$X$ Banach space. In $B(X)$, we can define a Lie product $[ , ]:[T_1,T_2]=T_1T_2-T_2T_1$ for any $T_1,T_2 \in B(X).$

Let $\mathcal{L}$ Lie Algebra. $\mathcal{L}^1=\mathcal{L}$ , $ \mathcal{L}^2=[\mathcal{L},\mathcal{L}]$, $\mathcal{L}^3=[\mathcal{L}^2,\mathcal{L}]$, $\mathcal{L}^n=[\mathcal{L}^{n-1},\mathcal{L}]$. Then;

$\mathcal{L} \supset \mathcal{L}^2 \supset \mathcal{L}^3 \supset ... \supset \mathcal{L}^n \supset ...$

If $\mathcal{L}$ in nilpotent then $\mathcal{L} \supset \mathcal{L}^2 \supset \mathcal{L}^3 \supset ... \supset \mathcal{L}^n= \left\{ 0 \right\}$

What does "Algebra $\mathcal{A}$ is generated by nilpotent Lie Algebra $\mathcal{L}$" mean?

I think this means that $\mathcal{A}= \bigcup \limits_{n=1} {\mathcal{L}^n}$. Is this true? If my idea is true, then $\mathcal{A}=\mathcal{L}$. I do not understand it.

  • $\begingroup$ Firstly, do you know what a Banach algebra is? $\endgroup$ – Cameron Williams Mar 30 '15 at 18:35
  • $\begingroup$ I know, of course, $\endgroup$ – ersanbatu Mar 30 '15 at 18:48
  • $\begingroup$ Based on the information given, I think your definition for $\mathcal{A}$ is reasonable. However, there is a little bit of ambiguity to me here as to what the notion of "product" should be in $\mathcal{A}$. Based on the fact that you want to build $\mathcal{A}$ from a Lie algebra, you might be tempted to use the Lie bracket as your product in $\mathcal{A}$. However I'm not $100\%$ convinced that this product will play nicely with the operator norm structure on $B(X)$ and I don't think it would give you a Banach algebra. $\mathcal{A}$ would probably inherit the usual product on $B(X)$ then. $\endgroup$ – Cameron Williams Mar 30 '15 at 18:52
  • $\begingroup$ Ok. I fix the question. $\endgroup$ – ersanbatu Mar 30 '15 at 19:07

Without more context I cannot be certain, but I don't think your interpretation is the one your author has in mind.

Given an associative algebra (like a Banach algebra), say $\mathcal{A}$, one gets a Lie algebra using the commutator bracket: $[x,y]=xy-yx$ where $x,y \in \mathcal{A}$. Here "$xy$" denotes $\mathcal{A}$'s (associative) multiplication.

So $\mathcal{A}$ can be viewed either as an associative algebra (with $(x,y) \mapsto xy$) or a Lie algebra (with $(x,y) \mapsto [x,y]=xy-yx$).

This is where it gets a little confusing. Suppose $\mathcal{L}$ is a subalgebra of $\mathcal{A}$. What is meant? a sub-associative-algebra or a sub-Lie-algebra? These are not the same thing.

A subalgebra (in the associative sense) is a subspace which is closed under the associative multiplication. A subalgebra (in the Lie algebra sense) is a subspace which is closed under the Lie bracket.

Every subalgebra (in the associative sense) is a subalgebra (in the Lie sense), but the converse is not true.

All that said, I would interpret "$\mathcal{A}$ is generated by a nilpotent Lie algebra $\mathcal{L}$" as follows:

There exists a Lie subalgebra $\mathcal{L}$ of $\mathcal{A}$ ($\mathcal{L}$ is a subspace closed under the bracket) such that $\mathcal{L}$ is nilpotent (as a Lie algebra) where $\mathcal{L}$ generates $\mathcal{A}$ as an associative algebra.

[Here $\mathcal{L}$ may or may not be a subalgebra of $\mathcal{A}$ in the associative sense.]

"Generated" here means: The smallest associative subalgebra of $\mathcal{A}$ containing $\mathcal{L}$ is $\mathcal{A}$ itself. Or, in other words, for each $a \in \mathcal{A}$ there exists some multivariate polynomial $f$ (in non-commuting variables) and elements $x_1,\dots,x_n \in \mathcal{L}$ such that $f(x_1,\dots,x_n)=a$. So every element in $\mathcal{A}$ can be expressed as a linear combination of words over $\mathcal{L}$.

  • $\begingroup$ Ok. Thank you very much. I understand. This was very helpful. $\endgroup$ – ersanbatu Mar 30 '15 at 19:51
  • $\begingroup$ Glad to help! :) $\endgroup$ – Bill Cook Mar 30 '15 at 20:06
  • 1
    $\begingroup$ @Bill Cook, small improvement: make precise that your multivariate polynomial $f$ is noncommutative which means that its monomials are words in $x_1,\dots,x_n$. $\endgroup$ – Duchamp Gérard H. E. Mar 30 '15 at 21:14
  • $\begingroup$ @DuchampGérardH.E. definitely. I should be more careful. $\endgroup$ – Bill Cook Mar 31 '15 at 1:14
  • $\begingroup$ Sure. Every associative algebra is a Lie algebra when given the commutator bracket. However, not all Lie algebras are themselves associative algebras (they can be embedded in an associative algebra but this associative algebra may be strictly larger). $\endgroup$ – Bill Cook Apr 7 '15 at 22:08

The use of nilpotent Lie algebra is classic when you want to pass from the Lie algebra (i.e. guiding vector fields in the context of analysis) to the corresponding group with a finite CBH (Campbell-Baker-Hausdorff) formula. This formula states that $$exp(A)exp(B)=exp(C)$$ where $C$ is a Lie series in $A$ and $B$, so in the general case, one has to do exta assumptions on the Lie algebra $\mathcal{L}$. However, in case $\mathcal{L}$ is nilpotent of order $n$ ($\mathcal{L}^{(n)}=(0)$), the development of $C=log(exp(A)exp(B))$ terminates in finitely many steps, because all terms with more than $n$ Lie brackets are zero. Then you get a nice $log<->exp$ correspondence between the Lie algebra and its corresponding Lie Group. In a Banach algebra, all exponentials converge, so the exponentials of your Lie algebra i.e. the set $$\{exp(X)\}_{X\in \mathcal{L}}$$ form a group. This is not the case in general for sub-Lie algebras of $B(X)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.