Geodesics on $SU(4)$

Question

Are the geodesics of the following metrics on $SU(4)$ known or easy (in a way not known to me!) to find?

In the adjoint representation, one can express the Killing form as a matrix and consider it as an inner product on $\mathfrak{su}(4)$. This matrix is some multiple of the identity (i.e. a scalar matrix). The geodesics of this metric are well known to be the one parameter sub-groups of $SU(4)$.

Consider the basis of $\mathfrak{su}(4)$ given by $\{i\sigma^n \otimes \sigma^m\}$ where $n,m =0, x,y,z$ but not both are $0$. $\sigma^0$ is the identity matrix and the others are the usual Pauli matrices.

I want to know the geodesics of the metrics formed by right translating an inner product at the identity which is given by diagonal matrix (it would have dimension $15$ by $15$) in this basis for $\mathfrak{su}(4)$. Ideally, I'd like to know the geodesics in terms of the 15 diagonal elements of the matrix defining the inner product on $\mathfrak{su}(4)$.

A way to represent the solution as a matrix exponential would be especially helpful!

Answer 1

In the OP's particular case, the situation is somehwat simpler than the general case that José discusses. That's because the family of left-invariant metrics on $\mathrm{SU}(4)$ that the OP wants to consider has special properties, although just how special does not become apparent until one looks at the problem from a rather different viewpoint, using the fact that $\mathrm{SU}(4)$ is $\mathrm{Spin}(6)$. (In fact, one has $\mathrm{SU}(4)/\{\pm I_4\}=\mathrm{SO}(6)$, and the problem is much easier to describe and treat as a problem on $\mathrm{SO}(6)$, as will be seen.)

First, though, a quick review of the geodesic equations for a left-invariant metric on a compact, semi-simple Lie group $G$: If $\kappa:{\frak{g}}\times{\frak{g}}\to\mathbb{R}$ is the Killing form on ${\frak{g}} = T_eG$, and $\omega:TG\to{\frak{g}}$ is the canonical left-invariant form, then the standard bi-invariant metric on $G$ is given by $\mathrm{d}s^2 = -\kappa(\omega,\omega)$. Any other left-invariant metric on $G$ can be written uniquely in the form $\mathrm{d}\bar s^2 = -\kappa(B\omega,\omega)$, where $B:{\frak{g}}\to{\frak{g}}$ is a positive definite $\kappa$-symmetric linear isomorphism. To find the $\mathrm{d}\bar s^2$-geodesic passing through $g_0\in G$ with initial velocity $L'_{g_0}(v_0)\in T_{g_0}G = L'_{g_0}\bigl({\frak{g}}\bigr)$, one has a $2$-step procedure: First, one finds the curve $v:\mathbb{R}\to{\frak{g}}$ that satisfies the Euler equation (a nonlinear ODE initial value problem) $$ v'(t) = B^{-1}\bigl[v(t),Bv(t)\bigr],\qquad v(0) = v_0 $$ and then the curve $g:\mathbb{R}\to G$ satisfying the Lie equation $$ \omega\bigl(g'(t)\bigr) = v(t),\qquad g(0) = g_0\,. $$ (When $G$ is a matrix group, this latter equation is just $g'(t) = g(t) v(t)$, with initial value $g(0) = g_0$.)

Note that, when $v_0$ is an eigenvector of $B$, the solution of the Euler equation is $v(t) = v_0$, and so the geodesic is just $g(t) = g_0 \exp(tv_0)$ (i.e., the left-translation of a $1$-parameter subgroup). More generally, if $B$ preserves a subalgebra ${\frak{s}}\subset {\frak{g}}$ that contains $v_0$, then the problem reduces to finding the geodesic in the corresponding subgroup $S\subset G$ (which is totally geodesic in $G$ with respect to the metric $\mathrm{d}\bar s^2$).

Next, in the OP's specific case, one has ${\frak{g}} = {\frak{su}}(4) = {\frak{so}}(6)$ and the OP has prescribed an orthogonal basis $\mathbf{b}$ consisting of 15 elements in ${\frak{su}}(4)$ and wants to consider, all together, the $15$-dimensional cone of metrics determined by the set of positive definite symmetric transformations $B:{\frak{su}}(4)\to {\frak{su}}(4)$ that preserve the $15$ lines spanned by the elements of $\mathbf{b}$. What is not apparent in the OP's description is the great deal of symmetry that the basis $\mathbf{b}$ possesses.

This is much more apparent when one, instead, uses the alternative form ${\frak{so}}(6)$, i.e., the skew-symmetric linear transformations of $\mathbb{R}^6$ with its standard inner product. In this form, one can describe the OP's basis $\mathbf{b}$ as follows: Let $e_1,\dots,e_6$ be an orthonormal basis of $\mathbb{R}^6$ and let $E_{ij}\in {\frak{so}}(6)$ for $1\le i<j\le 6$ be the rank $2$ linear transformation that satisfies $E_{ij}(e_i) = e_j$ and $E_{ij}(e_j) = - e_i$. Then the basis $\mathbf{b} = \bigl(E_{ij}\bigr)_{i<j}$ is orthonormal with respect to the Killing form of ${\frak{so}}(6)$, and it corresponds, under an appropriate isomorphism, to the OP's prescribed basis of ${\frak{su}}(4)$, at least up to signs (which are immaterial to the problem). (Verifying this is an interesting exercise for the reader.)

Now, a subalgebra ${\frak{s}}\subset {\frak{so}}(6)$ is invariant under all of the positive definite linear transformations $B:{\frak{so}}(6)\to {\frak{so}}(6)$ that preserve $\mathbf{b}$ up to multiples if and only if it has a basis that is a subset of $\mathbf{b}$. There are many such subspaces, and this makes it easy to compute the geodesics for many initial values $v_0$:

1. There are $15$ such maximal tori ${\frak{t}}\subset {\frak{so}}(6)$: For any permutation $\pi = \bigl(\pi(1),\ldots,\pi(6)\bigr)$ let ${\frak{t}}_\pi$ be spanned by the three elements $E_{\pi(1)\pi(2)}$, $E_{\pi(3)\pi(4)}$, and $E_{\pi(5)\pi(6)}$. Then, for $v_0\in {\frak{t}}_\pi$, the solution to the Euler equation is $v(t) = v_0$, so the corresponding geodesics for all of the $15$-parameter family of left-invariant metrics are left-translates of $1$-parameter subgroups.

2. There are $20$ such copies of ${\frak{so}}(3)\subset {\frak{so}}(6)$: For any triple $(i,j,k)$ with $1\le i<j<k\le 6$, let ${\frak{so}}(3)_{ijk}$ be spanned by the elements $E_{ij}$, $E_{ik}$, and $E_{jk}$. Then this defines a subgroup of $\mathrm{SO}(6)$ that is totally geodesic for all of the metrics in the OP's class, and each of these metrics restricts to be a left-invariant metric on each such $\mathrm{SO}(3)$. (Unfortunately, these include the general left-invariant metrics on $\mathrm{SO}(3)$, and, as is well-known, the geodesic equations for the generic such metric on $\mathrm{SO}(3)$ can only be integrated using the Jacobian elliptic functions. [See any good book on mechanics for this integration, where it is described as solving the rigid body problem. Also, note the onset of chaos already in this simple case.] As a result, it follows that it is hopeless to expect a general solution in any explicit form, even for the Euler equation.) Note, by the way, that these $20$ copies of totally geodesic $\mathrm{SO}(3)$s in $\mathrm{SO}(6)$ can be grouped into $10$ pairs that commute with each other, which generates $10$ totally geodesic copies of $\mathrm{SO}(3)\times\mathrm{SO}(3)$ on which the geodesic equations for all the metrics in the family reduce to solving independent pairs of $3$-dimensional rigid body problems.

3. There are, of course, other subgroups that are totally geodesic for the entire $15$-dimensional cone of metrics, such as $15$ copies of $\mathrm{SO}(2)\times\mathrm{SO}(4)$, and $6$ copies of $\mathrm{SO}(5)$. But the Euler equations become progressively harder to solve, and, as far as I know, there is no general solution known for this family of left-invariant metrics on $\mathrm{SO}(5)$ and maybe not even for $\mathrm{SO}(4)$. (Even the $\mathrm{SO}(2)\times\mathrm{SO}(4)$ case is not easy: Even though the Lie algebra of $\mathrm{SO}(4)$ splits as the direct sum of two subalgebras, this splitting is not preserved by the generic linear transformation $B$ in the $15$-dimensional family, and, as a result, the Euler equations do not usually uncouple to simpler equations.)

My conclusion is that, while one can compute the geodesics for this family of left-invariant metrics on $\mathrm{SO}(6)$ for special subspaces of initial conditions for the Euler equations, to get the general solution in any explicit form is probably not possible.

Answer 2

The geodesics you seek are the so-called homogeneous geodesics. Not all geodesics will be of this form, but there certainly exist. In the literature, for some reason, people consider left-invariant metrics and not right-invariant metrics, but by considering the "opposite group" you can use results for left-invariant metrics to your situation.

By a result of Kajzer (Kajzer V.V., Conjugate points of left-invariant metrics on Lie groups, Sov. Math., 34 (1990), translation from Izv. Vyssh. Uchebn. Zaved. Mat., 342 (1990), 27–37.) there is at least one homogeneous geodesic for left-invariant metrics in a compact semi-simple Lie group; that is, one which is given by the orbit of a one-parameter subgroup.

By a result of Szenthe (Szenthe J., Homogeneous geodesics of left-invariant metrics, Univ. Iagel. Acta Math., 38 (2000), 99–103.) [MathSciNet link] if in addition group has rank ≥ 2 then there are an infinite number of homogeneous geodesics.

A later result of Szenthe (Szenthe J., On the set of homogeneous geodesic of a left-invariant metric, Univ. Iagel. Acta Math. No. 40 (2002), 170–181.) [MathSciNet link] characterises those geodesic vectors in the Lie algebra.

Let $\left<-,-\right>$ denote the negative of Killing form, which is a positive-definite inner product on the Lie algebra $\mathfrak{g}$ and let $B$ denote the positive-definite inner product on the Lie algebra which agrees with the left-invariant metric at the identity. Let $\hat B: \mathfrak{g}\to \mathfrak{g}$ be the associated endomorphism defined for all $X,Y\in\mathfrak{g}$ by $$ \left<X, \hat B Y \right> = B(X,Y) $$

Then $X \in \mathfrak{g}$ is geodesic if and only if $$ [X, \hat B X] = 0 $$ i.e., $\hat B X$ belongs to the isotropy subalgebra of $X$ in $\mathfrak{g}$. This is quadratic equation on $X$, so effectively computing this set might not be trivial.