Let $E\to M$ be a real vector bundle over a differentiable manifold $M$ and let $p_{1}(E)$ denote its first Pontryagin class. I would like to know if there is any formula allowing to write $p_{1}(\Lambda^2 E)$ in terms of $p_{1}(E)$. I am mostly interested in the case where the dimension of $M$ is less or equal than seven and $E=TM$ is the tangent bundle of the manifold. In particular, $p_{1}(TM)\in H^{4}(M,\mathbb{Z})$, and for the case that I am interested in, $H^{4}(M,\mathbb{Z}) = \mathbb{Z}_{10}$ and thus $p_{1}(TM) = k\, u$, where $k=0,\dots,9,$ and $u$ is the generator of $\mathbb{Z}_{10}$.
Thanks.
I think, it is true that $\Lambda ^2(E\otimes \mathbb C)=(\Lambda^2E)\otimes \mathbb C$.
And for $n$-dimensional complex vector bundle $E'$ splitting principle gives us $c_2(\Lambda^2E')=c_1(E')^2\cdot\frac{(n-1)(n-2)}{2}+c_2(E')\cdot(3n-3)$. So, maybe, $p(E)$ is not sufficient to compute $p_1(\Lambda^2E)$, because you have to know $c_1(E\otimes\mathbb C)$.
EDIT: for example, if $E$ is orientable, then $c_1(E\otimes \mathbb C)=0$, because we can induce $E$ from tautological bundle over oriented grassmannian $G_+(n,\infty)$, and $H^2(G_+(n,\infty))=0$.