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.


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:

  • $c^k_{ij} = - c^k_{ji}$
  • $\sum_k c^k_{ij} c^m_{kl} + c^k_{jl}c^m_{kl} + c^k_{li}c^m_{kj} = 0$
  • $\begingroup$ 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

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