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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!

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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.

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  • $\begingroup$ Beautiful answer, as usual! $\endgroup$ – José Figueroa-O'Farrill Aug 5 '14 at 10:11
  • $\begingroup$ My only other comment is that I actually said "right translating" not left. I've often been confused about how to translate results about left invariant metrics into ones about right invariant metrics. $\endgroup$ – Benjamin Aug 21 '14 at 16:45
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    $\begingroup$ Fortunately, that's easy; just compose with the inverse map of the Lie group, which converts a left-invariant metric $\lambda$ on a Lie group $G$ to the right-invariant one $\rho$ on $G$ that has the same inner product at the identity as $\lambda$. Of course, you should also reverse the parametrization, if you want the velocity to stay the same. I.e., if $c:\mathbb{R}\to G$ is a geodesic of a left-invariant metric with initial velocity $v\in{\frak{g}} = T_eG$, then $\hat c(t) = c(-t)^{-1}$ defines the geodesic of the corresponding right-invariant metric with the same initial velocity. $\endgroup$ – Robert Bryant Aug 21 '14 at 16:56
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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.

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  • $\begingroup$ Thanks, that is a helpful ref, however I didn't mean the exponential of a constant matrix. $\endgroup$ – Benjamin Aug 1 '14 at 19:20
  • $\begingroup$ I see. But $SU(4)$ is compact and connected, so the exponential map is surjective. Any geodesic can be written as the exponential of some curve in the Lie algebra. So are you simply asking to find the geodesics? If the geodesic is homogeneous, the geodesic equation localises to the algebraic equation in my answer. Otherwise, I'm afraid that nothing saves you from having to solve the geodesic equation. $\endgroup$ – José Figueroa-O'Farrill Aug 1 '14 at 23:18
  • $\begingroup$ I realise that every curve on the group can be written as an exponential. However, I was hoping to express the time dependent generator of a geodesic in terms of the 15 numbers defining the inner product $\endgroup$ – Benjamin Aug 1 '14 at 23:21
  • $\begingroup$ Did you try solving the geodesic equation? $\endgroup$ – José Figueroa-O'Farrill Aug 2 '14 at 0:10
  • $\begingroup$ Yes. I only fiddled for a while though, to no avail. I was hoping that the euler poincare equations would solve it and let me avoid calculating cristofel symbols. $\endgroup$ – Benjamin Aug 2 '14 at 0:17

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