Lie Bracket of Bivectors

Question

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.

Answer 1

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: $$\begin{aligned}\ [v_1 \odot v_2, w_1 \odot w_2] &= (v_1 \odot v_2)(w_1)\odot w_2 + w_1 \odot(v_1 \odot v_2)(w_2) \\ &= \omega(v_1,w_1)v_2\odot w_2 + \omega(v_2,w_1)v_1\odot w_2 + \omega(v_1,w_2)w_1 \odot v_2 + \omega(v_2,w_2)w_1 \odot v_1\end{aligned}$$

Answer 2

(expanding on comments)

This is the cross product, or commutator product, in Clifford algebra.

$$X\times Y=\frac{XY-YX}{2}$$

It's straightforward to verify that, in any associative algebra, the commutator is a derivation:

$$X\times(YZ)=(X\times Y)Z+Y(X\times Z)$$ $$X\times(Y\times Z)=(X\times Y)\times Z+Y\times(X\times Z)$$

(In what follows, lowercase letters denote vectors.)

Suppose $X$ and $Y$ are two simple (i.e. wedge-factorable) bivectors. The wedge product of vectors is (isomorphic to) the antisymmetric part of the Clifford product:

$$X=a\wedge b=\frac{ab-ba}{2}=a\times b$$ $$Y=c\wedge d=\frac{cd-dc}{2}=c\times d$$

And by definition of the Clifford algebra, your bilinear form (which I'll write as a dot product) is the symmetric part of the Clifford product:

$$B(a,b)=a\cdot b=\frac{ab+ba}{2}=\frac{(a+b)^2-a^2-b^2}{2}$$

It's also straightforward to verify the bac-cab formula (note that we're not using the 3D vector cross product, which would give the opposite sign):

$$a\times(b\times c)=(a\cdot b)c-b(a\cdot c)=\frac{(a\cdot b)c+c(a\cdot b)}{2}-\frac{(a\cdot c)b+b(a\cdot c)}{2}$$

Putting all these pieces together, we can derive your Lie bracket formula:

$$X\times Y=X\times(c\times d)$$ $$=(X\times c)\times d+c\times(X\times d)$$ $$=\big((a\times b)\times c\big)\times d+c\times\big((a\times b)\times d\big)$$ $$=\big(a(b\cdot c)-(a\cdot c)b\big)\times d+c\times\big(a(b\cdot d)-(a\cdot d)b\big)$$ $$=(b\cdot c)(a\times d)-(a\cdot c)(b\times d)+(b\cdot d)(c\times a)-(a\cdot d)(c\times b)$$ $$=(b\cdot c)(a\wedge d)-(a\cdot c)(b\wedge d)-(b\cdot d)(a\wedge c)+(a\cdot d)(b\wedge c)$$