# triviality of Whitney sums of a vector bundle

Let a $3$-dimensional subspace $V$ of $\mathbb{R}^4$ be $$V=\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4\mid\sum_{i=1}^4x_i=0\}.$$ The alternating group $A_4$ acts on $V$ by $$\sigma(x_1,x_2,x_3,x_4)=(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)} ,x_{\sigma(4)})$$ for any $\sigma\in A_4$.

Since $V$ is linearly isometric to $\mathbb{R}^3$, $SO(3)$ acts on $V$ canonically. Moreover, $A_4$ acts on $V$ as a subgroup of $SO(3)$. Hence we have a covering map $$A_4\to SO(3)\to SO(3)/A_4.$$ Letting $A_4$ act on $\mathbb{R}^4$ by permuting basis and attaching $\mathbb{R}^4$ as fibres, we have a vector bundle associated to the covering map $$\xi: \mathbb{R}^4\to SO(3)\times _{A_4}\mathbb{R}^4\to SO(3)/A_4.$$

Question: (1). Is $\xi$ a trivial vector bundle?

(2). Is $\xi^{\oplus 2}$ (2-fold Whitney sum) a trivial vector bundle?

(3). What is the smallest integer $n$ such that $\xi^{\oplus n}$ is a trivial vector bundle?

My attempt: I plan to compute the Stiefel-Whitney class. But I am not sure whether all Stiefel-Whitney classes of $\xi$ are trivial or not?

• Real vector bundles over CW-complexes of dimension $\leq 3$ are completely classified by the Stiefel-Whitney classes (seems to be an old result of Whitney). So if the classes are trivial, then so is the bundle. – Matthias Wendt Oct 9 '15 at 13:40
• @MatthiasWendt: the comment is not quite right. You forgot the Euler class even though it is zero in the case at hand. The reference is "Classification of Oriented Sphere Bundles Over A 4-Complex" by A. Dold and H. Whitney, see maths.ed.ac.uk/~aar/papers/doldwhit.pdf – Igor Belegradek Oct 9 '15 at 20:28
• 3-dimensional real vector bundles over CW-complexes of dimension $\leq 3$ are completely classified by the first two Stiefel-Whitney classes. The obstruction for triviality over the 1-skeleton is $w_1$, and the obstruction for the existence of a 2-frame over the 2-skeleton reduces to $w_2$, so there exists a trivial 2-plane bundle inside of your vector bundle restricted to the 2-skeleton. This 2-plane bundle extends over the 3-skeleton as $\pi_2(V_2(\Bbb R^3))=0$. Now endow your bundle with a metric to find a complement to this 2-plane bundle. – PVAL Oct 9 '15 at 21:27
• The four dimensional vector bundle has a trivial one dimensional sub bundle induced by the invariant vector (1,1,1,1). That means you have to look at the 3 dimensional sub bundle $SO(3)×_{A_4}V$. – Asghar Ghorbanpour Oct 10 '15 at 6:04

There is a simple general argument showing that $\xi$ is trivial: Take any homogeneous space $G/H$ and any representation $V$ of $G$ and restrict the representation to $H$. Then the homogeneous vector bundle $G\times_H V\to G/P$ is a trivial as a vector bundle. A trivialization can be written down explicitly. It is induced by the map $G\times V\to (G/H)\times V$ defined by $(g,v)\mapsto (gH,g\cdot v)$. This is evidently $H$-invariant and factorizes to an isomorphism $G\times_H V\to (G/H)\times V$ of vector bundles.
Edit (following the comments by @Sebastian_Goette and @Asghar_Ghorbanpour): This argument can be applied both to the three dimensional representation of $A_4$ called $V$ in the question and to the representation on $\mathbb R^4$ by permutation of coordinates. The latter representation also extends to $SO(3)$ as the direct sum of the representation on $V$ and a trivial representation.
It follows from the answer to your methane molecule question that the first two Stiefel-Whitney classes are trivial, because the corresponding cohomology groups are trivial. Then $w_3$, which is the mod $2$ reduction of $W_3=\beta w_2$, is also trivial.
As in my comment, the triviality of the Stiefel-Whitney classes implies triviality of the bundle $\xi$ since $\dim SO(3)= 3$, by a result of Whitney in: