Let $V$ be an n dimensional real vector space and $V^*$ be the dual vector space.
We have a non degenerate inner product $(\centerdot,\centerdot)$ in $V\oplus V^*$ such that
$(v+\xi , u+\eta)=-\frac{1}{2}\{<v,\eta >+<u, \xi>\}$ where $v,u \in V, \xi, \eta \in V^* and <\centerdot,\centerdot>$ is a canonical paring $V \times V^* \rightarrow \mathbb{R}$.
I cannot understand following the decomposition $\mathfrak{so}(V \oplus V^*)=$End$V \oplus \Lambda^2 V^* \oplus \Lambda^2 V$.
Please help me.
Recall that for any inner product space $(\Bbb W, g)$, we get a natural $SO(g)$-module isomorphism $$\Phi: \mathfrak{so}(g) \cong \Bbb \Lambda^2 \Bbb W^*$$ given by identifying $\mathfrak{so}(g)$ as a subalgebra of $\mathfrak{gl}(\Bbb W) \cong \Bbb W \otimes \Bbb W^*$ and lowering an index using the inner product $g$; explicitly, this isomorphism is $$\Phi(X)(v, w) := (X \cdot v, w),$$ where $\cdot$ denotes the standard (Lie algebra) action. Then, the identification follows from applying the familiar decomposition $$\Lambda^2 (\Bbb A \oplus \Bbb B) \cong \Lambda^2 \Bbb A \oplus (\Bbb A \otimes \Bbb B) \oplus \Lambda^2 \Bbb B$$ to $\Bbb A = V$ and $\Bbb B = V^*$ (and the canonical isomorphism $$\text{End}(V) \cong V \otimes V^* ).$$
(NB that this result does not depend on the inner product having the particular form given in the question, i.e., on being the natural extension of the canonical pairing between $V$ and $V^*$.)