I'm trying to calculate the gradient (wrt to the bi-invariant metric) of the following functions $F\_1, F\_2 : SU(n) \rightarrow \mathbb{R}$ defined by $F\_1(U) = | Tr (G^{\dagger} U) |^2$, $F\_2(U) = \Re (Tr (G^{\dagger} U)) $ where $G \in SU(n)$ is given.


Can there be a coordinate free expression for $\nabla F$?


(Posted here after no answer on MSE).



Gradient on $SU(n)$
Here is a calculation for the derivative $DF$ (or the differential $dF$), which does not depend on the metric.


Define $f:SU(n)\to\mathbb C$ by $f(U)=Tr(G^\dagger U)$.
If we think of $f$ being defined on all of $\mathbb C^{n\times n}$, then it is linear and $Df(U)(H)=f(H)$ for any matrix $H$.
If we restrict our attention to $SU(n)$, we need to choose $H$ from the tangent space at $U$.
If we identify this with the Lie algebra $\mathfrak{su}(n)$ by left multiplication ($H\in\mathfrak{su}(n)$ is identified with $UH\in U\mathfrak{su}(n)=T\_USU(n)$), we get the derivative $Df(U):\mathfrak{su}(n)\to\mathbb C$ defined by $Df(U)H=f(UH)=Tr(G^\dagger UH)$.


The function $F\_2$ is just the real part of $f$, so its derivative is the real part of the derivative: $DF\_2(U)=\Re(Tr(G^\dagger U \cdot))$.
The function $F\_1$ is $f$ composed with $\phi:\mathbb C\to\mathbb C$, given by $\phi(z)=|z|^2$.
Since $D\phi(z)w=2\Re(\bar wz)$, we get
$$
DF\_1(U)(H)
=
D\phi(f(U))Df(U)H
=
2\Re(\overline{Tr(G^\dagger U)}Tr(G^\dagger UH)).
$$
In other words, the derivative is
$$
DF\_1(U)=2\Re(\overline{Tr(G^\dagger U)}Tr(G^\dagger U\cdot)).
$$


