In String Theory it is well known that a string can propagate on backgrounds such as a $T^2$ fibred over a circle. This fibration can be non-trivial in the following sense:
Given $T^2$ generators $J^i, \quad i=1,2$ (i.e. $U(1)^2$ generators) with $[J^1,J^2]=0$ and $S^1$ generator $J^y$, we have a 'twist matrix' $N$ such that $[J^y, J^i]=N^i_jJ^j$. The non-triviality of the fibration is encoded in $N$, and normally we take $N\in \mathcal{L} (SL(2, \mathbb{Z}))$.
Now, noting that the cartan torus of $SU(3)$ is $T^2$, we would like to extend this fibration to $SU(3)$, i.e. fibre $SU(3)$ non-trivially over a circle in the same way. The generators of the fibre are $J^i, J^a$, $a=1,...,6$, and these obey the usual $SU(3)$ commutation relations. We now have some twist for the $J^a$ as well, $[J^y, J^a]=M^a_bJ^b$, where $M$ should probably be determined by $N$, but I haven't completely verified this yet.
The issue is whether this is actually possible, i.e. can we preserve the $SU(3)$ algebra in the fibre whilst still having a non-trivial fibration over the circle? I'm not sure if this is something which is known or if there are known methods for dealing with this sort of thing.
First, if you want a principal $SU(3)$ bundle over $S^1$, you only get the trivial. "Principal" means that the structure group of the bundle is $SU(3)$ acting each of the fibers (which are all diffeo to $SU(3)$) acts simply transitively, by either left or right multiplication.
The proof I'm thinking of works with $SU(n)$ replaced by any connected compact Lie group $G$. It's a fact (due to Milnor?) that for every such Lie group, there is a contractible topological space $EG$ which admits a free action by $G$. The quotient is denoted $BG = EG/G$. It turns out that for any reasonable space $X$, there is a natural equivalence between prinicpal $G$-bundles over $X$ and homotopy classes of maps from $X$ into $BG$.
Now, from the long exact sequence in homotopy groups associated to the fibration $G\rightarrow EG\rightarrow BG$ (together with the fact that $EG$ is contractible), we learn that $\pi_k(BG)\cong \pi_{k-1}(G)$ for all $k\geq 1$. In particular, the fact that $G$ is connected implies that $BG$ is simply connected. But then this implies that all maps from $X= S^1$ into $BG$ are homotopically trivial. Thus, for $X = S^1$ and $G$ a connected compact Lie group, there is exactly one principal $G$ bundle over $S^1$. Since the product $G\times S^1$ is one such example, it must be the only one. In other words, all principal $G$-bundles (with $G$ connected) over $S^1$ are trivial.
But what about non-principal bundles? In other words, can we find a fiber bundle $SU(3)\rightarrow E\rightarrow S^1$ which is non-trivial, and where the fiber just happens to be diffeomorphic to $SU(3)$?
I claim that we can for any $SU(n)$. The idea is to find a diffeomorphism $f:SU(n)\rightarrow SU(n)$ which is not isotopic to the identity. For $n = 2$, $SU(2)$ is diffeomorphic to $S^3$. So, for $f$, we can use your favorite map which is not isotopic to the identity. Reflections are wonderful examples.
For $SU(n)$ with $n\geq 3$, we will pick $f:SU(n)\rightarrow SU(n)$ with $f(A) = \overline{A}$. (Curiously, for $n=2$, this map is isotopic to the identity. In fact, thinking of $S^3$ as unit quaternions, $f$ is just conjugation by $j\in S^3$. Then a path from $j$ to $1\in S^3$ gives an isotopy to the identity.)
Proposition: The map $f$ above is not isotopic to the identity for $n\geq 3$.
Proof: We will show that $f_\ast:H^5(SU(n))\rightarrow H^5(SU(n))$ (which is really a map from $\mathbb{Z}$ to itself for $n\geq 3$) is multiplication by $-1$. Believing this, it follows that $f$ is not even homotopic to the identity.
We prove the claim by induction, beginning with $n = 3$. Since $SU(3)$ acts transitively on $S^5$ with stabilizer $SU(2)$, we get a fiber bundle $SU(2)\rightarrow SU(3)\xrightarrow{\pi} S^5$. As $SU(2) = S^3$ we can use the Gysin sequence to see that $\pi^\ast:H^5(S^5)\rightarrow H^5(SU(3))$ is an isomorphism.
I'm not sure how to draw commutative diagrams on MSE, but by using complex conjugation as a map from $SU(2)$ to itself, from $SU(3)$ to itself, and from $S^5\subseteq \mathbb{C}^3$ to itself, we get a map from the bundle $SU(2)\rightarrow SU(3)\rightarrow S^5$ to itself. Complex conjugation fixes the point $(1,0,0)\in S^5$ and acts as a composition of 3 reflections on $T_{(1,0,0)}S^5$, so is orientation reversing. In particular, the induced map $H^5(S^5)\rightarrow H^5(S^5)$ is multplication by $-1$. Now, using the fact that $\pi^ast$ is an isomorphism on $H^5$, it follows that the map $H^5(SU(3))\rightarrow SU(3)$ is multiplication by $-1$. This finishes the proof of the base case.
Now, for the inductive case, we assume the induced map on $H^5{SU(n)}$ is multiplication by $-1$ for some $n\geq 3$. Now, $SU(n+1)$ acts transitively on $S^{2n+1}$ with stabilizer subgroup $SU(n)$, so we get a fibration $SU(n)\xrightarrow{i} SU(n+1)\rightarrow S^{2n+1}$. Now a quick spectral sequence argument (together with the fact that $2n+1 \geq 7 > 5$ shows that $i^\ast H^5(SU(n+1))\rightarrow H^5(SU(n))$ is an isomorphism. Now you create the same comutative diagram as in the base case and use the inductive hypothesis to conclude. $\square$
Great, so we have this weird $f$ to play with. Here's how we make a non-trivial bundle over $S^1$. First, consider the trivial bundle $SU(n)\times [0,1]$. We define an equivalence relation on this space by saying $(A,0)\sim (f(A),1)$ for any $A\in SU(n)$. (Remember, $f$ means complex conjugation for $n\geq 3$ and a reflection for $n=2$.) I'll denote the quotient of $SU(n)\times [0,1]/\sim$ by $E$. Then there is a natural map $p:E\rightarrow S^1$ mapping $[A,t]$ to $[t]$, where we are thinking of $S^1 = [0,1]/0\sim 1$.
This map is a fiber bundle with fiber $SU(n)$.
Propsition: This bundle is non-trivial. In fact, $E$ isn't even homotopy equivalent to $SU(n)\times S^1$
Proof: For $n=2$, $E$ isn't even orientable because reflections are orientation reversing. (In fact, for $n\geq 3$, $f$ is often orientation reversing, but, e.g. when $n = 5,6$ it's orientation preserving.)
So, let's assume $n\geq 3$. Let's determine $H_5(E)$. Note that $H_5(S^1\times SU(n)) = \mathbb{Z}$ using Kunneth together with proof of the first proposition. Now, following Ryan Budney's answer at https://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle together with the fact that $f_\ast $ is $-1$ on $H_5$, shows that $H_5(E)\cong \mathbb{Z}/2\mathbb{Z}.$ $\square$