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

[Kreizhn]

Hello all,

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

My overall goal is to express P as an exponential product
$$P = \exp\left( \alpha_1 H_1 \right) \cdots \exp\left( \alpha_d H_d \right)$$
for $\alpha_i \in \mathbb R$. 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
$$P = \exp\left( - \mu G \right)$$
for some $G \in \mathfrak u(N)$. 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 $H_i$, how would I do this?

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

 

[Ben Niehoff]

Unfortunately, what you are trying to do is not possible, unless your Lie group happens to be Abelian. What you need is this:

http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula

 

[Kreizhn]

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 $S=\left\{H_1, \ldots, H_d \right\}$. 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 $\alpha_i$ for fixed index sets $I_j$ so long as the Lie algebra generated by $\left\{ H_1, \ldots, H_d \right\}$ is the entire algebra $\mathfrak{su}(N)$. 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 $\mathfrak{su}(N)$. 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.

 

[Kreizhn]

With regard to decomposing G: The only thing I can think of is computationally generating a basis for $\mathfrak{su}(N)$ 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.

 

[Kreizhn]

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
$$S=\left\{ H_1, \ldots, H_d \right\}$$
to create a basis for $\mathfrak{su}(N)$. That is, if S does not generate the entire Lie Algebra, $\exists j,k \in \left\{1, \ldots, d \right\}$ such that $[H_j, H_k]$ is linearly independent of S. Thus $H_{d+1} = e^{-iH_j \tilde t} H_k e^{iH_j \tilde t}$ is linearly independent of S for arbitrarily small $\tilde t$.

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

Now we can use successive Cartan decompositions of the $\mathfrak{su}(N)$ (where I forgot to mention that $N=2^n$ for some n ) into $\mathfrak{su}(2)$ as a vector space such that the map
$$\phi(X) = \exp(X_1) \cdots \exp(X_2)$$
for $X_q \in \mathfrak{su}(2), \forall q$ is a diffeomorphism in a local neighbourhood V of the group identity. We can project any operator $G \in \mathfrak{SU}(N)$ into this neighbourhood by considering sufficiently large exponentials of a scaled principle logarithm. That is, we can write $G^{1/m} = \exp(A/m) \in V$. Then the map $\phi$ 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,
$$s> d \quad \Rightarrow \quad \exists 0<j,k<s, H_s = e^{-iH_jt} H_k e^{iH_jt} \quad \Rightarrow \quad e^{-iH_s \tau} = e^{-iH_jt} e^{-iH_k \tau} e^{-i H_j t}$$
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.