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In differential geometry, tangent vectors are considered operators. At point p, the local tangent space is defined as $$ T_p(M)=\{X^i\partial_i|X\in R^n\} $$ This is quite easy to understand for me.

However, I study information geometry recently and get stuck with the tangent vectors defined on statistical manifold.

As http://en.wikipedia.org/wiki/Information_geometry points out, the tangent vectors defined at point $p_\xi$ are $\partial_ip_\xi$ in mixture representation. I really cannot understand why tangent spaces can be defined like this! It does not make any sense to me. And I cannot digest the explanations on wikipedia.

Can anyone help me understand it? Thanks in advance!

I can understand the previous problem now. Is there anyone who has studied information geometry before. I have a new question.

on http://en.wikipedia.org/wiki/Information_geometry $D[\partial_i\partial_j||\cdot]= D[\cdot||\partial_i\partial_j]=-D[\partial_i||\partial_j]$. I think they should all equal to 0. Here is my reason:

because $D[\partial_j||\cdot]=0$, we have $$ 0=\partial_iD[\partial_j||\cdot]=\partial_iD((\partial_j)_p||p)=\partial_i\partial_jD(p||p)=D((\partial_i\partial_j)_p||p)=D[\partial_i\partial_j||\cdot] $$

But the true result seems to support such equation: $$ \partial_iD((\partial_j)_p||p)=D((\partial_i\partial_j)_p||p)+D((\partial_i)_p||(\partial_j)_p) $$

Why?

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    $\begingroup$ The wikipedia article needs work. Perhaps the book referenced there could help. But I think it is clear that if you have a family of probability distributions, which you think of as functions, and they depend smoothly on a parameter, and you differentiate in that parameter, you get a function. A tangent vector is the derivative of a 1-parameter family of points. $\endgroup$ – Ben McKay Apr 13 '14 at 7:27
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    $\begingroup$ The book Methods of Information Geometry seems to be well written (after a glance) and looks much easier to digest than the wikipedia article. $\endgroup$ – Ben McKay Apr 13 '14 at 7:29
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There are situations where you think of a function space as an infinite-dimensional manifold. Then the tangent space is also infinite-dimensional, and also naturally a function space. The tangent space to a product manifold is the direct sum of the tangent spaces, and a function space is like a (contiuum) product, so its tangent space should be a product over the same index set, i.e. consist of functions on the same domain.

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  • $\begingroup$ Yeah! I get it. $\endgroup$ – user49550 Apr 16 '14 at 7:11

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