
> 
> **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](https://mathoverflow.net/a/267427/20052)).
Geodesics satisfy the Euler equation (see again [Bryant's comment](https://mathoverflow.net/a/267427/20052) or this [Bryant's answer](https://mathoverflow.net/a/177758/20052)).
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](https://mathoverflow.net/a/177758/20052)). 
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
$$
\mathrm{diam}(\mathrm{SU}(2), g\_{(a,b,c)})=\frac{\pi}{\max\{a,b,c\}}.
$$



Diameter of $\mathrm{SU}(2)$ endowed with a left-invariant metric
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. 


