My exact problem is a Riemannian manifold defined on $SU(2^n)$, where the metric is defined as follows: If $U(t)$ is a curve so that $U′(t)=−iH(t)U(t)$ (so just a unitary evolution of a quantum system of n qubits with a time dependent Hamiltonian $H(t)$), we can defined a metric at $U(t)$ as $$\langle H(t),H(t)\rangle =w_P(P(H(t)),P(H(t)))+w_Q(Q(H(t)),Q(H(t)))$$ where P,Q are orthogonal projections of H(t) into two orthogonal subspaces and $(P(H),P(H)),(Q(H),Q(H))$ are inner products on these subspaces,$w_P,w_Q$ are weighting factors. It has been indicated that for some parameter selections of $(P(H),P(H)),(Q(H),Q(H))$ and $w_P,w_Q$, the manifold shows a negative curvature almost everywhere.
To make things clear, obviously $H(t)$ is a vector of the Lie algebra of $SU(2^n)$. What the projectors $P$ and $Q$ did is to project $H(t)$ into two orthogonal subspaces of the Lie algebra, let's say $PH,QH$, and then we define the metric so that we have different weights $w_P,w_Q$ on this two subspaces. Here more specifically, $PH$ is the subspace of all Hamiltonian that only involves no more than 2 qubits (local or simple). Or in another way, we know every $U$ can be reached by an EXP mapping $\operatorname{Exp}(-iH(U))$ since we always have $H(U)=i\operatorname{Log}(U)$, so that we have a diffeomorphism between the Lie algebra and the Lie group. Instead, the Riemannian manifold defined above will try to reach each $U$ from $I$ by a geodesic curve $U(t)$. But the metric is in fact setting constraints on the possible 'velocity' direction $U'(t)$ so that we prefer that $H(t)\sim U'(t)$ can only falls in a subspace $PH$ of the Lie algebra. If in an extreme case, the wight $w_Q$ goes to infinity, then we only allow a velocity in the subspace $PH$ but not in $QH$. For me it seems quite possible that not every $U$ will be attainable from $I$. That's to say, there might exist a point $U$ so that there will be no geodesic/or a general curve with a finite length that can connect $I$ and $U$ because we are not allowed to travel along any direction in $QH$ (just imagine from point (0,0) you can not reach an arbitrary point $(x,y)$ in a unit square if you are only allowed to travel in x direction).
I was told by M. Miller(thanks) that "compact Riemannian manifold always has a finite diameter". Since $SU(N)$ is compact, so definitely the diameter of the Riemannian manifold is finite. This seems to close the question. But my intuition does not see it. It seems there might exist some points $U$ that can not be reached from $I$ with a finite length. I think the contradiction might arise from the fact that I am thinking the case where $w_Q\rightarrow\infty$. So does this mean if both $w_P,w_Q$ are finite, we have a manifold with a finite diameter. But if $w_Q\rightarrow\infty$, then we have a different manifold so that "compact Riemannian manifold always has a finite diameter" does not apply? Is the manifold still a Riemannian manifold with $w_Q\rightarrow\infty$?
P.S.: Since I am working on $SU(2^n)$, I was posted by Norbert Schuch that there is a way to decompose any $U \in SU(2^n)$ into simple operations involving no more than 2 qubits(let's say the unitary operators that can be achieved by a selected subspace $PH$) with a complexity of $n^3e^n$. This is exactly a special case of $w_Q\rightarrow\infty$ since we only allow simple/local operations in $PH$.
Does this mean the answer highly depends on the selection of $PH, QH$?