Understand Riemannian cross-derivative on product manifolds

Question

Suppose we have a smooth function $f:\mathcal{M}\times\mathcal{N}\rightarrow\mathbb{R}$ where the domain is a product of two Riemannian manifolds. The Riemannian cross-derivative ([1], section 2) is defined as: $$ \mathrm{grad}_{xy}^2:= \mathrm{D}_x\mathrm{grad}_y f(x,y): \mathrm{T}_x\mathcal{M}\rightarrow \mathrm{T}_y\mathcal{N} $$

I'm trying to understand this operator. I have two specific questions:

First, to me, if we fix $y$ then $\mathrm{grad}_y f(x,y):\mathcal{M}\rightarrow\mathrm{T}\mathcal{N}$, and the differential on $x$ would result in an operator $\mathrm{D}_x\mathrm{grad}_y f(x,y):\mathrm{T}\mathcal{M}\rightarrow\mathrm{T}\mathrm{T}\mathcal{N}$, which seems not the same as the definition in the paper. Is there any point I missed here?

Second, I'm trying to understand it similar to its Euclidean counterparts. Under what curcumstances do we know that $\mathrm{grad}_{xy}^2$ and $\mathrm{grad}_{yx}^2$ are adjoints? That's to say, do we have: $$ \langle \eta,\mathrm{grad}_{xy}^2(\xi) \rangle_{y} = \langle\mathrm{grad}_{yx}^2(\eta),\xi \rangle_{x}, \forall \xi\in \mathrm{T}_{x}\mathcal{M}\text{ and }\forall \eta\in \mathrm{T}_{y}\mathcal{N} $$

Any comments or references are appreciated, thanks!

References:

[1] Han, Andi, et al. "Riemannian Hamiltonian methods for min-max optimization on manifolds." SIAM Journal on Optimization 33.3 (2023): 1797-1827.

Answer 1

1

If you fix $y$, the function $x \mapsto \mathrm{grad}_y f(x,y)$ is a function mapping $M \to T_y N$; note that the codomain is a single, fixed vector space. So the differential in $x$ is a mapping $TM \to T(T_yN)$, but as $T_yN$ is a linear space its tangent space is canonically isomorphic to itself.

2

They should always be adjoints. You have, extending $\eta$ to a constant map $M \to T_yN$, $$ \langle \eta, \mathrm{grad}_{xy}^2(\xi)\rangle_y = \xi( \langle \eta, \mathrm{grad}_y f(x,y) \rangle_y) = \xi(\eta(f)) $$ So: given a vector $\eta\in T_yN$ and a vector $\xi$ in $T_xM$, extend them to a vector field $\eta$ on $N$ and a vector field $\xi$ on $M$, and then extend them trivially to vector fields on $\tilde{\eta}(x,y) = (0,\eta(y))$ and $\tilde{\xi}(x,y) = (\xi(x),0)$ on $M\times N$. Then our computation above shows that the difference between the two expression you are asking about is exactly $[\tilde{\xi},\tilde{\eta}]f$, but it is easy to check that this Lie Bracket vanishes.