Help with intuitively understanding why the space of bivectors from 4D Euclidean space is 6 dimensional

Question

The wedge product of two 4D Euclidean space vectors is of dimension 6. I can understand mathematically that the six basis vectors are linearly independent, from the math itself. But I fail to see this intuitively. My confusion is the following:

Each basis vector represents an oriented area inside 4D space, perpendicular to the two basis vectors used in the wedge product. We can add two of these areas and obtain another area. I understand that the orientation of the areas add in a different way that 4D vector do, but I cannot see how or in what way this can result in a vector space of area orientations that is 6 dimensional, rather than 4 dimensional. Is there an easy way to visualize why this is so? (or, did I state something wrong?)

Answer 1

It is the fact that there are six planes in 4D which are pair-wise partially orthogonal to each other. Partially orthogonal means one of the planes contains at least one vector which is orthogonal to the other plane. Total orthogonality (meaning all pairs of vectors, one from each plane, are orthogonal) implies partial orthogonality.

We can see this directly from the standard basis. First, $e_1\wedge e_2$ and $e_3\wedge e_4$ are totally orthogonal. Then we just take one vector from each in every way we can: $$ e_1\wedge e_3,\quad e_1\wedge e_4,\quad e_2\wedge e_3,\quad e_2\wedge e_4. $$ Convince yourself that all six of these planes are pair-wise partially orthogonal.

Suppose there is a seventh plane $B$. There cannot be a nonzero vector in $B$ orthogonal to all the other planes, since then that vector would be orthogonal to all $e_i$ and hence would be zero. So for at least one of the standard planes $E$, there is no $v \in B$ orthogonal to $E$, but there is $v \in E$ orthogonal to $B$. This is impossible for two planes, so $B$ does not exist.

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You should also convince yourself that the geometric algebra dot product of two bivectors $B_1\cdot B_2 = 0$ iff $B_1$ and $B_2$ are partially orthogonal.

Answer 2

A basis for space of bivectors consists of pairs of vectors from a basis for the the original space, and $$ \binom{4}{2} = 6. $$ We can list them explicitly: \begin{array}{cccc} v_1 \wedge v_2 & v_1 \wedge v_3 & v_1 \wedge v_4 \\ & v_2 \wedge v_3 & v_2 \wedge v_4 \\ & & v_3 \wedge v_4 \end{array} and see that there are $6$ of them.

Each of these $v_i \wedge v_j$ describes a plane $\Pi_{ij} = \operatorname{span}\{v_i, v_j\} \subset \mathbb{R}^4$, together with an orientation.

Answer 3

The geometric aspect provides motivation. Consider projective 3-space over an arbitrary field (There is nothing special about $\mathbb R$ in this context.) where the points are given by 4 homogeneous coordinates, i.e. ratios (x:y:z:t). Two points $P_1=(x_1:y_1:z_1:t_1),P_2=(x_2:y_2:z_2:t_2)$ are distinct iff the rank of $\begin{bmatrix}x_1&y_1&z_1&t_1 \\x_2&y_2&z_2&t_2\end{bmatrix}$ is 2. The Grassman-Plucker coordinates of the line $L$ joining $P_1$ and $P_2$ are the point with 6 homogeneous coordinates$(\ell:m:n:\ell^{\prime}:m^{\prime}:n^{\prime})$ in projective 5-space where $$\ell=\begin{vmatrix}x_1&x_2\\t_1&t_2 \end{vmatrix},m=\begin{vmatrix} y_1&y_2\\t_1&t_2\end{vmatrix},n=\begin{vmatrix}z_1&z_2\\t_1&t_2 \end{vmatrix},$$ $$\ell^{\prime}=\begin{vmatrix}y_1&y_2\\z_1&z_2 \end{vmatrix},m^{\prime}=-\begin{vmatrix} x_1&x_2\\z_1&z_2\end{vmatrix},n^{\prime}=\begin{vmatrix}x_1&x_2\\y_1&y_2 \end{vmatrix}.$$ Then the point $(\ell:m:n:\ell^{\prime}:m^{\prime}:n^{\prime})$ is independent of which two points on $L$ are used. Moreover, $$\ell \ell^{\prime}+mm^{\prime}+nn^{\prime}=0$$ and there is a bijection between lines in projective 3-space and points on this quadric surface in projective 5-space.

Answer 4

The reason your intuition fails is that in 3D to every plane there is one normal vector. This means you automatically associate a bivector in 3D with its normal vector. 3 axes = 3 bivectors right? Well, in 4D there can be 2 linearly independent normal vectors to every plane. I made a diagram of this to make it visually more clear. The color of each bivector is a mix of the colors of the basis vectors it is composed of. For example cyan = blue + green. I also show each vector that the plane is normal to.

In 3D:

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In 4D: