# Why tangent vector of statistical manifold is a function?

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?

• 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. – Ben McKay Apr 13 '14 at 7:27
• The book Methods of Information Geometry seems to be well written (after a glance) and looks much easier to digest than the wikipedia article. – Ben McKay Apr 13 '14 at 7:29