Non-trivial fibration of $SU(3)$ over $S^1$? 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. 
 A: 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 $H_5(E)\cong \mathbb{Z}/2\mathbb{Z}.$ $\square$
