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