Can we construct a Lie algebra from the squares of SU(1,1)

[Buddha_the_Scientist]

I am trying to decompose some exponential operators in quantum optics. The interesting thing is that the operators includes operators from Su(1,1) algebra $$ [K_+,K_-]=-2K_z \quad,\quad [K_z,K_\pm]=\pm K_\pm.$$
For example this one: $$ (K_++K_-)^2.$$ But as you can see they are squares of it.
I wonder if it is possible to construct algebra for such operators for example an algebra includes $$K_+^2, K_-^2,...$$ I tried, it doesn't seem possible but still wanted to ask. There exists many operators in a similar fashion. Up to now I could only resort to approximation techniques. The operators are bosonic mode operators so squares aren't equal to identity.

 

[fresh_42]

A normal (say e.g. associative) algebra product as the square is, usually (unless it is trivially zero) breaches the Jacobi identity, so the answer in general is no. But for every associative algebra, the setting $[X,Y]:=XY-YX$ gives you a Lie algerbra, but this requires an associative multiplication first. Another possibility is, to systematically add all needed products as new basis elements. E.g. in your example we have
$$
(K_++K_-)^2 = K_+^2 +K_+K_-+K_-K_++K_-^2
$$
and we can consider all these new products as new basis vectors: $B_1=K_+,B_2=K_-,B_3=K_+^2,B_4=K_+K_-,B_5=K_-K_+,B_6=K_-^2$ and so on. Without any properties about finiteness, esp. for the mixed products, we will get an infinite dimensional Lie algebra, freely generated by those products.

The main difficulty is, that the normal products alone don't belong to the Lie algebra anymore, they have to be artificially added. E.g. if we look at matrices with trivial trace, which form a Lie algebra, this property doesn't hold any longer for their products:
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}\cdot \begin{bmatrix}0&0\\1&0\end{bmatrix} = \begin{bmatrix}1&0\\0&0\end{bmatrix}
$$
The reason is, as already mentioned, the Jacobi Identity. This is basically the product or Leibniz rule of differentiation:
$$
\operatorname{ad}([X,Y])=[\operatorname{ad}(X),\operatorname{ad}(Y)]=\operatorname{ad}(X)\circ \operatorname{ad}(Y) - \operatorname{ad}(Y) \circ \operatorname{ad}(X)
$$
with the Lie multiplication $\operatorname{ad}\, : \,Z \mapsto [X,Z]\,$.

 

[Buddha_the_Scientist]

Thank you for the reply. My point was if it is possible to construct a finite algebra easily from the squares of operators that constitutes a finite Lie algebra already. For example consider the following algebra:
$$[L_1,L_2]=I$$
Where I represents identity. Now basically these operators form a group:
$$ L_1, L_2, I$$
Now if we take squares and check the commutation relations we see that:
$$L_1^2, L_2^2, (L_1L_2+L_2L_1)$$
also form a group. The first group is sort of simple because of the identity operator, but I was wondering if there are generalizations for these sort of constructions.

 

[fresh_42]

Not sure what you mean by group in this context. It looks more as if we're still talking about algebras.

Thank you for the reply. My point was if it is possible to construct a finite algebra easily from the squares of operators that constitutes a finite Lie algebra already. For example consider the following algebra:
$$[L_1,L_2]=I$$
Where I represents identity. Now basically these operators form a group:
$$ L_1, L_2, I$$

This is the smallest Heisenberg algebra. It can be represented by $(3\times 3)-$matrices with $L_1=\delta_{12}\, , \,L_2=\delta_{23}\; , \;I=\delta_{13}\,.$

Now if we take squares and check the commutation relations we see that:
$$L_1^2, L_2^2, (L_1L_2+L_2L_1)$$
also form a group. The first group is sort of simple because of the identity operator, but I was wondering if there are generalizations for these sort of constructions.

With the representation above, we have $L_1^2=L_2^2=L_2L_1=0$ and $L_2L_1=I\,.$ So in case you meant something different, you'll have to define it completely. It is not clear what $[X,Y]$ should be for $X,Y \in \{L_1^2, L_2^2, (L_1L_2+L_2L_1)\} \cup \{L_1,L_2,I\}$not to mention even more general elements as e.g. $L_1^{n_1}L_2^{n_2}L_1^{n_3}L_2^{n_4} \ldots$ and so on.

 

[Buddha_the_Scientist]

As I noted in my first post I am talking about bosonic mode operators, not matrices, the squares doesn't equal to I. Consider the annihilation and creation operators:
$$a,a^{\dagger}$$
Now they constitutes the Heisenberg algebra and as I pointed in my post:
$$a^2,a^{\dagger^2},a^{\dagger} a+aa^{\dagger}$$
also constitutes a group. This is the SU(2) group.
That's what I meant.

 

[fresh_42]

As I noted in my first post I am talking about bosonic mode operators, not matrices, the squares doesn't equal to I. Consider the annihilation and creation operators:
$$a,a^{\dagger}$$

Every finite dimensional complex Lie algebra can be viewed as a matrix algebra. (Ado)

Whether it is represented by $I=\sum \delta_{ii}$ or not is irrelevant. You haven't defined what $[L_i,I]$ should have been, so I assumed $[L_i,I]=0$ because what you have given was incomplete to define a Lie algebra. The representation of a Heisenberg algebra by the maximal nilpotent subalgebra of $\mathfrak{gl}_3$ is the standard representation. And it is what it says: a representation by choosing basis vectors and defining a multiplication table. $I=\delta_{13}$ satisfies the definition, what it looks like doesn't matter. If you meant $[L_i,I]=L_i$ instead, you should have mentioned, as usually the identity commutates with all other operators. With this new definition we basically get $\mathfrak{sl_2}\cong \mathfrak{su}_2$. Neither of them contains elements as $L_i^2$

Now they constitutes the Heisenberg algebra and as I pointed in my post:
$$a^2,a^{\dagger^2},a^{\dagger} a+aa^{\dagger}$$
also constitutes a group. This is the SU(2) group.
That's what I meant.

In a group, you have one operation, which you can either write as multiplication or addition, but not both. Furthermore, I don't see any inverses here. So what we really have is an algebra, but in this case, the question is: which one? This means, can you write down a complete basis for the underlying vector space, whether it is an operator space or not, and the multiplication table for this algebra? Both is needed to determine, what it is. If you insist on a basis $\{A,B,I,A^2,B^2,AB+BA\}$ chances are that it is neither an associative nor a Lie algebra, but I don't want to guess all possible multiplication tables. In any case, it is never a group.

 

[Buddha_the_Scientist]

Ok, sorry for my ignorance.