Why are the fundamental and anti-fundamental representation in $text{SL}(2,mathbb{C})$ not equivalent?












2












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I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $text{SL}(2,mathbb{C})$, $2 times 2$ matrix with determinant of $1$, are not equivalent. This means that no similarity transformation can map one of them to the other.



My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.



Can anyone please explain the explanation of my professor, or perhaps give another explanation?






group-theory representation-theory lie-groups







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  • 3




    $begingroup$
    I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
    $endgroup$
    – Tobias Kildetoft
    Nov 25 '18 at 7:28






  • 6




    $begingroup$
    Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
    $endgroup$
    – Joppy
    Nov 26 '18 at 10:57










  • $begingroup$
    Crossposted from physics.stackexchange.com/q/443092/2451
    $endgroup$
    – Qmechanic
    Dec 1 '18 at 12:39










  • $begingroup$
    ...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
    $endgroup$
    – Qmechanic
    Jan 28 at 16:44
















2












$begingroup$


I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $text{SL}(2,mathbb{C})$, $2 times 2$ matrix with determinant of $1$, are not equivalent. This means that no similarity transformation can map one of them to the other.



My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.



Can anyone please explain the explanation of my professor, or perhaps give another explanation?






group-theory representation-theory lie-groups







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
    $endgroup$
    – Tobias Kildetoft
    Nov 25 '18 at 7:28






  • 6




    $begingroup$
    Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
    $endgroup$
    – Joppy
    Nov 26 '18 at 10:57










  • $begingroup$
    Crossposted from physics.stackexchange.com/q/443092/2451
    $endgroup$
    – Qmechanic
    Dec 1 '18 at 12:39










  • $begingroup$
    ...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
    $endgroup$
    – Qmechanic
    Jan 28 at 16:44














2












2








2


1



$begingroup$


I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $text{SL}(2,mathbb{C})$, $2 times 2$ matrix with determinant of $1$, are not equivalent. This means that no similarity transformation can map one of them to the other.



My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.



Can anyone please explain the explanation of my professor, or perhaps give another explanation?






group-theory representation-theory lie-groups







share|cite|improve this question











$endgroup$




I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $text{SL}(2,mathbb{C})$, $2 times 2$ matrix with determinant of $1$, are not equivalent. This means that no similarity transformation can map one of them to the other.



My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.



Can anyone please explain the explanation of my professor, or perhaps give another explanation?






group-theory representation-theory lie-groups




group-theory representation-theory lie-groups






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 5 '18 at 3:31









Andrews

3901317




3901317










asked Nov 25 '18 at 1:23









foxielmaofoxielmao

1319




1319








  • 3




    $begingroup$
    I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
    $endgroup$
    – Tobias Kildetoft
    Nov 25 '18 at 7:28






  • 6




    $begingroup$
    Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
    $endgroup$
    – Joppy
    Nov 26 '18 at 10:57










  • $begingroup$
    Crossposted from physics.stackexchange.com/q/443092/2451
    $endgroup$
    – Qmechanic
    Dec 1 '18 at 12:39










  • $begingroup$
    ...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
    $endgroup$
    – Qmechanic
    Jan 28 at 16:44














  • 3




    $begingroup$
    I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
    $endgroup$
    – Tobias Kildetoft
    Nov 25 '18 at 7:28






  • 6




    $begingroup$
    Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
    $endgroup$
    – Joppy
    Nov 26 '18 at 10:57










  • $begingroup$
    Crossposted from physics.stackexchange.com/q/443092/2451
    $endgroup$
    – Qmechanic
    Dec 1 '18 at 12:39










  • $begingroup$
    ...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
    $endgroup$
    – Qmechanic
    Jan 28 at 16:44








3




3




$begingroup$
I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
$endgroup$
– Tobias Kildetoft
Nov 25 '18 at 7:28




$begingroup$
I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism.
$endgroup$
– Tobias Kildetoft
Nov 25 '18 at 7:28




6




6




$begingroup$
Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
$endgroup$
– Joppy
Nov 26 '18 at 10:57




$begingroup$
Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, mathbb{C})$. Perhaps this helps.
$endgroup$
– Joppy
Nov 26 '18 at 10:57












$begingroup$
Crossposted from physics.stackexchange.com/q/443092/2451
$endgroup$
– Qmechanic
Dec 1 '18 at 12:39




$begingroup$
Crossposted from physics.stackexchange.com/q/443092/2451
$endgroup$
– Qmechanic
Dec 1 '18 at 12:39












$begingroup$
...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
$endgroup$
– Qmechanic
Jan 28 at 16:44




$begingroup$
...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge.
$endgroup$
– Qmechanic
Jan 28 at 16:44










1 Answer
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  1. For $$G~:=~SL(2,mathbb{C})~:=~{gin {rm Mat}_{2times 2}(mathbb{C})mid det g = 1 }tag{1}$$
    viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $jin frac{1}{2}mathbb{N}_0,$ and of complex dimension $2j+1$.


  2. For the same group $$G~:=~SL(2,mathbb{C})~cong~ Spin(1,3,mathbb{R})tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation
    $$rho: Gto GL(2,mathbb{C}), qquad rho(g)~=~bar{g}, qquad g~in~ G, tag{3}$$
    of the defining representation (1) are not equivalent, i.e. there does not exist an element $Min GL(2,mathbb{C})$ such that
    $$forall gin G: Mg=bar{g}M, tag{4}$$
    cf. e.g. user Dan Yand's Math.SE answer.


  3. One complexification of $G$ is $$G_{mathbb{C}}~cong~Spin(1,3,mathbb{C})cong SL(2,mathbb{C})times SL(2,mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $j_L,j_Rin frac{1}{2}mathbb{N}_0$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $(1/2,0)$ and $(0,1/2)$, respectively.







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    1 Answer
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    active

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    0












    $begingroup$


    1. For $$G~:=~SL(2,mathbb{C})~:=~{gin {rm Mat}_{2times 2}(mathbb{C})mid det g = 1 }tag{1}$$
      viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $jin frac{1}{2}mathbb{N}_0,$ and of complex dimension $2j+1$.


    2. For the same group $$G~:=~SL(2,mathbb{C})~cong~ Spin(1,3,mathbb{R})tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation
      $$rho: Gto GL(2,mathbb{C}), qquad rho(g)~=~bar{g}, qquad g~in~ G, tag{3}$$
      of the defining representation (1) are not equivalent, i.e. there does not exist an element $Min GL(2,mathbb{C})$ such that
      $$forall gin G: Mg=bar{g}M, tag{4}$$
      cf. e.g. user Dan Yand's Math.SE answer.


    3. One complexification of $G$ is $$G_{mathbb{C}}~cong~Spin(1,3,mathbb{C})cong SL(2,mathbb{C})times SL(2,mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $j_L,j_Rin frac{1}{2}mathbb{N}_0$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $(1/2,0)$ and $(0,1/2)$, respectively.







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$


      1. For $$G~:=~SL(2,mathbb{C})~:=~{gin {rm Mat}_{2times 2}(mathbb{C})mid det g = 1 }tag{1}$$
        viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $jin frac{1}{2}mathbb{N}_0,$ and of complex dimension $2j+1$.


      2. For the same group $$G~:=~SL(2,mathbb{C})~cong~ Spin(1,3,mathbb{R})tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation
        $$rho: Gto GL(2,mathbb{C}), qquad rho(g)~=~bar{g}, qquad g~in~ G, tag{3}$$
        of the defining representation (1) are not equivalent, i.e. there does not exist an element $Min GL(2,mathbb{C})$ such that
        $$forall gin G: Mg=bar{g}M, tag{4}$$
        cf. e.g. user Dan Yand's Math.SE answer.


      3. One complexification of $G$ is $$G_{mathbb{C}}~cong~Spin(1,3,mathbb{C})cong SL(2,mathbb{C})times SL(2,mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $j_L,j_Rin frac{1}{2}mathbb{N}_0$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $(1/2,0)$ and $(0,1/2)$, respectively.







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$


        1. For $$G~:=~SL(2,mathbb{C})~:=~{gin {rm Mat}_{2times 2}(mathbb{C})mid det g = 1 }tag{1}$$
          viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $jin frac{1}{2}mathbb{N}_0,$ and of complex dimension $2j+1$.


        2. For the same group $$G~:=~SL(2,mathbb{C})~cong~ Spin(1,3,mathbb{R})tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation
          $$rho: Gto GL(2,mathbb{C}), qquad rho(g)~=~bar{g}, qquad g~in~ G, tag{3}$$
          of the defining representation (1) are not equivalent, i.e. there does not exist an element $Min GL(2,mathbb{C})$ such that
          $$forall gin G: Mg=bar{g}M, tag{4}$$
          cf. e.g. user Dan Yand's Math.SE answer.


        3. One complexification of $G$ is $$G_{mathbb{C}}~cong~Spin(1,3,mathbb{C})cong SL(2,mathbb{C})times SL(2,mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $j_L,j_Rin frac{1}{2}mathbb{N}_0$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $(1/2,0)$ and $(0,1/2)$, respectively.







        share|cite|improve this answer











        $endgroup$




        1. For $$G~:=~SL(2,mathbb{C})~:=~{gin {rm Mat}_{2times 2}(mathbb{C})mid det g = 1 }tag{1}$$
          viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $jin frac{1}{2}mathbb{N}_0,$ and of complex dimension $2j+1$.


        2. For the same group $$G~:=~SL(2,mathbb{C})~cong~ Spin(1,3,mathbb{R})tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation
          $$rho: Gto GL(2,mathbb{C}), qquad rho(g)~=~bar{g}, qquad g~in~ G, tag{3}$$
          of the defining representation (1) are not equivalent, i.e. there does not exist an element $Min GL(2,mathbb{C})$ such that
          $$forall gin G: Mg=bar{g}M, tag{4}$$
          cf. e.g. user Dan Yand's Math.SE answer.


        3. One complexification of $G$ is $$G_{mathbb{C}}~cong~Spin(1,3,mathbb{C})cong SL(2,mathbb{C})times SL(2,mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $j_L,j_Rin frac{1}{2}mathbb{N}_0$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $(1/2,0)$ and $(0,1/2)$, respectively.








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        edited Dec 5 '18 at 18:52

























        answered Dec 5 '18 at 12:34









        QmechanicQmechanic

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