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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.

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  • $\begingroup$ 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
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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 $[X,Y]\in V$.

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  • $\begingroup$ 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

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