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Simple Lie algebras and ideals

  1. Dec 30, 2005 #1
    I understand the notation [tex][\mathcal{L},\mathcal{L}]=0[/tex] to be "the whole Lie algebra commutes with itself", but the precise meaning of [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex] I'm unsure of.

    Does it mean that every element of [tex]\mathcal{L}[/tex] can be generated by taking the commutator of two elements within [tex]\mathcal{L}[/tex]?

    I've been asked to show that if [tex]\mathcal{L}[/tex] is simple, and [tex][\mathcal{L},\mathcal{L}] \not= 0[/tex], then [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

    Am I right in thinking if [tex]I[/tex] is an ideal, then by definition [tex][I,\mathcal{L}] = I[/tex]. Since [tex]\mathcal{L}[/tex] is simple, then [tex]I = 0[/tex] or [tex]I =\mathcal{L}[/tex]. The only possible combinations of ideals are therefore [tex][0,\mathcal{L}]=0[/tex], [tex][\mathcal{L},\mathcal{L}]=0[/tex] and [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

    ThoughI don't feel it constitutes much of a proof. I'd be very grateful for any help given.
     
    Last edited: Dec 30, 2005
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  3. Dec 30, 2005 #2

    AKG

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    In the context of groups, [L,L] would be called the commutator subgroup, so I'm guessing that in this context, it would be called the commutator subalgebra. It is the subalgebra generated by the commutators, where a commutator is any element of the form xyx-1y-1, where x and y are in L.
    It depends on what you mean by "generated". It is not necessarily true that every element of L can be expressed in the form xyx-1y-1, BUT every element of L CAN be expressed as a product of commutators.
    By definition? No, I don't think so. Since I is an ideal, you can argue that [I,L] is a subalgebra of I, but I don't think you can say that, by definition, it is I. If so, then they wouldn't even need to tell you "If [L,L] is not 0..."
    Yes.
    No, why? Like I said, [L,L] could conceivably be any subalgebra of L.

    You're told that L is simple, so its only ideals are 0 and L. You're asked to show that [L,L] = L, given that [L,L] is not 0. Do you see that it suffices to prove something more general about [L,L]?
     
  4. Dec 31, 2005 #3
    Ah, I didn't get the distinction between a subalgebra and an ideal (/rereads notes).

    Saying I is an ideal is a stronger statement than saying I is a subalgebra. That gets rid of a few question marks I had in my head about some things, and also makes most of my first post's ideas fairly stupid, something I'm used to with this material. Being an applied mathematician and suddenly finding so much pure maths in particle physics has thrown me a bit....

    So while L doesn't have any proper ideals, it can have plenty of subalgebras within, such as L(SU(3)) having L(SU(2)) subalgebras within it.
    Given the typical method of proof in my notes and the fact [L,L] != 0, I'd hazard a guess at it being to show something about the Killing form for say [tex]\kappa (l_{1},[l_{2},l_{3}])[/tex] and degeneracy, but exactly what I'm not sure. Wish I'd paid more attention in Linear Algebra 2 years ago....
     
  5. Dec 31, 2005 #4

    AKG

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    I've studied groups, and have some knowledge of algebras, but none of Lie algebras. Also, I previously said that [L,L] is generated by elements of the form xyx-1y-1. This may be wrong, so I suggest you look up the definition of the subalgebra [L,L]. I don't know what the Killing form is, or why you'd use it. You know that L has only 0 and L as ideals, and you know [L,L] is not 0, so in order to show that [L,L] = L, it suffices to show that, in general, [L,L] is not just a subalgebra, but is in fact an ideal.

    Since I haven't studied Lie algebras, this may not be the way to go about doing it, but consider it.
     
  6. Dec 31, 2005 #5

    selfAdjoint

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    And a subset is an ideal if, whenever you multiply an element of the parent algebra by an element of the subset, the product is an element of the subset. This is the definition of an ideal; think of it in terms of the even numbers among the integers. Take any integer and multiply it by any even number and the result is another even number.

    So now what do you have to prove about [L,L]?
     
  7. Dec 31, 2005 #6

    AKG

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    I know what an ideal is, but I don't know much about Lie algebras.
     
  8. Jan 1, 2006 #7

    matt grime

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    For AKG and the OP:

    You do not need to use the killing form. Putting it in group terms, if S is a simple group then the commutator (derived) subgroup is either trivial or all of S: proof the commutator subgroup is normal, S is simple.

    So translating here (ie self adjoint's hint) your (AKG) proof is exactly correct.


    [LL] is the derived subalgebra, and is the (smallest) subalgebra containing the elements [ab], ie a typical element is a sum

    [tex] \sum [a_i b_i] [/tex]

    where a_i and b_i are in L.

    Note that just as in group theory the abelian simple objects are treated differently (a lie algebra is abelian if [LL]=0, which you can think of as saying that ab-ba=0, or ab=ba, for all a and b, though this of course isn't really allowed, since in formal terms a lie algebra does not actually have a product ab, but merely a lie bracket. however just like 'any' hilbert space is actually l_2, 'any' lie algebra is a matrix algebra, (has a faithful representation) so ab makes sense though it is not constrained to lie in the lie algebra, eg the matrix product of two elements in sl_n is not in sl_n (trace zero matrices))
     
    Last edited: Jan 1, 2006
  9. Jan 1, 2006 #8

    AKG

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    Treating this problem like a group theoretic one doesn't seem to work. If L is any Lie algebra, c is any commutator, and l is any element of L, then lcc-1 is in [L,L], hence so is l, thus [L,L] = L. Note that this holds even when L is not simple, so long as we can treat L as a group, and we can assume [L,L] is an ideal, it holds. The problem here must be in treating L as a group, in particular, saying that if c is in [L,L] then so is c-1. c-1 may not even be defined. A commutator in a group is xyx-1y-1. In a Lie algebra, it is apparently xy - yx? Also, instead of xy, the product is actually the Lie bracket [x,y], which is nonassociative? I don't see how one could do this problem without knowing anything about Lie algebras.

    If we have the commutator [x,y] - [y,x], and left-multiply it by some arbitrary element z of L, we get:

    [z,[x,y] - [y,x]]
    = -[[x,y] - [y,x],z]
    = [[y,x],z] - [[x,y],z]

    Is this right so far? What am I supposed to do with this?
     
  10. Jan 1, 2006 #9

    matt grime

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    You can't do the question without knowing the definitions, that's obvious, but I was explaining that your notional proof of 'show the derived algebra is an ideal hence either 0 or L if L is simple' is correct by *analogy*. It is by *analogy* the same as the derived subgroup of a simple group G is either 1 or G. I did say you had to translate it to the lie algebra case.

    In this case that the derived subalgebra is an ideal is quite trivial: anything in [L,[L,L]] is clearly in [L,L]. indeed the sequence of ideals by bracketing one more copy of L is a very important series in lie algebra theory, as is the one gotten by repeatedly taking derived subalgebras of derived subalgebras (note these are strictly different in general; when they are the same is also an important result in the abstract theory of lie algebras).

    Lie algebras do not have an algebra multiplication, they have a lie bracket composition: an anti-commutative bilinear map from LxL to L satisfying the jacobi identity. it is better not to think of it as a commutator of a group, since it isn't one, but as the commutator in the sense of matrices. the idea of multiplying elements xy and subtracting yx is one for you to forget since there is no need for xy to even be defined.
     
    Last edited: Jan 1, 2006
  11. Jan 2, 2006 #10
    As with so much in my Symmetries course, it seems the answer is so obvious and simple once you know it :redface:

    Thanks for your help guys. This is definitely one of those courses where just sitting there and thinking about it loads is the name of the game because solutions don't seem to require a lot of complex maths, just lateral thought. Though one question involved computing 8 8x8 matrices, then computing another 8x8 matrix from the trace of their multiples (ie the Killing form from the trace-ad definition :yuck: )

    AKG, as an example of how AB-BA can be in a Lie algebra, but not AB or BA we were given

    [tex]A = V^{i}(x)\frac{\partial}{\partial x^{i}}[/tex] (Einstein summation assumed)

    [tex]AB = V^{i}(x)\frac{\partial}{\partial x^{i}} \left( W^{j}(x)\frac{\partial}{\partial x^{j}}\right) = V^{i}(x)\frac{\partial}{\partial x^{i}}W^{j}(x)\frac{\partial}{\partial x^{j}} + V^{i}(x)W^{j}(x)\frac{\partial^{2}}{\partial x^{i}\partial x^{j}}[/tex]

    This clearly isn't the same type of operator as A, but AB-BA will be, because partial derivatives commute.
     
    Last edited: Jan 2, 2006
  12. Jan 2, 2006 #11

    matt grime

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    There are two reasons for this kind of question: one to make people realize that the avoidance of computational examples is actually quite a good thing; two to give those who are struggling with the theoretical stuff can at least get some marks.
     
  13. Jan 2, 2006 #12
    It was slightly annoying when the lecturer wrote on his answers, just above the Killing form matrix "Computed in Maple!!" and then the same again with the Killing form in a different basis showing it was compact.
     
  14. Jan 2, 2006 #13

    matt grime

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    That's a little unnecessary. But then I think of lie algebras as subsets of nxn matrices with [ab]=ab-ba (ie as algebriac gadgets; i don't even think of them as the tangent spaces of lie groups [though I do think of them in terms of algebraic groups]). Any sl_n would do (trace zero matrices): it is easy to find two matrices with trace zero but whose matrix composition doesn't, eg a diagonal matrix with real nonzero entries that sum to zero: its square is not trace zero.

    Harder is to think of a lie algebra that is not simply an associative algebra and where [ab]=ab-ba (ie one where [ab] is defined but ab is not)
     
    Last edited: Jan 2, 2006
  15. Jan 3, 2006 #14

    George Jones

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    In some sense, i.e., via its universal enveloping algebra, any Lie algebra can be thought of as arising from the commutator product of an associative algebra.

    Regards,
    George
     
  16. Jan 3, 2006 #15

    matt grime

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    How about one that is naturally occuring and not made to be an associative algebra by fiat? And, moreover one that does not require us to extend to an infinite dimensional algebra (if I recall the universal enveloping algbra is the quotient of the tensor algebra by some relation).


    Your answer is a little like saying that all (comutative) monoids arise from rings by adding formal inverses, but that still doesn't stop the natural numbers being interesting and defined within their own right.
     
    Last edited: Jan 3, 2006
  17. Jan 3, 2006 #16

    George Jones

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    I was fairly sure that you would find my example to be quite contrived.

    The appropriate ideal to mod out is the one generated by x tensor y - y tensor x - [x,y].

    Here's another example that also might be contrived. Every Lie algebra can (I think - can't find a reference) be considered to be a (bivector) subspace (closed under the commutator product) of some Clifford algebra.

    A Clifford algebra Cl(V,g) is generated by a vector space V and a bilinear g. One way of constructing Cl(V,g) is to take the quotient of the tensor algebra by the relation v tensor v - g(v,v), but there are other, possibly more concrete ways, of constructing Cl(V,g).

    Regards,
    George
     
  18. Jan 3, 2006 #17

    matt grime

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    Again, these are contrived, in my opinion. Just give me a naturally occuring lie algebra that is not gotten from simply considering an associative algebra or a subspace of some larger infinite dimensional in some cases associative algebra if possible. I can't think of one; this isn't a test; it's a genuine query; there is nothing that requires the product of two elements in a lie algebra to exist, merely their bracket, though trivially we can declare them to exist, just as we can formally declare lots of things to exist. For those with such a mind, what about vector fields and the lie derivative?
     
  19. Jan 3, 2006 #18

    George Jones

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    Isn't this example given in post #10?

    Regards,
    George
     
  20. Jan 3, 2006 #19

    matt grime

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    Is it? As I know next to nothing about differential geometry I can't tell if you are implying that AB is *not* a vector field, or if it as case where AB is defined but not of the correct type.
     
  21. Jan 3, 2006 #20

    George Jones

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    Let M be a differentiable manifold and C^{infinity}(M) be the ring of smooth functions from M into R. A vector field A is a smooth assignment of a tangent vector A_p (which, among other things, is a map from C^{infinity}(M) into R) at each p in M. Smooth means that the for every f in C^{infinity}(M), the function Af: M -> R defined by

    (Af)(p) = A_p (f)

    is also in C^{infinity}(M).

    Consequently, a vector field A can be viewed as a map A: C^{infinity}(M) -> C^{infinity}(M). From the properties of tangent vectors, this map is linear (hence the notation Af) and a derivation on the ring C^{infinity}(M).

    Let A and B be vector fields. AoB is defined as a linear map from C^{infinity}(M) into C^{infinity}(M), but AoB is not, in general, a derivation on C^{infinity}(M), and thus not a vector field. However, AoB - BoA is a derivation and a vector field.

    AoB(fg) = A(B(fg)) = A(gBf +fBg) = AgBf + gA(Bf) + AfBg +fA(Bg)

    The presence of the first and third terms on the right show that AoB is not a derivation, however, in the combination AoB - BoA, these offending cancel with similar terms in -BoA.

    Under this commutator, the space of vector fields is a Lie algebra, and the product (function composition) of Lie algebra elements, i.e., vector fields, also exists, but is not itself a vector field.

    Regards,
    George
     
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