Let $\mathfrak{g}$ be a Lie Algebra. Then isn't necessarily true that all vector spaces $V \subset \mathfrak{g}$ are a Lie subalgebra (it is easy to construct an example that this fails).

Now, weird things starting to happen. Define the linear map

$$\rho: \mathfrak{g}\to \{0\} $$ $$X \mapsto 0, $$

and the set $\tilde{V} = \{X \in V; \rho([X,Y]) = 0,$ $\forall$ $Y$ $\in$ $V$}.

I will prove that $\tilde{V} $ is a subalgebra, i.e. $\tilde{V}$ is a linear subspace of $\mathfrak{g}$, and $\{[X,Y]; X,Y \in \tilde{V}\} \subset \tilde{V}$.

Indeed, $\tilde{V}$ is a vector space, because, if $X,Y$ $\in$ $\tilde{V}$, then is it obvious that $X + \lambda Y$ $\in$ $\tilde{V}$.

Moreover if $X,Y \in \tilde{V} \Rightarrow [X,Y] \in \tilde{V},$ because $\rho([[X,Y],Z]) = 0$, $\forall$ $Z$ $\in$ $V$.

Then, we conclude that $\tilde{V}$ is a Lie subalgebra, but $V = \tilde{V}$, implying that $V$ is a Lie subalgebra. So we concluded that any subspace of $\mathfrak{g}$ is a Lie subalgebra!

Where is the error in my argumentation? I can not see what I'm confusing.

  • $\begingroup$ I don't think so... when I write $ρ([[X,Y],Z])$ I'm just using that $[X,Y]$ and $[[X,Y],Z]$ $\in$ $\mathfrak{g}$. $\endgroup$ – Matheus Manzatto Aug 28 '18 at 2:13
  • $\begingroup$ I see, but I think the issue is in the same spot. Your definition of $\tilde{V}$ is $\{ X \in V \mid \forall Y \in V (\rho([X,Y]) = 0)\}$. That $\rho([[X,Y],Z])=0$ for every $Z \in V$ only implies $[X,Y] \in \tilde{V}$ if we already assume that $[X,Y] \in V$. $\endgroup$ – Hayden Aug 28 '18 at 2:16

Your argument fails at the step

Moreover if $X,Y \in \tilde{V} \Rightarrow [X,Y] \in \tilde{V},$ because $\rho([[X,Y],Z]) = 0$, $\forall$ $Z$ $\in$ $V$.

By definition of $\tilde{V}$, the fact that $\rho([[X,Y],Z])=0,\;\forall$ $Z$ $\in$ $V$ does not, in and of itself, imply $[X,Y]\in \tilde{V}$, unless you already know $[X,Y]\in V$.

| cite | improve this answer | |
  • $\begingroup$ Of course, I'm sorry. I was stuck in it for a long time. $\endgroup$ – Matheus Manzatto Aug 28 '18 at 2:24
  • $\begingroup$ No problem. Sometimes the simplest errors are the hardest ones to spot, since the error, once made, gets a waiver on being scrutinized $\endgroup$ – quasi Aug 28 '18 at 2:34

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.