I have an exercise which starts saying that the Lie algebra of a two-dimensional group is $\left[X,Y\right]=X-Y$. Later, the exercise asks other questions. But, in my opinion, it is impossible that such a Lie bracket could define a Lie algebra. It obviously fulfills the anti-symmetry condition, i.e, $\left[X,Y\right]=-\left[Y,X\right]$. But it does not fulfill linearity $\left[\alpha X+\beta Y,Z\right]=\alpha\left[X,Z\right]+\beta\left[Y,Z\right]$, nor the Jacobi associativity $\left[X,\left[Y,Z\right]\right]+\left[Z,\left[X,Y\right]\right]+\left[Y,\left[Z,X\right]\right]=0$.
For instance, by applying the rule, I obtain the following results regarding linearity:
$\left[\alpha X+\beta Y,Z\right]=\alpha X+\beta Y-Z$, which is obviously different, in general, to $\alpha\left[X,Z\right]+\beta\left[Y,Z\right]=\alpha X +\beta Y-\left(\alpha+\beta\right)Z$.
In addition, for the Jacobi identity, I obtain the following partial results by applying the rule for the given Lie bracket:
$\left[X,\left[Y,Z\right]\right]=X-\left[Y,Z\right]=X-\left(Y-Z\right)=X-Y+Z$, and similarly for the other brackets, by rearrangement of the letters I obtain:
$\left[Z,\left[X,Y\right]\right]=Z-X+Y$, and $\left[Y,\left[Z,X\right]\right]=Y-Z+X$.
Therefore, after summing the three terms I finally obtain for the associativity rule of Jacobi that $X+Y+Z=0$, and the left-hand side of this equation is not zero in general…
Am I missing something about the given defining rule of $\left[X,Y\right]=X-Y$? I am asking this because I am teaching Mathematical Physics at Bachelor degree in an university, and this was an exercise of an examination in past years. Therefore, I don't know if I am very silly about the meaning of the rule or truly the exercise has no sense at all.
I am sorry if this is not the adequate forum to ask the question and I apologize for that in such a case.
Best Answer
It is better to come up with an explicit counterexample, if you claim that it is "obviously" different. Let $L=\Bbb R^2$ with basis $\{e_1,e_2\}$ and Lie bracket $[x,y]=x-y$. Then we have, by bilinearity, $$ e_1-e_2=[e_1,e_2]=[e_1+e_2,e_2]=(e_1+e_2)-e_2=e_1. $$ This is a contradiction.
However, if we only require $[x,y]=x-y$ for basis elements, then there is the nonabelian Lie algebra in dimension $2$ with $[e_1,e_2]=e_1-e_2$.