What Does [\mathcal{L},\mathcal{L}] = \mathcal{L} Imply in Simple Lie Algebras?

[AlphaNumeric]

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

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

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

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

ThoughI don't feel it constitutes much of a proof. I'd be very grateful for any help given.

 

[AKG]

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

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.

Does it mean that every element of \mathcal{L} can be generated by taking the commutator of two elements within \mathcal{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.

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

Am I right in thinking if I is an ideal, then by definition [I,\mathcal{L}] = I.

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..."

Since \mathcal{L} is simple, then I = 0 or I =\mathcal{L}.

Yes.

The only possible combinations of ideals are therefore [0,\mathcal{L}]=0, [\mathcal{L},\mathcal{L}]=0 and [\mathcal{L},\mathcal{L}]=\mathcal{L}.

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]?

 

[AlphaNumeric]

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.

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]?

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 \kappa (l_{1},[l_{2},l_{3}]) and degeneracy, but exactly what I'm not sure. Wish I'd paid more attention in Linear Algebra 2 years ago...

 

[AKG]

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.

 

[selfAdjoint]

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.

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]?

 

[AKG]

I know what an ideal is, but I don't know much about Lie algebras.

 

[matt grime]

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

\sum [a_i b_i]

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))

 

[AKG]

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?

 

[matt grime]

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.

 

[AlphaNumeric]

In this case that the derived subalgebra is an ideal is quite trivial: anything in [L,[L,L]] is clearly in [L,L]

As with so much in my Symmetries course, it seems the answer is so obvious and simple once you know it

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 )

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

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

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}}

This clearly isn't the same type of operator as A, but AB-BA will be, because partial derivatives commute.

 

[matt grime]

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 )

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.

 

[AlphaNumeric]

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.

 

[matt grime]

AKG, as an example of how AB-BA can be in a Lie algebra, but not AB or BA we were given
A = V^{i}(x)\frac{\partial}{\partial x^{i}} (Einstein summation assumed)
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}}
This clearly isn't the same type of operator as A, but AB-BA will be, because partial derivatives commute.

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)

 

[George Jones]

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)

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

 

[matt grime]

How about one that is naturally occurring 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.

 

[George Jones]

How about one that is naturally occurring and not made to be an associative algebra by fiat?

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

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).

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

 

[matt grime]

Again, these are contrived, in my opinion. Just give me a naturally occurring 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?

 

[George Jones]

For those with such a mind, what about vector fields and the lie derivative?

Isn't this example given in post #10?

Regards,
George

 

[matt grime]

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.

 

[George Jones]

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

 

[Don Aman]

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

And of course the same argument works with the space of derivations of any algebra. The commutator is again a derivation, but the product (which is defined by composition of maps) is not.

Or what about the tangent space to any Lie group? If the Lie group is not a matrix group, there is no product, so it is not simply the commutator in some associative algebra (though it can always be embedded in one as such).

 

[matt grime]

Again, that isn't an example of the behaviour I asked: for AoB is still defined but it just isn't a vector field. I asked is there a naturally occurring lie algebra such that the lie bracket [ab] is not defined as ab-ba for some composition of a and b in some sense.

I can produce some (infintely many, obviously) artificial examples: consider the vector space with basis e,f,h and with the bracket defined such as to make it isomorphic to sl_2, but is there a naturally occurring one? all naturally occurring ones I can think of inherit some extra structure, and beyond defining some merely by a list of rules then i don't know of a way of getting one that doesn't come with extra structure.

 

[George Jones]

Again, that isn't an example of the behaviour I asked

I made the post in response to your comment

I can't tell if you are implying that AB is *not* a vector field ...

I didn't think that I needed to state explicitly that this showed that a Lie algebra of vector fields was not the example for which you had earlier asked.

Also, a Lie algebra of vector fields is infinite-dimensional. Do you require that the example be a finite-dimensional Liealgebra? In any case, I know of no 'natural' example.

I think that any finite-dimensional Lie algebra is isomorphic to a Lie algebra of matrices, but maybe not in a natural way.

Regards,
George

 

[matt grime]

I made the post in response to your comment
I didn't think that I needed to state explicitly that this showed that a Lie algebra of vector fields was not the example for which you had earlier asked.

I didn't believe it was (since it was clearly using the product of two objects) and as such I was puzzled as to why you'd repeatedly post it in response to a request that asked about something else entirely.

Also, a Lie algebra of vector fields is infinite-dimensional. Do you require that the example be a finite-dimensional Liealgebra? In any case, I know of no 'natural' example.

nope, any 'natural' example, if anyone knows of one. it is purely a 'for my own knowledge'

I think that any finite-dimensional Lie algebra is isomorphic to a Lie algebra of matrices, but maybe not in a natural way.
Regards,
George

Ie is there an faithful representation of any (finite dimensional) lie algebra (any semisimple lie algebra certainly does)? Even if there is (which is eminently reasonablem though off the top of my head I don't know either) that wouldn't preclude it being a natural example of a space without composition, cf my unnatural example of the span of e,f,h = C^3 giving a copy of sl_2. There is no composition of elements e,f,h in C^3 given, though I am cheating by using a 'forgetful' construction.

Of course 'natural' is an undefined term.

 

[Don Aman]

It seems you consider the universal enveloping algebra "contrived", but as it is a universal functor from the category of Lie algebras, it is natural in the usual sense of the word.

But every Lie algebra has a multiplication, and the only way you'll find one without multiplication is if you turn a blind eye to it. That's why I mention the Lie algebras of Lie groups that aren't matrix groups (say, Spin(n) for a big enough n). The only way it'll have a multiplication is if you go out of your way to define one.

 

[matt grime]

It seems you consider the universal enveloping algebra "contrived", but as it is a universal functor from the category of Lie algebras, it is natural in the usual sense of the word.

it is universal in the usual sense of the word. but embedding a (small) lie algebra into an (infinite dimensional) quotient of a tensor algebra does not endow the original lie algebra with a 'naturally occurring' product, does it? the product so given is not (necessarily) defined on the lie algebra before you make this construction. the example i gave of C^3 with relations forcing it to be sl_2 shows that. C^3 comes with no 'natural' algebra structure for which [ef]=ef-fe.

we can embed (finitely generated) abelian monoids into (finitely generated abelian) groups (at least thati s my gut reaction, i can't say I've proved it, but it seems clear), that doesn't mean that there isn't a naturally occurring monoid that isn't a group, does it?

But every Lie algebra has a multiplication, and the only way you'll find one without multiplication is if you turn a blind eye to it.

and the question was if there was a naturally occurring situation when we *do* turn a blind eye to it, or more accurately when we plain don't see it to begin with (like N in Z). why is that so hard to grasp? i can't think of one, and you can't either, apparently, so why do you keep telling me you can't think of one?

That's why I mention the Lie algebras of Lie groups that aren't matrix groups (say, Spin(n) for a big enough n). The only way it'll have a multiplication is if you go out of your way to define one.

so now you can think of one? which is it?

As another example I know fundamental group of a space can be made arbitrary, that doesn't mean every group is a fundamental group, it is merely isomorphic to one. Is the distinction clear?

 

[Don Aman]

As another example I know fundamental group of a space can be made arbitrary, that doesn't mean every group is a fundamental group, it is merely isomorphic to one. Is the distinction clear?

Of course the distinction is clear. Every (fd) Lie algebra is isomorphic to a matrix algebra, which has a natural product. But that doesn't mean it is a matrix algebra, in exactly the same sense that not every group is a fundamental group. I mention now, for the third time, an example of a Lie algebra that is not a matrix algebra, and therefore does not have your product: the Lie algebra of a Lie group which is not a matrix group. It is of course isomorphic to a matrix algebra, and so in that sense has a product, but then again, so is every (fd) Lie algebra. It is of course embeddable in its universal enveloping algebra, and so in that sense has product, but then again, so is every Lie algebra.

Are there Lie algebras without products? Sure, simply don't include a product. Are there Lie algebras which are not isomorphic to algebras with product? No. As far as I can tell, the answer to your question is either "obviously yes" or "obviously no", depending on your view.

 

[matt grime]

Are there Lie algebras without products? Sure, simply don't include a product. Are there Lie algebras which are not isomorphic to algebras with product? No. As far as I can tell, the answer to your question is either "obviously yes" or "obviously no", depending on your view.

which is why i included the words 'naturally occurring', that was sort of the key point in the question that didn't make it vacuous. indeed i gave a formal example of a lie algebra that didn't come with a product

however, as for the example given two tangents at the identity what is [ab] of them? seriously, I'm an algebraist, know nothing of differential geometry, i thought the tangent space was the span of the operators \partial_x which do have a natural product since they are in the associative algebra of differential operators, and that the bracket was ab-ba in this space.

now, if as is entirely possible, there is another definition of tangent space that doesn't require you to evaluate things in this algebra, but which takes in a,b and spits out [ab] naturally then that'd be what the question wanted to know about.

 

[George Jones]

In brief: the Lie algebra of a Lie Group is the Lie algebra of left-invariant vector fields, and, since a left-invariant vector field is uniquely determined by its value in the tangent space of the identity, the commutator product of left-invariant vector can be used to give the tangent space of the identity Lie algbera structure.

Now the details.

Let M be an n-diemnsional differentiable manifold, and let f: M \rightarrow M be a diffeomorphism. f can be used to define for any p \in M a pushforward map (differential) f_{*} : T_{p} \rightarrow T_{f\left(p\right)} between tangent spaces as follows.

Suppose p \in M is in the image of a curve \gamma: I \rightarrow M, where I is an interval in \mathbb{R}. Then, \gamma' = f \circ \gamma is a curve that has f\left(p\right) in its image. Let v be the tangent vector to \gamma at p, and define f_* by taking f_{*} \left( v \right) to be the tangent vector to \gamma' at f \left( p \right).

This is the natural definition of f_* to use with the equivalence classes of curves definition of tangent vectors, and the equivalent definition of f_* for the derivation definition of tangent vectors shows that every h_* is linear.

Now let M also have Lie group structure. For any g \in M, define a diffeomorphism (action) l_{g} : M \rightarrow M by l_{g} h = gh for every h \in M.

Let X be a vector field on M, and X_h be its representative in the tangent space T_h at h \in M. Use l_{g*} to push forward to a vector l_{g*} X_{h} in T_{l_g h} = T_{gh}. But X[/tex] itself has a representativeX_{l_g h} in T_{l_g h} = T_{gh}.<br /> <br /> The vector field X is said to be left-invariant if l_{g*} X_{h} = X_{l_{g} h} for every h \in M. It is easy to show that any left-invariant vector field is determined by its representative at the identity. Also, if X and Y are 2 left-invariant vector fields, then so is the vector field \left[ X , Y \right].<br /> <br /> Let X_e and Y_e be arbitrary tangent vectors in the tangent space of the identity, and let X and Y be the unigue vector fields that they determine. Define \left[ X_{e}, Y_{e} \right] by<br /> <br /> \left[ X_{e}, Y_{e} \right] = \left[ X , Y \right]_{e}.<br /> <br /> This turns the tangent space at the indentity into a Lie algebra isomorphic to the Lie algebra of left-invariant vector fields. Members of the Lie algebra of left-invariant vector fields have natural compositions as vector fields (that aren&#039;t vector fields), but I don&#039;t see how to extend this to an associative product on the tangent space at the identity.<br /> <br /> So, it seems that for a general Lie group, the Lie algebra of left-invariant vector fields doesn&#039;t meet your criteria, but the Lie algebra of the tangent space at the identity might meet your criteria. ?<br /> <br /> Regards,<br /> George

 

[Don Aman]

however, as for the example given two tangents at the identity what is [ab] of them?

tangent vectors are first order differential operators. the product of two of those is a second order differential operator.