# Are all bivectors in three dimensions simple?

I want to show that all bivectors in three dimensions are simple.

If I understand correctly, a bivector is simply an element from the two-fold exterior product $$\bigwedge^2V$$ of a vector space $$V$$, right?

We can define $$\wedge(e_i\otimes e_j):=e_i\otimes e_j - e_j\otimes e_i$$. Now let $$T=t^{ij}e_i\wedge e_j\mapsto t^{ij}(e_i\otimes e_j - e_j\otimes e_i)=(t^{ij}-t^{ji})e_i\otimes e_j$$. This is injective. Because the wedge product as a linear map from the tensor product to the exterior product maps all symmetric tensors to 0.

We see that total antisymmetric tensors in this case are represented by skew-symmetric matrices. To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.

But the rank of a skew-symmetric matrix is never one.

I must have made a conceptual mistake somewhere again. Does anybody have a hint for me?

Geometrically, one can use the canonical isomorphism between the two-fold exterior product and $$\mathbb{R}^3$$ itself to show that any anyisymmetric tensor can be thought of as a vector in 3D, which in turn can be represented by the cross product of two vectors, that are not collinear to each other but orthogonal to that vector.

• $(e_1+e_2)\wedge(e_3-e_1)$ – user8268 Nov 13 '18 at 23:12
• Does "simple" mean "of the form $u \wedge v$ for some $u, v \in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n \geq 2$, then each element of $\Lambda^{n-1}\left(V\right)$ can be written as $v_1 \wedge v_2 \wedge \cdots \wedge v_{n-1}$ for some $v_1, v_2, \ldots, v_{n-1} \in V$. – darij grinberg Nov 14 '18 at 2:42
• Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $\mathbb{R} ^3$ itself. – Thomas Wening Nov 14 '18 at 11:05
• Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post. – Thomas Wening Nov 14 '18 at 14:24

To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.

This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $$1$$. You're conflating $$e_i \otimes e_j - e_j \otimes e_i$$, which as a tensor has rank $$2$$ or $$0$$, with $$e_i \wedge e_j$$. These are not the same object; one of them lives in $$V^{\otimes 2}$$ and the other one lives in $$\Lambda^2(V)$$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.

Anyway, here's a proof of darij's more general claim in the comments. Let $$v_1 \in \Lambda^{n-1}(V)$$ be a vector, where $$\dim V = n$$. Choose a nonzero element $$\omega \in \Lambda^n(V)$$, hence an identification of it with the ground field $$k$$. Then the exterior product

$$\wedge : V \times \Lambda^{n-1}(V) \to \Lambda^n(V) \cong k$$

is a nondegenerate bilinear pairing. Extend $$v_1$$ to a basis $$v_1, \dots v_n \in \Lambda^{n-1}(V)$$. Then it has a unique dual basis $$e_1, \dots e_n \in V$$ defined by the condition that

$$e_i \wedge v_j = \delta_{ij} \omega \in \Lambda^n(V).$$

Then the $$v_i$$ must also be the dual basis of the $$e_i$$ with respect to this pairing. But this dual basis in turn must be

$$v_i = (-1)^{i-1} \frac{\omega}{e_1 \wedge \dots \wedge e_n} e_1 \wedge \dots \wedge \widehat{e_i} \wedge \dots \wedge e_n$$

where the hat denotes that we omit $$e_i$$, and in particular

$$v_1 = \frac{\omega}{e_1 \wedge \dots \wedge e_n} e_2 \wedge \dots \wedge e_n.$$

• Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $\frac{\omega}{e^1\wedge\cdots\wedge e_n}$ mean? – Thomas Wening Nov 14 '18 at 21:41
• @Thomas: $\Lambda^n(V)$ is one-dimensional, and $\omega$ and $e_1 \wedge \dots \wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $\omega = c e_1 \wedge \dots \wedge e_n$. It refers to this scalar. – Qiaochu Yuan Nov 14 '18 at 21:44
• Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_i\wedge v_i$ to get all the indices in monotonously increasing order, right? – Thomas Wening Nov 14 '18 at 21:59
• Yes, that's right. – Qiaochu Yuan Nov 14 '18 at 22:00
• Good. The pairing allows for the definition of a dual basis, once I have fixed one in $\bigwedge^2\mathbb{R}^3$. But how does that allow me to decompose an arbitrary $\omega\in\bigwedge^{n-1}V$ into $\omega=v\wedge u$ with $u,v\in V$? – Thomas Wening Nov 14 '18 at 22:11