theory quantum-computation braid-groups
edited Mar 15 at 1:28
Juan Ospina
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
edited Mar 15 at 1:28
Juan Ospina
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
edited Mar 15 at 1:28
Juan Ospina
1,5841614
edited Mar 15 at 1:28
Juan Ospina
1,5841614
edited Mar 15 at 1:28
Juan Ospina
1,5841614
1,5841614
asked Feb 3 at 1:09
GKiuGKiu
917
asked Feb 3 at 1:09
GKiuGKiu
917
asked Feb 3 at 1:09
GKiuGKiu
917
917
$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
1
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
1
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
1
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
1
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27
|
show 2 more comments
$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
1
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
1
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
1
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
1
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27
$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
1
1
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
1
1
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
1
1
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
1
1
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
I'm going to make a guess that the motivation for your question is representations of $B_n$ that commute with representations of "quantum $mathfrak{sl}(2)$." In this case, $mathbb{C}^2$ would be the two-dimensional irreducible representation of $U_q(mathfrak{sl}(2))$, known as $V_1$ since it is the highest-weight $1$ representation.
One may conceptualize this as looking for maps from $B_n$ to the space of endomorphisms of $(mathbb{C}^2)^{otimes n}$ that commute with the $mathcal{U}_q(mathfrak{sl}(2))$-action, where this quantum group is acting on each $mathbb{C}^2$ factor simultaneously. That is, maps $B_nto operatorname{End}_{mathcal{U}_q(mathfrak{sl}(2))}(V_1^{otimes n})$.
The $q=1$ case is the representation Mike Miller came up with: $B_3$ acts on $V_1^{otimes 3}$ by permuting the tensor factors via the $B_3to S_3$ homomorphism, and this intertwines with $(v_1otimes v_2otimes v_3) x=xv_1otimes x v_2otimes x v_3$ for $xinmathfrak{sl}(2)$, where $xv_i$ is simply the matrix-vector product.
More generally, we need to know the "$R$-matrix" for how a two-strand twist of $B_3$ corresponds to a homomorphism $V_1otimes V_1to V_1otimes V_1$. The earlier conceptualization leads us to the Temperley-Lieb algebra, which for our purposes we will take $qinmathbb{C}^times$ and then take $mathbb{C}[B_n]$ and quotient it by (i.e., impose) the following relations:
One can check that these satisfy the relations for $B_n$, and so this is a well-defined quotient. So: given a braid in $B_n$, one may resolve all crossings and expand the word into a linear combination of crossing-free Temperley-Lieb diagrams. For $B_3$, every braid reduces to a linear combination of the following five diagrams:
At this point, we can make a representation by deciding what "cups" and "caps" should be. According to some notes I have, the following choice works, where ${e_1,e_2}$ forms a basis for $mathbb{C}^2$ and ${e^1,e^2}$ forms a corresponding dual basis for $(mathbb{C}^2)^*$:
If I calculated it correctly, this is the two-strand twist as a matrix (derived from an $R$-matrix), with basis ${e_{11},e_{12},e_{21},e_{22}}$:
To get the matrices for $B_3$, one can perform the Kronecker product with the $2times 2$ identity matrix, which has the effect of giving the matrix for a braid with an additional strand on one side. These were checked in Mathematica to satisfy $sigma_1sigma_2sigma_1=sigma_2sigma_1sigma_2$:
begin{align}
sigma_1 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right) \
sigma_2 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 \
0 & 0 & 0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right)
end{align}
This representation splits into irreducibles for generic values of $q$ (that is, except for a few roots of unity). In particular, this eight-dimensional representation splits as $2V_1oplus V_3$. The projection onto the $V_3$ summand is given by the third Jones-Wenzl projector, which graphically can be written as
and with some elbow grease this may be turned into an $8times 8$ projection matrix. (For $q=1$, this is the projection onto $operatorname{Sym}^3 V_1$.)
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
add a comment |
$begingroup$
Other representation, which is unitary, is given by
These matrices were checked in Maple in order to verify than they satisfy the braiding relation or the Yang-Baxter condition.
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other representation for three qbits:
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other possibility is the Jones representation:
with the condition
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
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4 Answers
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active
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4 Answers
4
active
oldest
votes
active
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active
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$begingroup$
I'm going to make a guess that the motivation for your question is representations of $B_n$ that commute with representations of "quantum $mathfrak{sl}(2)$." In this case, $mathbb{C}^2$ would be the two-dimensional irreducible representation of $U_q(mathfrak{sl}(2))$, known as $V_1$ since it is the highest-weight $1$ representation.
One may conceptualize this as looking for maps from $B_n$ to the space of endomorphisms of $(mathbb{C}^2)^{otimes n}$ that commute with the $mathcal{U}_q(mathfrak{sl}(2))$-action, where this quantum group is acting on each $mathbb{C}^2$ factor simultaneously. That is, maps $B_nto operatorname{End}_{mathcal{U}_q(mathfrak{sl}(2))}(V_1^{otimes n})$.
The $q=1$ case is the representation Mike Miller came up with: $B_3$ acts on $V_1^{otimes 3}$ by permuting the tensor factors via the $B_3to S_3$ homomorphism, and this intertwines with $(v_1otimes v_2otimes v_3) x=xv_1otimes x v_2otimes x v_3$ for $xinmathfrak{sl}(2)$, where $xv_i$ is simply the matrix-vector product.
More generally, we need to know the "$R$-matrix" for how a two-strand twist of $B_3$ corresponds to a homomorphism $V_1otimes V_1to V_1otimes V_1$. The earlier conceptualization leads us to the Temperley-Lieb algebra, which for our purposes we will take $qinmathbb{C}^times$ and then take $mathbb{C}[B_n]$ and quotient it by (i.e., impose) the following relations:
One can check that these satisfy the relations for $B_n$, and so this is a well-defined quotient. So: given a braid in $B_n$, one may resolve all crossings and expand the word into a linear combination of crossing-free Temperley-Lieb diagrams. For $B_3$, every braid reduces to a linear combination of the following five diagrams:
At this point, we can make a representation by deciding what "cups" and "caps" should be. According to some notes I have, the following choice works, where ${e_1,e_2}$ forms a basis for $mathbb{C}^2$ and ${e^1,e^2}$ forms a corresponding dual basis for $(mathbb{C}^2)^*$:
If I calculated it correctly, this is the two-strand twist as a matrix (derived from an $R$-matrix), with basis ${e_{11},e_{12},e_{21},e_{22}}$:
To get the matrices for $B_3$, one can perform the Kronecker product with the $2times 2$ identity matrix, which has the effect of giving the matrix for a braid with an additional strand on one side. These were checked in Mathematica to satisfy $sigma_1sigma_2sigma_1=sigma_2sigma_1sigma_2$:
begin{align}
sigma_1 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right) \
sigma_2 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 \
0 & 0 & 0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right)
end{align}
This representation splits into irreducibles for generic values of $q$ (that is, except for a few roots of unity). In particular, this eight-dimensional representation splits as $2V_1oplus V_3$. The projection onto the $V_3$ summand is given by the third Jones-Wenzl projector, which graphically can be written as
and with some elbow grease this may be turned into an $8times 8$ projection matrix. (For $q=1$, this is the projection onto $operatorname{Sym}^3 V_1$.)
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
add a comment |
$begingroup$
I'm going to make a guess that the motivation for your question is representations of $B_n$ that commute with representations of "quantum $mathfrak{sl}(2)$." In this case, $mathbb{C}^2$ would be the two-dimensional irreducible representation of $U_q(mathfrak{sl}(2))$, known as $V_1$ since it is the highest-weight $1$ representation.
One may conceptualize this as looking for maps from $B_n$ to the space of endomorphisms of $(mathbb{C}^2)^{otimes n}$ that commute with the $mathcal{U}_q(mathfrak{sl}(2))$-action, where this quantum group is acting on each $mathbb{C}^2$ factor simultaneously. That is, maps $B_nto operatorname{End}_{mathcal{U}_q(mathfrak{sl}(2))}(V_1^{otimes n})$.
The $q=1$ case is the representation Mike Miller came up with: $B_3$ acts on $V_1^{otimes 3}$ by permuting the tensor factors via the $B_3to S_3$ homomorphism, and this intertwines with $(v_1otimes v_2otimes v_3) x=xv_1otimes x v_2otimes x v_3$ for $xinmathfrak{sl}(2)$, where $xv_i$ is simply the matrix-vector product.
More generally, we need to know the "$R$-matrix" for how a two-strand twist of $B_3$ corresponds to a homomorphism $V_1otimes V_1to V_1otimes V_1$. The earlier conceptualization leads us to the Temperley-Lieb algebra, which for our purposes we will take $qinmathbb{C}^times$ and then take $mathbb{C}[B_n]$ and quotient it by (i.e., impose) the following relations:
One can check that these satisfy the relations for $B_n$, and so this is a well-defined quotient. So: given a braid in $B_n$, one may resolve all crossings and expand the word into a linear combination of crossing-free Temperley-Lieb diagrams. For $B_3$, every braid reduces to a linear combination of the following five diagrams:
At this point, we can make a representation by deciding what "cups" and "caps" should be. According to some notes I have, the following choice works, where ${e_1,e_2}$ forms a basis for $mathbb{C}^2$ and ${e^1,e^2}$ forms a corresponding dual basis for $(mathbb{C}^2)^*$:
If I calculated it correctly, this is the two-strand twist as a matrix (derived from an $R$-matrix), with basis ${e_{11},e_{12},e_{21},e_{22}}$:
To get the matrices for $B_3$, one can perform the Kronecker product with the $2times 2$ identity matrix, which has the effect of giving the matrix for a braid with an additional strand on one side. These were checked in Mathematica to satisfy $sigma_1sigma_2sigma_1=sigma_2sigma_1sigma_2$:
begin{align}
sigma_1 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right) \
sigma_2 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 \
0 & 0 & 0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right)
end{align}
This representation splits into irreducibles for generic values of $q$ (that is, except for a few roots of unity). In particular, this eight-dimensional representation splits as $2V_1oplus V_3$. The projection onto the $V_3$ summand is given by the third Jones-Wenzl projector, which graphically can be written as
and with some elbow grease this may be turned into an $8times 8$ projection matrix. (For $q=1$, this is the projection onto $operatorname{Sym}^3 V_1$.)
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
add a comment |
$begingroup$
I'm going to make a guess that the motivation for your question is representations of $B_n$ that commute with representations of "quantum $mathfrak{sl}(2)$." In this case, $mathbb{C}^2$ would be the two-dimensional irreducible representation of $U_q(mathfrak{sl}(2))$, known as $V_1$ since it is the highest-weight $1$ representation.
One may conceptualize this as looking for maps from $B_n$ to the space of endomorphisms of $(mathbb{C}^2)^{otimes n}$ that commute with the $mathcal{U}_q(mathfrak{sl}(2))$-action, where this quantum group is acting on each $mathbb{C}^2$ factor simultaneously. That is, maps $B_nto operatorname{End}_{mathcal{U}_q(mathfrak{sl}(2))}(V_1^{otimes n})$.
The $q=1$ case is the representation Mike Miller came up with: $B_3$ acts on $V_1^{otimes 3}$ by permuting the tensor factors via the $B_3to S_3$ homomorphism, and this intertwines with $(v_1otimes v_2otimes v_3) x=xv_1otimes x v_2otimes x v_3$ for $xinmathfrak{sl}(2)$, where $xv_i$ is simply the matrix-vector product.
More generally, we need to know the "$R$-matrix" for how a two-strand twist of $B_3$ corresponds to a homomorphism $V_1otimes V_1to V_1otimes V_1$. The earlier conceptualization leads us to the Temperley-Lieb algebra, which for our purposes we will take $qinmathbb{C}^times$ and then take $mathbb{C}[B_n]$ and quotient it by (i.e., impose) the following relations:
One can check that these satisfy the relations for $B_n$, and so this is a well-defined quotient. So: given a braid in $B_n$, one may resolve all crossings and expand the word into a linear combination of crossing-free Temperley-Lieb diagrams. For $B_3$, every braid reduces to a linear combination of the following five diagrams:
At this point, we can make a representation by deciding what "cups" and "caps" should be. According to some notes I have, the following choice works, where ${e_1,e_2}$ forms a basis for $mathbb{C}^2$ and ${e^1,e^2}$ forms a corresponding dual basis for $(mathbb{C}^2)^*$:
If I calculated it correctly, this is the two-strand twist as a matrix (derived from an $R$-matrix), with basis ${e_{11},e_{12},e_{21},e_{22}}$:
To get the matrices for $B_3$, one can perform the Kronecker product with the $2times 2$ identity matrix, which has the effect of giving the matrix for a braid with an additional strand on one side. These were checked in Mathematica to satisfy $sigma_1sigma_2sigma_1=sigma_2sigma_1sigma_2$:
begin{align}
sigma_1 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right) \
sigma_2 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 \
0 & 0 & 0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right)
end{align}
This representation splits into irreducibles for generic values of $q$ (that is, except for a few roots of unity). In particular, this eight-dimensional representation splits as $2V_1oplus V_3$. The projection onto the $V_3$ summand is given by the third Jones-Wenzl projector, which graphically can be written as
and with some elbow grease this may be turned into an $8times 8$ projection matrix. (For $q=1$, this is the projection onto $operatorname{Sym}^3 V_1$.)
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
$endgroup$
I'm going to make a guess that the motivation for your question is representations of $B_n$ that commute with representations of "quantum $mathfrak{sl}(2)$." In this case, $mathbb{C}^2$ would be the two-dimensional irreducible representation of $U_q(mathfrak{sl}(2))$, known as $V_1$ since it is the highest-weight $1$ representation.
One may conceptualize this as looking for maps from $B_n$ to the space of endomorphisms of $(mathbb{C}^2)^{otimes n}$ that commute with the $mathcal{U}_q(mathfrak{sl}(2))$-action, where this quantum group is acting on each $mathbb{C}^2$ factor simultaneously. That is, maps $B_nto operatorname{End}_{mathcal{U}_q(mathfrak{sl}(2))}(V_1^{otimes n})$.
The $q=1$ case is the representation Mike Miller came up with: $B_3$ acts on $V_1^{otimes 3}$ by permuting the tensor factors via the $B_3to S_3$ homomorphism, and this intertwines with $(v_1otimes v_2otimes v_3) x=xv_1otimes x v_2otimes x v_3$ for $xinmathfrak{sl}(2)$, where $xv_i$ is simply the matrix-vector product.
More generally, we need to know the "$R$-matrix" for how a two-strand twist of $B_3$ corresponds to a homomorphism $V_1otimes V_1to V_1otimes V_1$. The earlier conceptualization leads us to the Temperley-Lieb algebra, which for our purposes we will take $qinmathbb{C}^times$ and then take $mathbb{C}[B_n]$ and quotient it by (i.e., impose) the following relations:
One can check that these satisfy the relations for $B_n$, and so this is a well-defined quotient. So: given a braid in $B_n$, one may resolve all crossings and expand the word into a linear combination of crossing-free Temperley-Lieb diagrams. For $B_3$, every braid reduces to a linear combination of the following five diagrams:
At this point, we can make a representation by deciding what "cups" and "caps" should be. According to some notes I have, the following choice works, where ${e_1,e_2}$ forms a basis for $mathbb{C}^2$ and ${e^1,e^2}$ forms a corresponding dual basis for $(mathbb{C}^2)^*$:
If I calculated it correctly, this is the two-strand twist as a matrix (derived from an $R$-matrix), with basis ${e_{11},e_{12},e_{21},e_{22}}$:
To get the matrices for $B_3$, one can perform the Kronecker product with the $2times 2$ identity matrix, which has the effect of giving the matrix for a braid with an additional strand on one side. These were checked in Mathematica to satisfy $sigma_1sigma_2sigma_1=sigma_2sigma_1sigma_2$:
begin{align}
sigma_1 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right) \
sigma_2 &= left(
begin{array}{cccccccc}
sqrt{q} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 \
0 & frac{1}{sqrt{q}} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & sqrt{q} & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & sqrt{q}-frac{1}{q^{3/2}} & frac{1}{sqrt{q}} & 0 \
0 & 0 & 0 & 0 & 0 & frac{1}{sqrt{q}} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & sqrt{q} \
end{array}
right)
end{align}
This representation splits into irreducibles for generic values of $q$ (that is, except for a few roots of unity). In particular, this eight-dimensional representation splits as $2V_1oplus V_3$. The projection onto the $V_3$ summand is given by the third Jones-Wenzl projector, which graphically can be written as
and with some elbow grease this may be turned into an $8times 8$ projection matrix. (For $q=1$, this is the projection onto $operatorname{Sym}^3 V_1$.)
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
edited Feb 4 at 0:07
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
answered Feb 3 at 23:00
Kyle MillerKyle Miller
10.1k930
10.1k930
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
add a comment |
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
$begingroup$
This is exactly what I needed. Thank you very much for your amazing, clearly explained answer!
$endgroup$
– GKiu
Feb 4 at 11:20
add a comment |
$begingroup$
Other representation, which is unitary, is given by
These matrices were checked in Maple in order to verify than they satisfy the braiding relation or the Yang-Baxter condition.
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other representation, which is unitary, is given by
These matrices were checked in Maple in order to verify than they satisfy the braiding relation or the Yang-Baxter condition.
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other representation, which is unitary, is given by
These matrices were checked in Maple in order to verify than they satisfy the braiding relation or the Yang-Baxter condition.
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
$endgroup$
Other representation, which is unitary, is given by
These matrices were checked in Maple in order to verify than they satisfy the braiding relation or the Yang-Baxter condition.
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
edited Mar 14 at 13:23
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
answered Mar 14 at 12:37
Juan OspinaJuan Ospina
1,5841614
1,5841614
add a comment |
add a comment |
$begingroup$
Other representation for three qbits:
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other representation for three qbits:
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other representation for three qbits:
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
$endgroup$
Other representation for three qbits:
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
answered Mar 14 at 18:33
Juan OspinaJuan Ospina
1,5841614
1,5841614
add a comment |
add a comment |
$begingroup$
Other possibility is the Jones representation:
with the condition
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other possibility is the Jones representation:
with the condition
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
$endgroup$
add a comment |
$begingroup$
Other possibility is the Jones representation:
with the condition
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
$endgroup$
Other possibility is the Jones representation:
with the condition
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
answered Mar 15 at 19:49
Juan OspinaJuan Ospina
1,5841614
1,5841614
add a comment |
add a comment |
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$begingroup$
What book are you using?
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 1:41
1
$begingroup$
By now I've looked into a mixture of readable books and papers I could find (I'm a physicists, the majority of the hardcore mathematical literature is unreadable to me). Do you have some good suggestion?
$endgroup$
– GKiu
Feb 3 at 11:13
1
$begingroup$
There is a homomorphism $B_3 to S_3$, the symmetric group on 3 letters, and if $V$ is a vector space then $S_3$ acts on $V otimes V otimes V$, acting on simple tensors by permuting the factors: eg $(123) cdot a otimes b otimes c = c otimes a otimes b$. This gives a representation of $B_3$ using the above homomorphism. If you don't want any non-trivial elements of $B_3$ to give the identity matrix, you want to ask that your representation is "faithful".
$endgroup$
– user98602
Feb 3 at 14:10
1
$begingroup$
It is a vector space and is otherwise irrelevant. You could replace it with $Bbb R$. (This is why I think you should probably give some details about what you want to demand of this representation: I suspect this is not what you want.)
$endgroup$
– user98602
Feb 3 at 20:02
1
$begingroup$
If you wanted the individual factors to matter you could also use the projection suggested by @Mike and take the $2$-dimensional irreducible rep for $S_3$, then tensor that with itself twice. But once again, it is not clear if this would really be the sort of thing you are looking for.
$endgroup$
– Tobias Kildetoft
Feb 3 at 21:27