3
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ This is an abuse of terminology.. Sometimes the metric you're seeing is meant to be an inner product or pseudo-inner product. $\endgroup$ – Cameron Williams Jan 15 '20 at 2:36
  • $\begingroup$ @CameronWilliams are you saying the two words are synonyms? This is quickly getting more confusing ... $\endgroup$ – Allure Jan 15 '20 at 2:45
  • 2
    $\begingroup$ An inner product induces a norm $\langle v,v\rangle= \|v\|^2$ The norm satisfies the definitions of a metric (i.e. triangle inequality, etc.). $\endgroup$ – Doug M Jan 15 '20 at 3:13
  • $\begingroup$ @Allure A metric at a point $p$ in a manifold is a $(0,2)$ tensor and so is an inner product on the tangent space at $p$. The inner product they've defined is on the manifold and so isn't a metric. $\endgroup$ – Oliver Jones Jan 15 '20 at 6:57
  • 1
    $\begingroup$ There are two different meanings of the word "metric" in mathematics. One definition of "metric" (sometimes called a "distance function") is the one in your Wikipedia link. The other definition is that of a "Riemannian metric" (which is sometimes just called a "metric" for short). By "metric," this paper means "Riemannian metric." The four concepts of "distance function," "inner product," "norm," and "Riemannian metric" are related to one another, with some of these relationships being more elementary than others. $\endgroup$ – Jesse Madnick Jan 15 '20 at 7:41
4
$\begingroup$

(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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ One day I'll look back at your answer and think "oh yeah, it's easy to see!". Unfortunately, today is not that day xD $\endgroup$ – Allure Jan 16 '20 at 5:51
  • $\begingroup$ @Allure It sounds like you need to learn the fundamentals of manifold theory. $\endgroup$ – Oliver Jones Jan 16 '20 at 5:54
  • $\begingroup$ If I have to (fortunately I think I understand this particular proof and might not need more), at least I'll know what the name of the theory is! One question: what's $x$ in your answer? The elements of the symmetric matrix? $\endgroup$ – Allure Jan 16 '20 at 6:05
  • $\begingroup$ @Allure Yes, the $x_{i,j}$'s are the entries of the matrix. They give us the chart from which the $dx_{i,j}$'s are defined. $\endgroup$ – Oliver Jones Jan 16 '20 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.