0

Consider the three dimensional Lie algebra with basis $x,y,z$ and commutator relations $$[x,y]=y-z$$ $$[y,z]=0$$ $$[z,x]=y-z$$ I want to find a faithful matrix representation of this Lie algebra, by Ado's theorem one exists. The adjoint representation is not faithful so it needs to be some other set of matrices which give this.

I have tried to find a three $3\times3$ matrices which satisfy the above by brute force but the complexity and runtime of such a programme is extremely high. If anyone knows a solution or a practical way of finding one that would be great.

Matthew
  • 1,354
  • 1
    How about using the basis ${x , y, y+z}$ instead of ${x , y, z}$? The benefit being that $y+z$ is central. – Nate Feb 17 '20 at 18:57
  • @Nate I'm not quite sure how that helps me find the matrices – Matthew Feb 18 '20 at 13:54
  • @Matthew with that transformation, your Lie algebra becomes $[x,y] = w$, $[y, w] = 0$, and $[w,x] = 0$. – Cameron Williams Feb 18 '20 at 14:06
  • 1
    This is the Heisenberg Lie algebra, which has a faithful linear representation of degree $3$. It is given here. – Dietrich Burde Feb 18 '20 at 14:15
  • @Nate I realised there was an error in my question, thanks for your help though that was an interesting way of seeing the equivalence of that Lie algebra to the Heisenberg. – Matthew Feb 18 '20 at 18:03
  • @DietrichBurde as per my response to Nate I made an error in my question, thanks for the comment. – Matthew Feb 18 '20 at 18:04
  • 1
    Mathew, no problem. I have computed faithful representations for all $3$-dimensional Lie algebras (which was not so difficult as you think. You can try it with upper-triangular matrices). Also here we can do a base change, see this post. – Dietrich Burde Feb 18 '20 at 18:34
  • @DietrichBurde I am trying to do exactly the same, in particular I am looking at Bianchi III in the Bianchi Classification. Did you have any method for doing so, other than ingenuity and guesswork? – Matthew Feb 18 '20 at 18:38
  • 1
    One method to try first is to take upper-triangular matrices. This is much easier and gives you in most cases already a faithful representation. One can also say something on the minimal degree (for the Heisenberg it is $3$), so one knows where to start - see my article here. – Dietrich Burde Feb 18 '20 at 18:41
  • @DietrichBurde your article is great, exactly what I was looking for. Thank you! – Matthew Feb 18 '20 at 18:51

0 Answers0