- #1
Kreizhn
- 743
- 1
Hopefully this is a simple enough question.
Let (M,g) be a matrix Riemannian manifold and [itex] f: M \to \mathbb R[/itex] a smooth function. Take [itex] p \in M [/itex] and let [itex] \{ X_1,\ldots, X_n \} [/itex] be a local orthonormal frame for a neighbourhood of p. We can define a gradient of f in a neighbourhood of p as
[tex] \nabla f = \sum_{i=1}^n X_i f X_i [/tex]
Now in my situation, I'm actually trying to compute this is a local coordinate system. Let [itex] \{ x^{jk} \} [/itex] be a local coordinate system and write [itex] X_i = a_i^{jk} \frac{\partial}{\partial x^{jk}} [/itex]. Then our gradient, evaluated at (for simplicity) p becomes
[tex] \begin{align*}
(\nabla f)_p &= \sum_{i=1}^n \underbrace{\left( a_i^{jk} \left.\frac{\partial}{\partial x^{jk}}\right|_p \right)}_{X_i} f \underbrace{\left(a_i^{mn} \left.\frac{\partial}{\partial x^{jk}}\right|_p\right)}_{X_i} \\
&= \sum_{i=1}^n a_i^{jk} a_i^{mn} \frac{\partial f}{\partial x^{jk}}(p) \left.\frac{\partial}{\partial x^{jk}}\right|_p
\end{align*}
[/tex]
Now we know that [itex] (X_i f)(p) = X_i(p) f \in \mathbb R [/itex] if we use the derivation definition of tangent vectors. But this term simply corresponds to
[tex] a_i^{jk} \frac{\partial f}{\partial x^{jk}}(p) \in \mathbb R[/tex]
Now here is my problem. Let's take [itex] M = U(N) [/itex] the unitary group of dimension N, and [itex] f(p) = \text{tr}[y^\dagger p]\text{tr}[p^\dagger y ] [/itex]
for some [itex] y \in U(N)[/itex]. We know that [itex] f(p) \in \mathbb R [/itex] since [itex] \langle y,p \rangle = \text{tr}[y^\dagger p ] [/itex] is just the Hilbert-Schmidt inner-product, so f(p) is nothing more that [itex] f(p) = |\langle y,p \rangle|^2. [/itex]. But when I compute the terms
[tex] a_i^{jk} \frac{\partial f}{\partial x^{jk}}(p) [/tex]
I get values in [itex] \mathbb C [/itex]. In fact, I can't see any reason why this HAS to be in [itex] \mathbb R [/itex]. Did I just make a silly mistake?
Let (M,g) be a matrix Riemannian manifold and [itex] f: M \to \mathbb R[/itex] a smooth function. Take [itex] p \in M [/itex] and let [itex] \{ X_1,\ldots, X_n \} [/itex] be a local orthonormal frame for a neighbourhood of p. We can define a gradient of f in a neighbourhood of p as
[tex] \nabla f = \sum_{i=1}^n X_i f X_i [/tex]
Now in my situation, I'm actually trying to compute this is a local coordinate system. Let [itex] \{ x^{jk} \} [/itex] be a local coordinate system and write [itex] X_i = a_i^{jk} \frac{\partial}{\partial x^{jk}} [/itex]. Then our gradient, evaluated at (for simplicity) p becomes
[tex] \begin{align*}
(\nabla f)_p &= \sum_{i=1}^n \underbrace{\left( a_i^{jk} \left.\frac{\partial}{\partial x^{jk}}\right|_p \right)}_{X_i} f \underbrace{\left(a_i^{mn} \left.\frac{\partial}{\partial x^{jk}}\right|_p\right)}_{X_i} \\
&= \sum_{i=1}^n a_i^{jk} a_i^{mn} \frac{\partial f}{\partial x^{jk}}(p) \left.\frac{\partial}{\partial x^{jk}}\right|_p
\end{align*}
[/tex]
Now we know that [itex] (X_i f)(p) = X_i(p) f \in \mathbb R [/itex] if we use the derivation definition of tangent vectors. But this term simply corresponds to
[tex] a_i^{jk} \frac{\partial f}{\partial x^{jk}}(p) \in \mathbb R[/tex]
Now here is my problem. Let's take [itex] M = U(N) [/itex] the unitary group of dimension N, and [itex] f(p) = \text{tr}[y^\dagger p]\text{tr}[p^\dagger y ] [/itex]
for some [itex] y \in U(N)[/itex]. We know that [itex] f(p) \in \mathbb R [/itex] since [itex] \langle y,p \rangle = \text{tr}[y^\dagger p ] [/itex] is just the Hilbert-Schmidt inner-product, so f(p) is nothing more that [itex] f(p) = |\langle y,p \rangle|^2. [/itex]. But when I compute the terms
[tex] a_i^{jk} \frac{\partial f}{\partial x^{jk}}(p) [/tex]
I get values in [itex] \mathbb C [/itex]. In fact, I can't see any reason why this HAS to be in [itex] \mathbb R [/itex]. Did I just make a silly mistake?