I would like to see an example of a fiber bundle $\mathbb{T}^2 \to X \xrightarrow{\pi} B$ whose fibers are tori $\mathbb{T}^2:=(S^1)^2$ and whose vertical tangent bundle $\ker(d\pi) \to X$ is not decomposable? Which homotopy groups does one need to look at to construct such an example?


(This problem has been edited using inputs from the comments.)



A torus bundle whose vertical tangent bundle is indecomposable
Put $T=\{(z\_0,z\_1,z\_2)\in(S^1)^3:z\_0z\_1z\_2=1\}$, so $T$ is homeomorphic to $S^1\times S^1$ and has an obvious action of the symmetric group $\Sigma\_3$. The action of $\Sigma\_3$ on $H\_1(T;\mathbb{R})$ is indecomposable, and this homology group can also be identified with the tangent space to $T$ at the identity element.


Now put $F=\{(u\_0,u\_1,u\_2)\in\mathbb{C}^3:u\_i\neq u\_j\text{ for } i\neq j\}$. This also has a natural action of $\Sigma\_3$, which is free. Put $E=(T\times F)/\Sigma\_3$ and $B=F/\Sigma\_3$, so we have a fibre bundle $T\to E\to B$. This will have the required indecomposability. Note that the identity element of $T$ gives a section of the projection $E\to B$.


The cohomology of $F$ is a well-known example. For $i\neq j$ we can define $\pi\_{ij}\colon F\to\mathbb{C}\setminus\{0\}$ by $\pi\_{ij}(z)=z\_j-z\_i$, and by pulling back the generator of $H^1(\mathbb{C}\setminus\{0\})$ we get an element $a\_{ij}\in H^1(F)$. These satisfy $a\_{ij}=a\_{ji}$ and $a\_{ij}^2=0$ and the Arnol'd relation $a\_{ij}a\_{jk}+a\_{jk}a\_{ki}+a\_{ki}a\_{ij}=0$. It follows that $\{1,a\_{01},a\_{02},a\_{12},a\_{01}a\_{02},a\_{01}a\_{12}\}$ is a basis for $H^\*(F)$. From this description you can calculate $H^\*(B)$ and $H^\*(E)$ using the Serre spectral sequence. With rational coefficients this just degenerates to $H^\*(B;\mathbb{Q})=H^\*(F;\mathbb{Q})^{\Sigma\_3}=\mathbb{Q}$ and $H^\*(E;\mathbb{Q})=H^\*(T\times F;\mathbb{Q})^{\Sigma\_3}$. It can't be too hard to calculate that last ring explicitly, but I have not done so.


If you prefer to work with homotopy groups, it is standard that $\pi\_1(B)$ is the braid group $Br\_3$. This has a standard surjective homomorphism to $\Sigma\_3$, whose kernel is the pure braid group $PBr\_3$, which can be identified with $\pi\_1(F)$. It is also known that $\pi\_i(F)=\pi\_i(B)=\pi\_i(T)=0$ for $i>1$. The map $PBr\_3\to\Sigma\_3$ gives an action of $PBr\_3$ on $\pi\_1(T)\simeq\mathbb{Z}^2$, using which we can form the semidirect product $PBr\_3\ltimes\mathbb{Z}^2$. Using the split fibre bundle $T\to E\to B$ we can identify $\pi\_1(E)$ with this semidirect product, and also see that $\pi\_i(E)=0$ for $i>1$.


