Lie algebra generator relation $T^a T^b propto T^c $ valid for any $a,b$?












1












$begingroup$


Given a Lie Algebra (such as $su(n), so(n))$ can I always find a set of generators + identity ${T^a}cup {id}$ such that there exists a $c$ for any given $a,b$ such that $T^a T^b = C(a,b) T^c $ for a $a,b$-dependent function $C$ into the real/complex numbers?






lie-algebras







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$endgroup$












  • $begingroup$
    What does $propto$ stand for.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:44






  • 1




    $begingroup$
    @JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:51










  • $begingroup$
    are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:56










  • $begingroup$
    sorry, let's add ${ id }$ to the set of generators
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:58






  • 1




    $begingroup$
    Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 17:26


















1












$begingroup$


Given a Lie Algebra (such as $su(n), so(n))$ can I always find a set of generators + identity ${T^a}cup {id}$ such that there exists a $c$ for any given $a,b$ such that $T^a T^b = C(a,b) T^c $ for a $a,b$-dependent function $C$ into the real/complex numbers?






lie-algebras







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $propto$ stand for.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:44






  • 1




    $begingroup$
    @JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:51










  • $begingroup$
    are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:56










  • $begingroup$
    sorry, let's add ${ id }$ to the set of generators
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:58






  • 1




    $begingroup$
    Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 17:26
















1












1








1





$begingroup$


Given a Lie Algebra (such as $su(n), so(n))$ can I always find a set of generators + identity ${T^a}cup {id}$ such that there exists a $c$ for any given $a,b$ such that $T^a T^b = C(a,b) T^c $ for a $a,b$-dependent function $C$ into the real/complex numbers?






lie-algebras







share|cite|improve this question











$endgroup$




Given a Lie Algebra (such as $su(n), so(n))$ can I always find a set of generators + identity ${T^a}cup {id}$ such that there exists a $c$ for any given $a,b$ such that $T^a T^b = C(a,b) T^c $ for a $a,b$-dependent function $C$ into the real/complex numbers?






lie-algebras




lie-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 13:59







tonydo

















asked Dec 19 '18 at 13:41









tonydotonydo

1497




1497












  • $begingroup$
    What does $propto$ stand for.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:44






  • 1




    $begingroup$
    @JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:51










  • $begingroup$
    are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:56










  • $begingroup$
    sorry, let's add ${ id }$ to the set of generators
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:58






  • 1




    $begingroup$
    Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 17:26




















  • $begingroup$
    What does $propto$ stand for.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:44






  • 1




    $begingroup$
    @JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:51










  • $begingroup$
    are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
    $endgroup$
    – ziggurism
    Dec 19 '18 at 13:56










  • $begingroup$
    sorry, let's add ${ id }$ to the set of generators
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:58






  • 1




    $begingroup$
    Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 17:26


















$begingroup$
What does $propto$ stand for.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 13:44




$begingroup$
What does $propto$ stand for.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 13:44




1




1




$begingroup$
@JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
$endgroup$
– ziggurism
Dec 19 '18 at 13:51




$begingroup$
@JoséCarlosSantos: "proportional to". So OP is asking whether there exist constants $k^{ab}_c$ such that $T^aT^b=k^{ab}_c T^c$ (with no Einstein summation, just a single constant per equation)
$endgroup$
– ziggurism
Dec 19 '18 at 13:51












$begingroup$
are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
$endgroup$
– ziggurism
Dec 19 '18 at 13:56




$begingroup$
are $mathfrak{su}(n)$ or $mathfrak{so}(n)$ even closed under multiplication?
$endgroup$
– ziggurism
Dec 19 '18 at 13:56












$begingroup$
sorry, let's add ${ id }$ to the set of generators
$endgroup$
– tonydo
Dec 19 '18 at 13:58




$begingroup$
sorry, let's add ${ id }$ to the set of generators
$endgroup$
– tonydo
Dec 19 '18 at 13:58




1




1




$begingroup$
Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 17:26






$begingroup$
Why would such identities be of interest? A representation of the Lie algebra need not preserve such product relations, only commutators.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 17:26












1 Answer
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$begingroup$

The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.



You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample:
$$
S_z=operatorname {diag} ~ (1,0,-1), qquad Longrightarrow qquad S_z S_z = operatorname {diag} (1,0,1).
$$
Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.






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    $begingroup$

    The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.



    You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample:
    $$
    S_z=operatorname {diag} ~ (1,0,-1), qquad Longrightarrow qquad S_z S_z = operatorname {diag} (1,0,1).
    $$
    Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.



      You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample:
      $$
      S_z=operatorname {diag} ~ (1,0,-1), qquad Longrightarrow qquad S_z S_z = operatorname {diag} (1,0,1).
      $$
      Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.



        You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample:
        $$
        S_z=operatorname {diag} ~ (1,0,-1), qquad Longrightarrow qquad S_z S_z = operatorname {diag} (1,0,1).
        $$
        Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.






        share|cite|improve this answer











        $endgroup$



        The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.



        You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample:
        $$
        S_z=operatorname {diag} ~ (1,0,-1), qquad Longrightarrow qquad S_z S_z = operatorname {diag} (1,0,1).
        $$
        Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 21:15

























        answered Dec 20 '18 at 15:14









        Cosmas ZachosCosmas Zachos

        1,810522




        1,810522






























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