Adjoint representation of $\mathfrak{so}(n)$

Question

So I know $\mathfrak{so}(n)$ consists of anti-symmetric matrices, but I don't see why the adjoint representation is isomorphic to the anti-symmetric tensor square of the vector representation. I see that we can associate each matrix with an anti-symmetric tensor, but I don't get how it relates with the adjoint representation. Can someone please help me clarify some of this confusion. Thanks.

Answer 1

The adjoint representation $\mathfrak{so}(n)$ is isomorphic to the second exterior power and not to the second symmetric power. The isomorphisms are given as follows: For a matrix $X\in\mathfrak{so}(n)$, the map $(v,w)\mapsto \langle Xv,w\rangle$ is bilinear and skew symmetric, thus defining an element of $\Lambda^2\mathbb R^{n*}\cong \Lambda^2\mathbb R^n$. Conversely, you associate to $w_1\wedge w_2$ to the map $f:\mathbb R^n\to\mathbb R^n$ defined by $v\mapsto \langle v,w_1\rangle w_2-\langle v,w_2\rangle w_1$. One immediately verifies that this map lies in $\mathfrak{so}(n)$ and that the two constructions are inverse to each other.