3

From Humphreys' Introduction to Lie Algebras and Representation Theory:

We can determine (up to isomorphism) all Lie algebra of dimension $2$. Start with a basis $x,y$ of $L$. Clearly, all products in $L$ yield scalar multiples of $[xy]$. If these are all $0$, then $L$ is abelian. Otherwise, we can replace $x$ in the basis by a vector spanning the one-dimensional space of multiples of the original $[xy]$, and take $y$ to be any other vector independent of the new $x$. Then $[xy]=ax$ $(a\ne 0)$. Replacing $y$ by $a^{-1}y$, we finally get $[xy]=x$. Abstractly, therefore, at most one nonabelian $L$ exists (the reader should check that $[xy]=x$ actually defines a Lie algebra.)

(1) If we replace $y$ by $a^{-1}y$, we should get $a^{-1}[xy]=ax\rightarrow [xy]=a^2x$, shouldn't we?

(2) Why does $[xy]=x$ define a Lie algebra? If we take $y=x$, we get $[xx]=x$. But aren't we supposed to have $[xx]=0$?

PJ Miller
  • 8,193
  • 2
    By "Replacing $y$ by $a^{-1}y$", the author means you let $z = a^{-1}y$, and then take the basis $(x,z)$ instead of $(x,y)$. Then you have $[xz] = [x(a^{-1}y)] = a^{-1}[xy] = a^{-1}(ax) = x$. – Daniel Fischer Aug 10 '13 at 13:42

2 Answers2

3

Re 1): By "Replacing $y$ by $a^{-1}y$", the author means you let $z = a^{-1}y$, and then take the basis $(x,z)$ instead of $(x,y)$. Then you have $$[xz] = [x(a^{-1}y)] = a^{-1}[xy] = a^{-1}(ax) = x.$$

Re 2): $[xy] = x$ means you define the product for the two basis elements, in general you have

$$[(ax+by)(cx+dy)] = [(ax)(cx+dy)] + [(by)(cx+dy)] = ac[xx]+ ad[xy] + bc[yx] + bd[yy]$$

and use $[xx] = 0 = [yy]$ and $[yx] = -[xy]$.

Daniel Fischer
  • 206,697
  • 1
    I see, thank you. Looks like I'm quite inexperienced with mathematical writing. At least I'm learning! – PJ Miller Aug 10 '13 at 13:55
2

The answer to your first question is in the comment by Daniel.

When the author says $[xy]=x$ he requires $x,y$ to for a basis. So you can't take $y=x$ in this equality.

Sasha Patotski
  • 7,161