Question
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**Is [Bregman divergence](https://en.wikipedia.org/wiki/Bregman_divergence) free of coordinates?**


Although it is invariant w.r.t. which local affine coordinate you take, is it possible to prove that it does not change w.r.t. an arbitrary change of coordinates?


(I am reading "Information Geometry and Its Applications" (Amari, 2016).)



Is Bregman divergence independent of coordinates?
No. Away from a critical point, any function $\psi$ becomes linear in some coordinates. In such coordinates, the Bregman divergence $D\_{\psi}$ of that function $\psi$, as defined in your book (p. 14 equation (1.44)), vanishes. So the Bregman divergence of a function, as defined in your book, is not coordinate invariant. It is, however, invariant under affine changes of coordinates. 


On the other hand, consider the Bregman divergence to be defined not using affine coordinates and some function (as it is in your book), but instead to be defined as using a flat affine connection $\nabla$ with trivial monodromy and some differentiable function $\psi$, by
$$D\_{\psi,\nabla}(\xi,\eta)=\psi(\xi)-\psi(\eta)-d\psi(\xi)v$$
where $\exp\_{\xi}v=\eta$ for the exponential map $\exp$ of $\nabla$.
This Bregman divergence is coordinate independent, but dependent on not just the function: it depends also on the choice of the flat affine connection $\nabla$. 


The dual connection is defined, in that book, using the Legendre transformation applied to $\psi$, so depends on $\psi$. Also, the Legendre transformation depends on the affine structure, which depends on the choice of the flat affine connection $\nabla$. So it doesn't make sense to pose a set of dual connections, without a function $\psi$. The notion of duality is not defined independently of $\psi$. But once you have $\psi$ and $\nabla$, you have the Bregman divergence $D\_{\psi,\nabla}$ defined as above.


