6
$\begingroup$

Edit: 28 January 2023 I just realized that this metric is frequently used in this paper https://hal.science/hal-01382281/document

Let $$M=\{(x_1,x_2,\ldots,x_n)\in \mathbb{R}^n\mid x_i>0,\;i=1,2,\ldots,n\}$$ For $X=(x_1,x_2,\ldots,x_n)\in M$ put $|X|=\sum_{i=1}^n x_i$.

We consider the Shahshahani Riemannian metric $g$ on $M$ with diagonal tensor metric $g_{ii}=\frac{|X|}{x_i}$.

What is the dimension and the precise structure of the group of all isometries of $(M,g)$?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Are you asking about global isometries rather than infinitesimal isometries? In the former case, it's just the symmetric group on $n$-letters acting as permutations of the $x_i$ (when $n>1$). The Lie algebra of Killing vector fields, though, has dimension $\tfrac12n(n{-}1)$ (when $n>1$) and is isomorphic to the Lie algebra of $\mathrm{SO}(n)$. $\endgroup$
    – Robert Bryant
    Commented Jun 13, 2018 at 13:53
  • $\begingroup$ @RobertBryant Yes I was asking for global isometries however your answer about Lie algebra of Killing vector field is very helpful. I try to understand the details of both part. Thanks for this very helpfull comment. $\endgroup$
    – Ali Taghavi
    Commented Jun 13, 2018 at 19:14
  • $\begingroup$ @RobertBryant I can not understand why Killing vector fields is isometric to the Lie algebra of $SO(n)$. For $n=2$, what is a precise Killing vector field for that metric? This may help me to find all Killing vector fields. $\endgroup$
    – Ali Taghavi
    Commented Jun 13, 2018 at 20:50
  • 1
    $\begingroup$ Actually, what I said above in the case $n=2$ is not correct because, in the case $n=2$, there are more Killing fields. (See the note in my answer below.) However, when $n>2$ there are no more Killing fields than those generated by the obvious $\mathrm{SO}(n)$-action. $\endgroup$
    – Robert Bryant
    Commented Jun 14, 2018 at 10:24

1 Answer 1

Reset to default
6
$\begingroup$

The point is that, if you set $x_i = {u_i}^2$ where $u_i>0$, this becomes a diffeomorphism of $M$ with itself with the property that, in the $u$-coordinates, the Shahshahani metric becomes $$ g = 4({u_1}^2+\cdots+{u_n}^2)\bigl({\mathrm{d}u_1}^2+\cdots + {\mathrm{d}u_n}^2\bigr). $$ Clearly, this metric is just the flat metric in the $u$-coordinates on the positive orthant times the squared '$u$-distance' from the origin. When $n\ge2$, any rotation in the $u$-coordinates would preserve the metric on the entire $\mathbb{R}^n$, but it wouldn't preserve the positive orthant, which is $M$.

When $n=2$, this is a flat metric: Set $z = u_1 {+} i\,u_2$. then $g = 4 z\bar z\, \mathrm{d}z\circ\mathrm{d}\bar z =\mathrm{d}w\circ\mathrm{d}\bar w$ where $w = z^2 = (u_1 + i\,u_2)^2 = ({u_1}^2{-}{u_2}^2)+ i\,(2u_1u_2)$. Thus, $w:M\to\mathbb{C}$ isometrically embeds $(M,g)$ as the upper half-plane in $\mathbb{C}$ when $\mathbb{C}$ is given its standard metric. The global isometries are the translations by a real number in the $w$-coordinate together with reflection in the imaginary axis. Meanwhile, the Lie algebra of Killing fields has dimension $3$ instead of $1$.

However, when $n>2$, the metric is not flat, and the Killing fields are the infinitesimal generators of the obvious $\mathrm{SO}(n)$-action, as can be seen by direct calculation or conversion to 'polar coordinates'. Meanwhile, globally on $M$, you only have the rotations and reflections (in the $u$-coordinates) that preserve the positive orthant, and this is just the permutations in the $u_i$, which is the permutations in the $x_i$.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I wanted to write something similar, but I was missing one argument. You write, correctly, that any rotation preserves the metric. But the question remains: are there additional transformations that preserve the metric? Without the factor $(u_1^2 + \cdots + u_n^2)$, in the Euclidean case, the obvious answer would be no. $\endgroup$
    – jarauh
    Commented Jun 14, 2018 at 9:24
  • 1
    $\begingroup$ @jarauh: I have added a note to address your concerns. You are right that the case $n=2$ is special. $\endgroup$
    – Robert Bryant
    Commented Jun 14, 2018 at 10:23
  • $\begingroup$ @RobertBryant Thank you for extending your comment to this very helpful answer. $\endgroup$
    – Ali Taghavi
    Commented Jun 15, 2018 at 8:44
  • $\begingroup$ @RobertBryant Is there a uniform upper bound for the cardinality of the isometric group of a Riemannian metric on the plane whose isometric group is a finite group. According to your answer the Shahshani metric, which can be counted as a metric on the whole plane(after a diffeomorphism), has an order 2 isometric group. Now, can one imagine metrics whith arbitrary large order of finite isometry group? $\endgroup$
    – Ali Taghavi
    Commented Jun 20, 2018 at 7:31
  • 1
    $\begingroup$ @AliTaghavi: Yes, there are metrics with arbitrarily large finite group of isometries. For example, for any $n\ge 1$ you can have a metric on $\mathbb{R}^2$ whose only isometries are the rotations by $2k\pi/n$ about the origin, a cyclic group of order $n$. $\endgroup$
    – Robert Bryant
    Commented Jun 20, 2018 at 13:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .