# Distance of parametrized skew hermitian exponentials

Consider two skew-adjoint matrices $$A$$ and $$A'$$, i.e. $$A^*=-A$$ and $$A'^*=-A'$$. It is well-known that

$$e^{-tA}$$ and $$e^{-tA'}$$ are unitary operators.

I would like to know:

Is it true that $$\sup_{t \in \mathbb{R}} \Vert e^{-tA}-e^{-tA'} \Vert = 2(1-\delta_{A,A'})?$$

• What is $\delta_{A, A'}$? – Nik Weaver May 25 at 19:45
• the kronecker delta that is 1 if $A=A'$ and zero otherwise. – Kung Yao May 25 at 19:47
• I see, thank you. – Nik Weaver May 25 at 19:50
• So the question is: if two strongly continuous unitary groups are not identical, then must their pointwise distance be 2? – Nik Weaver May 25 at 19:51
• Please don't delete your question after it's been given an answer, this is disrespectful. – YCor May 25 at 20:57

This is not true in general. E.g., let $$A:=\left( \begin{array}{cc} -i & 0 \\ 0 & 0 \\ \end{array} \right),\quad A':=\frac i2\, \left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \\ \end{array} \right).$$ Then, for real $$t$$, $$e^{-tA}=\left( \begin{array}{cc} e^{i t} & 0 \\ 0 & 1 \\ \end{array} \right),\quad e^{-tA'}=\frac12\,\left( \begin{array}{cc} 1+e^{i t} & 1-e^{i t} \\ 1-e^{i t} & 1+e^{i t} \\ \end{array} \right),$$ and $$\|e^{-tA}-e^{-tA'}\|=\sqrt{1-\cos t},$$ so that $$\sup_{t\in\mathbb R}\|e^{-tA}-e^{-tA'}\|=\sqrt2\ne2(1-\delta_{A,A'}).$$