Suppose $V$ is a (finite dimensional) vector space and $B:V\otimes V\to\mathbf{C}$ a non-degenerate symmetric bilinear form (not necessarily definite). The Lie algebra $\mathfrak{so}(V,B)=\{T:V\to V| B(Tv, w) + B(v, Tw) = 0 \text{ and } Tr(T)=0\}$ is naturally isomorphic as a vector space to the space of bivectors $\Lambda^2V$ by $\phi:\Lambda^2V\to \mathfrak{so}(V,B)$ where $\phi(x\wedge y)(v) = B(y,v)x – B(x, v)y$. One can work out that $\phi$ is a Lie algebra isomorphism when $\mathfrak{so}(V,B)$ has the usual commutator bracket, and $\Lambda^2V$ is endowed with the bracket
$$[x_1\wedge x_2, y_1\wedge y_2] = -B(x_1, y_1)x_2 \wedge y_2 + B(x_1, y_2)x_2 \wedge y_1 + B(x_2, y_1)x_1 \wedge y_2 – B(x_2, y_2)x_1 \wedge y_1.$$
(It's entirely possible the signs in the above are off.)
What is this bracket on $\Lambda^2V$? Surely it has a name or some geometric meaning.
Best Answer
As mr_e_man notes it happens to coincide with the cross product (possibly up to scale) under the identification $\Lambda^2V \cong \mathbb{C}^3$ when $V$ is 3-dimensional. For higher dimensions, I think simply thinking of it as a commutator of endomorphisms $[A,B] = AB - BA$ is the most sensible geometrical interpretation.
Indeed, remember $\mathrm{End}(V) = V^* \otimes V$ and the natural Lie algebra on this looks like:
$$ [v_1^* \otimes v_2,w_1^* \otimes w_2] = v_1^*(w_2) w_1^* \otimes v_2 -w_1^*(v_2)v_1^* \otimes w_2. $$
Then in the presence of a nondegenerate symmetric bilinear form we have a natural isomorphism $V^* \cong V$ which gives the identification you mention: $\mathfrak{so}(V) \cong \Lambda^2V$ (viewing $\Lambda^2V$ as a subspace of $V^* \otimes V \cong V \otimes V$ e.g. as spans of elements like $v_1^* \otimes w_1 - w_1^* \otimes v_1$). Hopefully it clear that this more general formula reduces to yours under these conditions.
Indeed note that the symmetric property wasn't vital here so we get a similar story for symplectic forms as well: $\mathfrak{sp}(V) \cong S^2V$ (for $V$ even dimensional) via $(v \odot w)(x) = \omega(v,x)w + \omega(w,x)v$ and the Lie bracket looks like: $$\begin{aligned}\ [v_1 \odot v_2, w_1 \odot w_2] &= (v_1 \odot v_2)(w_1)\odot w_2 + w_1 \odot(v_1 \odot v_2)(w_2) \\ &= \omega(v_1,w_1)v_2\odot w_2 + \omega(v_2,w_1)v_1\odot w_2 + \omega(v_1,w_2)w_1 \odot v_2 + \omega(v_2,w_2)w_1 \odot v_1\end{aligned}$$