# Conditions for a subspace to be Lie algebra with projected bracket

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:

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