# Cayley Transform, Exponential Mapping and more…

Assume a self-adjoint operator, represented as hermitian matrix $H=H^\dagger$. To my knowledge there are at least 2 mappings of $H$ onto unitary matrices:

1. Cayley's Transformation with $U_1=\frac{H+i1}{H-i1}$ ($1$ is a unit matrix of appropriate dimension)
2. Exponential Mapping with $U_2=e^{iH}$.

Are there other mappings with a different structure and if not can one prove that?

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Assume that a mapping $H\mapsto U$ is of such a form that $U$ can be expressed as a power series in $H$ (I don't think this is a huge loss of generality to assume this, perhaps none.) Then $$U=c_{0}I+c_{1}H+c_{2}H^{2} + c_{3}H^{3} + \cdots$$ Since $U$ unitary implies $e^{i\theta}U$ is unitary, then WLOG we may take $c_{0}$ to be real. Moreover, if $U$ is unitary, then $U^{-1}= U^{*}$, so (assuming $U^{-1}$ can be represented as a power series in $H$) $$U^{-1}=c_{0}I+c_{1}^{*}H+c_{2}^{*}H^{2} + c_{3}^{*}H^{3}+ \cdots$$ The condition $UU^{-1} = I$ gives $$I=UU^{-1}=c_{0}^{2}I+c_{0}(c_{1}+c_{1}^{*})H+\cdots$$ whereupon we immediately determine that $c_{0}=1$ and that $c_{1}+c_{1}^{*}=0$. Hence $c_{1}$ is purely imaginary, and we write $c_{1}=ib_{1}$. With this in hand, we expand the products of these power series again: $$I=UU^{-1}=I+(c_{2}+c_{2}^{*}+b_{1}^{2})H^{2}+\cdots$$ so that $2\mathfrak{R}(c_{2})=-b_{1}^{2}$ but the imaginary part of $c_{2}$ is a free parameter. So we've already found 2 free parameters, each of which generate infinite families of unitary maps $H\mapsto U$. So there are a ton of these maps.

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+1 for a ton. Thx. – draks ... Feb 6 '12 at 7:32
Do you know, which of these mappings $H \mapsto U$ are invertible, like mentioned in this answer? – draks ... Apr 5 '12 at 17:41
No, I have no clue. – Nick Thompson Apr 12 '12 at 9:03
It's been a while, but is it right to say that, the infinity of mappings comes from the fact that the $c_k$'s are complex? Since we always pay the $\Re / \Im$ part for the condition $I=UU^{-1}$ and get the other for free? So are there no power series with real $c_k$, for general $H$? What if I restrict to symmetric $H$? – draks ... Jul 12 '12 at 18:55

Suppose that $H\mapsto U$ and $Hv=\lambda v$. Then for 1) $Uv = \frac{\lambda+i}{\lambda-i}v$ and for 2) $Uv = \exp(i\lambda)v$. Let $Uv=\kappa v$.

For 1) as $\lambda \rightarrow \infty$, $\kappa \rightarrow 1$. However, $\kappa$ never attains the value of $1$ and so the Cayley Transform does not map Hermitian matrices onto the unitary matrices (it is, however, one-to-one).

For 2) $\exp(i(\lambda + 2n\pi))=exp(i\lambda))$. So the matrix exponential is not one-to-one (it is, however, onto).

In the comments it is mentioned that a bijection is desired. We limit ourselves to a class of functions that preserve eigenvectors while mapping real eigenvalues to complex ones of modulus 1. This would include the power series in $H$, as assumed in Nick's answer. Thus, we can simply look for bijective maps from $\mathbb{R}\rightarrow S^1$.

Since $\mathbb{R}$ and $S^1$ are not homeomorphic, there are no bi-continuous bijections between the two.

There are, however, discontinuous bijections between the two sets (see this question for bijection from $(0,1) \rightarrow (0,1]$, which gives us a bijection $\mathbb{R}\rightarrow S^1$. However, in this example, there are countably infinite discontinuities, so it is not a particularly nice bijection).

Without rigor, I would say that none of the power series mappings would be invertible on the basis that the series would need to converge everywhere (or the mapping is ill-defined) and therefore will be continuous everywhere. Since our mapping is defined by the power series, it is analytic, hence we can find a power series for the inverse function by the Lagrange inversion theorem (there is possibly an issue if the derivative at a point is 0, but since the function is invertible it hopefully isn't a problem). Therefore, the inverse is continuous. Therefore, if a power series is a bijection it is bicontinous. This is a contradiction (as $\mathbb{R}$ and $S^1$ are not homeomorphic). Therefore, no power series in $H$ gives a bijection from the Hermitian to unitary matrices.

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+1 Thanks for your thoughts... – draks ... Feb 7 '14 at 11:28