8
$\begingroup$

Basic question: What is the diameter of $\mathrm{SU}(2)$ endowed with a left-invariant metric?

Now, let me give more information. Set $$ X_1= \begin{pmatrix} i &\\ &-i \end{pmatrix},\; X_2= \begin{pmatrix} &1\\ -1& \end{pmatrix},\; X_3= \begin{pmatrix} &i\\ i& \end{pmatrix}. $$ It is sufficient to consider the left-invariant metrics with inner product on $\mathfrak{su}(2)$ satisfying that $\{aX_1,bX_2,cX_3\}$ is an orthonormal basis, where $a,b,c$ are positive real numbers. Indeed, any permutation of these numbers will return an isometric metric. Let us denote by $g_{(a,b,c)}$ the left-invariant metric (and also the inner product on $\mathfrak{su}(2)$) mentioned above. The goal is to write an expression for $\mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)})$ in terms of $a$, $b$ and $c$.

It is well known that $\mathrm{SU}(2)$ is diffeomorphic to the $3$-dimensional sphere. For example, a map is given by $$ \begin{pmatrix} u&-\bar v \\ v&\bar u \end{pmatrix} \mapsto \begin{pmatrix} u\\ v \end{pmatrix}, $$ for $u,v\in\mathbb{C}$ satisfying that $|u|^2+|v|^2=1$. Since the Riemannian manifolds at hand are homogeneous, the diamenter is attained at a pair of points $(I_2,h)$, where $I_2$ denotes the $2\times 2$ identity matrix and $h$ is in $\mathrm{SU}(2)$. Since $I_2$ and $-I_2$ are opposite points in $S^3$, the next question seems to be affirmative.

Question 1: $\mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)}) = \mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2)$?

Any idea to prove it (in case is affirmative)? Let's move on, assuming this is true. The geodesics on $(\mathrm{SU}(2), g_{(a,b,c)})$ starting from $I_2$ are not in general one-parameter subgroups (see Robert Bryant's comment in Tsemo Aristide's answer). Geodesics satisfy the Euler equation (see again Bryant's comment or this Bryant's answer). Let $\dot\gamma (t) = d L_{\gamma(t)}(X_{\gamma}(t))$, and for $X\in\mathfrak{su}(2)$, set $\textrm{ad}_{g_{(a,b,c)}}^*(X):\mathfrak{su}(2) \to\mathfrak{su}(2)$ defined by $$ g_{(a,b,c)}(\textrm{ad}_{g_{(a,b,c)}}^*(X)Y,Z)= g_{(a,b,c)}(Y,[X,Z]). $$ The Euler equation is $$ \dot X_{\gamma}(t) = \textrm{ad}_{g_{(a,b,c)}}^*(X_\gamma(t)) X_{\gamma}(t). $$ For $X_\gamma(t)= a_1(t)X_1+a_2(t)X_2+a_3(t)X_3$, it reduces to $$ \begin{array}{rcl} \dot a_1(t) &=&2a^2\, a_2(t)a_3(t) (\frac{1}{b^2}-\frac{1}{c^2}),\\[1mm] \dot a_2(t) &=&2b^2\, a_1(t)a_3(t) (\frac{1}{c^2}-\frac{1}{a^2}),\\[1mm] \dot a_3(t) &=&2c^2\, a_1(t)a_2(t) (\frac{1}{a^2}-\frac{1}{b^2}). \end{array} $$

Example 1: When $a=b=c$ (round metric), it follows immediately that $a_j(t)\equiv a_j(0)$ is constant for all $j$ and then $\gamma(t)=\exp(tX_\gamma(0))$ is an one-parameter subgroup. Moreover, one can check (as expected) that $\mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) = \pi/a$ since every geodesic from $I_2$ to $-I_2$ has this length.

We go back to the arbitrary case. Although in general not every geodesic is an one-parameter subgroup, it is for some particular cases (remember Bryant's answer). Take the initial velocity vector as $X_1$, thus $a_1(0)=1$ and $a_2(0)=a_3(0)=0$. Euler equation immediately implies that $\dot a_j(0)=0$ for all $j$, thus again $X_\gamma(t)=X_1$ for all $t$ and $\gamma(t)=\exp(tX_1)$. Consequently, since $\gamma(\pi)=-I_2$, then the length of $\gamma$ on $[0,\pi]$ is $$ \int_0^\pi g_{(a,b,c)}(X_\gamma(t),X_\gamma(t))^{1/2} dt = \pi\, g_{(a,b,c)}(X_1,X_1)^{1/2} = \frac{\pi}{a}. $$ Similarly, taking $X_2$ or $X_3$ as initial velocity vector, we obtain that the corresponding geodesics from $I_2$ to $-I_2$ have length $\pi/b$ and $\pi/c$ respectively. Hence $$ \mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) \leq \frac{\pi}{\max\{a,b,c\}}. $$

Question 2: $\mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) = {\pi}/{\max\{a,b,c\}}$?

In case Questions 1 and 2 are affirmative, then we conclude that $$ \mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)})=\frac{\pi}{\max\{a,b,c\}}. $$

$\endgroup$
7
$\begingroup$

Let's suppose without loss of generality that $a \ge b \ge c$. In a recent preprint "Left-invariant geometries on $\mathrm{SU}(2)$ are uniformly doubling" (arXiv:1708.03021) with Maria Gordina and Laurent Saloff-Coste, we show that (in your notation) the diameter of the left-invariant metric $g_{(a,b,c)}$ is comparable to $1/b$, uniformly over all left-invariant metrics. See Proposition 7.1 and note that our parameters $a_1, a_2, a_3$ are the reciprocals of your $a,b,c$ (up to some factors of 2). So in fact it is the middle coefficient which controls the diameter, not the largest one.

Consider what happens as $a \to \infty$ and the $X_1$ direction becomes very "cheap". It is true that the distance between $I_2$ and $-I_2$ becomes very small, as you say. But the diameter of this metric is not small. It "collapses" to the two-dimensional quotient of $\mathrm{SU}(2)$ by a circle, but not to a point. This is the content of our Lemma 5.4. So the answer to your Q1 is negative.

I think the answer to your Q2 is yes, certainly at least up to a constant, but I'd have to think about how to make sure.

If $a,b$ both become very large, then the manifold does collapse to a point, even if $c$ remains moderate. This is because of the relation $[X_1, X_2] = 2X_3$, which allows you to "replicate" motion in the $X_3$ direction by making a loop in the $X_1, X_2$ "plane" (at cost $1/(ab)$ instead of $1/c$), so if they are both cheap then motion in $X_3$ becomes cheap as well.

On the other hand, if $c$ becomes small while $a,b$ remain of moderate size, the manifold's diameter does not blow up, again because of $[X_1, X_2]$. Instead, it becomes a compact non-degenerate sub-Riemannian manifold, whose diameter is still finite. If $b,c$ both become small then the diameter does blow up; you can't replicate $X_2, X_3$ with $X_1$ alone, since it commutes with itself.

$\endgroup$
  • $\begingroup$ Nice answer! Do you think it is possible to give a closed explicit expression for the diameter in term of $a$, $b$ and $c$? $\endgroup$ – emiliocba Sep 19 '17 at 17:49
  • 1
    $\begingroup$ @emiliocba: I don't know of such an expression, nor any particularly promising approaches to obtaining one. Sorry. $\endgroup$ – Nate Eldredge Sep 19 '17 at 17:54
  • $\begingroup$ @emiliocba: I am sure such an expresson is possible, since the corresponding dynamical system - the ``free rigid body'' is integrable -(in terms of elliptic function). $\endgroup$ – Richard Montgomery Sep 28 '17 at 16:38
5
$\begingroup$

Write $A = 1/a, B = 1/b, C = 1/c$ so that, in the problem solver's notation we have, for example, $<X_1, X_1 > = A^2$, and the metric is
$$ds^2 _{a,b,c} = A^2 \sigma_1 ^2 + B^2 \sigma_2 ^2 + C^2 \sigma_3 ^2, $$ the $\sigma_i$ forming the basis for $Lie(SU(2))^*$ dual to the $X_i$. Write $diam(A,B,C)$ for the diameter you are interested in. This function is positive, homogeneous of degree 1 and invariant under permutations. Your ordering convention becomes $A \le B \le C$. Here is what I can show :

(1) $\pi A \le diam(A,B,C) \le \pi B$

(2) $diam(A,A,C) = \pi A$.

Proof of (1) assuming (2). Use monotonicity:
$$A \le A' , B \le B' , C \le C' \Longrightarrow diam(A, B,C) \le diam(A', B', C').$$ This monontonicity inequality follows directly from the same inequalities for the corresponding Riemannian metrics, viewed as a quadratic function on the tangent bundle. Then $diam(A,A, C) \le diam(A,B,C) \le diam(B,B, C)$. Now use (2).

Proof of (2). As the problem poser stated, $diam(A,A,A) = \pi A$. By monotonicity $diam(A,A,A) \le diam (A,A, C) \le diam(A, A, \infty) := lim_{C \to \infty} diam(A,A, C)$. So, it suffices to prove $$diam(A,A, \infty) = \pi A.$$

The trick to computing this limit is to realize that the limiting geometric object is a subRiemannian [sR] metric: a distribution on the tangent space equiped with a metric on the distribution. The distribution is the span of $X_1, X_2$ -- the infinite coefficient for $C$ forbids finite length paths from travelling transverse to the distribution. This sR metric happens to be my personal favorite, the one associated to the Hopf fibration $S^3 \to S^2$. In my book, `A Tour of subRiemannian Geometry'', p. 154, example 11.3.1, I compute all the geodesics for this sR metric in the case $A = 1$ and the diameter computation follows from this understanding. (Theorem 11.8 p. 153 is the idea which drives the computations.)

Here goes. The Hopf fibration is a principal circle bundle and this distribution form the horizontal space for the canonical (eg $U(2)$-invariant) connection on that bundle. So we call curves tangent to the distribution ``horizontal''. The Hopf projection maps horizontal curves to curves on $S^2$ in a way that preserves their length, provided the metric on the two-sphere is the one it receives as a sphere of radius $R = 1/2$ in Euclidean 3-space. The sR geodesics are precisely the horizontal lifts of circles -- great or small -- on this two-sphere. By rotational symmetry, looking at all geodesics passing through the identity, so all circles passing through, say, the North pole, we see that every sR geodesic has a conjugate point by the time its associated circle closes up (the sR geodesic returns to its initial fiber). From this it follows that the longest minimizing geodesics possible are realized by great circles going once around the equator. These have lengths $2 \pi R = \pi$. Their associated holonomy is $\pi$, meaning that their horizontal lift connects $I$ to $-I$, as the problem proposer suggested. All this shows that $diam(1,1, \infty) = \pi$. Scaling (homogeneity) now yields $diam(A,A, \infty) = \pi A$. QED

End note. I am confident that I can also compute $diam(A,C,C)$. Surprisingly, this turns out to be more difficult than $diam(A,A,C)$. I will report back if I get something pleasing.

$\endgroup$
  • $\begingroup$ Thanks for the answer! I had found a similar solution with more or less the same ideas as in your answer, which is again similar to the proof of Prop. 7.1 mentioned by Nate Eldredge. I am sorry for the delay in communicating here... The inequalities are (in your notation): $\pi B < diam(A,B,C)\leq 2\pi B $, as Nate Eldredge predicted. It follows from the inequalities for the diameter plus the explicit calculation of them when two parameters are equal (i.e., a Berger's sphere). This can be obtained from [Sakai, Cut loci of Berger's spheres] (see also arxiv.org/abs/1504.05472v3). $\endgroup$ – emiliocba Sep 29 '17 at 12:38
  • 1
    $\begingroup$ Is there some error in my upper bound of $diam(A,B,C) \le \pi B$ in my (1)? $\endgroup$ – Richard Montgomery Sep 30 '17 at 19:59
  • $\begingroup$ Sorry, I was missing the factor $2$ in both sides. Thank you since I figured out because of your comment. I have just posted my full answer. $\endgroup$ – emiliocba Oct 2 '17 at 12:20
  • $\begingroup$ Wonderful!! Nice formula. Thanks. $\endgroup$ – Richard Montgomery Oct 3 '17 at 16:18
2
$\begingroup$

Here I propose the following answer to my question. It uses the same ideas as in the two previous answers by Nate Eldredge and Richard Montgomery. Comments are welcome.

The basic question is not known. So, it would be useful to have some bounds for $\mathrm{diam}(\mathrm{SU}(2),g_{(a,b,c)})$ in terms of $a$, $b$ and $c$. From now on we assume $a\geq b\geq c>0$. We will prove (I will avoid writing $\mathrm{SU}(2)$ in the diameter function.)

$$\frac{\pi}{2b}\leq \mathrm{diam}(g_{(a,b,c)})\leq \frac{\pi}{b}.$$

Moreover, there will be a more precise lower bound in terms of $a$ and $b$. Nate Eldredge had predicted that $\mathrm{diam}(g_{(a,b,c)})$ and $b$ are comparable.

We have that $$ g_{(t,b,b)}(X,X)\leq g_{(a,b,b)}(X,X)\leq g_{(a,b,c)}(X,X)\leq g_{(b,b,c)} (X,X) $$ for all left-invariant vector fields $X$, if $t\geq a$. Hence $$ \mathrm{diam}(g_{(t,b,b)})\leq \mathrm{diam}(g_{(a,b,b)})\leq \mathrm{diam}(g_{(a,b,c)})\leq \mathrm{diam}(g_{(b,b,c)}). $$

It is well known that the diameter is equal to the maximum distance between a fixed point and the points in the corresponding cut locus. The cut locus of any Berger's sphere (i.e.\ $(\mathrm{SU}(2),g_{(a,b,c)})$ with $a=b$ or $b=c$) is computed. Sakai did it for $a=b\geq c$, and recently Podobryaev-Sachkov in PS gave the full answer.

Although the diameter of a Berger's sphere is not stated in PS, the following expression can be obtained by the same procedure used in PS to obtain the diameter of $\mathrm{SO}(3)$ endowed with a left-invariant metric (in its notation, the multisets $\{\!\{I_1,I_2,I_3\}\!\}$ and $\{\!\{1/(2a)^2,1/(2b)^2,1/(2c)^2\}\!\}$ coincide): $$ \textrm{diam}(\textrm{SU}(2),g_{(a,b,c)}) = \begin{cases} \pi/b &\quad\text{if } a=b\geq c,\\ \pi/a &\quad\text{if } a>b=c\geq a/\sqrt{2},\\ \dfrac{\pi}{2b\sqrt{1-b^2/a^2}} &\quad\text{if } a>\sqrt{2}b=\sqrt{2}c. \end{cases} $$ This expression was kindly communicated by the first named author of PS.

It follows immediately that $\mathrm{diam}(g_{(a,b,c)})\leq {\pi}/{b}$ as Richard Montgomery asserted. Furthermore, $$ \mathrm{diam}(g_{(a,b,c)})\geq \mathrm{diam}(g_{(a,b,b)}) = \begin{cases} \pi/a &\quad\text{if } a\leq \sqrt{2}b,\\ \dfrac{\pi}{2b\sqrt{1-b^2/a^2}} &\quad\text{if } a>\sqrt{2}b. \end{cases} $$ Since the above function is decreasing, then $\mathrm{diam}(g_{(a,b,c)})\geq \lim_{t\to\infty} \mathrm{diam}(g_{(t_1,b,b)})={\pi}/{2b}$.

$\endgroup$

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.