# Lie Bracket of Bivectors

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.
• It is exactly the cross product in geometric algebra. en.wikipedia.org/wiki/Bivector#Product_of_two_bivectors Nov 1, 2022 at 23:24
• (Though, in GA usually the scalars are real, not complex. But much of it generalizes to arbitrary fields (char$\neq2$) and arbitrary symmetric bilinear forms.) Nov 1, 2022 at 23:35
• I think saying "cross product" can be confusing, so let me elaborate a bit more. What @mr_e_man is saying is you: (1) Form the Clifford algebra $Cl(V, B)$ associated with $B$ when it is symmetric. (2) Use the canonical linear isomorphism $Cl(V, B) \cong {\bigwedge} V$ to define bivectors. (3) Then the Lie bracket on bivectors $X, Y \in Cl(V, B)$ is the commutator product $X\times Y = \tfrac12(XY - YX)$ using the Clifford product. It may still be possible to use Clifford algebras when $B$ is not symmetric, but I am unsure and that is too much to get into in this comment. Nov 2, 2022 at 15:38
• @Nicholas This is maybe more of another question than a comment, but it is not at all obvious to me that the commutator of two bivectors in the Clifford algebra is again a bivector. Nov 2, 2022 at 15:43
• Well, you could just take the bivector part of the Clifford product of the two bivectors. Nov 2, 2022 at 15:44

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}

• What @mr_e_man is talking about here isn't exactly the vector cross product, it's the commutator product $A\times B = \tfrac12(AB - BA)$ where $AB$ and $BA$ are Clifford products. This works for all Lie algebras and all dimensions. It happens to be the same as the vector cross product in three dimensions. Nov 2, 2022 at 15:17
• Thanks; this certainly puts the bracket in a much more natural light. I still wonder if there is a geometric interpretation; it looks something like the interior product of a bivector, though a cursory search didn't yield a formal definition of any such interior product. Nov 2, 2022 at 15:40
• @NicholasTodoroff Good point, I misunderstood what he was saying Nov 2, 2022 at 16:25
• @Steve well that's (sort of) what is going in my more general formula. The interior product is just contraction of tensors and $[v_1^* \otimes v_2,w_1^* \otimes w_2]$ could be thought of as the difference between contracting $v_1^* \otimes v_2 \otimes w_1^* \otimes w_2$ in two different ways: the inner pair $v_1^* \otimes w_1^*(v_2) \otimes w_2$ or the outer pair $v_1^*(w_2) \otimes v_2 \otimes w_1^*$. Nov 2, 2022 at 16:54
• @Steve The Clifford algebra formulation is entirely geometric. At least in the real Euclidean case, If $A, B$ are bivectors blades whose planes $[A], [B]$ are orthogonal or parallel subspaces then $A$ and $B$ commute so $A\times B = 0$. Otherwise, the planes intersect in a line $L$, and $C = A\times B$ is the unique bivector with orientation given by the right-hand-rule such that $[C]$ is orthogonal to $L$ but intersects $[A], [B]$ non-trivially, and $C^2 = -A^2B^2\sin^2\theta$ where $\theta$ is the angle between $[A], [B]$ using $L$ as an axis... Nov 2, 2022 at 18:56

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:

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