0

Consider a three dimensional Lie algebra $L$ with basis $a_1, a_2$ and $a_3$ which satisfy following commutation relation. \begin{align} [a_1,a_2]&=0\\ [a_1,a_3]&=a_1\\ [a_2,a_3]&=-a_2 \end{align}

Is it possible to find a representation $\rho: L\to \mathrm{End}(V)$ with finite dimensional V such that $\rho(a_1)=(\rho(a_2))^{-1}$ as elements of $GL(V)$?

For context: this algebra describes a quantum circuit as described in this paper (https://journals.aps.org/prb/pdf/10.1103/PhysRevB.53.4027?casa_token=kelWUXKoPVoAAAAA%3AHF236bMnQ_GbDKWMdySqluoDE_9PZ3Cbxb-lN31u0FxcGnWrtwgqfaonQ-mvy17mZi2m3iioarigfA)

What I see so far is that there can not be two-dimensional matrix representation because, from the commutation relation, we see that $a_1$ and $a_2$ are traceless. Moreover, $a_1$ and $a_2$ are inverses of each other. But the inverse of a $2\times 2$ traceless matrix is just a constant multiplied by itself. Thus, we run into inconsistent commutation relations.

But for 3D, I do not see any ways to argue whether a faithful representation exists or not. And the same with higher dimensions.

Any hints on how to approach the problem for $3*3$ case?

user824530
  • 713
  • Sorry, the commutator is the commutator of the matrices. – user824530 May 06 '23 at 01:25
  • And in a matrix representation a1 and a2 are the matrix inverse of one another – user824530 May 06 '23 at 01:26
  • 2
    What you wrote does not make sense: if "the commutator in the Lie algebra is the matrix commutator" then that means the elements of the algebras are matrices, and therefore you already have a finite dimensional faithful representation. – Mariano Suárez-Álvarez May 06 '23 at 03:29
  • My mind reading machine tells me that what you mean is: consider the free associative algebra generated by three letters x, y and z, and divide by the ideal generated by xy-1, yx-1, xz-zx-x and yz-zy+y. Now let L be the lie algebra constructed from that associative algebra (or the Lie subalgebras generated inside that one by the three letters) – Mariano Suárez-Álvarez May 06 '23 at 03:32
  • Morally the enveloping algebra of that Lie algebra can be constructed by starting with the enveloping algebra of the solvable 2-dimensional Lie algebra spanned by u and v subject to [u,v]=v, and then localizing at u. – Mariano Suárez-Álvarez May 06 '23 at 03:35
  • 1
    Or do you maybe mean: Given a three-dimensional Lie algebra $L$ with basis $a_1, a_2, a_3$ satisfying the given commutator relations. Then is there a representation $\rho: L \rightarrow End(V)$ with finite-dimensional $V$ such that $\rho(a_1) = (\rho(a_2))^{-1}$ as elements of $GL(V)$? – Torsten Schoeneberg May 06 '23 at 14:58
  • 1
    @Torsten, Yes that I what I mean. Is there a finite dimensional representation where in the representation $\rho(a_1)=(\rho(a_2))^{-1}$.

    If $[a_1,a_2]=1$ and $a_1a_2=a_3$ this is the Harmonic oscillator algebra which do not have finite dimensional representations. But now here $[a_1,a_2]=0$ so such finite dimensional representations could exist.

     – user824530 May 06 '23 at 15:24
  • Please edit the question accordingly then. I think the answer is no at least in characteristic $0$ then, and if you edit the question and I have time I might write an answer later. – Torsten Schoeneberg May 06 '23 at 18:24
  • Ok. I'll edit the question. Thank you for your help . – user824530 May 07 '23 at 02:12

1 Answers1

1

Thanks for revising the question.

If we are working over a field of characteristic $0$, then the answer is no. Because e.g. equation 2 now forces

$$\rho(a_1) \rho(a_3) - \rho(a_3) \rho(a_1) = \rho(a_1)$$

but it is a very useful result of Jacobson's that for matrices over characteristic $0$ fields, $AB-BA = A$ forces $A$ to be nilpotent. (More generally, this is already true as soon as $A$ commutes with $AB-BA$.)

So if there were such a finite-dimensional representation, the matrix representing $a_1$ (and also the one representing $a_2$) would have to be nilpotent, and thus cannot even have an inverse.

I do not know if one can construct such a representation over a field of positive characteristic. I would not be surprised if one can. The above result certainly does not hold any more.

Update: There is a somewhat "natural", three-dimensional representation over any field of characteristic $3$:

$\rho(a_1) = \pmatrix{0&1&0\\0&0&1\\1&0&0}$, $\rho(a_2) = \pmatrix{0&0&1\\1&0&0\\0&1&0}$, $\rho(a_3) = \pmatrix{0&0&0\\0&1&0\\0&0&-1}$.

And a four-dimensional representation of this Lie algebra over a field of characteristic $2$: Let the field be $\mathbb F_4 = \{0,1, \zeta, \zeta^2=\zeta^{-1}=\zeta+1 \}$ and set

$\rho(a_1) = \pmatrix{0&\zeta&0&0\\ 1&0&0&0\\0&0&0&1\\0&0&1&0}$, $\rho(a_2) = \pmatrix{0&1&0&0\\ \zeta^{-1} &0&0&0\\0&0&0&1\\0&0&1&0}$, $\rho(a_3) = \pmatrix{1&0&0&0\\ 0&0&0&0\\0&0&1&0\\0&0&0&0}$

Torsten Schoeneberg
  • 26,051