# gradient flow on $SU(n)$

Define the following cost functions $f_1, f_2 :SU(n) \rightarrow \mathbb{R}$ by $f_1(U) = Re \left( \text{Tr}\left(G^{\dagger} U \right) \right)$ and $f_2(U) = \left| \left( \text{Tr}\left(G^{\dagger} U \right) \right) \right|^2$ for some fixed $G \in SU(n)$.

What are the gradients $\nabla f_1, \nabla f_2$ of these functions and how can they be expressed?

• Where is $W$ used? – Robin Goodfellow Jul 11 '15 at 16:25
• Fixed it, thanks – Benjamin Jul 11 '15 at 17:44
• Your first sentence gets an endorsement from the Redundancy Department of Redundancy :-) – Mariano Suárez-Álvarez Jul 11 '15 at 17:49
• Is part of your question what the gradient of a function on $SU(n)$ is? – Mariano Suárez-Álvarez Jul 11 '15 at 17:53

So let's do $f_2$. What you need to do is to first work out the derivative in $M_n$, and then project onto $SU(n)$. That is, we identify $\mathfrak{su}(n)$ with its dual via the inner product $\langle U,V \rangle = \text{Re}\bigl(\text{Tr}(U^\dagger V)\bigr)$. (You need to do this identification if you want to figure out gradient flow, because $U^{-1} \frac{dU}{dt}$ has to be in $\mathfrak{su}(n)$.) Note that I am thinking of $SU(n)$ and $\mathfrak{su}(n)$ as a Lie group or Lie algebra respectively over the real numbers, not the complex numbers.
So if $V \in \mathfrak{su}(n)$, then at $U \in SU(n)$ we have $$f_2(U(I + \epsilon V)) = f_2(U) + \epsilon \langle \nabla f_2,V \rangle + O(\epsilon^2) .$$ Multiplying out, we obtain $$\langle \nabla f_2,V \rangle = 2 \text{Re} \bigl(\overline{\text{Tr}(G^\dagger U)} \, \text{Tr}(G^\dagger U V) \bigr) = 2 \text{Re} \bigl(\text{Tr}(H^\dagger U V) \bigr) ,$$ where $$K = \text{Tr}(G^\dagger U) \, G .$$ So we see that $$\nabla f_2 = 2 U^\dagger K + \Lambda ,$$ where $\Lambda \in \mathfrak{su}(n)^\perp$ is the "Lagrange multiplier" such that $\nabla f_2 \in \mathfrak{su}(n)$. Here $$\mathfrak{su}(n)^\perp = \{\Lambda \in M_n : \text{\langle \Lambda,W\rangle = 0 for all W \in \mathfrak{su}(n)}\} ,$$ After some reflection, you realize that $\mathfrak{su}(n)^\perp$ is the space of Hermitian matrices, and that $\nabla f_2$ is the anti-Hermitian part of $2 U^\dagger K$, that is $$\nabla f_2 = U^\dagger K - K^\dagger U .$$ So the gradient flow is $$\frac{dU}{dt} = U \, \nabla f_2 = K - U K^\dagger U .$$
• $\nabla f_2$ is going to be an element of the dual space of $\mathfrak{su}(n)$. One construction of the dual of $\mathfrak{su}(n)$ is to think of it as a quotient of $M_n$ via the duality $\langle \nabla f_2, V\rangle = \text{Tr}((\nabla f_2)^T V) = \sum_{ij} (\nabla f_2)_{ij} V_{ij}$. – Stephen Montgomery-Smith Jul 12 '15 at 18:21
• Except actually I am looking at the dual of the lie algebra translated by left multiplication of $U$. The whole thing is so much easier if you think of it in terms of matrices than abstract objects. – Stephen Montgomery-Smith Jul 12 '15 at 18:24