TITLE: Diameter of $\mathrm{SU}(2)$ endowed with a left-invariant metric
QUESTION [9 upvotes]: 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
$$
\mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)})=\frac{\pi}{\max\{a,b,c\}}.
$$

REPLY [2 votes]: 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
$\mathrm{diam}(g_{(a,b,c)})\geq \lim_{t\to\infty} \mathrm{diam}(g_{(t_1,b,b)})={\pi}/{2b}$.