Let's say that I have a curve in $\mathbb{C}^N$ given by the action of the unitary group: $$x(t) = e^{Ht}x_0,~ H \in \mathfrak{u}(N),~ ||x_0||=1$$ Here, $H$ is an NxN skew-Hermitian matrix (for very large $N$). I can approximate this to first order as: $$\tilde x(t) \approx x_0 + t(Hx_0) + O(t^2)$$ However, this map isn't unitary; $||\tilde x(t)|| \neq 1$. A better first-order approximation seems to be a great circle on the unit sphere in $\mathbb{C}^N$: $$\tilde x(t)' \approx \cos(\alpha t)x_0 + \alpha^{-1}\sin(\alpha t)Hx_0,~\alpha = ||Hx_0||$$ My question is: Is there a simple generalization of this to higher order terms? Specifically, I'm looking for a family of curves: $$\gamma_M(t) : \mathbb{R} \rightarrow S_C$$ that satisfy: $$\frac{d^k}{dt^k} \gamma_M(t)|_{t=0} = \begin{cases}(H^k)x_0 & (k \le M) \\ 0 & (k > M)\end{cases}$$ where $S_C$ is the unit sphere in $\mathbb{C}^N$. It should be obvious that: $$\lim_{M\rightarrow\infty} \gamma_M(t) = \exp(Ht)x_0$$

For context, $H$ represents a complicated linear combination of elements of $\mathfrak{u}(N)$, and is intractable to compute explicitly (thus making the true exponential map nearly impossible to compute). However, I can compute the action of $H$ on vectors, so I want the closest easy-to-compute unitary transform as a function of some number of iterated applications of $H$.


It does not verify your condition on the derivatives, but this looks like a job for the Cayley transform: take $\tilde{x}(t) = (I-\frac{1}{2}Ht)^{-1}(I+\frac{1}{2}Ht)x_0$.

The scalar map $z\mapsto \frac{1+\frac{1}{2}tz}{1-\frac{1}{2}tz}$ sends the imaginary axis (including $\infty) $onto the unit circle, so this transform produces unitary matrices from skew-Hermitian ones, exactly like the exponential. Moreover, it is a first-order approximation of $\exp(Ht)$.


Something is apparently broken. If for any $k > M\ :\ \frac{d^k}{dt^k} \gamma_M(t) = 0$, then $\gamma_M$ must be a polynomial. But polynomials (except degree-0) that take values on a sphere do not exist. For a function where the codomain is the unit sphere, is an approximation by polynomials a good idea? Or maybe I just misunderstand what means $\frac{d^k}{dt^k}$ on a sphere?

  • $\begingroup$ I believe the OP wants $(d^k/dt^k)\gamma_M$ as specified only at $t=0$. $\endgroup$ – Christian Remling Aug 9 '14 at 17:12
  • $\begingroup$ Yes, that's correct - I meant at t=0 $\endgroup$ – Jarred Aug 11 '14 at 22:31
  • $\begingroup$ This will still run you into trouble: Suppose you had such a curve $\gamma_M$ as you describe it. Since $\langle \gamma_M(t),\gamma_M(t)\rangle \equiv 1$, differentiating this relation $2M$ times and setting $t=0$ will give you $$\bigl\langle \gamma_M^{(M)}(0),\gamma_M^{(M)}(0)\bigr\rangle = 0,$$ so you'll have to have $$\langle H^kx_0,H^kx_0\rangle = 0.$$ I don't think you want this. $\endgroup$ – Robert Bryant Aug 12 '14 at 11:16
  • $\begingroup$ That's a good point. I'm not sure what the exact conditions should be; I just know that they should be a function of $\{H^kx_0\}_{k\le M}$ and they should capture the exponential map in the limit as $M\rightarrow \infty$. $\endgroup$ – Jarred Aug 12 '14 at 16:31

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