How is Manopt's gradient descent on manifolds connected to Amari's natural gradient?

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Oleg Kachan

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Mar 13, 2021, 5:04:03 AMMar 13
to Manopt
Dear all,

As far as I understand, Amari's natural gradient learning rule converts the Euclidean gradient \nabla to Riemannian one \tilde{\nabla} by multiplying it by the inverse of the manifold's metric tensor matrix G:

w^{(k+1)} = w^{(k)} - \alpha \tilde{\nabla} L,
w^{(k+1)} = w^{(k)} - \alpha G^{-1} \nabla L.

How that connects to "make a step in the tangent space-retract to the manifold" way described in Edelman, Absil, and Boumal? Could someone explain?

Also reading Nicolas's book I understand the purpose of retraction, but now catching the "conversion" egrad2rgrad, and it seems is exactly what Amari's natural gradient does but without (maybe explicit) retraction.

Thanks.

Nicolas Boumal

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Mar 14, 2021, 4:14:55 AMMar 14
to Manopt
Hello,

In Amari's setting, the manifold (that is, the set over which we optimize) is a linear space. For example, it is just R^n. That manifold is turned into a Riemannian manifold by defining a Riemannian metric (an inner product which can vary as a function of the point x on the manifold). That metric is described by a positive definite matrix G(x).

Since Amari's manifold is linear, we don't really need a retraction: we can move away from x along any chosen direction v, and x+tv will remain on the manifold. Contrast this to the situation where the manifold is nonlinear (e.g., a sphere, and x is a point on the sphere and v is a tangent vector to the sphere at x).

This should clarify why there are retractions in one setting and not in the other.

Now, for the G(x)^{-1} in the expression for the gradient: that is exactly the same in Amari's setting and in the more general Riemannian optimization setting. (It takes a bit of effort to see it, but it's  the same in the end.)

Best,
Nicolas
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