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I'm sure I'm missing something really obvious here. This seems too stupid.

On page 47 of Erdmann & Wildon's Introduction to Lie Algebras, we have the following set up. Let $L$ be a Lie subalgebra of $\mathfrak{gl}(V)$ of dimension $\geq 1$, and let $A \leq L$ be a maximal Lie subalgebra and let $\overline{L} = L/A$. Then they define $\varphi: A \to \mathfrak{gl}(\overline{L})$ by $\varphi(a)(x+A) := [a,x] + A$.

My question is: why do we not have $\varphi = 0$? Or do we have this? I mean, if $a \in A, x+A \in \overline{L}$, then $\varphi(a)(x+A) = [a,x] + A = [a+A,x+A]=[0,x+A]=0$, right?

Maybe I'm not understanding how the braket is defined on the quotient, or something.

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2 Answers 2

up vote 1 down vote accepted

Let's have an example. Let $L = span_{\mathbb F} \{x, y\}$, and $[x, y] = y$ (Note that every two dimensional Lie algebra are of this form). Let $A = span\{x\}$. Then the map $$\phi: A \to gl(\bar L)$$

is $\phi(cx)(y + A) = [cx, y] + A = cy+ A$. This is not a zero map.

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Ok, but why don't we have $[cx,y] + A = [cx+A,y+A] = [0,y+A]=0$? –  nigelvr Oct 29 '13 at 5:50
    
Your first equality is wrong. As the bracket is bilinear in both entries, $[cx+A, y+A] = [cx, y] + [A, y] + [cx, A] + [A, A] = [cx, y]+ [A, y] +A \neq [cx, y] + A$ –  John Oct 29 '13 at 5:55

The issue here is that $A$ is assumed to be a subalgebra, but not necessary an ideal. Actually $[a,x] + A = [a + A, x + A]$ is well defined if and only if $A$ is an ideal. Hence we cannot consider the quotient $L/A$ as a Lie algebra, just as a vector space.

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