Prove that the acceleration of $ghg^{-1}h^{-1}$ is $2[X, Y]$.

Question

This problem has been driving me up the wall!

Let $G$ be a matrix group and let $g(t)$ and $h(t)$ be paths in $G$ with $g(0) = I$, $g^\prime(0) = X$, $h(0) = I$, and $h^\prime(0) = Y$. Prove that the group commutator $g(t)h(t)g^{-1}(t)h^{-1}(t)$ is equal to $I + [X, Y]t^2 + \mathcal{O}(t^3)$.

My problem: My calculation keeps showing that the group commutator's acceleration vanishes...

To simplify notation, I will write $g$ instead of $g(t)$.

(PLEASE: Before leaving the correct solution, please tell me where my solution is going wrong..)

First, I take for granted that \begin{equation} \tag{1} (g^{-1})^\prime(0) = -g^\prime(0) \end{equation}
and \begin{equation} \tag{2} \left(ghg^{-1}h^{-1}\right)^\prime(0) = 0. \end{equation} I now show that \begin{equation} \tag{3} (g^{-1})^{\prime \prime}(0) = 2(g^\prime)^2(0) - g^{\prime \prime}(0). \end{equation} $I = g^{-1}g$ implies that $0 = (g^{-1})^\prime g + g^{-1}g^\prime$, so that \begin{equation} 0 = (g^{-1})^{\prime \prime} g + (g^{-1})^\prime g^\prime + (g^{-1})^\prime g^\prime + g^{-1} g^{\prime \prime}. \end{equation} Evaluating at $0$ and rearranging, we obtain the desired result for $(g^{-1})^{\prime \prime}(0)$.

From Eqs. $(1)$ and $(2)$, we conclude that \begin{align} \left( (hg)^{-1} \right)^{\prime \prime}(0) &= \left\{ \left( \left( g^{-1} \right)^\prime h^{-1} + g^{-1} \left( h^{-1} \right)^\prime \right)^\prime \right\} \bigg|_{t=0} \\ &= \left\{ \left( g^{-1} \right)^{\prime \prime} + 2 g^\prime h^\prime + \left( h^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0} \\ &= \left\{ 2 \left( g^\prime \right)^2 - g^{\prime \prime} +2 g^\prime h^\prime + 2 \left( h^\prime \right)^2 - h^{\prime \prime} \right\} \bigg|_{t=0}. \tag{4} \end{align}

Now, for the main result:

Taking the first derivative of the group commutator, we obtain \begin{equation} (ghg^{-1}h^{-1})^\prime = (gh)^\prime (hg)^{-1} + gh \left( (hg)^{-1} \right)^\prime. \end{equation}

Taking the second derivative and evaluating at $0$, we have (using Eq. $(3)$) \begin{align} (ghg^{-1}h^{-1})^{\prime \prime}(0) & = \left\{ (gh)^{\prime \prime}(hg)^{-1} + (gh)^\prime \left( (hg)^{-1} \right)^\prime + (gh)^\prime \left( (hg)^{-1} \right)^\prime + gh \left( (hg)^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0} \\ & = \left\{ (gh)^{\prime \prime} - 2(gh)^\prime (hg)^\prime + \left( (hg)^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0} \\ &= \left\{ g^{\prime \prime} + 2 g^\prime h^\prime + h^{\prime \prime} \right\} \bigg|_{t=0} \\ &- \left\{ 2 \left( g^\prime + h^\prime \right) \left( h^\prime + g^\prime \right) \right\} \bigg|_{t=0} \\ &+ \left\{ 2 \left( g^\prime \right)^2 - g^{\prime \prime} +2 g^\prime h^\prime + 2 \left( h^\prime \right)^2 - h^{\prime \prime} \right\} \bigg|_{t=0} \\ &= 0. \left(\text{Edit as of }8/21/13: \text{This is not equal to } 0, \text{but }2[X,Y]! \right) \end{align}

Obviously, something has gone terribly wrong...

Thanks for your help!

Answer 1

Well, shucks, if non-commutativity won't be the death of me...

The last line of my proof is not equal to $0$, but $2[g^\prime(0), h^\prime(0)] = 2[X,Y]$.

Sorry about that everyone! :) :)

Answer 2

I would do it this way: if $g = I + X t + g''(0) t^2/2 + O(t^3)$ and $h = I + Y t + h''(0) t^2/2 + O(t^3)$, $$g h = I + (X + Y) t + (g''(0)/2 + X Y + h''(0)/2) t^2 + O(t^2)$$ Similarly, $g^{-1} = I - X t + (X^2 - g''(0)/2) t^2 + O(t^3)$ and $h^{-1} = I - Y t + (Y^2 - h''(0)/2) t^2 + O(t^3)$ so $$g^{-1} h^{-1} = I - (X + Y) t + (X^2 - g''(0)/2 + X Y + Y^2 - h''(0)/2) t^2 + O(t^3)$$ Multiply $gh$ and $g^{-1} h^{-1}$: $$ I + \left(\frac{g''(0)}{2} + X Y + \frac{h''(0)}{2} - (X+Y)^2 + X^2 - \frac{g''(0)}{2} + X Y + Y^2 - \frac{h''(0)}{2}\right) t^2 + O(t^3)$$ Now note that the $g''$ and $h''$ terms cancel and $(X+Y)^2 = X^2 + X Y + Y X + Y^2$.