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I am just starting learn about Lie algebras and I saw it defined that a Lie algebra is a vector space with a commutator operation that is 1) bilinear, 2) satisfies the Jacobi identity, 3) [x,x]=0. Over a one dimensional real inner-product space, is it the possible to form a Lie algebra by defining the commutator operation to be the inner product? Thanks.

Duncan Ramage
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Patrick Fraser
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  • No, because the values of the Lie operation on a vector space $V$ lie in $V$, not in $\Bbb R$. – Angina Seng Aug 18 '17 at 03:18
  • Ah, okay that makes sense, but in the one dimensional case, since $V=\mathbb{R}$, how does one distinguish between the field and the vector space (after all, as sets, they are identical)? – Patrick Fraser Aug 18 '17 at 03:24
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    Even in the case that $V= \Bbb R$ (on which any inner product is just a scalar multiple of ordinary multiplication), is it the case that $x^2=0$ for all $x$? Or that $3xyz=0$ for all $x,y,z \in \Bbb R$? – shalop Aug 18 '17 at 03:25

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No. You can show that for one dimension real inner product vector spaces, that is, $\mathbb{R}$, that $\langle x, x\rangle = 0$ if and only if $x = 0$. You can see how this is going to be a problem for axiom 3 when we ask that $\langle 1, 1 \rangle = 0$.

Duncan Ramage
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