Scalar multiplication of automorphism

Question

Let $\phi$ be an automorphism from a Lie algebra $L$ to itself. Is it necessary that $\phi(-x)=-\phi(x)$ for an element $x\in L$.

The automorphism must satisfy $[\phi(x),\phi(y)]=\phi([x,y])$. We may substitute $y$ by $-y$ to get $[\phi(x),\phi(-y)]=\phi(-[x,y])$. Still, I don't think we can get the relation $\phi(-x)=-\phi(x)$.

Which part of the definition implies that the map is linear? — PJ Miller, Aug 12 '13 at 15:03

score 2 · Accepted Answer · answered Aug 12 '13 at 09:58

2

By definition, an automorphism of a Lie algebra L is a vector space automorphism $\varphi: L \rightarrow L$ such that $\varphi([x,y])=[\varphi(x),\varphi(y)]$. So one always has $\varphi(-x)=-\varphi(x)$.

answered Aug 12 '13 at 09:58

user68316

461

Exactly which part of the definition implies $\varphi(-x)=-\varphi(x)$? – PJ Miller Aug 12 '13 at 15:02
It follows from the "vector space automorphism" part, not the $\varphi([x,y])=[\varphi(x),\varphi(y)]$ part, is that right? – PJ Miller Aug 12 '13 at 15:47

Scalar multiplication of automorphism

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