Can Lie Algebraic Decomposition Simplify Matrix Exponentials in \mathfrak{U}(N)?

In summary, the conversation discusses the problem of expressing a unitary matrix P as an exponential product using the Lie group of unitary matrices and a set of generating matrices, as well as the possibility of reducing the problem to finding a decomposition of an element G in terms of the generating set. The conversation also mentions the possibility of working in the special unitary group and the potential use of numerical algorithms for finding the values of \alpha_i. Ultimately, the conversation ends with a solution involving a similarity transformation and successive Cartan decompositions to generate a basis for the Lie algebra and express all elements in terms of the generating set.
  • #1
Kreizhn
743
1
Hello all,

I have a small problem that I would appreciate some assistance with. I'm working with the Lie group [itex] \mathfrak U(N) [/itex] of unitary matrices and I have an element [itex] P \in \mathfrak U(N) [/itex]. Furthermore, I have a set
[tex] \left\{ H_1, \ldots, H_d \right\} [/tex]
that generate the Lie algebra [itex] \mathfrak u(N) [/itex] of skew-Hermitian matrices under a Lie bracket given by a commutator [itex] [A,B] = AB-BA [/itex].

My overall goal is to express P as an exponential product
[tex] P = \exp\left( \alpha_1 H_1 \right) \cdots \exp\left( \alpha_d H_d \right) [/tex]
for [itex] \alpha_i \in \mathbb R [/itex]. I know in general that this decomposition is probably not a simple one. On the other hand, I do know that I can express P as
[tex] P = \exp\left( - \mu G \right) [/tex]
for some [itex] G \in \mathfrak u(N) [/itex]. It seems to me then that maybe I can reduce this to simply finding a decomposition of G in terms of the generating set of the Lie algebra. Even if the decomposition is in terms of the Lie bracket, I should be able to exploit properties of the matrix exponential to get what I desire.

In any case, I am not certain if there is an easier way of doing this. Or if we can indeed just focus on decomposing G in terms of [itex] H_i [/itex], how would I do this?

P.S. If it helps, I can instead choose to work in the special unitary group [itex] \mathfrak{SU}(N) [/itex] and hence [itex] \mathfrak{su}(N) [/itex] the Lie algebra of traceless skew-Hermitian matrices. Also, I can write
[tex] P = \exp\left[ \alpha_1 \sum_{i \in I_1} H_i \right] \cdots \exp\left[\alpha_k \sum_{i \in I_k} H_i \right] [/tex]
For [itex] I_j \subseteq \left\{1, \ldots, d \right\} [/itex]. That is, the argument of the exponentials need not be only a single generating element, but it can also be a sum of generators.
 
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  • #3
Hi Ben,

Thanks for your reply. Luckily, this is for application in a numerical algorithm, and hence so long as the infinite product gradually has less and less contribution then truncation on the order of machine-epsilon will not be noticeable.

By chance, do you know how to "decompose" the element G mentioned above in terms of the set [itex] S=\left\{H_1, \ldots, H_d \right\} [/itex]. It seems like this should certainly be possible, but I'm uncertain as how to do it.

While analytically it may not be a well posed problem, it is actually possible to create numerical algorithms to find the values of [itex] \alpha_i [/itex] for fixed index sets [itex] I_j [/itex] so long as the Lie algebra generated by [itex] \left\{ H_1, \ldots, H_d \right\} [/itex] is the entire algebra [itex] \mathfrak{su}(N) [/itex]. While not presented in the exact same context, you may be interested in the result of the following paper:

C.Altafini. Controllability of quantum mechanical systems by root space decomposition of [itex] \mathfrak{su}(N) [/itex]. Journal of Mathematical Physics, 43(5): 2051-2062, 2002

Also, the Khaneja-Glaser decomposition offers an interesting way of writing special unitary matrices in terms of an exponential product using recursive Cartan decompositions of the space. Though I don't think this is really applicable to the current problem.
 
  • #4
With regard to decomposing G: The only thing I can think of is computationally generating a basis for [itex] \mathfrak{su}(N) [/itex] by computing commutators of S and then adding linearly independent elements until I have a spanning set. At that point I just vectorize all the components and solve it like any other vector space problem. This works, but is not very efficient or elegant. If anyone has a better idea, it would be much appreciated.
 
  • #5
In case anyone ever stumbles upon this problem in the future, I have solved it. One can use a similarity transformation on the original set of generators
[tex] S=\left\{ H_1, \ldots, H_d \right\} [/tex]
to create a basis for [itex] \mathfrak{su}(N) [/itex]. That is, if S does not generate the entire Lie Algebra, [itex] \exists j,k \in \left\{1, \ldots, d \right\} [/itex] such that [itex] [H_j, H_k] [/itex] is linearly independent of S. Thus [itex] H_{d+1} = e^{-iH_j \tilde t} H_k e^{iH_j \tilde t} [/itex] is linearly independent of S for arbitrarily small [itex] \tilde t[/itex].

For the sake of contradiction, assume this is not true. Then we can write
[tex] H_{d+1} = \sum_{r=1}^d a_r(\tilde t) H_r [/tex]
But differentiating with respect to [itex] \tilde t[/itex] and evaluating at zero will then give
[tex] [H_j, H_k] = \sum_{r=1}^d \frac{da_r}{dt}(0) H_r [/tex]
which is a contradiction to the assumption that S doesn't span [itex] \mathfrak{su}(N) [/itex]. We can do this until we have generated a basis for the Lie Algebra.

Now we can use successive Cartan decompositions of the [itex] \mathfrak{su}(N) [/itex] (where I forgot to mention that [itex] N=2^n [/itex] for some n ) into [itex] \mathfrak{su}(2) [/itex] as a vector space such that the map
[tex] \phi(X) = \exp(X_1) \cdots \exp(X_2) [/tex]
for [itex] X_q \in \mathfrak{su}(2), \forall q [/itex] is a diffeomorphism in a local neighbourhood V of the group identity. We can project any operator [itex] G \in \mathfrak{SU}(N) [/itex] into this neighbourhood by considering sufficiently large exponentials of a scaled principle logarithm. That is, we can write [itex] G^{1/m} = \exp(A/m) \in V[/itex]. Then the map [itex] \phi [/itex] gives us a parameterization in terms of the basis from S. Furthermore, note that elements generated by S are expressible in S since they are derived via similarity transform. That is,
[tex] s> d \quad \Rightarrow \quad \exists 0<j,k<s, H_s = e^{-iH_jt} H_k e^{iH_jt} \quad \Rightarrow \quad e^{-iH_s \tau} = e^{-iH_jt} e^{-iH_k \tau} e^{-i H_j t} [/tex]
which can be recursively done until each element is in S.

Thus every element can be written in terms of the matrix exponential of the generating set.
 

Related to Can Lie Algebraic Decomposition Simplify Matrix Exponentials in \mathfrak{U}(N)?

What is Lie algebraic decomposition?

Lie algebraic decomposition is a mathematical concept that involves breaking down a complex Lie algebra into smaller, simpler parts. It is a way of analyzing and understanding the structure of a Lie algebra, which is a mathematical object that describes the properties of a group of transformations.

Why is Lie algebraic decomposition important?

Lie algebraic decomposition is important because it allows us to study the structure of a Lie algebra and make connections between different parts of the algebra. This can help us to understand the underlying symmetry and geometry of a system, and can also be used in applications such as physics and engineering.

What are the different types of Lie algebraic decomposition?

There are several types of Lie algebraic decomposition, including Cartan decomposition, Iwasawa decomposition, and Jordan decomposition. Each type involves breaking down the algebra in a different way, depending on the specific properties of the algebra.

How is Lie algebraic decomposition used in physics?

Lie algebraic decomposition is used in physics to study the symmetries and transformations of physical systems. It can be applied to various areas of physics, such as quantum mechanics, particle physics, and general relativity, and is especially useful in understanding the behavior of particles and fields in a given system.

Are there any real-world applications of Lie algebraic decomposition?

Yes, Lie algebraic decomposition has many real-world applications in addition to its use in physics. It is used in data analysis and pattern recognition, as well as in cryptography and coding theory. It is also used in engineering and robotics to understand the underlying structure and behavior of systems.

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