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.

  • $\begingroup$ What is the source for this exercise? (Or, could you post the full exercise?) I agree with you that for the reasons you state, this does not define a Lie algebra. $\endgroup$ Commented May 22 at 18:59
  • $\begingroup$ @NoOneIsHere The source of this exercise is the collection of examinations from past years given by one of the previous professor teaching the subject. Additionally, the exercise was asking, after giving that Lie bracket about finding all the finite-dimensional irreducible representations of the group with such a Lie algebra. $\endgroup$ Commented May 22 at 21:33

1 Answer 1


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 $[e_1,e_2]=e_1-e_2$.

  • $\begingroup$ Thank you very much for the clarification. In fact, at some point I also thought that the Lie bracket was being written only for the basis elements of that two-dimensional Lie algebra. The problem is that the subject I am teaching is for physicists, and normally we are much less formal and clear in the way we enunciate the problems than mathematicians do. But I now think that your explanation is the right one to understand the meaning of the wording. Therefore, thank you very much. $\endgroup$ Commented May 22 at 21:45

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