1
$\begingroup$

I am working on my bachelor's thesis in mathematical physics, and I have stumbled across a problem that I cannot seem to solve. Since it seems a quite natural question, I am hoping that someone has studied this kind of problem before, even if I cannot seem to find any article on it.

Let $\mathfrak{g}$ be a (possibly infinite) Lie algebra with bracket $[ \cdot\, , \cdot ]$. What are the conditions that a (finite-dimensional) subspace $S\subset \mathfrak{g}$ must satisfy in order to be a Lie Algebra with bracket $\Pi^S[ \cdot\, , \cdot]$ , where $\Pi^S$ is the orthogonal projector on $S$ with respect to a given scalar product?

(Excluding the trivial case in which S is closed with respect to $[ \cdot\, , \cdot ]$)

In my particular case $\mathfrak{g}$ is the infinite dimensional algebra of divergence-free vector fields on the 3D torus $\mathbb T^3$ and $S$ is a finite Fourier truncation.

$\endgroup$
1
$\begingroup$

Say that $S$ is $n$-dimensional. Pick a basis $\{e_i\}$ for $\mathfrak{g}$ so that $e_1,\dots,e_n$ is a basis for $S$. Then the projection $\pi_S$, as a matrix, looks like

$$ \left( \begin{array}{cc} \mathrm{Id}_n & 0 \\ 0 & 0 \end{array} \right) $$

Let $c^k_{ij}$ be the structure constants of $\mathfrak{g}$ in this basis; i.e. $[e_i,e_j] = \sum\limits_{k=1}^{\mathrm{dim}(\mathfrak{g})} c^k_{ij} e_k$

Then the structure constants for the new bracket $\pi_S[-,-]$ are the same $c^k_{ij}$, but just restricting to $k=1,\dots,n$.

So if you know the structure constants for $\mathfrak{g}$ in a nice basis like this, maybe you can check that $c^k_{ij}$ for just $k\leq n$ also satisfy the requirements to be a Lie algebra:

  • $c^k_{ij} = - c^k_{ji}$
  • $\sum_k c^k_{ij} c^m_{kl} + c^k_{jl}c^m_{kl} + c^k_{li}c^m_{kj} = 0$
$\endgroup$
  • $\begingroup$ Hi Nick, thank you for the answer. That is exactly what I have done! I am glad to see someone agrees. $\endgroup$ – Angelo Brillante Romeo Sep 27 '18 at 19:27

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.