What is the relation between "inner product" and "metric"? I'm reading this paper. In the second paragraph of page 53, they write:

We define the inner product of two symmetric matrices $S_1$ and $S_2$ to be $\mathrm{Tr}(S_1S_2)$. The metric in the space of symmetric matrices is $ds^2=\mathrm{Tr}[(dS)^2]$, where $S$ is a symmetric matrix.

I'm trying to understand these two sentences. Based on what I've found out, "inner product" is something which you are free to define as long as it satisfies three conditions (conjugate symmetry, linearity in the first element, and positive-definiteness). The trace of two symmetric matrices satisfies all three, so it's a valid inner product, and the first sentence makes sense. 
That is as far as I got however. I'm guessing that "metric" refers to this definition from Wikipedia, but I don't see where the expression $ds^2=\mathrm{Tr}[(dS)^2]$ came from or how it's related to the inner product. I get the feeling there's something elementary I'm missing here, but I'm too unfamiliar with the jargon to tell what it is. Can someone explain?
 A: (1) If $V$ is a real vector space, then on $V$ one can choose an inner product $\langle \cdot, \cdot \rangle$.  Once an inner product is chosen, one automatically gets a norm on $V$ by defining $\Vert v \Vert^2 = \langle v, v\rangle$.  Consequently, one can define a distance function on $V$ by setting $d(v,w) = \Vert v - w \Vert$.
(2) If $M$ is a differentiable manifold, then its tangent spaces $T_pM$ are vector spaces.  Consequently, we can (smoothly) choose an inner product on each tangent space $T_pM$, and such a choice is called a Riemannian metric on $M$ (even though the choice is really on all the $T_pM$'s).  A Riemannian metric on $M$ gives rise to a distance function on $M$.
(Side note: Once one chooses a Riemannian metric, $M$ is called a Riemannian manifold, but no matter.)
(3) In your case, the set of $3 \times 3$ symmetric matrices --- let's call this set $S^2(\mathbb{R}^3)$ --- can be thought of as both a vector space and a differentiable manifold.
Thinking of $S^2(\mathbb{R}^3)$ as a vector space, the authors choose an inner product for it.  Thinking of $S^2(\mathbb{R}^3)$ as a differentiable manifold, the authors choose a Riemannian metric for it (meaning an inner product for all its tangent spaces).
The authors chose the inner product and the Riemannian metric in such a way that the two distance functions on $S^2(\mathbb{R}^3)$ --- one coming from the inner product, and the other coming from the Riemannian metric --- are the same.
A: You have an obvious chart on the set of symmetric matrices from which you can construct a basis for the tangent space and its dual. In the paper, $dS$ refers to the matrix of differentials and it's easy to see that 
$$
\begin{align*}
ds^2 &= \text{Tr}(dS^2)\\\\
&=\sum_{i=1}^ndx_{ii}^2+2\sum_{i<j}dx_{ij}^2.
\end{align*}
$$
The set of $n\times n$ symmetric matrices can be identified with ${\Bbb R}^N$, with $N=n(n+1)/2$, and so can be identified with its tangent space at any point. Under this identification, the inner product on the matrices agrees with the inner product on the tangent space coming from the metric.
